(PDF) Chapter 23 Solutions - DOKUMEN.TIPS (2023)

Chapter 23 Solutions

23.1 (a) N = 10.0 grams

107.87 grams mol

6.02 × 1023 atomsmol

47.0

electronsatom

= 2.62 × 1024

(b) # electrons added = Q

e= 1.00 × 10−3 C

1.60 × 10-19 C electron= 6.25 × 1015

or 2.38 electrons for every 109 already present

23.2 (a) Fe =

ke q1q2

r2 =8.99 × 109 N ⋅ m2/ C2( ) 1.60 × 10−19 C( )2

(3.80 × 10− 10 m)2 = 1.59 × 10−9 N (repulsion)

(b) Fg =

Gm1m2

r2 =6.67 × 10−11 N ⋅ m2 kg2( )(1.67 × 10−27 kg)2

(3.80 × 10−10 m)2 = 1.29 × 10−45 N

The electric force is larger by 1.24 × 1036 times

q1q2

r2 = Gm1m2

r2 with q1 = q2 = q and m1 = m2 = m, then

=Gke

=6.67 × 10−11 N ⋅ m2 / kg2

8.99 × 109 N ⋅ m2 / C2 = 8.61 × 10−11 C / kg

23.3 If each person has a mass of ≈ 70 kg and is (almost) composed of water, then each personcontains

N ≈

70,000 grams18 grams mol

6.02 × 1023 moleculesmol

protonsmolecule

≈2.3 × 1028 protons

With an excess of 1% electrons over protons, each person has a charge

q = (0.01)(1.6 × 10−19 C)(2.3 × 1028) = 3.7 × 107 C

So F = ke

q1q2

r2 = (9 × 109)(3.7 × 107 )2

0.62 N = 4 × 1025 N ~ 1026 N

This force is almost enough to lift a "weight" equal to that of the Earth:

Mg = (6 × 1024 kg)(9.8 m s2) = 6 × 1025 N~ 1026 N

2 Chapter 23 Solutions

23.4 We find the equal-magnitude charges on both spheres:

F = ke q1q2

r 2 = ke

r 2 so

q = r

Fke

= 1.00 m( ) 1.00 × 104 N8.99 × 109 N ⋅ m2/ C2 = 1.05 × 10−3 C

The number of electron transferred is then

Nxfer = 1.05 × 10−3 C( ) 1.60 × 10−19 C / e−( ) = 6.59 × 1015 electrons

The whole number of electrons in each sphere is

Ntot = 10.0 g

107.87 g / mol

6.02 × 1023 atoms / mol( ) 47 e− / atom( ) = 2.62 × 1024 e−

The fraction transferred is then

f =

Nxfer

Ntot=

6.59 × 1015

2.62 × 1024 = 2.51 × 10–9 = 2.51 charges in every billion

23.5

F = keq1q2

r2 =8.99 × 109 N ⋅ m2 C2( ) 1.60 × 10−19 C( )2

6.02 × 1023( )2

2(6.37 × 106 m)[ ] 2 = 514 kN

*23.6 (a) The force is one of attraction. The distance r in Coulomb's law is the distance betweencenters. The magnitude of the force is

F = keq1q2

r2 = 8.99 × 109 N ⋅ m2

12.0 × 10−9 C( ) 18.0 × 10−9 C( )

(0.300 m)2 = 2.16 × 10−5 N

(b) The net charge of − × −6 00 10 9. C will be equally split between the two spheres, or

−3.00 × 10−9 C on each. The force is one of repulsion, and its magnitude is

F = keq1q2

r2 = 8.99 × 109 N ⋅ m2

3.00 × 10−9 C( ) 3.00 × 10−9 C( )

(0.300 m)2 = 8.99 × 10−7 N

Chapter 23 Solutions 3

23.7 F1 = ke

q1q2

r2 = (8.99 × 109 N ⋅ m2/ C2 )(7.00 × 10−6 C)(2.00 × 10−6 C)(0.500 m)2 = 0.503 N

F2 = ke

q1q2

r2 = (8.99 × 109 N ⋅ m2 / C2 )(7.00 × 10−6 C)(4.00 × 10−6 C)(0.500 m)2 = 1.01 N

Fx = (0.503 + 1.01) cos 60.0°= 0.755 N

Fy = (0.503 − 1.01) sin 60.0°= −0.436 N

F = (0.755 N)i − (0.436 N)j = 0.872 N at an angle of 330°

Goal Solution Three point charges are located at the corners of an equilateral triangle as shown in Figure P23.7.Calculate the net electric force on the 7.00−µC charge.

G : Gather Information: The 7.00−µC charge experiences a repulsive force F1 due to the 2.00−µCcharge, and an attractive force F2 due to the −4.00−µC charge, where F2 = 2F1. If we sketch theseforce vectors, we find that the resultant appears to be about the same magnitude as F2 and isdirected to the right about 30.0° below the horizontal.

O : Organize : We can find the net electric force by adding the two separate forces acting on the

7.00−µC charge. These individual forces can be found by applying Coulomb’s law to each pair ofcharges.

A : Analyze: The force on the 7.00−µC charge by the 2.00−µC charge is

F1 =

8.99 × 109 N ⋅ m2/ C2( ) 7.00 × 10−6 C( ) 2.00 × 10−6 C( )0.500 m( )2 cos60°i + sin60° j( ) = F1 = 0.252i + 0.436j( ) N

Similarly, the force on the 7.00−µC by the −4.00−µC charge is

F2 = − 8.99 × 109

N ⋅ m2

7.00 × 10−6 C( ) −4.00 × 10−6 C( )

0.500 m( )2 cos60°i − sin60° j( ) = 0.503i − 0.872j( ) N

Thus, the total force on the 7.00−µC, expressed as a set of components, is

F = F1 + F2 = 0.755 i − 0.436 j( ) N = 0.872 N at 30.0° below the +x axis

L : Learn: Our calculated answer agrees with our initial estimate. An equivalent approach to thisproblem would be to find the net electric field due to the two lower charges and apply F=qE to findthe force on the upper charge in this electric field.

4 Chapter 23 Solutions

*23.8 Let the third bead have charge Q and be located distance x from the left end of the rod. Thisbead will experience a net force given by

F =

ke 3q( )Qx2 i +

ke q( )Qd − x( )2 −i( )

The net force will be zero if

3x2 = 1

d − x( )2 , or d − x = x

This gives an equilibrium position of the third bead of x = 0.634d

The equilibrium is stable if the third bead has positive charge .

*23.9 (a) F = kee2

r 2 = (8.99 × 109 N ⋅ m2/C 2)

(1.60 × 10–19 C)2

(0.529 × 10–10 m)2 = 8.22 × 10–8 N

(b) We have F = mv2

r from which v = Fr

8.22 × 10−8 N( ) 0.529 × 10−10 m( )9.11× 10−31 kg

= 2.19 × 106 m/s

23.10 The top charge exerts a force on the negative charge

keqQ

d 2( )2 + x2 which is directed upward and

to the left, at an angle of tan−1 d / 2x( ) to the x-axis. The two positive charges together exertforce

2 keqQ

d2 4 + x2( )

(− x)i

d2 4 + x2( )1/2

= ma or for x << d 2, a ≈ − 2keqQ

md 3 / 8x

(a) The acceleration is equal to a negative constant times the excursion from equilibrium, as i n

a = −ω2x, so we have Simple Harmonic Motion with ω2 =

16keqQmd 3 .

T = 2π

ω=

π2

md3

keqQ , where m is the mass of the object with charge −Q .

(b) vmax = ωA = 4a

keqQmd 3

Chapter 23 Solutions 5

23.11 For equilibrium, Fe = −Fg , or qE = −mg −j( ). Thus, E = mg

qj.

(a)

E = mgq

j = (9.11× 10−31 kg)(9.80 m s2 )

−1.60 × 10−19 C( ) j = − 5.58 × 10−11 N C( )j

(b)

E = mgq

j =1.67 × 10−27 kg( ) 9.80 m s2( )

1.60 × 10−19 C( ) j = 1.02 × 10−7 N C( )j

23.12 Fy = 0:∑ QE j + mg(− j) = 0

∴ m =

QEg

(24.0 × 10-6 C)(610 N / C)9.80 m / s2 = 1.49 grams

*23.13 The point is designated in the sketch. The magnitudes of theelectric fields, E1, (due to the –2.50 × 10–6 C charge) and E2 (due tothe 6.00 × 10–6 C charge) are

E1 = keqr 2

= (8.99 × 109 N · m2/C 2)(2.50 × 10–6 C)

d2 (1)

E2 = keqr 2

= (8.99 × 109 N · m2/C 2)(6.00 × 10–6 C)

(d + 1.00 m)2 (2)

Equate the right sides of (1) and (2) to get (d + 1.00 m)2 = 2.40d 2

or d + 1.00 m = ±1.55d

which yields d = 1.82 m or d = – 0.392 m

The negative value for d is unsatisfactory because that locates a point between the charges

where both fields are in the same direction. Thus, d = 1.82 m to the left of the -2.50 µC charge.

23.14 If we treat the concentrations as point charges,

E+ = ke

qr2

= 8.99 × 109

N ⋅ m2

40.0 C( )

1000 m( )2 −j( ) = 3.60 × 105 N / C −j( ) (downward)

E− = ke

qr2

= 8.99 × 109

N ⋅ m2

40.0 C( )

1000 m( )2 −j( ) = 3.60 × 105 N / C −j( ) (downward)

E = E+ + E− = 7.20 × 105 N / C downward

6 Chapter 23 Solutions

*23.15 (a) E1 = keq

r2 =8.99 × 109( ) 7.00 × 10−6( )

0.500( )2 = 2.52 × 105 N C

E2 = keq

r2 =8.99 × 109( ) 4.00 × 10−6( )

0.500( )2 = 1.44 × 105 N C

Ex = E2 − E1 cos 60°= 1.44 × 105 − 2.52 × 105 cos 60.0°= 18.0 × 103 N C

Ey = −E1 sin 60.0°= −2.52 × 105 sin 60.0°= −218 × 103 N C

E = [18.0i − 218 j] × 103 N C = [18.0i − 218 j] kN C

(b) F = qE = 2.00 × 10−6 C( ) 18.0i − 218 j( ) × 103 N C = 36.0i − 436 j( ) × 10−3 N = 36.0i − 436 j( )mN

*23.16 (a) E1 =

ke q1

r12 − j( ) =

8.99 × 109( ) 3.00 × 10−9( )0.100( )2 − j( )

= − 2.70 × 103 N C( )j

E2 =

ke q2

r22 −i( ) =

8.99 × 109( ) 6.00 × 10−9( )0.300( )2 −i( ) = − 5.99 × 102 N C( )i

E = E2 + E1 = − 5.99 × 102 N C( )i − 2.70 × 103 N C( )j

(b) F = qE = 5.00 × 10−9 C( ) −599i − 2700 j( )N C

F = −3.00 × 10−6 i − 13.5 × 10−6 j( )N = − −( )3 00 13 5. .i j µN

23.17 (a) The electric field has the general appearance shown. It is zero

at the center , where (by symmetry) one can see that the threecharges individually produce fields that cancel out.

(b) You may need to review vector addition in Chapter Three.

The magnitude of the field at point P due to each of the chargesalong the base of the triangle is E = ke q a2 . The direction of the fieldin each case is along the line joining the charge in question to point P as shown in the diagram at the right. The x components add tozero, leaving

E = keq

a2 sin60.0°( )j + keqa2 sin60.0°( )j =

keqa2 j

Chapter 23 Solutions 7

Goal Solution Three equal positive charges q are at the corners of an equilateral triangle of side a, as shown in FigureP23.17. (a) Assume that the three charges together create an electric field. Find the location of a point(other than ∞) where the electric field is zero. (Hint: Sketch the field lines in the plane of the charges.)(b) What are the magnitude and direction of the electric field at P due to the two charges at the base?

G : The electric field has the general appearance shown by the black arrows in the figure to the right.This drawing indicates that E = 0 at the center of the triangle, since a small positive charge placed atthe center of this triangle will be pushed away from each corner equally strongly. This fact could beverified by vector addition as in part (b) below.

The electric field at point P should be directed upwards and about twice the magnitude of the electricfield due to just one of the lower charges as shown in Figure P23.17. For part (b), we must ignore theeffect of the charge at point P , because a charge cannot exert a force on itself.

O : The electric field at point P can be found by adding the electric field vectors due to each of the twolower point charges: E = E1 + E2

A : (b) The electric field from a point charge is

As shown in the solution figure above, E1 = ke

qa2 to the right and upward at 60°

E2 = ke

qa2 to the left and upward at 60°

E = E1 + E2 = ke

qa2 cos60°i + sin60° j( ) + −cos60°i + sin60° j( )[ ]

= ke

qa2 2 sin60° j( )[ ] = 1.73ke

qa2 j

L :

The net electric field at point P is indeed nearly twice the magnitude due to a singlecharge and is entirely vertical as expected from the symmetry of the configuration. Inaddition to the center of the triangle, the electric field lines in the figure to the rightindicate three other points near the middle of each leg of the triangle where E = 0 , butthey are more difficult to find mathematically.

23.18 (a) E = ke q

r2 = (8.99 × 109)(2.00 × 10−6 )(1.12)2 = 14, 400 N C

Ex = 0 and Ey = 2(14, 400) sin 26.6°= 1.29 × 104 N C

so E == 1.29 × 104 j N C

(b) F = Eq = (1.29 × 104 j)(−3.00 × 10−6 ) = −3.86 × 10−2 j N

8 Chapter 23 Solutions

23.19 (a) E = ke q1

r12 ~1 + ke q2

r 22 ~2 + ke q3

r32 ~3

ke 2q( )a2 i +

ke 3q( )2a2 (i cos 45.0° + j sin 45.0°)

ke 4q( )a2 j

E = 3.06

keqa2 i + 5.06

keqa2 j =

5.91

keqa2 at 58.8°

(b) F E= =q 5.91

keq2

a2 at 58.8°

23.20 The magnitude of the field at (x, y) due to charge q at (x0 , y0 )is given by E = keq r2 where r is the distance from (x0 , y0 ) to

(x, y). Observe the geometry in the diagram at the right.From triangle ABC , r

2 = (x − x0 )2 + (y − y0 )2 , or

r = (x − x0 )2 + (y − y0 )2 , sinθ = (y − y0 )

r, and

cosθ = (x − x0 )

Thus, Ex = Ecosθ = ke q

r2(x − x0 )

ke q(x − x0 )[(x − x0 )2 + (y − y0 )2]3/2

and Ey = Esinθ = ke q

r2(y − y0 )

ke q(y − y0 )[(x − x0 )2 + (y − y0 )2]3/2

23.21 The electric field at any point x is E = keq

(x – a)2 – keq

(x – (–a))2 = keq(4ax)(x2 – a2)2

When x is much, much greater than a, we find E ≈ (4a)(keq)

x 3

23.22 (a) One of the charges creates at P a field E = ke Q/nR2 + x2

at an angle θ to the x-axis as shown.

When all the charges produce field, for n > 1, the componentsperpendicular to the x-axis add to zero.

The total field is nke (Q/n)i

R2 + x2 cos θ = keQxi

(R2 + x2)3/2

(b) A circle of charge corresponds to letting n grow beyond all bounds, but the result does notdepend on n. Smearing the charge around the circle does not change its amount or itsdistance from the field point, so it does not change the field. .

Chapter 23 Solutions 9

23.23 E = keq

r2 ~∑ = keqa2 (− i)+ keq

(2a)2 (− i) + keq(3a)2 (− i) + . . .

= − keqi

a2 1 + 122 + 1

33 + . . .

− π2keq

6 a2 i

23.24 E = keλl

d l+ d( ) =ke Q /l( )l

d l+ d( ) = keQd l+ d( ) =

(8.99 × 109)(22.0 × 10–6)(0.290)(0.140 + 0.290)

E = 1.59 × 106 N/C , directed toward the rod .

23.25 E = ∫ ke dqx 2

where dq = λ0 dx

E = ke λ0 ⌡⌠x0

∞

dxx 2

= ke

–

∞

= keλ 0

x0 The direction is –i or left for λ0 > 0

23.26 E = dE = keλ 0x0 dx −i( )

∞∫∫ = − keλ 0x0 i x−3 dx

∞∫

= − keλ 0x0 i − 12x2

∞

keλ 0

2x0

−i( )

23.27 E = kexQ

(x2 + a2)3/2 = (8.99 × 109)(75.0 × 10–6)x

(x 2 + 0.1002)3/2 = 6.74 × 105 x

(x 2 + 0.0100)3/2

(a) At x = 0.0100 m, E = 6.64 × 106 i N/C = 6.64 i MN/C

(b) At x = 0.0500 m, E = 2.41 × 107 i N/C = 24.1 i MN/C

(d) At x = 1.00 m, E = 6.64 × 105 i N/C = 0.664 i MN/C

10 Chapter 23 Solutions

23.28 E =

ke Q x(x 2 + a2 )3/2

For a maximum,

dEdx

= Qke1

(x2 + a2 )3/2 − 3x2

(x2 + a2 )5/2

= 0

x2 + a2 − 3x2 = 0 or

x = a

Substituting into the expression for E gives

E = ke Qa

2( 32 a2 )3/2 = ke Q

3 32 a2

2ke Q3 3 a2 =

Q6 3πe0a2

23.29 E = 2πke σ 1 − x

x2 + R2

= 2π 8.99 × 109( ) 7.90 × 10−3( ) 1 − x

x2 + 0.350( )2

= 4.46 × 108 1 − x

x2 + 0.123

(a) At x = 0.0500 m, E = 3.83 × 108 N C = 383 MN C

(b) At x = 0.100 m, E = 3.24 × 108 N C = 324 MN C

(d) At x = 2.00 m, E = 6.68 × 108 N C = 6.68 MN C

23.30 (a) From Example 23.9: E = 2π ke σ 1 − x

x2 + R2

σ = Q

πR2 = 1.84 × 10−3 C m2

E = (1.04 × 108 N C)(0.900) = 9.36 × 107 N C = 93.6 MN/C

appx: E = 2πke σ = 104 MN/C (about 11% high)

(b) E = (1.04 × 108 N / C) 1 − 30.0 cm

30.02 + 3.002 cm

= (1.04 × 108 N C)(0.00496) = 0.516 MN/C

appx: E = ke

Qr2 = (8.99 × 109)

5.20 × 10−6

(0.30)2 = 0.519 MN/C (about 0.6% high)

Chapter 23 Solutions 11

23.31 The electric field at a distance x is Ex = 2πkeσ 1 − x

x2 + R 2

This is equivalent to

Ex = 2πkeσ 1 − 1

1+ R 2 x2

For large x, R2 x2 << 1 and

R 2

x2 ≈ 1+R 2

2x2

Ex = 2πkeσ 1 − 1

1+ R2 (2x2 )[ ]

= 2πkeσ1+ R 2 (2x2 ) − 1( )

1+ R 2 (2x2 )[ ]

Substitute σ = Q/π R2,

Ex =keQ 1 x 2( )

1+ R 2 (2x 2 )[ ] = keQ x 2 + R 2

But for x > > R,

1x2 + R2 2

≈ 1x2 , so

Ex ≈ keQ

x2 for a disk at large distances

23.32 The sheet must have negative charge to repel the negative charge on the Styrofoam. Themagnitude of the upward electric force must equal the magnitude of the downwardgravitational force for the Styrofoam to "float" (i.e., Fe = Fg ).

Thus, −qE = mg , or −q

σ2e0

= mg which gives σ = − 2e0mg

23.33 Due to symmetry Ey = ∫ dEy = 0, and Ex = ∫ dE sin θ = ke ∫ dq sin θ

r 2

where dq = λ ds = λr dθ, so that, Ex = keλ

rsinθ dθ

π∫ = keλ

r(−cosθ)

π= 2keλ

where λ = qL and r =

π . Thus, Ex =

2ke qπL2 =

2(8.99 × 109 N · m2/C 2)(7.50 × 10–6 C)π(0.140 m)2

Solving, E = Ex = 2.16 × 107 N/C

Since the rod has a negative charge, E = (–2.16 × 107 i) N/C = –21.6 i MN/C

12 Chapter 23 Solutions

23.34 (a) We define x = 0 at the point where we are to find the field. One ring, with thickness dx, hascharge Qdx/h and produces, at the chosen point, a field

dE =

ke x(x2 + R 2 )3/2

Q dxh

The total field is

E = dEall charge

∫ =

keQ x dxh(x2 + R 2 )3/2 i

d + h∫

keQ i2h

(x2 + R 2 )− 3/2 2x dxx = d

d + h∫

E =

keQ i2 h

(x2 + R 2 )− 1/2

(− 1/ 2) x = d

d + h =

keQ ih

1(d2 + R 2 )1/2 − 1

(d + h)2 + R 2( )1/2

(b) Think of the cylinder as a stack of disks, each with thickness dx, charge Q dx/h, and charge-per-area σ = Q dx / πR 2h. One disk produces a field

dE =

2πkeQ dxπR2h

1 − x(x2 + R2 )1/2

So, E = dE

all charge∫ =

x = d

d + h∫

2keQ dxR2 h

1 − x(x2 + R2 )1/2

E =

2 keQ iR 2h

dxd

d + h∫ − 1

2 (x2 + R 2 )− 1/2 2x dxx = d

d + h∫

= 2keQ i

R2hx

d+ h− 1

2(x2 + R2 )1/2

1/ 2 d

d + h

E = 2 keQ i

R 2hd + h − d − (d + h)2 + R 2( )1/2

+ (d2 + R 2 )1/2

2 keQ iR 2h

h + (d2 + R 2 )1/2 − (d + h)2 + R 2( )1/2

23.35 (a) The electric field at point P due to each element of length dx, is dE = kedq

(x2 + y2 ) and is directed

along the line joining the element of length to point P . By symmetry,

Ex = dEx = 0 ∫ and since dq = λ dx,

E = Ey = dEy = dE cos θ∫∫ where

cos θ =

y(x2 + y2 )1 2

Therefore,

dx(x2 + y2 )3 2 =

2keλ sinθ0

(b) For a bar of infinite length, θ → 90° and Ey =

2keλy

Chapter 23 Solutions 13

*23.36 (a) The whole surface area of the cylinder is A = 2πr2 + 2πrL = 2πr r + L( ).

Q = σA = 15.0 × 10−9 C m2( )2π 0.0250 m( ) 0.0250 m + 0.0600 m[ ] = 2.00 × 10−10 C

(b) For the curved lateral surface only, A = 2πrL.

Q = σA = 15.0 × 10−9 C m2( )2π 0.0250 m( ) 0.0600 m( ) = 1.41× 10−10 C

*23.37 (a) Every object has the same volume, V = 8 0.0300 m( )3 = 2.16 × 10−4 m3 .

For each, Q = ρV = 400 × 10−9 C m3( ) 2.16 × 10−4 m3( ) = 8.64 × 10−11 C

(b) We must count the 9.00 cm2 squares painted with charge:

(i) 6 × 4 = 24 squares

Q = σA = 15.0 × 10−9 C m2( )24.0 9.00 × 10−4 m2( ) = 3.24 × 10−10 C

(ii) 34 squares exposed

Q = σA = 15.0 × 10−9 C m2( )34.0 9.00 × 10−4 m2( ) = 4.59 × 10−10 C

(iii) 34 squares

Q = σA = 15.0 × 10−9 C m2( )34.0 9.00 × 10−4 m2( ) = 4.59 × 10−10 C

(iv) 32 squares

Q = σA = 15.0 × 10−9 C m2( )32.0 9.00 × 10−4 m2( ) = 4.32 × 10−10 C

Q = λ = 80.0 × 10−12 C m( )24 × 0.0300 m( ) = 5.76 × 10−11 C

(ii) Q = λ = 80.0 × 10−12 C m( )44 × 0.0300 m( ) = 1.06 × 10−10 C

(iii) Q = λ = 80.0 × 10−12 C m( )64 × 0.0300 m( ) = 1.54 × 10−10 C

14 Chapter 23 Solutions

(iv) Q = λ = 80.0 × 10−12 C m( )40 × 0.0300 m( ) = 0.960 × 10−10 C

Chapter 23 Solutions 15

22.38

22.39

23.40 (a)

q2= −6

18=

− 1

(b) q1 is negative, q2 is positive

23.41 F = qE = ma a = qE

v = vi + at v = qEt

electron: ve = (1.602 × 10−19)(520)(48.0 × 10−9)

9.11× 10−31 = 4.39 × 106 m/s

in a direction opposite to the field

proton: vp = (1.602 × 10−19)(520)(48.0 × 10−9)

1.67 × 10−27 = 2.39 × 103 m/s

in the same direction as the field

23.42 (a) a = qE

m= (1.602 × 10−19)(6.00 × 105)

(1.67 × 10−27 )= 5.76 × 1013 m s so a = −5.76 × 1013 i m s2

(b) v = vi + 2a(x − xi )

0 = vi2 + 2(−5.76 × 1013)(0.0700) vi = 2.84 × 106 i m s

0 = 2.84 × 106 + (−5.76 × 1013)t t = 4.93 × 10−8 s

16 Chapter 23 Solutions

23.43 (a)

a = qEm

=1.602 × 10−19( ) 640( )

1.67 × 10−27( ) = 6.14 × 1010 m/s2

(b) v = vi + at

1.20 × 106 = (6.14 × 1010)t

t = 1.95 × 10-5 s

x = 1

2 1.20 × 106( ) 1.95 × 10−5( ) = 11.7 m

(d) K = 12 mv 2 = 1

2 (1.67 × 10−27 kg)(1.20 × 106 m / s)2 = 1.20 × 10-15 J

23.44 The required electric field will be in the direction of motion . We know that Work = ∆K

So, –Fd = – 12 m v 2

i (since the final velocity = 0)

This becomes Eed = 1

2mvi

2 or E =

12 m v 2

e d

E = 1.60 × 10–17 J

(1.60 × 10–19 C)(0.100 m) = 1.00 × 103 N/C (in direction of electron's motion)

23.45 The required electric field will be in the direction of motion .

Work done = ∆K so, –Fd = – 12 m v 2

i (since the final velocity = 0)

which becomes eEd = K and E = Ke d

Chapter 23 Solutions 17

Goal Solution The electrons in a particle beam each have a kinetic energy K . What are the magnitude and direction ofthe electric field that stops these electrons in a distance of d?

G : We should expect that a larger electric field would be required to stop electrons with greater kineticenergy. Likewise, E must be greater for a shorter stopping distance, d . The electric field should be i nthe same direction as the motion of the negatively charged electrons in order to exert an opposingforce that will slow them down.

O : The electrons will experience an electrostatic force F = qE. Therefore, the work done by the electricfield can be equated with the initial kinetic energy since energy should be conserved.

A : The work done on the charge is W = F ⋅d = qE ⋅dand Ki +W = Kf = 0Assuming v is in the + x direction, K + −e( )E ⋅ di = 0

eE ⋅ di( ) = K

E is therefore in the direction of the electron beam: E = K

edi

L : As expected, the electric field is proportional to K , and inversely proportional to d . The direction ofthe electric field is important; if it were otherwise the electron would speed up instead of slowingdown! If the particles were protons instead of electrons, the electric field would need to be directedopposite to v in order for the particles to slow down.

23.46 The acceleration is given by v2 = v2i + 2a(x – xi)or v2 = 0 + 2a(–h)

Solving, a = – v2

Now ∑ F = ma: –mgj + qE = – mv2 j

Therefore qE =

– m v 2

2h + m g j

(a) Gravity alone would give the bead downward impact velocity

2 9.80 m / s2( ) 5.00 m( ) = 9.90 m / s

To change this to 21.0 m/s down, a downward electric field must exert a downward electricforce.

(b) q = mE

v 2

2h – g = 1.00 × 10–3 kg1.00 × 104 N/C

N · s2

kg · m

(21.0 m/s)2

2(5.00 m) – 9.80 m/s2 = 3.43 µC

18 Chapter 23 Solutions

23.47 (a) t = x

v= 0.0500

4.50 × 105 = 1.11 × 10-7 s = 111 ns

(b) ay = qE

m= (1.602 × 10−19)(9.60 × 103)

(1.67 × 10−27 )= 9.21× 1011 m / s2

y − yi = vyit + 12 ayt2

y = 12 (9.21× 1011)(1.11× 10−7 )2 = 5.67 × 10-3 m = 5.67 mm

vy = vy i + ay = (9.21 × 1011)(1.11 × 10-7) = 1.02 × 105 m/s

23.48 ay = qE

(1.602 × 10−19)(390)(9.11× 10−31)

= 6.86 × 1013 m / s2

(a) t =

2vi sinθay

from projectile motion equations

t =

2(8.20 × 105)sin 30.0°6.86 × 1013 = 1.20 × 10-8 s) = 12.0 ns

(b) h =

vi2 sin2 θ2ay

=(8.20 × 105)2 sin2 30.0°

2(6.86 × 1013)= 1.23 mm

vi2 sin 2θ2ay

=(8.20 × 105)2 sin 60.0°

2(6.86 × 1013)= 4.24 mm

23.49 vi = 9.55 × 103 m/s

(a) ay = eE

m= (1.60 × 10−19)(720)

(1.67 × 10−27 )= 6.90 × 1010 m s2

R = vi

2 sin 2θay

= 1.27 × 10-3 m so that

(9.55 × 103)2 sin 2θ6.90 × 1010 = 1.27 × 10−3

sin 2θ = 0.961 θ = 36.9° 90.0° – θ = 53.1°

(b) t = R

vix= R

vi cosθIf θ = 36.9°, t = 167 ns If θ = 53.1°, t = 221 ns

Chapter 23 Solutions 19

*23.50 (a) The field, E1, due to the 4.00 × 10–9 C charge is in the –x direction.

E1 = ke qr 2

= (8.99 × 109 N · m2/C 2)(− 4.00 × 10–9 C)

(2.50 m)2 i = −5.75i N/C

Likewise, E2 and E3, due to the 5.00 × 10–9 C charge and the 3.00 × 10–9 C charge are

E2 = ke qr 2

= (8.99 × 109 N · m2/C 2)(5.00 × 10–9 C)

(2.00 m)2 i = 11.2

N/C

E3 = (8.99 × 109 N · m2/C 2)(3.00 × 10–9 C)

(1.20 m)2 i = 18.7 N/C

ER = E1 + E2 + E3 = 24.2 N/C in +x direction.

(b) E1 = ke qr 2

= −8.46 N / C( ) 0.243i + 0.970j( )

E2 = ke qr 2

= 11.2 N / C( ) +j( )

E3 = ke qr 2

= 5.81 N / C( ) −0.371i + 0.928j( )

Ex = E1x + E3x = – 4.21i N/C Ey = E1y + E2y + E3y = 8.43j N/C

ER = 9.42 N/C θ = 63.4° above –x axis

23.51 The proton moves with acceleration ap = qE

1.60 × 10−19 C( ) 640 N / C( )1.67 × 10−27 kg

= 6.13 × 1010 m s2

while the e− has acceleration ae =

1.60 × 10−19 C( ) 640 N/C( )9.11× 10−31 kg

= 1.12 × 1014 m s2 = 1836 ap

(a) We want to find the distance traveled by the proton (i.e., d = 12 apt2 ), knowing:

4.00 cm = 1

2 apt2 + 12 aet

2 = 1837 12 apt2( )

Thus, d = 1

2 apt2 = 4.00 cm1837

= 21.8 µm

20 Chapter 23 Solutions

(b) The distance from the positive plate to where the meeting occurs equals the distance thesodium ion travels (i.e., dNa = 1

2 aNat2). This is found from:

4.00 cm = 12 aNat2 + 1

2 aClt2:

4.00 cm = 1

2eE

22.99 u

t2 + 1

2eE

35.45 u

This may be written as 4.00 cm = 1

2 aNat2 + 12 0.649aNa( )t2 = 1.65 1

2 aNat2( )

so dNa = 1

2 aNat2 = 4.00 cm1.65

= 2.43 cm

23.52 From the free-body diagram shown, ∑Fy = 0

and T cos 15.0° = 1.96 × 10–2 N

So T = 2.03 × 10–2 N

From ∑Fx = 0, we have qE = T sin 15.0°

or q = T sin 15.0°

E = (2.03 × 10–2 N) sin 15.0°

1.00 × 103 N/C = 5.25 × 10–6 C = 5.25 µC

23.53 (a) Let us sum force components to find

∑Fx = qEx – T sin θ = 0, and ∑Fy = qEy + T cos θ – mg = 0

Combining these two equations, we get

q = m g

(Ex cot θ + Ey) =

(1.00 × 10-3)(9.80)(3.00 cot 37.0° + 5.00) × 105 = 1.09 × 10–8 C = 10.9 nC

(b) From the two equations for ∑Fx and ∑Fy we also find

T = qEx

sin 37.0° = 5.44 × 10–3 N = 5.44 mN

Free Body Diagramfor Goal Solution

Chapter 23 Solutions 21

Goal Solution A charged cork ball of mass 1.00 g is suspended on a light string in the presence of a uniform electric field,as shown in Fig. P23.53. When E = 3.00i + 5.00j( ) × 105 N / C, the ball is in equilibrium at θ = 37.0°. Find(a) the charge on the ball and (b) the tension in the string.

G : (a) Since the electric force must be in the same direction as E, the ball must be positively charged. Ifwe examine the free body diagram that shows the three forces acting on the ball, the sum of whichmust be zero, we can see that the tension is about half the magnitude of the weight.

O : The tension can be found from applying Newton's second law to this statics problem (electrostatics,in this case!). Since the force vectors are in two dimensions, we must apply ΣF = ma to both the xand y directions.

A : Applying Newton's Second Law in the x and y directions, and noting that ΣF = T + qE + Fg = 0,

ΣFx = qEx − T sin 37.0°= 0 (1)

ΣFy = qEy + T cos 37.0° − mg = 0 (2)

We are given Ex = 3.00 × 105 N / C and Ey = 5.00 × 105 N / C; substituting T from (1) into (2):

q = mg

Ey + Ex

tan 37.0°

= (1.00 × 10−3 kg)(9.80 m / s2 )

5.00 + 3.00tan 37.0°

× 105 N / C= 1.09 × 10−8 C

(b) Using this result for q in Equation (1), we find that the tension is T = qEx

sin 37.0°= 5.44 × 10−3 N

L : The tension is slightly more than half the weight of the ball ( Fg = 9.80 × 10−3 N) so our result seemsreasonable based on our initial prediction.

23.54 (a) Applying the first condition of equilibrium to the ball gives:

ΣFx = qEx − T sinθ = 0 or T = qEx

sinθ= qA

sinθand ΣFy = qEy + T cosθ − mg = 0 or qB + T cosθ = mg

Substituting from the first equation into the second gives:

q Acotθ + B( ) = mg , or q =

mgAcotθ + B( )

(b) Substituting the charge into the equation obtained from ΣFx yields

T = mg

Acotθ + B( )A

sinθ

mgAAcosθ + Bsinθ

22 Chapter 23 Solutions

Goal Solution A charged cork ball of mass m is suspended on a light string in the presence of a uniform electric field, asshown in Figure P23.53. When E = Ai + Bj( ) N / C, where A and B are positive numbers, the ball is i nequilibrium at the angle θ . Find (a) the charge on the ball and (b) the tension in the string.

G : This is the general version of the preceding problem. The known quantities are A , B , m, g , and θ .The unknowns are q and T .

O : The approach to this problem should be the same as for the last problem, but without numbers tosubstitute for the variables. Likewise, we can use the free body diagram given in the solution toproblem 53.

A : Again, Newton's second law: −T sin θ + qA = 0 (1)

and + T cosθ + qB − mg = 0 (2)

(a) Substituting T = qA

sinθ, into Eq. (2),

qAcosθsinθ

+ qB = mg

Isolating q on the left, q = mg

Acot θ + B( )

(b) Substituting this value into Eq. (1), T = mgA

Acos θ + Bsinθ( )

L : If we had solved this general problem first, we would only need to substitute the appropriate valuesin the equations for q and T to find the numerical results needed for problem 53. If you find thisproblem more difficult than problem 53, the little list at the Gather step is useful. It shows whatsymbols to think of as known data, and what to consider unknown. The list is a guide for decidingwhat to solve for in the Analysis step, and for recognizing when we have an answer.

23.55 F = ke q1q2

r 2 tan θ =

15.060.0 θ = 14.0°

F1 = (8.99 × 109)(10.0 × 10–6)2

(0.150)2 = 40.0 N

F3 = (8.99 × 109)(10.0 × 10–6)2

(0.600)2 = 2.50 N

F2 = (8.99 × 109)(10.0 × 10–6)2

(0.619)2 = 2.35 N

Fx = –F3 – F2 cos 14.0° = –2.50 – 2.35 cos 14.0° = – 4.78 N

Fy = –F1 – F2 sin 14.0° = – 40.0 – 2.35 sin 14.0° = – 40.6 N

Fnet = F 2x + F 2y = (– 4.78)2 + (– 40.6)2 = 40.9 N

tan φ = FyFx

= – 40.6– 4.78 φ = 263°

Chapter 23 Solutions 23

23.56 From Fig. A: dcos . .30 0 15 0° = cm, or d =

°15 0

30 0.

cos . cm

From Fig. B: θ = sin−1 d

50.0 cm

= sin−1 15.0 cm

50.0 cm cos 30.0°( )

= 20.3°

mg= tanθ

or Fq = mg tan 20.3° (1)

From Fig. C: Fq = 2F cos 30.0° = 2

keq2

0.300 m( )2

cos 30.0° (2)

Equating equations (1) and (2), 2

keq2

0.300 m( )2

cos 30.0° = mg tan 20.3°

q2 = mg 0.300 m( )2 tan 20.3°

2ke cos 30.0°

q2 =2.00 × 10−3 kg( ) 9.80 m s2( ) 0.300 m( )2 tan 20.3°

2 8.99 × 109 N ⋅ m2 C2( )cos 30.0°

q = 4.20 × 10−14 C2 = 2.05 × 10−7 C= 0.205 µC

Figure A

Figure B

Figure C

23.57 Charge Q/2 resides on each block, which repel as point charges:

F = ke(Q/2)(Q/2)

L 2 = k(L – L i)

Q = 2Lk L − Li( )

ke= 2 0.400 m( ) 100 N / m( ) 0.100 m( )

8.99 × 109 N ⋅ m2 / C2( ) = 26.7 µC

23.58 Charge Q/2 resides on each block, which repel as point charges: F =

ke Q 2( ) Q 2( )L2 = k L − Li( )

Solving for Q , Q = 2L

k L − Li( )ke

24 Chapter 23 Solutions

*23.59 According to the result of Example 23.7, the lefthand rodcreates this field at a distance d from its righthand end:

E = keQ

d(2a + d)

dF = keQQ

2a dx

d(d + 2a)

F = keQ 2

2a ∫

x = b – 2a dx

x(x + 2a) = keQ 2

–

12a ln

2a + xx

b – 2a

F = +keQ 2

4a2

– ln

2a + bb + ln

bb – 2a =

keQ 2

4a2 ln b2

(b – 2a)(b + 2a) =

keQ 2

4a2 ln

b2 – 4a2

*23.60 The charge moves with acceleration of magnitude a given by ∑F = ma = q E

(a) a = q E

m = 1.60 × 10–19 C (1.00 N/C)

9.11 × 10–31 kg = 1.76 × 1011 m/s2

Then v = vi + at = 0 + at gives t = va =

3.00 × 107 m/s1.76 × 1011 m/s2 = 171 µs

(b) t = va =

vmqE =

(3.00 × 107 m/s)(1.67 × 10–27 kg)(1.60 × 10–19 C)(1.00 N/C)

= 0.313 s

23.61 Q = ∫ λdl = ∫ 90.0°–90.0° λ 0 cos θ Rdθ = λ 0 R sin θ 90.0°

–90.0° = λ 0 R [1 – (–1)] = 2λ 0R

Q = 12.0 µC = (2λ 0 )(0.600) m = 12.0 µC so λ 0 = 10.0 µC/m

dFy = 14πe0

3.00 µC( ) λdl( )R2

cosθ = 14πe0

3.00 µC( ) λ 0 cos2 θRdθ( )R2

Fy = ∫

90.0°

–90.0°

8.99 × 109 N · m2

C 2 (3.00 × 10–6 C)(10.0 × 10–6 C/m)

(0.600 m) cos2 θdθ

Fy = 8.99 30.0( )

0.60010−3 N( ) 1

2+ 1

2cos2θ( )dθ

−π/2

π/2

∫

Fy = 0.450 N( ) 1

2π + 1

4sin 2θ −π/2

π/2( ) = 0.707 N Downward.

Since the leftward and rightward forces due to the two halves of thesemicircle cancel out, Fx = 0.

Chapter 23 Solutions 25

23.62 At equilibrium, the distance between the charges is r = 2 0.100 m( )sin10.0°= 3.47 × 10−2 m

Now consider the forces on the sphere with charge +q , and use ΣFy = 0:

ΣFy = 0: T cos 10.0°= mg, or T = mg

cos 10.0°(1)

ΣFx = 0: Fnet = F2 − F1 = T sin10.0° (2)

Fnet is the net electrical force on the charged sphere. Eliminate Tfrom (2) by use of (1).

Fnet = mgsin 10.0°

cos 10.0°= mg tan 10.0°

= 2.00 × 10−3 kg( ) 9.80 m / s2( )tan 10.0°= 3.46 × 10−3 N

Fnet is the resultant of two forces, F1 and F2 . F1 is the attractive forceon +q exerted by –q, and F2 is the force exerted on +q by the externalelectric field.

Fnet = F2 – F1 or F2 = Fnet + F1

F1 = 8.99 × 109 N ⋅ m2 / C2( ) 5.00 × 10−8 C( ) 5.00 × 10−8 C( )3.47 × 10−3 m( )2 = 1.87 × 10−2 N

Thus, F2 = Fnet + F1 yields F2 = 3.46 × 10−3 N + 1.87 × 10−2 N = 2.21× 10−2 N

and F2 = qE , or E = F2

q= 2.21× 10−2 N

5.00 × 10−8 C = 4.43 × 105 N/C = 443 kN/C

23.63 (a) From the 2Q charge we have Fe − T2 sinθ2 = 0 and mg − T2 cosθ2 = 0

Combining these we find

mg= T2 sinθ2

T2 cosθ2= tanθ2

From the Q charge we have Fe − T1 sinθ1 = 0 and mg − T1 cosθ1 = 0

Combining these we find

mg= T1 sinθ1

T1 cosθ1= tanθ1 or θ2 = θ1

(b) Fe = ke 2QQ

r2 = 2keQ2

If we assume θ is small then . Substitute expressions for Fe and tan θ into either

equation found in part (a) and solve for r.

mg= tanθ then and solving for r we find

26 Chapter 23 Solutions

23.64 At an equilibrium position, the net force on the charge Q is zero. The equilibrium positioncan be located by determining the angle θ corresponding to equilibrium. In terms of lengths s,

12 a 3, and r, shown in Figure P23.64, the charge at the origin exerts an attractive force

keQq (s + 12 a 3)2 . The other two charges exert equal repulsive forces of magnitude keQq r2.

The horizontal components of the two repulsive forces add, balancing the attractive force,

Fnet = keQq

2 cos θr2 − 1

(s + 12 a 3)2

= 0

From Figure P23.64, r =

12 a

sin θ s = 1

2 a cot θ

The equilibrium condition, in terms of θ, is Fnet = 4

keQq 2 cos θ sin2θ − 1

( 3 + cot θ)2

= 0

Thus the equilibrium value of θ is 2 cos θ sin2 θ( 3 + cot θ)2 = 1.

One method for solving for θ is to tabulate the left side. To three significant figures the valueof θ corresponding to equilibrium is 81.7°. The distance from the origin to the equilibriumposition is x = 1

2 a( 3 + cot 81.7°) = 0.939a

θ 2 cosθ sin2 θ( 3 + cotθ)2

60°70°80°90°81°81.5°81.7°

42.6541.22601.0911.0240.997

23.65 (a) The distance from each corner to the center of the square is

L 2( )2 + L 2( )2 = L 2

The distance from each positive charge to −Q is then z2 + L2 2 .Each positive charge exerts a force directed along the line joining

q and −Q , of magnitude

keQqz2 + L2 2

The line of force makes an angle with the z-axis whose cosine is

z2 + L2 2

The four charges together exert forces whose x and y componentsadd to zero, while the z-components add to F =

− 4keQq z

z2 + L2 2( )3 2 k

Chapter 23 Solutions 27

(b) For z << L, the magnitude of this force is

Fz ≈ − 4keQqz

L2 2( )3 2 = − 4 2( )3 2 keQqL3

z = maz

Therefore, the object’s vertical acceleration is of the form az = −ω2z

with ω2 = 4 2( )3 2 keQq

mL3 = keQq 128mL3

Since the acceleration of the object is always oppositely directed to its excursion fromequilibrium and in magnitude proportional to it, the object will execute simple harmonicmotion with a period given by

T = 2π

ω= 2π

128( )1 4mL3

keQq=

π8( )1 4

mL3

keQq

23.66 (a) The total non-contact force on the cork ball is: F = qE + mg = m g +

qEm

which is constant and directed downward. Therefore, it behaves like a simple pendulum i nthe presence of a modified uniform gravitational field with a period given by:

T = 2π L

g + qEm

= 2π 0.500 m

9.80 m / s2 +2.00 × 10−6 C( ) 1.00 × 105 N / C( )

1.00 × 10−3 kg

= 0.307 s

(b) Yes . Without gravity in part (a), we get T = 2π L

qE m

T = 2π 0.500 m

2.00 × 10−6 C( ) 1.00 × 105 N / C( ) 1.00 × 10−3 kg= 0.314 s (a 2.28% difference).

23.67 (a) Due to symmetry the field contribution from each negative charge isequal and opposite to each other. Therefore, their contribution tothe net field is zero. The field contribution of the +q charge is

E = keq

r2 = keq3 a2 4( ) = 4keq

3a2

in the negative y direction, i.e., E = − 4ke q

3a2 j

28 Chapter 23 Solutions

(b) If Fe = 0, then E at P must equal zero. In order for the field to cancel at P , the − 4q must beabove + q on the y-axis.

Then, E = 0 = − keq

1.00 m( )2 + ke(4q)y2 , which reduces to y

2 = 4.00 m2 .

Thus, y = ±2.00 m . Only the positive answer is acceptable since the − 4q must be located

above + q. Therefore, the − 4q must be placed 2.00 meters above point P along the + y − axis .

23.68 The bowl exerts a normal force on each bead, directed alongthe radius line or at 60.0° above the horizontal. Consider thefree-body diagram of the bead on the left:

ΣFy = nsin 60.0°−mg = 0 ,

or n = mg

sin 60.0°

Also, ΣFx = −Fe + ncos 60.0° = 0,

ke q2

R2 = ncos 60.0° = mgtan 60.0°

= mg3

Thus, q = R

mgke 3

1 2

60.0˚

23.69 (a) There are 7 terms which contribute:

3 are s away (along sides)

3 are 2 s away (face diagonals) and sin θ =

= cos θ

1 is 3 s away (body diagonal) and sinφ = 1

The component in each direction is the same by symmetry.

F = keq

s2 1+2

2 2+ 1

3 3

(i + j + k) =

keq2

s2 (1.90)(i + j + k)

(b) F = F x

2 + F y2 + F z

2 = 3.29

ke q2

s2 away from the origin

Chapter 23 Solutions 29

23.70 (a) Zero contribution from the same face due to symmetry, oppositeface contributes

keqr2 sin φ

where

r = s

+ s2

+ s2 = 1.5 s = 1.22 s

E = 4

keqsr3 = 4

(1.22)3keqs2 =

2.18

keqs2

sin φ = s/r

(b) The direction is the k direction.

*23.71

dE = ke dq

x2 + 0.150 m( )2−x i + 0.150 m j

x2 + 0.150 m( )2

=keλ −x i + 0.150 m j( )dx

x2 + 0.150 m( )2[ ] 3 2

E = dEall charge∫ = keλ

−x i + 0.150 m j( )dx

x2 + 0.150 m( )2[ ] 3 2x=0

0.400 m∫

dqx

0.150 m

E = keλ+i

x2 + 0.150 m( )20

0.400 m

+ 0.150 m( )j x

0.150 m( )2 x2 + 0.150 m( )20

0.400 m

E = 8.99 × 109

N ⋅ m2

35.0 × 10−9

i 2.34 − 6.67( ) m + j 6.24 − 0( ) m[ ]

E = −1.36i + 1.96 j( ) × 103 N C = −1.36i + 1.96 j( ) kN C

23.72 By symmetry Ex∑ = 0. Using the distances as labeled,

Ey = ke

q(a2 + y2 )

sin θ +q

(a2 + y2 ) sin θ − 2q

∑

But sin θ =

(a2+y2 ), so

E = Ey = 2keq

y(a2 + y2 )3 2 − 1

∑

Expand (a2 +y2)− 3 2 as (a2 + y2 )− 3 2 = y−3 − (3 2)a2y−5 + . . .

Therefore, for a << y, we can ignore terms in powers higher than 2,

and we have E = 2keq

1y2 − 3

y4 − 1y2

E = − ke 3qa2

30 Chapter 23 Solutions

23.73 The field on the axis of the ring is calculated in Example 23.8, E = Ex = kexQ

(x2 + a2 )3/2

The force experienced by a charge –q placed along the axis of the ring is

F = −keQq

x(x2 + a2 )3/2

and when x << a , this becomes F = keQq

This expression for the force is in the form of Hooke's law,

with an effective spring constant of k = keQq a3

Since ω = 2πf = k m , we have f =

12π

keQqma3

23.74 The electrostatic forces exerted on the two charges result in anet torque τ = −2Fa sin θ = −2Eqa sin θ.

For small θ, sin θ ≈ θ and using p = 2qa, we have τ = –Epθ.

The torque produces an angular acceleration given by τ = Iα = I

d2θdt2

Combining these two expressions for torque, we have

d2θdt2 + Ep

θ = 0

This equation can be written in the form

d2θdt2 = − ω2θ where

ω2 = Ep

This is the same form as Equation 13.17 and thefrequency of oscillation is found by comparisonwith Equation 13.19, or

f = 1

2πpEI

12π

2qaEI

Chapter 24 Solutions

24.1 (a) ΦE = EA cos θ = (3.50 × 103)(0.350 × 0.700) cos 0° = 858 N · m2/C

(b) θ = 90.0° ΦE = 0

24.2 ΦE = EA cos θ = (2.00 × 104 N/C)(18.0 m2)cos 10.0° = 355 kN · m2/C

24.3 ΦE = EA cos θ

A = π r 2 = π (0.200)2 = 0.126 m2

5.20 × 105 = E (0.126) cos 0°

E = 4.14 × 106 N/C = 4.14 MN/C

24.4 The uniform field enters the shell on one side and exits on the other so the total flux is zero .

24.5 (a) ′ = ( )( )A 10 0 30 0. . cm cm

′A = 300 cm2 = 0.0300 m2

ΦE, ′A = E ′A cosθ

ΦE, ′A = 7.80 × 104( ) 0.0300( )cos 180°

ΦE, ′A = −2.34 kN ⋅ m2 C

0.0 cm

3 0.0 cm

0.0˚

(b) ΦE, A = EA cosθ = 7.80 × 104( ) A( )cos 60.0°

A = 30.0 cm( ) w( ) = 30.0 cm( ) 10.0 cm

cos 60.0°

= 600 cm2 = 0.0600 m2

ΦE, A = 7.80 × 104( ) 0.0600( )cos 60°= + 2.34 kN ⋅ m2 C

Chapter 24 Solutions 33

ΦE, total = − 2.34 kN ⋅ m2 C + 2.34 kN ⋅ m2 C + 0 + 0 + 0 = 0

34 Chapter 24 Solutions

24.6 (a) ΦE = E ⋅ A = (ai + b j) ⋅ A i = aA

(b) ΦE = (ai + bj) ⋅ Aj = bA

24.7 Only the charge inside radius R contributes to the total flux.

ΦE = q /e0

24.8 ΦE = EA cosθ through the base

ΦE = 52.0( ) 36.0( )cos 180°= –1.87 kN · m2/C

Note the same number of electric field lines go through the baseas go through the pyramid's surface (not counting the base).

For the slanting surfaces, ΦE = +1.87 kN ⋅ m2 / C

24.9 The flux entering the closed surface equals the flux exiting the surface. The flux entering theleft side of the cone is

ΦE = E ⋅ dA =∫ ERh . This is the same as the flux that exits the right

side of the cone. Note that for a uniform field only the cross sectional area matters, not shape.

*24.10 (a) E = keQr 2

8.90 × 102 = (8.99 × 109)Q

(0.750)2 , But Q is negative since E points inward.

Q = – 5.56 × 10–8 C = – 55.6 nC

(b) The negative charge has a spherically symmetric charge distribution.

24.11 (a) ΦE = qin

e0=

+5.00 µC − 9.00 µC + 27.0 µC − 84.0 µC( )8.85 × 10−12 C2 / N ⋅ m2 = – 6.89 × 106 N · m2/C = – 6.89 MN · m2/C

(b) Since the net electric flux is negative, more lines enter than leave the surface.

Chapter 24 Solutions 35

24.12 ΦE = qin

Through S1 ΦE = −2Q + Q

e0=

− Qe0

Through S2 ΦE = + Q − Q

e0= 0

Through S3 ΦE = −2Q + Q − Q

e0=

− 2Q

Through S4 ΦE = 0

24.13 (a) One-half of the total flux created by the charge q goes through the plane. Thus,

ΦE, plane = 1

2ΦE, total = 1

2qe0

q2e0

(b) The square looks like an infinite plane to a charge very close to the surface. Hence,

ΦE, square ≈ ΦE, plane =

q2e0

24.14 The flux through the curved surface is equal to the flux through the flat circle, E0 πr 2 .

24.15 (a)+Q2 e0

Simply consider half of a closed sphere.

(b)–Q2 e0

(from ΦΕ, total = ΦΕ, dome + ΦΕ, flat = 0)

36 Chapter 24 Solutions

Goal Solution A point charge Q is located just above the center of the flat face of a hemisphere of radius R, as shown i nFigure P24.15. What is the electric flux (a) through the curved surface and (b) through the flat face?

G : From Gauss’s law, the flux through a sphere with a point charge in it should be Q e0 , so we shouldexpect the electric flux through a hemisphere to be half this value: Φcurved = Q 2e0 . Since the flatsection appears like an infinite plane to a point just above its surface so that half of all the field linesfrom the point charge are intercepted by the flat surface, the flux through this section should alsoequal Q 2e0 .

O : We can apply the definition of electric flux directly for part (a) and then use Gauss’s law to find theflux for part (b).

A : (a) With δ very small, all points on the hemisphere are nearly at distance R from the charge, so thefield everywhere on the curved surface is keQ / R2 radially outward (normal to the surface).Therefore, the flux is this field strength times the area of half a sphere:

Φcurved = E ⋅ dA∫ = ElocalAhemisphere

= ke

QR2

12( ) 4πR2( ) = 1

4πe0Q 2π( ) = Q

2e0

(b) The closed surface encloses zero charge so Gauss's law gives

Φcurved + Φflat = 0 or Φflat = −Φcurved = −Q

2e0

L : The direct calculations of the electric flux agree with our predictions, except for the negative sign i npart (b), which comes from the fact that the area unit vector is defined as pointing outward from anenclosed surface, and in this case, the electric field has a component in the opposite direction (down).

24.16 (a) ΦE, shell = qin

e0= 12.0 × 10−6

8.85 × 10−12 = 1.36 × 106 N ⋅ m2 / C = 1.36 MN · m2/C

(b) ΦE, half shell = 12 (1.36 × 106 N ⋅ m2 / C) = 6.78 × 105 N ⋅ m2 / C = 678 kN · m2/C

24.17 From Gauss's Law, ΦE = E ⋅ dA∫ = qin

e0.

Thus, ΦE = Q

e0= 0.0462 × 10−6 C

8.85 × 10-12 C2 N ⋅ m2 = 5.22 kN ⋅ m2 C

Chapter 24 Solutions 37

24.18 If R ≤ d, the sphere encloses no charge and ΦE = qin /e0 = 0

If R > d, the length of line falling within the sphere is 2 R 2 − d2

so ΦΕ = 2λ R2 − d2 e0

24.19 The total charge is Q − 6 q . The total outward flux from the cube is Q − 6 q( )/e0 , of whichone-sixth goes through each face:

ΦE( )one face =

Q − 6 q6e0

ΦE( )one face =

Q − 6 q6e0

= (5.00 − 6.00) × 10−6 C ⋅ N ⋅ m2

6 × 8.85 × 10−12 C2 = − ⋅18.8 kN m /C2

24.20 The total charge is Q − 6 q . The total outward flux from the cube is Q − 6 q( )/e0 , of whichone-sixth goes through each face:

ΦE( )one face =

Q − 6 q6e0

24.21 When R < d, the cylinder contains no charge and ΦΕ = 0 .

When R > d, ΦE = qin

e0=

λLe0

24.22 ΦE, hole = E ⋅ Ahole = keQ

πr2( )

=8.99 × 109 N ⋅ m2 C2( ) 10.0 × 10−6 C( )

0.100 m( )2

π 1.00 × 10−3 m( )2

ΦE, hole = 28.2 N ⋅ m2 C

38 Chapter 24 Solutions

Chapter 24 Solutions 39

24.23 ΦE = qin

e0= 170 × 10−6 C

8.85 × 10-12 C2 N ⋅ m2 = 1.92 × 107 N ⋅ m2 C

(a) ΦE( )one face = 1

6 ΦE = 1.92 × 107 N ⋅ m2 C6

ΦE( )one face = 3 20. MN m C2⋅

(b) ΦE = 19.2 MN ⋅ m2 C

(c) The answer to (a) would change because the flux through each face of the cube would not beequal with an unsymmetrical charge distribution. The sides of the cube nearer the chargewould have more flux and the ones farther away would have less. The answer to (b) wouldremain the same, since the overall flux would remain the same.

24.24 (a) ΦE = qin

8.60 × 104 = qin

8.85 × 10−12

qin = 7.61× 10−7 C = 761 nC

(b) Since the net flux is positive, the net charge must be positive . It can have any distribution.

24.25 No charge is inside the cube. The net flux through the cube is zero. Positive flux comes outthrough the three faces meeting at g. These three faces together fill solid angle equal to one-eighth of a sphere as seen from q, and together pass flux

q e0( ) . Each face containing a

intercepts equal flux going into the cube:

0 = ΦE, net = 3ΦE, abcd + q / 8e0

ΦE, abcd = −q / 24e0

40 Chapter 24 Solutions

24.26 The charge distributed through the nucleus creates a field at the surface equal to that of a pointcharge at its center: E = keq r2

E = (8.99 × 109 Nm2/C 2)(82 × 1.60 × 10–19 C)

[(208)1/3 1.20 × 10–15 m] 2

E = 2.33 × 1021 N/C away from the nucleus

24.27 (a) E = ke Qr

a3 = 0

(b) E = ke Qr

a3 = (8.99 × 109)(26.0 × 10–6)(0.100)

(0.400)3 = 365 kN/C

(8.99 × 109)(26.0 × 10–6)(0.400)2 = 1.46 MN/C

(d) E = ke Qr 2

= (8.99 × 109)(26.0 × 10–6)

(0.600)2 = 649 kN/C

The direction for each electric field is radially outward.

*24.28 (a) E = 2ke λ

3.60 × 104 = 2(8.99 × 109)(Q/2.40)

(0.190)

Q = + 9.13 × 10–7 C = +913 nC

(b) E = 0

24.29 ∫ο E · dA = qin

e0 =

∫ ρ dV

e0 =

e0 l π r 2

E2π rl = ρ

e0 l π r 2

Chapter 24 Solutions 41

E = ρ

2 e0 r away from the axis

Goal Solution Consider a long cylindrical charge distribution of radius R with a uniform charge density ρ . Find theelectric field at distance r from the axis where r < R.

G : According to Gauss’s law, only the charge enclosed within the gaussian surface of radius r needs to beconsidered. The amount of charge within the gaussian surface will certainly increase as ρ and rincrease, but the area of this gaussian surface will also increase, so it is difficult to predict which ofthese two competing factors will more strongly affect the electric field strength.

O : We can find the general equation for E from Gauss’s law.

A : If ρ is positive, the field must be radially outward. Choose as the gaussian surface a cylinder of length

L and radius r , contained inside the charged rod. Its volume is πr2L and it encloses charge ρπr2L.The circular end caps have no electric flux through them; there E ⋅ dA = EdAcos90.0°= 0. The curvedsurface has E ⋅ dA = EdAcos0° , and E must be the same strength everywhere over the curved surface.

Gauss’s law,

E ⋅ dA∫ = qe0

, becomes

E dA∫CurvedSurface

= ρπr2Le0

Now the lateral surface area of the cylinder is 2πrL : E 2πr( )L = ρπr2L

Thus, E = ρ r

2e0 radially away from the cylinder axis

L : As we expected, the electric field will increase as ρ increases, and we can now see that E is alsoproportional to r . For the region outside the cylinder ( r > R), we should expect the electric field todecrease as r increases, just like for a line of charge.

24.30 σ = 8.60 × 10−6 C / cm2( ) 100 cm

= 8.60 × 10−2 C / m2

E = σ2e0

= 8.60 × 10−2

2 8.85 × 10−12( ) = 4.86 × 109 N / C

The field is essentially uniform as long as the distance from the center of the wall to the fieldpoint is much less than the dimensions of the wall.

24.31 (a) E = 0

42 Chapter 24 Solutions

(b) E = keQ

r2 = (8.99 × 109)(32.0 × 10−6 )(0.200)2 = 7.19 MN/C

Chapter 24 Solutions 43

24.32 The distance between centers is 2 × 5.90 × 10–15 m. Each produces a field as if it were a pointcharge at its center, and each feels a force as if all its charge were a point at its center.

F = keq1q2

r 2 =

8.99 × 109 N · m2

C 2 (46)2 (1.60 × 10–19 C)2

(2 × 5.90 × 10–15 m)2 = 3.50 × 103 N = 3.50 kN

*24.33 Consider two balloons of diameter 0.2 m, each with mass 1 g , hangingapart with a 0.05 m separation on the ends of strings making angles of10˚ with the vertical.

(a) ΣFy = T cos 10° − mg = 0 ⇒ T = mg

cos 10°

ΣFx = T sin 10° − Fe = 0 ⇒ Fe = T sin 10° , so

Fe = mg

cos 10°

sin 10° = mg tan 10°= 0.001 kg( ) 9.8 m s2( )tan 10°

Fe ≈ 2 × 10−3 N ~10-3 N or 1 mN

(b) Fe = keq

2 × 10−3 N ≈

8.99 × 109 N ⋅ m2 C2( )q2

0.25 m( )2

q ≈ 1.2 × 10−7 C ~10−7 C or 100 nC

r2 ≈8.99 × 109 N ⋅ m2 C2( ) 1.2 × 10−7 C( )

0.25 m( )2 ≈ 1.7 × 104 N C ~10 kN C

(d) ΦE = q

e0≈ 1.2 × 10−7 C

8.85 × 10−12 C2 N ⋅ m2 = 1.4 × 104 N ⋅ m2 C ~ 10 kN ⋅ m2 C

24.34 (a)

ρ = Q43

πa3= 5.70 × 10−6

43 π(0.0400)3 = 2.13 × 10−2 C / m3

qin = ρ 4

3 πr3( ) = 2.13 × 10−2( ) 43 π( ) 0.0200( )3 = 7.13 × 10−7 C = 713 nC

(b) qin = ρ 4

3 πr3( ) = 2.13 × 10−2( ) 43 π( ) 0.0400( )3 = 5.70 µC

44 Chapter 24 Solutions

24.35 (a) E = 2ke λ

2 8.99 × 109 N ⋅ m2 C2( ) 2.00 × 10−6 C( ) 7.00 m[ ]0.100 m

E = 51.4 kN/C, radially outward

(b) ΦE = EAcosθ = E(2πr )cos 0˚

ΦE = 5.14 × 104 N C( )2π 0.100 m( ) 0.0200 m( ) 1.00( ) = 646 N ⋅ m2 C

24.36 Note that the electric field in each case is directed radially inward, toward the filament.

(a) E = 2keλ

2 8.99 × 109 N ⋅ m2 C2( ) 90.0 × 10−6 C( )0.100 m

= 16.2 MN C

(b) E = 2keλ

2 8.99 × 109 N ⋅ m2 C2( ) 90.0 × 10−6 C( )0.200 m

= 8.09 MN C

2 8.99 × 109 N ⋅ m2 C2( ) 90.0 × 10−6 C( )1.00 m

= 1.62 MN C

24.37 E =

σ2e0

=9.00 × 10−6 C / m2

2(8.85 × 10−12 C2 / N ⋅ m2)= 508 kN/C, upward

24.38 From Gauss's Law, EA = Q

σ = QA = e0 E = (8.85 × 10-12)(130)= 1.15 × 10-9 C/m2 = 1.15 nC/m2

24.39 ∫ο E dA = E(2π rl ) = qin

e0 E =

qin/l

2π e0 r =

λ2π e0 r

(a) r = 3.00 cm E = 0 inside the conductor

(b) r = 10.0 cm E = 30.0 × 10–9

2π (8.85 × 10–12)(0.100) = 5400 N/C, outward

2π (8.85 × 10–12)(1.00) = 540 N/C, outward

Chapter 24 Solutions 45

*24.40 Just above the aluminum plate (a conductor), the electric field is E = ′σ e0 where the charge Qis divided equally between the upper and lower surfaces of the plate:

Thus

′ = ( ) =σQ

AQA

and E = Q

2e0A

For the glass plate (an insulator), E = σ / 2e0 where σ = Q / A since the entire charge Q is onthe upper surface.

Therefore, E = Q

2e0A

The electric field at a point just above the center of the upper surface is the same for each ofthe plates.

E = Q

2e0A, vertically upward in each case (assuming Q > 0)

*24.41 (a) E = σ e0 σ = (8.00 × 104)(8.85 × 10–12) = 7.08 × 10-7 C/m2

σ = 708 nC/m2 , positive on one face and negative on the other.

(b) σ = QA Q = σA = (7.08 × 10–7) (0.500)2 C

Q = 1.77 × 10–7 C = 177 nC , positive on one face and negative on the other.

24.42 Use Gauss's Law to evaluate the electric field in each region, recalling that the electric field iszero everywhere within conducting materials. The results are:

E = 0 inside the sphere and inside the shell

E = ke

Qr2 between sphere and shell, directed radially inward

E = ke

2Qr2 outside the shell, directed radially inward

Charge –Q is on the outer surface of the sphere .

Charge +Q is on the inner surface of the shell ,

46 Chapter 24 Solutions

and +2Q is on the outer surface of the shell.

Chapter 24 Solutions 47

24.43 The charge divides equally between the identical spheres, with charge Q/2 on each. Then theyrepel like point charges at their centers:

F = ke(Q/2)(Q/2)

(L + R + R)2 = ke Q 2

4(L + 2R)2 = 8.99 × 109 N · m2(60.0 × 10-6 C)2

4 C 2(2.01 m)2 = 2.00 N

*24.44 The electric field on the surface of a conductor varies inversely with the radius of curvature ofthe surface. Thus, the field is most intense where the radius of curvature is smallest and vise-versa. The local charge density and the electric field intensity are related by

E = σ

e0 or σ = e0E

(a) Where the radius of curvature is the greatest,

σ =e0Emin = 8.85 × 10−12 C2 N ⋅ m2( ) 2.80 × 104 N C( ) = 248 nC m2

(b) Where the radius of curvature is the smallest,

σ =e0Emax = 8.85 × 10−12 C2 N ⋅ m2( ) 5.60 × 104 N C( ) = 496 nC m2

24.45 (a) Inside surface: consider a cylindrical surface within the metal. Since E inside the conductingshell is zero, the total charge inside the gaussian surface must be zero, so the insidecharge/length = – λ.

0 = λ + qin ⇒ qin = –λ

Outside surface: The total charge on the metal cylinder is 2λl = qin + qout .

qout = 2λl+ λl

so the outside charge/length = 3λ

(b) E = 2ke (3λ)

r = 6ke λ

r =

3λ2πe0r

48 Chapter 24 Solutions

24.46 (a) E = keQ

r2 =8.99 × 109( ) 6.40 × 10−6( )

0.150( )2 = 2.56 MN/C, radially inward

(b) E = 0

Chapter 24 Solutions 49

24.47 (a) The charge density on each of the surfaces (upper and lower) of the plate is:

σ = 1

2qA

= 1

2(4.00 × 10−8 C)

(0.500 m)2 = 8.00 × 10−8 C / m2 = 80.0 nC / m2

(b) E =

σe0

k =8.00 × 10−8 C / m2

8.85 × 10−12 C2 / N ⋅ m2

k = 9.04 kN / C( )k

24. 48 (a) The charge +q at the center induces charge −q on the inner surface of the conductor, where itssurface density is:

σa =

−q4πa2

(b) The outer surface carries charge Q + q with density

σb =

Q + q4πb2

24.49 (a) E = 0

(b) E = keQ

r2 =8.99 × 109( ) 8.00 × 10−6( )

0.0300( )2 = 7.99 × 107 N / C = 79.9 MN/C

(d) E = keQ

r2 =8.99 × 109( ) 4.00 × 10−6( )

0.0700( )2 = 7.34 × 106 N / C = 7.34 MN/C

24.50 An approximate sketch is given at the right. Notethat the electric field lines should be perpendicularto the conductor both inside and outside.

50 Chapter 24 Solutions

24.51 (a) Uniform E, pointing radially outward, so ΦE = EA. The arc length is ds = Rd θ , and the circumference is 2π r = 2π R sin θ

A = 2πrds = (2πRsinθ)Rdθ = 2πR2 sinθdθ

∫0

∫∫ = 2πR2(−cosθ)0θ = 2πR2(1 − cosθ)

ΦE = 1

4πe0

QR2 ⋅ 2πR2(1 − cosθ) =

Q2e0

(1 − cosθ) [independent of R!]

(b) For θ = 90.0° (hemisphere): ΦE = Q

2e0(1 − cos 90°) =

Q2e0

2e0(1 − cos 180°) =

Qe0

[Gauss's Law]

*24.52 In general, E = ay i + bz j + cxk

In the xy plane, z = 0 and E = ay i + cxk

ΦE = E ⋅ dA = ay i + cxk( )∫∫ ⋅ k dA

ΦE = ch x dxx=0

w∫ = ch

2x=0

c hw2

x = 0

x = w

y = 0 y = h

dA = hdx

*24.53 (a) qin = +3Q − Q = +2Q

(b) The charge distribution is spherically symmetric and qin > 0 . Thus, the field is directed

radially outward .

r2 =

2keQr2 for r ≥ c

(d) Since all points within this region are located inside conducting material, E = 0 for b < r < c .

(e) ΦE = E ⋅ dA∫ = 0 ⇒ qin =e0ΦE = 0

(f) qin = +3Q

(g) E = keqin

r2 =

3keQr2 (radially outward) for a ≤ r < b

Chapter 24 Solutions 51

(h) qin = ρV = +3Q

43 πa3

πr3

+3Q

(i) E = keqin

r2 = ke

r2 +3Qr3

3keQ

ra3 (radially outward) for 0 ≤ r ≤ a

(j) From part (d), E = 0 for b < r < c . Thus, for a spherical gaussian surface with b < r < c ,

qin = +3Q + qinner = 0 where qinner is the charge on the inner surface of the conducting shell.

This yields qinner = −3Q

(k) Since the total charge on the conducting shell is

qnet = qouter + qinner = −Q , we have

qouter = −Q − qinner = −Q − −3Q( ) = +2Q

(l) This is shown in the figure to the right.

ra b c

24.54 The sphere with large charge creates a strong field to polarize the other sphere. That means itpushes the excess charge over to the far side, leaving charge of the opposite sign on the nearside. This patch of opposite charge is smaller in amount but located in a stronger externalfield, so it can feel a force of attraction that is larger than the repelling force felt by the largercharge in the weaker field on the other side.

24.55 (a)

E ⋅ dA = E 4πr2( )∫ = qin e0

For r < a, qin = ρ 4

3πr3( ) so

E = pr

3e0

For a < r < b and c < r, qin = Q so that E = Q

4πr2e0

For b ≤ r ≤ c, E = 0, since E = 0 inside a conductor.

(b) Let q1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside theconductor, the total charge enclosed by a spherical surface of radius b ≤ r ≤ c must be zero.

Therefore, q1 + Q = 0 and σ1 = q1

4π b 2 =

– Q4π b 2

Let q2 = induced charge on the outside surface of the hollow sphere. Since the hollow sphereis uncharged, we require q1 + q2 = 0

and σ2 = q1

4πc2 =

Q4πc2

52 Chapter 24 Solutions

24.56

E ⋅ dA∫ = E 4πr2( ) = qin

(a) −3.60 × 103 N C( )4π 0.100 m( )2 = Q

8.85 × 10−12 C2 N ⋅ m2 ( a < r < b)

Q = −4.00 × 10−9 C = −4.00 nC

(b) We take ′Q to be the net charge on the hollow sphere. Outside c,

+2.00 × 102 N C( )4π 0.500 m( )2 = Q + ′Q

8.85 × 10−12 C2 N ⋅ m2 ( r > c)

Q + ′Q = +5.56 × 10−9 C, so ′Q = +9.56 × 10−9 C = +9.56 nC

hollow sphere. Thus, Q1 = −Q = +4.00 nC

Then, if Q2 is the total charge on the outer surface of the hollow sphere,

Q2 = ′Q − Q1 = 9.56 nC − 4.00 nC = +5.56 nC

24.57 The field direction is radially outward perpendicular to the axis. The field strength dependson r but not on the other cylindrical coordinates θ or z. Choose a Gaussian cylinder of radius rand length L. If r < a ,

ΦE = qin

e0and

E 2πrL( ) = λL

E = λ

2πre0or

E = λ

2πre0 (r < a)

If a < r < b, E 2πrL( ) =

λL + ρπ r2 − a2( )Le0

E =

λ + ρπ r2 − a2( )2πre0

a < r < b( )

If r > b , E 2πrL( ) =

λL + ρπ b2 − a2( )Le0

Chapter 24 Solutions 53

E =

λ + ρπ b2 − a2( )2πre0

( r > b )

54 Chapter 24 Solutions

24.58 Consider the field due to a single sheet and let E+and E– represent the fields due to the positive andnegative sheets. The field at any distance from eachsheet has a magnitude given by Equation 24.8:

E+ = E– =

σ2e0

(a) To the left of the positive sheet, E+ is directedtoward the left and E– toward the right and the netfield over this region is E = 0 .

(b) In the region between the sheets, E+ and E– are both directed toward the right and the net fieldis

E =

σe 0

toward the right

24.59 The magnitude of the field due to each sheet given by Equation 24.8is

E = σ

2e0 directed perpendicular to the sheet.

(a) In the region to the left of the pair of sheets, both fields are directedtoward the left and the net field is

E =

σe0

to the left

(b) In the region between the sheets, the fields due to the individual sheets are oppositely directedand the net field is

E = 0

E =

σe0

to the right

Chapter 24 Solutions 55

Goal Solution Repeat the calculations for Problem 58 when both sheets have positive uniform charge densities of valueσ. Note: The new problem statement would be as follows: Two infinite, nonconducting sheets of chargeare parallel to each other, as shown in Figure P24.58. Both sheets have positive uniform charge densitiesσ. Calculate the value of the electric field at points (a) to the left of, (b) in between, and (c) to the right ofthe two sheets.

G : When both sheets have the same charge density, a positive test charge at a point midway betweenthem will experience the same force in opposite directions from each sheet. Therefore, the electricfield here will be zero. (We should ask: can we also conclude that the electron will experience equaland oppositely directed forces everywhere in the region between the plates?)

Outside the sheets the electric field will point away and should be twice the strength due to one sheetof charge, so E = σ /e0 in these regions.

O : The principle of superposition can be applied to add the electric field vectors due to each sheet ofcharge.

A : For each sheet, the electric field at any point is E = σ (2e0 ) directed away from the sheet.

(a) At a point to the left of the two parallel sheets E i i i= − + − = −E E E1 2 2( ) ( ) ( ) = − σ

e0i

(b) At a point between the two sheets E i i= + − =E E1 2 0( )

L : We essentially solved this problem in the Gather information step, so it is no surprise that theseresults are what we expected. A better check is to confirm that the results are complementary to thecase where the plates are oppositely charged (Problem 58).

24.60 The resultant field within the cavity is the superposition oftwo fields, one E+ due to a uniform sphere of positive chargeof radius 2a , and the other E− due to a sphere of negativecharge of radius a centered within the cavity.

πr3ρe0

= 4πr2E+ so E+ = ρ r

3e0 = ρr

3e0

–

πr13ρ

e0= 4πr1

2E− so E− = ρ r1

3e0(− 1

) = −ρ

3e0r1

Since r = a + r1 , E− = −ρ(r − a)

3e0

E = E+ + E− = ρr

3e0− ρr

3e0+ ρa

3e0= ρa

3e0= 0i + ρa

3e0j

Thus, Ex = 0 and Ey = ρa

3e0 at all points within the cavity.

56 Chapter 24 Solutions

24.61 First, consider the field at distance r < R from the center of a uniform sphere of positivecharge Q = +e( ) with radius R .

4πr2( )E = qin

e0= ρVe0

= +e43 πR3

43 πr3

e0so

E = e

4πe0R3

r directed outward

(a) The force exerted on a point charge q = −e located at distance r from the center is then

F = qE = −e

e4πe0R3

r = − e2

4πe0R3

r = −Kr

(b) K = e2

4πe0R3 =

kee2

r , so

ar = − kee

meR3

r = −ω2r

Thus, the motion is simple harmonic with frequency f = ω

2π=

12π

kee2

meR3

(d)

f = 2.47 × 1015 Hz = 12π

8.99 × 109 N ⋅ m2 C2( ) 1.60 × 10−19 C( )2

9.11× 10−31 kg( )R3

which yields R3 = 1.05 × 10−30 m3, or R = 1.02 × 10−10 m = 102 pm

24.62 The electric field throughout the region is directed along x;therefore, E will be perpendicular to dA over the four faces ofthe surface which are perpendicular to the yz plane, and E willbe parallel to dA over the two faces which are parallel to the yzplane. Therefore,

ΦE = − Ex x=a( )A + Ex x=a+c( )A

= − 3 + 2a2( )ab + 3 + 2(a + c)2( )ab = 2abc(2a + c)

Substituting the given values for a, b, and c, we find ΦE = 0.269 N · m2/C

Q = ∈ 0ΦE = 2.38 × 10-12 C = 2.38 pC

24.63

E ⋅ dA∫ = E(4πr2 ) = qin

(a) For r > R, qin = Ar2

∫ (4πr2 )dr = 4π AR5

5 and E =

AR5

5e0r2

(b) For r < R, qin = Ar2

∫ (4πr2 )dr = 4πAr5

5 and E =

Ar3

5e0

Chapter 24 Solutions 57

24.64 The total flux through a surface enclosing the charge Q is Q/ e0 . The flux through the disk is

Φdisk = E ⋅ dA∫where the integration covers the area of the disk. We must evaluate this integral and set itequal to

14 Q/ e0 to find how b and R are related. In the figure, take dA to be the area of an

annular ring of radius s and width ds. The flux through dA is

E · dA = E dA cos θ = E (2π sds) cos θ

The magnitude of the electric field has the same value at all points withinthe annular ring,

E = 1

4πe0

Qr2 = 1

4πe0

Qs2 + b2 and

cosθ = b

r= b

(s2 + b2 )1/2

Integrate from s = 0 to s = R to get the flux through the entire disk.

ΦE, disk = Qb

2e0

s ds(s2 + b2 )3/20

R∫ = Qb

2e0−(s2 + b2 )1/2[ ]

R= Q

2e01 − b

(R2 + b2 )1/2

The flux through the disk equals Q/4 e0 provided that

b(R2 + b2 )1/2 = 1

This is satisfied if R = 3 b .

24.65

E ⋅ dA∫ = qin

e0= 1e0

4πr2dr0

∫

E4πr2 = 4πa

e0r dr

∫ = 4πae0

E = a

2e0 = constant magnitude

(The direction is radially outward from center for positive a; radially inward for negative a.)

58 Chapter 24 Solutions

24.66 In this case the charge density is not uniform, and Gauss's law is written as

E ⋅ dA = 1e0

ρ dV∫∫ .

We use a gaussian surface which is a cylinder of radius r, length , and is coaxial with thecharge distribution.

(a) When r < R, this becomes E( 2πrl) =

ρ0

e0a − r

∫ dV . The element of volume is a cylindrical

shell of radius r, length l, and thickness dr so that dV = 2πrl dr.

` E 2πrl( ) = 2πr2lρ0

− r3b

so inside the cylinder, E =

ρ0r2e0

a − 2r3b

(b) When r > R, Gauss's law becomes

E 2πrl( ) = ρ0

e0a − r

∫ 2πrldr( ) or outside the cylinder, E =

ρ0R2

2e0ra − 2R

24.67 (a) Consider a cylindrical shaped gaussian surface perpendicular tothe yz plane with one end in the yz plane and the other endcontaining the point x :

Use Gauss's law:

E ⋅ dA∫ = qin

By symmetry, the electric field is zero in the yz plane and isperpendicular to dA over the wall of the gaussian cylinder.Therefore, the only contribution to the integral is over the end capcontaining the point x :

E ⋅ dA∫ = qin

e0 or

EA = ρ Ax( )

so that at distance x from the mid-line of the slab, E = ρx

z x

gaussiansurface

(b) a = F

me= −e( )E

me= − ρe

mee0

The acceleration of the electron is of the form a = −ω2x with ω = ρe

mee0

Thus, the motion is simple harmonic with frequency f = ω

2π=

12π

ρemee0

Chapter 24 Solutions 59

24.68 Consider the gaussian surface described in the solution to problem 67.

(a) For x > d

2, dq = ρ dV = ρA dx = C Ax2 dx

E ⋅ dA = 1

e0dq∫∫

EA = CA

e0 x2 dx

d/2

∫ = 1

3CAe0

E = Cd3

24e0or

E = Cd3

24e0i for

d>2

; E = − Cd3

24e0i for

d< −2

(b) For − < <d

2 2

E ⋅ dA = 1e0

dq∫∫ = C Ae0

x2 dx = C Ax3

3e00

∫

E = C x3

3e0i for x > 0;

E = − Cx3

3e0i for x < 0

24.69 (a) A point mass m creates a gravitational acceleration g = − Gm

r2 at a distance r.

The flux of this field through a sphere is

g∫ ⋅ dA = − Gmr2 4πr2( ) = − 4πGm

Since the r has divided out, we can visualize the field as unbroken field lines. The same fluxwould go through any other closed surface around the mass. If there are several or no massesinside a closed surface, each creates field to make its own contribution to the net fluxaccording to

g ⋅ dA = − 4πGmin∫

(b) Take a spherical gaussian surface of radius r. The field is inward so

g∫ ⋅ dA = g 4πr2 cos 180° = − g 4πr 2

and − 4πGmin = −4G 43 πr3ρ

Then, − g 4πr2 = − 4πG 43 πr3ρ and g = 4

3 πrρG

Or, since ρ = ME / 43 πRE

3, g =

MEGr

RE3 or

g = MEGr

RE3 inward

Chapter 25 Solutions

25.1 ∆V = –14.0 V and

Q = –NA e = – (6.02 × 1023)(1.60 × 10–19 C) = – 9.63 × 104 C

∆V = WQ , so W = Q(∆V) = (– 9.63 × 104 C)(–14.0 J/C) = 1.35 ΜJ

25.2 ∆K = q∆ V 7.37 × 10-17 = q(115)

q = 6.41 × 10-19 C

25.3 W = ∆K = q∆ V

12 mv2 = e(120 V) = 1.92 × 10–17 J

Thus, v = 3.84 × 10−17 J

(a) For a proton, this becomes v = 3.84 × 10−17 J

1.67 × 10−27 kg = 1.52 × 105 m/s = 152 km/s

(b) If an electron, v = 3.84 × 10−17 J

9.11× 10−31 kg = 6.49 × 106 m/s = 6.49 Mm/s

Goal Solution (a) Calculate the speed of a proton that is accelerated from rest through a potential difference of 120 V.(b) Calculate the speed of an electron that is accelerated through the same potential difference.

G : Since 120 V is only a modest potential difference, we might expect that the final speed of the particleswill be substantially less than the speed of light. We should also expect the speed of the electron to besignificantly greater than the proton because, with me << mp , an equal force on both particles willresult in a much greater acceleration for the electron.

O : Conservation of energy can be applied to this problem to find the final speed from the kinetic energyof the particles. (Review this work-energy theory of motion from Chapter 8 if necessary.)

Chapter 25 Solutions 57

A : (a) Energy is conserved as the proton moves from high to low potential, which can be defined forthis problem as moving from 120 V down to 0 V:

Ki +Ui + ∆Enc = Kf +U f

0 + qV + 0 = 12 mvp

2 + 0

(1.60 × 10−19 C)(120 V)

1 J1 V ⋅ C

= 1

2 (1.67 × 10−27 kg)vp2

vp = 1.52 × 105 m / s

(b) The electron will gain speed in moving the other way, from Vi = 0 to Vf = 120 V:

Ki +Ui + ∆Enc = Kf +U f

0 + 0 + 0 = 12 mve

2 + qV

0 = 12 (9.11× 10−31 kg)ve

2 + (−1.60 × 10−19 C)(120 J / C)

ve = 6.49 × 106 m / s

L : Both of these speeds are significantly less than the speed of light as expected, which also means thatwe were justified in not using the relativistic kinetic energy formula. (For precision to threesignificant digits, the relativistic formula is only needed if v is greater than about 0.1 c.)

25.4 For speeds larger than one-tenth the speed of light, 12 mv2 gives noticeably wrong answers for

kinetic energy, so we use

K = mc2 1

1 − v2 / c2− 1

= 9.11× 10−31 kg( ) 3.00 × 108 m / s( )2 1

1 − 0.4002− 1

= 7.47 × 10–15 J

Energy is conserved during acceleration: Ki + Ui + ∆E = Kf + Uf

0 + qVi + 0 = 7.47 × 10–15 J + qVf

The change in potential is Vf – Vi : Vf – Vi = –7.47 × 10–15 J

q = – 7.47 × 10–15 J–1.60 × 10–19 C

= + 46.7 kV

The positive answer means that the electron speeds up in moving toward higher potential.

25.5 W = ∆K = − q∆V

0 − 1

2 9.11× 10−31 kg( ) 4.20 × 105 m / s( )2 =

− − 1.60 × 10−19 C( )∆V

From which, ∆V = – 0.502 V

58 Chapter 25 Solutions

*25.6 (a) We follow the path from (0, 0) to (20.0 cm, 0) to (20.0 cm, 50.0 cm).

∆U = – (work done)

∆U = − work from origin to (20.0 cm,0) ( ) − work from (20.0 cm,0) to (20.0 cm,50.0 cm) ( )

Note that the last term is equal to 0 because the force is perpendicular to the displacement.

∆U = – (qEx)(∆x) = – (12.0 × 10–6 C)(250 V/m)(0.200 m) = – 6.00 × 10–4 J

(b) ∆V = ∆Uq = –

6.00 × 10–4 J12.0 × 10–6 C

= – 50.0 J/C = – 50.0 V

*25.7 E =

∆ Vd

= 25.0 × 103 J/C1.50 × 10–2 m

= 1.67 × 106 N/C = 1.67 MN/C

*25.8 (a) ∆ V = Ed = (5.90 × 103 V/m)(0.0100 m) = 59.0 V

(b)12 mvf

2 = q(∆V) ; 12 (9.11 × 10–31) vf

2 = (1.60 × 10–19)(59.0)

vf = 4.55 × 106 m/s

25.9 ∆U = − 1

2m vf

2 − vi2( ) = − 1

29.11× 10−31 kg

1.40 × 105 m / s( )2

− 3.70 × 106 m / s( )2

= 6.23 × 10−18 J

∆U = q ∆V: + 6.23 × 10–18 = (–1.60 × 10–19)∆V

∆V = – 38.9 V The origin is at higher potential.

Chapter 25 Solutions 59

*25.10 VB − VA = −

∫ E ⋅ ds = −A

∫ E ⋅ ds −C

∫ E ⋅ ds

VB − VA = (−Ecos180°)

−0.300

0.500

∫ dy − (Ecos90.0°)−0.200

0.400

∫ dx

VB – VA = (325)(0.800) = + 260 V

25.11 (a) Arbitrarily choose V = 0 at x = 0. Then at other points,

V = − Ex and Ue = QV = −QEx . Between the endpoints of themotion,

(K + Us + Ue )i = (K + Us + Ue ) f

0 + 0 + 0 = 0 + 12 kxmax

2 − QExmax

so the block comes to rest when the spring is stretched by an amount

xmax = 2QE

2 50.0 × 10−6 C( ) 5.00 × 105 V m( )100 N m

= 0.500 m

(b) At equilibrium, ΣFx = − Fs + Fe = 0 or kx = QE. Thus, the equilibrium position is at

x = QE

50.0 × 10−6 C( ) 5.00 × 105 N C( )100 N m

= 0.250 m

d2xdt2 . Let

′x = x − QE

k, or x = ′x + QE

kso the equation of motion becomes:

−k ′x + QE

+ QE = m

d2 ′x + QE k( )dt2 , or

d2 ′xdt2 = − k

′x

This is the equation for simple harmonic motion a ′x = −ω2 ′x( ), with ω = k m . The period of

the motion is then

T = 2π

ω= 2π m

k= 2π 4.00 kg

100 N m= 1.26 s

(d) (K + Us + Ue )i + ∆E = (K + Us + Ue ) f

0 + 0 + 0 − µkmgxmax = 0 + 12 kxmax

2 − QExmax

60 Chapter 25 Solutions

xmax = 2(QE − µkmg)

2 50.0 × 10−6 C( ) 5.00 × 105 N C( ) − 0.200 4.00 kg( ) 9.80 m s2( )[ ]100 N m

= 0.343 m

25.12 (a) Arbitrarily choose V = 0 at 0. Then at other points V = − Ex and

Ue = QV = −QEx . Between the endpoints of the motion,

(K + Us + Ue )i = (K + Us + Ue ) f

0 + 0 + 0 = 0 + 12 kxmax

2 − QExmax so xmax =

2QEk

(b) At equilibrium, ΣFx = − Fs + Fe = 0 or kx = QE. So the equilibrium position is at x =

QEk

d2xdt2 . Let

′x = x − QE

k, or

x = ′x + QE

k, so

the equation of motion becomes:

−k ′x + QE

+ QE = m

d2 ′x + QE k( )dt2 , or

d2 ′xdt2 = − k

′x

This is the equation for simple harmonic motion a ′x = −ω2 ′x( ), with ω = k m

The period of the motion is then T = 2π

ω=

2π m

(d) (K + Us + Ue )i + ∆E = (K + Us + Ue ) f

0 + 0 + 0 − µkmgxmax = 0 + 12 kxmax

2 − QExmax

xmax =

2(QE − µkmg)k

25.13 For the entire motion, y − yi = vyit + 12 ayt2

0 − 0 = vit + 12 ayt2 so

ay = − 2vi

∑ Fy = may : −mg − qE = − 2mvi

E = m

q2vi

t− g

and

E = − m

q2vi

t− g

For the upward flight: vyf2 = vyi

2 + 2ay(y − yi )

0 = vi

2 + 2 − 2vi

(ymax − 0) and ymax = 1

4 vit

Chapter 25 Solutions 61

∆V = E ⋅ dy = + mq

2vi

t− g

ymax∫0

ymax

= mq

2vi

t− g

14 vit( )

∆V = 2.00 kg

5.00 × 10−6 C2(20.1 m s)

4.10 s− 9.80 m s2

14 (20.1 m s) 4.10 s( )[ ] = 40.2 kV

62 Chapter 25 Solutions

25.14 Arbitrarily take V = 0 at the initial point. Then at distance d downfield, where L is the rodlength, V = − Ed and Ue = − λLEd

(a) (K + U)i = (K + U) f

0 + 0 = 12 µLv 2 − λLEd

v =

2λEdµ

2(40.0 × 10−6 C / m)(100 N / C)(2.00 m)(0.100 kg / m)

= 0.400 m/s

(b) The same.

25.15 Arbitrarily take V = 0 at point P. Then (from Equation 25.8) the potential at the originalposition of the charge is – E · s = –EL cos θ. At the final point a, V = –EL. Suppose the table isfrictionless: (K + U)i = (K + U) f

0 − qEL cosθ = 12 mv 2 − qEL

v = = 2qEL(1 − cosθ)

m= 2(2.00 × 10−6 C)(300 N / C)(1.50 m)(1 − cos 60.0°)

0.0100 kg = 0.300 m/s

*25.16 (a) The potential at 1.00 cm is

V1 = ke qr =

(8.99 × 109 N · m2/C 2)(1.60 × 10–19 C)1.00 × 10–2 m

= 1.44 × 10–7 V

(b) The potential at 2.00 cm is

V2 = ke qr =

(8.99 × 109 N · m2/C 2)(1.60 × 10–19 C)2.00 × 10–2 m

= 0.719 × 10–7 V

Thus, the difference in potential between the two points is

∆V = V2 – V1 = –7.19 × 10–8 V

(c) The approach is the same as above except the charge is – 1.60 × 10–19 C. This changes the signof all the answers, with the magnitudes remaining the same.

That is, the potential at 1.00 cm is –1.44 × 10–7 V

The potential at 2.00 cm is – 0.719 × 10–7 V, so ∆V = V2 – V1 = 7.19 × 10–8 V .

Chapter 25 Solutions 63

25.17 (a) Since the charges are equal and placed symmetrically, F = 0

(b) Since F = qE = 0, E = 0

8.99 × 109 N · m2

C 2

2.00 × 10–6 C

0.800 m

V = 4.50 × 104 V = 45.0 kV

25.18 (a) Ex = ke q1

x2 + ke q2

(x – 2.00)2 = 0 becomes Ex = ke

x2 + –2q

(x – 2.00)2 = 0

Dividing by ke, 2qx2 = q(x – 2.00)2

x2 + 4.00x – 4.00 = 0

Therefore E = 0 when x = −4.00 ± 16.0 + 16.0

2= – 4.83 m

(Note that the positive root does not correspond to a physically valid situation.)

(b) V = ke q1

x + ke q2

(2.00 – x) = 0 or V = ke

x – 2q

(2.00 – x) = 0

Again solving for x, 2qx = q(2.00 – x)

For 0 ≤ x ≤ 2.00 V = 0 when x = 0.667 m

andq

x =

–2q

2 – x

For x < 0 x = –2.00 m

25.19 (a) U = keq1q2

r= −(8.99 × 109)(1.60 × 10−19)2

0.0529 × 10−9 = – 4.35 × 10–18 J = –27.2 eV

(b) U = ke q1q2

r = – (8.99 × 109)(1.60 × 10–19)2

22(0.0529 × 10–9) = – 6.80 eV

64 Chapter 25 Solutions

r = – ke e 2

∞ = 0

Goal Solution The Bohr model of the hydrogen atom states that the single electron can exist only in certain allowedorbits around the proton. The radius of each Bohr orbit is r = n2(0.0529 nm) where n = 1, 2, 3, . . . .Calculate the electric potential energy of a hydrogen atom when the electron is in the (a) first allowedorbit, n = 1; (b) second allowed orbit, n = 2; and (c) when the electron has escaped from the atom ( r = ∞).Express your answers in electron volts.

G : We may remember from chemistry that the lowest energy level for hydrogen is E1 = −13.6 eV, andhigher energy levels can be found from En = E1 / n2 , so that E2 = −3.40 eV and E∞ = 0 eV. (see section42.2) Since these are the total energies (potential plus kinetic), the electric potential energy aloneshould be lower (more negative) because the kinetic energy of the electron must be positive.

O : The electric potential energy is given by U = ke

q1q2

A : (a) For the first allowed Bohr orbit,

U = 8.99 × 109

N ⋅ m2

(−1.60 × 10−19 C)(1.60 × 10−19 C)

(0.0529 × 10−9 m) = −4.35 × 10−18 J = −4.35 × 10−18 J

1.60 × 10−19 J / eV= −27.2 eV

(b) For the second allowed orbit,

U = (8.99 × 109 N ⋅ m2 / C2 )

(−1.60 × 10−19 C)(1.60 × 10−19 C)22(0.0529 × 10−9 m) = −1.088 × 10−18 J = −6.80 eV

U = 8.99 × 109 N ⋅ m2/ C2( ) −1.60 × 10−19 C( ) 1.60 × 10−19 C( )

∞ = 0 J

L : The potential energies appear to be twice the magnitude of the total energy values, so apparently thekinetic energy of the electron has the same absolute magnitude as the total energy.

*25.20 (a) U = qQ

4πe0r =

(5.00 × 10–9 C)(– 3.00 × 10–9 C)(8.99 × 109 V · m/C)(0.350 m) = – 3.86 × 10–7 J

The minus sign means it takes 3.86 × 10–7 J to pull the two charges apart from 35 cm to a muchlarger separation.

(b) V = Q1

4πe0r1+ Q2

4πe0r2 = (5.00 × 10−9 C)(8.99 × 109 V ⋅ m / C)

0.175 m+ (−3.00 × 10−9 C)(8.99 × 109 V ⋅ m / C)

0.175 m

V = 103 V

Chapter 25 Solutions 65

25.21 V = ∑i k

qiri

V = (8.99 × 109)(7.00 × 10–6)

–1

0.0100 – 1

0.0100 + 1

0.0387

V = –1.10 × 107 C = –11.0 MV

66 Chapter 25 Solutions

*25.22 Ue = q4V1 + q4V2 + q4V3 = q4 1

4π ∈ 0

r1 +

q2r2

+ q3r3

= 10.0 × 10−6 C( )28.99 × 109 N ⋅ m2/ C2( ) 1

0.600 m+ 1

0.150 m+ 1

0.600 m( )2 + 0.150 m( )2

Ue= 8.95 J

25.23 U = U1 + U2 + U3 + U4

U = 0 + U12 + (U13 + U23) + (U14 + U24 + U34)

U = 0 + keQ

s+ keQ

s12

+ 1

+ keQ2

s1 + 1

2+ 1

U = keQ

s4 + 2

= 5.41 ke Q2

An alternate way to get the term 4 + 2 2( ) is to recognize that there are 4 side pairs and 2 facediagonal pairs.

*25.24 (a) V = keq1

r1+ keq2

r2= 2

keqr

= 28.99 × 109 N ⋅ m2 C2( ) 2.00 × 10−6 C( )

1.00 m( )2 + 0.500 m( )2

V = 3.22 × 104 V = 32.2 kV

(b) U = qV = −3.00 × 10−6 C( ) 3.22 × 104

= −9.65 × 10−2 J

2.00 µC 2.00 µC

P (0, 0.500 m)

(1.00 m, 0)(-1.00 m, 0)

*25.25 Each charge creates equal potential at the center. The total potential is:

V = 5

ke −q( )R

− 5keq

Chapter 25 Solutions 67

*25.26 (a) Each charge separately creates positive potential everywhere. The total potential produced bythe three charges together is then the sum of three positive terms. There is no point located

at a finite distance from the charges, where this total potential is zero.

(b) V = ke q

a+ ke q

2keqa

25.27 (a) Conservation of momentum: 0 = m1v1 i + m2v2 (–i) or v2 = m1v1m 2

By conservation of energy, 0 + ke (–q1)q2

d = 12 m 1v2

1 + 12 m 2v2

2 + ke(–q1)q2(r1 + r2)

and ke q1q2r1 + r2

– ke q1q2

d = 12 m 1v2

1 + 12 m 2

1v21

m 2

v1 = 2m2keq1q2

m1 m1 + m2( )1

r1 + r2− 1

v1 =

2 0.700 kg( ) 8.99 × 109 N ⋅ m2/ C2( ) 2 × 10−6 C( ) 3 × 10−6 C( )0.100 kg( ) 0.800 kg( )

18 × 10−3 m

− 11.00 m

= 10.8 m/s

v2 = m 1v1m 2

= (0.100 kg)(10.8 m/s)

0.700 kg = 1.55 m/s

(b) If the spheres are metal, electrons will move around on them with negligible energy loss toplace the centers of excess charge on the insides of the spheres. Then just before they touch,the effective distance between charges will be less than r1 + r2 and the spheres will really bemoving faster than calculated in (a) .

25.28 (a) Conservation of momentum: 0 = m1v1 i + m2v2 (− i) or v2 = m1v1 / m2

By conservation of energy, 0 +

ke(− q1)q2

d= 1

2 m1v 12 + 1

2 m2v 22 +

ke(− q1)q2

(r1 + r2 )

and

keq1q2

r1 + r2− keq1q2

m1v 12 +

m12v1

v1 =

2m2keq1q2

m1(m1 + m2 )1

r1 + r2− 1

v2 = m1

v1 =

2m1keq1q2

m2(m1 + m2 )1

r1 + r2− 1

moving faster than calculated in (a) .

68 Chapter 25 Solutions

25.29 V = keQ

rso

r = keQ

8.99 × 109 N ⋅ m2 C2( ) 8.00 × 10−9 C( )V

= 72.0 V ⋅ mV

For V = 100 V, 50.0 V, and 25.0 V, r = 0.720 m, 1.44 m, and 2.88 m

The radii are inversely proportional to the potential.

25.30 (a)

V x( ) = keQ1

r1+ keQ2

r2=

ke +Q( )x2 + a2

+ke +Q( )

x2 + −a( )2

V x( ) = 2keQ

x2 + a2= keQ

x a( )2 + 1

V x( )keQ a( ) =

x a( )2 + 1

(b) V y( ) = keQ1

r1+ keQ2

r2=

ke +Q( )y − a

+ ke −Q( )y + a

V y( ) = keQ

y a − 1− 1

y a + 1

V y( )keQ a( ) =

1y a − 1

− 1y a + 1

25.31 Using conservation of energy, we have Kf +U f = Ki +Ui .

But Ui =

keqα qgold

ri , and ri ≈ ∞ . Thus, Ui = 0.

Also Kf = 0 ( vf = 0 at turning point), so U f = Ki , or

keqα qgold

rmin= 1

2 mα vα2

rmin =

2keqα qgold

mα vα2 =

2(8.99 × 109 N ⋅ m2 / C2 )(2)(79) 1.60 × 10-19 C( )2

(6.64 × 10−27 kg)(2.00 × 107 m / s)2 = 2.74 × 10−14 m = 27.4 fm

Chapter 25 Solutions 69

25.32 Using conservation of energy

we have:

keeQr1

keeQr2

+ 12 mv2

which gives: v =

2keeQm

1r1

− 1r2

or v =

(2)(8.99 × 109 N ⋅ m2 / C2)(−1.60 × 10−19 C)(10−9 C)9.11× 10-31 kg

10.0300 m

− 1

0.0200 m

Thus, v = 7.26 × 106 m / s

25.33 U =

keqiqj

ri j∑ , summed over all pairs of i, j( ) where i ≠ j

U = ke

q −2q( )b

+−2q( ) 3q( )

2q( ) 3q( )b

+q 2q( )

q 3q( )a2 + b2

+2q −2q( )

a2 + b2

U = keq

2 −20.400

− 60.200

+ 60.400

+ 20.200

+ 30.447

− 40.447

U = 8.99 × 109( ) 6.00 × 10−6( )2 4

0.400− 4

0.200− 1

0.447

= – 3.96 J

25.34 Each charge moves off on its diagonal line. All charges have equal speeds.

(K + U)i∑ = (K + U) f∑

0 +

4 keq2

L+

2 keq2

2 L= 4 1

2 mv 2( ) +4 keq

2L+

2keq2

2 2 L

2 +

ke q2

L= 2mv 2

v = 1 +

keq2

70 Chapter 25 Solutions

25.35 A cube has 12 edges and 6 faces. Consequently, there are 12 edge pairs separated by s, 2 × 6 = 12

face diagonal pairs separated by 2 s, and 4 interior diagonal pairs separated 3 s.

U = ke q2

s12 + 12

2+ 4

= 22.8

ke q2

25.36 V = a + bx = 10.0 V + −7.00 V m( )x

(a) At x = 0, V = 10.0 V

At x = 3.00 m, V = − 11.0 V

At x = 6.00 m, V = − 32.0 V

(b) E = − dV

dx= −b = − −7.00 V m( ) = 7.00 N C in + x direction

25.37 V = 5x – 3x2y + 2yz2 Evaluate E at (1, 0 – 2)

Ex = – ∂V

∂x = – 5 + 6xy = – 5 + 6(1)(0) = – 5

Ey = – ∂V

∂y = +3x2 – 2z2 = 3(1)2 – 2(–2)2 = – 5

Ez = – ∂V

∂ z = – 4yz = – 4(0)(–2) = 0

E = Ex

2 + Ey2 + Ez

2 = −5( )2 + −5( )2 + 02 = 7.07 N/C

25.38 (a) For r < R V = keQ

Er = − dV

dr= 0

(b) For r ≥ R V = keQ

Er = − dV

dr= − − keQ

keQr2

Chapter 25 Solutions 71

25.39

Ey = − ∂V∂y

= − ∂∂y

keQl

lnl+ l2 + y2

Ey = keQly

1 − y2

l2 + y2 +l l2 + y2

keQ

y l2 + y2

25.40 Inside the sphere, Ex = Ey = Ez = 0 .

Outside, Ex = − ∂V

∂x= − ∂

∂xV0 − E0z + E0 a3z(x 2 + y2 + z 2 )− 3/2( )

So Ex = − 0 + 0 + E0 a3z(− 3 / 2)(x 2 + y2 + z2 )− 5/2 (2x)[ ] = 3E0 a3xz(x 2 + y2 + z2 )− 5/2

Ey = − ∂V

∂y= − ∂

∂yV0 − E0 z + E0a3z(x 2 + y2 + z2 )− 3/2( )

Ey= − E0a3z(− 3 / 2)(x 2 + y2 + z2 )− 5/2 2y = 3E0a3yz(x 2 + y2 + z2 )− 5/2

Ez = − ∂V

∂ z = E0 − E0a3z(−3 / 2)(x 2 +y 2 +z 2 )− 5/2 (2z) −E0a3(x 2 +y2 +z 2 )− 3/2

Ez = E0 + E0a3(2z 2 −x 2 −y 2 )(x 2 +y 2 +z 2 )− 5/2

*25.41

∆V = V2R − V0 = keQ

R2 + 2R( )2− keQ

R= keQ

R15

− 1

= – 0.553 keQR

*25.42 V = dV∫ = 1

4πe0

dqr∫

All bits of charge are at the same distance from O, so

V = 1

4πe0

= 8.99 × 109 N ⋅ m2

−7.50 × 10−6 C( )

0.140 m / π( ) = –1.51 MV

72 Chapter 25 Solutions

25.43 (a) [α] =

x = Cm ⋅

m = C

m 2

(b) V = ke ⌡⌠

dqr = ke ⌡

⌠

λ dxr

= keα

x dx(d + x)

∫ = keα L − d ln 1 + L

25.44

V = ke dqr∫ = ke

α x dx

b2 + L 2 − x( )2∫

Let z = L

2− x . Then

x = L

2− z , and dx = −dz

V = keα

L 2 − z( )(−dz)

b2 + z2∫ = − keα L2

b2 + z2∫ + keαzdz

b2 + z2∫ = − keαL

2ln(z + z2 + b2 ) + keα z2 + b2

V = − keα L

2ln L 2 − x( ) + L 2 − x( )2 + b2

+ keα L 2 − x( )2 + b2

V = − keα L2

lnL 2 − L + L 2( )2 + b2

L 2 + L 2( )2 + b2

+ keα L 2 − L( )2 + b2 − L 2( )2 + b2

V =

− keα L2

lnb2 + (L2 4) − L 2

b2 + (L2 4) + L 2

25.45

dV = ke dq

r2 + x2where dq = σ dA = σ2πr dr

V = 2πσke

r dr

r2 + x2a

∫ = 2πkeσ x2 + b2 − x2 + a2

25.46 V = ke

dqrall charge∫

= ke

λ dx−x

+ keλ ds

R+ ke

λ dxxR

3R∫semicircle∫−3R

−R∫

V = − ke λ ln(− x)

− 3R

− R+

ke λR

πR + ke λ ln xR

V = ke λ ln

3RR

+ ke λπ + ke λ ln 3 = ke λ (π + 2 ln 3)

Chapter 25 Solutions 73

25.47 Substituting given values into V = ke qr , 7.50 × 103 V =

(8.99 × 109 N·m2/C2) q(0.300 m)

Substituting q = 2.50 × 10−7 C, N = 2.50 × 10-7 C

1.60 × 10-19 C/e− = 1.56 × 1012 electrons

25.48 q1 + q2 = 20.0 µC so q1 = 20.0 µC – q2

q1q2

= r1r2

so20.0 µC – q2

q2 =

4.00 cm6.00 cm

Therefore 6.00(20.0 µC – q2) = 4.00q2 ;

Solving, q2 = 12.0 µC and q1 = 20.0 µC – 12.0 µC = 8.00 µC

(a) E1 = keq1

r12 =

8.99 × 109( ) 8.00 × 10−6( )0.0400( )2 = 4.50 × 107 V / m = 45.0 MV/m

E2 = keq2

r22 =

8.99 × 109( ) 12.0 × 10−6( )0.0600( )2 = 3.00 × 107 V / m = 30.0 MV/m

(b) V1 = V2 = keq2

r2= 1.80 MV

25.49 (a) E = 0 ; V = keq

R= (8.99 × 109)(26.0 × 10−6 )

0.140= 1.67 MV

(b) E = keq

r2 = (8.99 × 109)(26.0 × 10−6 )(0.200)2 = 5.84 MN/C away

V = keq

r= (8.99 × 109)(26.0 × 10−6 )

(0.200)= 1.17 MV

R2 = (8.99 × 109)(26.0 × 10−6 )(0.140)2 = 11.9 MN/C away

V = keq

R= 1.67 MV

25.50 No charge stays on the inner sphere in equilibrium. If there were any, it would create anelectric field in the wire to push more charge to the outer sphere. Charge Q is on the outer

sphere. Therefore, zero charge is on the inner sphere and 10.0 µC is on the outer sphere .

74 Chapter 25 Solutions

25.51 (a) Emax = 3.00 × 106 V / m = keQ

r2 = keQr

= Vmax1r

Vmax = Emaxr = 3.00 × 106(0.150) = 450 kV

(b)

keQmax

r2 = Emax or keQmax

r= Vmax

Qmax = Emaxr2

ke= 3.00 × 106(0.150)2

8.99 × 109 = 7.51 µC

Goal Solution Consider a Van de Graaff generator with a 30.0-cm-diameter dome operating in dry air. (a) What is themaximum potential of the dome? (b) What is the maximum charge on the dome?

G : Van de Graaff generators produce voltages that can make your hair stand on end, somewhere on theorder of about 100 kV (see the Puzzler at beginning of Chapter 25). With these high voltages, themaximum charge on the dome is probably more than typical point charge values of about 1 µC.

The maximum potential and charge will be limited by the electric field strength at which the airsurrounding the dome will ionize. This critical value is determined by the dielectric strength of airwhich, from page 789 or from Table 26.1, is Ecritical = 3 × 106 V / m. An electric field stronger than thiswill cause the air to act like a conductor instead of an insulator. This process is called dielectricbreakdown and may be seen as a spark.

O : From the maximum allowed electric field, we can find the charge and potential that would create thissituation. Since we are only given the diameter of the dome, we will assume that the conductor isspherical, which allows us to use the electric field and potential equations for a spherical conductor.With these equations, it will be easier to do part (b) first and use the result for part (a).

A : (b) For a spherical conductor with total charge Q , E = keQ

Q = Er2

ke=

3.00 × 106 V / m( ) 0.150 m( )2

8.99 × 109 N ⋅ m2 / C2 1 N ⋅ m / V ⋅ C( ) = 7.51 µC

(a) V = keQ

r= (8.99 × 109 N ⋅ m2 / C2 )(7.51× 10−6 C)

0.150 m= 450 kV

L : These calculated results seem reasonable based on our predictions. The voltage is about 4000 timeslarger than the 120 V found from common electrical outlets, but the charge is similar in magnitude tomany of the static charge problems we have solved earlier. This implies that most of these chargeconfigurations would have to be in a vacuum because the electric field near these point chargeswould be strong enough to cause sparking in air. (Example: A charged ball with Q = 1 µC and r = 1 mm would have an electric field near its surface of

E = keQ

r2 =9 × 109 N ⋅ m2/ C2( ) 1× 10−6 C( )

0.001 m( )2 = 9 × 109 V / m

which is well beyond the dielectric breakdown of air!)

Chapter 25 Solutions 75

25.52 V = ke q

r and E = ke qr2 Since E =

Vr ,

(b) r = VE =

6.00 × 105 V3.00 × 106 V/m

= 0.200 m and

(a) q = Vrke

= 13.3 µC

25.53 U = qV = ke q1q2r12

= (8.99 × 109) (38)(54)(1.60 × 10–19)2

(5.50 + 6.20) × 10–15 = 4.04 × 10–11 J = 253 MeV

*25.54 (a) To make a spark 5 mm long in dry air between flat metal plates requires potential difference

V = Ed = 3.0 × 106 V m( ) 5.0 × 10−3 m( ) = 1.5 × 104 V ~104 V

(b) Suppose your surface area is like that of a 70-kg cylinder with the density of water and radius12 cm. Its length would be given by

70 × 103 cm3 = π 12 cm( )2l l = 1.6 m

The lateral surface area is A = 2πr l = 2π 0.12 m( ) 1.6 m( ) = 1.2 m2

The electric field close to your skin is described by E = σ

e0= Q

Ae0, so

Q = EA ∈ 0 = 3.0 × 106

1.2 m2( ) 8.85 × 10−12

N ⋅ m2

~10−5 C

25.55 (a) V = ke Q

x + a – 2x +

1x – a

V = ke Q

x(x – a) – 2(x + a)(x – a) + x(x + a)

x(x + a)(x – a) = 2ke Qa2

x3 – xa2

(b) V = 2ke Qa2

x3 for ax << 1

76 Chapter 25 Solutions

25.56 (a) Ex = − dV

dx= − d

dx2keQa2

x3 − xa2

(2keQa2 )(3x2 − a2 )(x3 − xa2 )2 and Ey = Ez = 0

(b) Ex =

2(8.99 × 109 N ⋅ m2/ C2 )(3 × 10−6 C)(2 × 10−3 m)2 3(6 × 10−3 m)2 − (2 × 10−3 m)2[ ](6 × 10−3 m)3 − (6 × 10−3 m)(2 × 10−3 m)2[ ]2

Ex = 609 × 106 N/C = 609 MN/C

25.57 (a) E = Q

4π∈ 0 r2

V = Q

4π∈ 0 r

r =

= 3000 V500 V / m

= 6.00 m

(b) V = −3000 V = Q

4π∈ 0 (6.00 m)

Q = −3000 V

(8.99 × 109 V ⋅ m / C)(6.00 m) = – 2.00 µC

25.58 From Example 25.5, the potential created by the ring at the electron's starting point is

Vi = keQ

xi2 + a2

=ke 2π λ a( )

xi2 + a2

while at the center, it is Vf = 2πkeλ . From conservation of energy,

0 + −eVi( ) = 1

2 mevf2 + −eVf( )

vf2 = 2e

meVf − Vi( ) = 4πekeλ

me1 − a

xi2 + a2

vf2 =

4π 1.60 × 10−19( ) 8.99 × 109( ) 1.00 × 10−7( )9.11× 10−31 1 − 0.200

0.100( )2 + 0.200( )2

vf = 1.45 × 107 m/s

Chapter 25 Solutions 77

25.59 (a) Take the origin at the point where we will find the potential. One ring, of width dx, hascharge Q dx/h and, according to Example 25.5, creates potential

dV =

keQ dx

h x 2 + R 2

The whole stack of rings creates potential

V = dV

all charge∫ =keQ dx

h x 2 + R 2d

d + h∫

ke Qh

ln x + x 2 + R2

d + h =

ke Qh

lnd + h + (d + h)2 + R2

d + d2 + R2

(b) A disk of thickness dx has charge Q dx/h and charge-per-area Q dx/π R 2h. According toExample 25.6, it creates potential

dV = 2πke

Q dxπR2h

x 2 + R2 − x

Integrating,

V =

2keQR 2hd

d+h∫ x 2 + R 2 dx − x dx

2ke QR2 h

x x 2 + R2 +R2

2ln x + x 2 + R2

− x 2

d + h

V =

keQR 2h

(d + h) (d + h)2 + R2 − d d2 + R2 − 2dh − h2

+ R2 lnd + h + (d + h)2 + R2

d + d2 + R2

25.60 The positive plate by itself creates a field E = σ

2 ∈ 0 =

36.0 × 10–9 C/m2

2(8.85 × 10–12 C2/N · m2) = 2.03

kNC

away from the + plate. The negative plate by itself creates the same size field and between theplates it is in the same direction. Together the plates create a uniform field 4.07 kN/C in thespace between.

(a) Take V = 0 at the negative plate. The potential at the positive plate is then

V − 0 = − −4.07 kN / C( )dx

12.0 cm∫

The potential difference between the plates is V = (4.07 × 103 N/C)(0.120 m) = 488 V

(b)

2 mv2 + q V i =

2 mv2 + q V f

qV = (1.60 × 10–19 C)(488 V) = 12 m v 2

f = 7.81 × 10–17 J

78 Chapter 25 Solutions

(d) vf 2 = v 2i + 2a(x – xi)

(3.06 × 105 m/s)2 = 0 + 2a(0.120 m)

a = 3.90 × 1011 m/s2

(e) ∑F = ma = (1.67 × 10–27 kg)(3.90 × 1011 m/s2) = 6.51 × 10–16 N

(f) E = Fq =

6.51 × 10–16 N1.60 × 10–19 C

= 4.07 kN/C

25.61 W = V dq

∫ where V = ke qR ; Therefore, W =

keQ 2

25.62 (a) VB − VA = − E ⋅ ds

B∫ and the field at distance r from a uniformly

charged rod (where r > radius of charged rod) is

E = λ

2πe0r= 2keλ

In this case, the field between the central wire and the coaxialcylinder is directed perpendicular to the line of charge so that

VB − VA = − 2keλ

rdr

rb∫ = 2keλ lnra

, or ∆V = 2keλ ln

− λ

(b) From part (a), when the outer cylinder is considered to be at zero potential, the potential at adistance r from the axis is

V = 2keλ ln

The field at r is given by E = − ∂V

∂r= −2keλ

rra

− ra

= 2keλ

But, from part (a), 2keλ = ∆V

ln ra rb( ) . Therefore, E = ∆V

ln ra rb( )1r

Chapter 25 Solutions 79

25.63 V2 − V1 = − E ⋅ dr

r2∫ = − λ2πε0rr1

r2∫ dr

V2 – V1 =

−λ2π∈ 0

lnr2

25.64 For the given charge distribution, V x, y, z( ) =

ke q( )r1

+ke −2q( )

where r1 = x + R( )2 + y2 + z2 and r2 = x2 + y2 + z2

The surface on which V x, y, z( ) = 0

is given by keq

1r1

− 2r2

= 0, or 2r1 = r2

This gives: 4 x + R( )2 + 4y2 + 4z2 = x2 + y2 + z2

which may be written in the form: x2 + y2 + z2 + 8

x + 0( )y + 0( )z + 4

3R2

= 0 [1]

The general equation for a sphere of radius a centered at x0 , y0 , z0( ) is:

x − x0( )2 + y − y0( )2 + z − z0( )2 − a2 = 0

or x2 + y2 + z2 + −2x0( )x + −2y0( )y + −2z0( )z + x0

2 + y02 + z0

2 − a2( ) = 0 [2]

Comparing equations [1] and [2], it is seen that the equipotential surface for which V = 0 isindeed a sphere and that:

−2x0 = 8

3R ; −2y0 = 0; −2z0 = 0;

2 + y02 + z0

2 − a2 = 43

Thus, x0 = − 4

3R, y0 = z0 = 0, and

a2 = 16

9− 4

R2 = 4

9R2.

The equipotential surface is therefore a sphere centered at

− 43

R,0,0

, having a radius

80 Chapter 25 Solutions

25.65 (a) From Gauss's law, EA = 0 (no charge within)

EB = ke

r 2 = (8.99 × 109)(1.00 × 10−8 )

r 2 =

89.9r 2

V / m

EC = ke

(qA + qB)r2 = (8.99 × 109)

(− 5.00 × 10−9)r2 =

− 45.0

V / m

(b) VC = ke

(qA + qB)r

= (8.99 × 109)(− 5.00 × 10−9)

− 45.0

∴ At r2 , V = − 45.0

0.300= − 150V

Inside r2 , VB = − 150 V +

89.9r2 dr

r∫

= − 150 + 89.9

− 10.300

− 450 +

89.9r

∴ At r1, V = − 450 +

89.90.150

= + 150V so VA = + 150 V

25.66 From Example 25.5, the potential at the center of the ring is

Vi = keQ R and the potential at an infinite distance from the ringis Vf = 0. Thus, the initial and final potential energies of thepoint charge are:

Ui = QVi = keQ

R and U f = QVf = 0

From conservation of energy, Kf +U f = Ki +Ui

or 12 Mvf

2 + 0 = 0 + keQ2

Rgiving vf =

2keQ2

25.67 The sheet creates a field E1 =

σ2 ∈ 0

i for x > 0. Along the x − axis, the line of charge creates a

field

E2 =

λ2πr ∈ 0

away =λ

2π ∈ 0 (3.00 m − x)(− i) for x < 3.00 m

The total field along the x − axis in the region 0 < x < 3.00 m is then

E = E1 + E2 = σ

2 ∈ 0− λ

2π ∈ 0 3.00 − x( )

Chapter 25 Solutions 81

(a) The potential at point x follows from

V − V0 = − E ⋅ idx =

x∫ − σ

2 ∈ 0− λ

2π ∈ 0 3.00 − x( )

x∫

V = V0 − σ x

2 ∈ 0− λ

2π ∈ 0ln 1 − x

3.00

V = 1.00 kV − (25.0 × 10−9 C / m2)x

2(8.85 × 10−12 C2 / N ⋅ m2 )− 80.0 × 10−9 C / m

2π(8.85 × 10−12 C2 / N ⋅ m2 )ln 1 − x

3.00

V = 1.00 kV − 1.41

kVm

x − (1.44 kV) ln 1.00 − x

3.00 m

(b) At x = 0.800 m, V = 316 V

and U = QV = 2.00 × 10−9 C( ) 316 J C( ) = 6.33 × 10−7 J = 633 nJ

25.68 V = ke

λ dx

x2 + b2= keλ ln

a+L

∫ x + (x2 + b2 )

a+L =

keλ lna + L + a + L( )2 + b2

a + a2 + b2

25.69 (a) Er = − ∂V

∂r=

2kepcosθr3

In spherical coordinates, the θ component of the gradient is

∂∂θ

Therefore, Eθ = −1

r∂V∂θ

kepsinθr3

For r >> a, Er (0°) = 2kep

r3 and Er (90°) = 0 , Eθ (0°) = 0 and Eθ (90°) = kep

These results are reasonable for r >> a .

However, for r → 0, E(0) → ∞ .

(b) V =

kepy(x2 + y2 )3/2 and

Ex = − ∂V

∂x= 3kepxy

(x2 + y2 )5/2

Ey = − ∂V

∂y=

kep(2y2 − x2 )(x2 + y2 )5/2

82 Chapter 25 Solutions

25.70 (a) EA > EB since E =

∆∆Vs

(b) EB = – ∆V

∆s = –

(6 – 2) V2 cm = 200 N/C down

25.71 For an element of area which is a ring of radius r and width dr, dV = ke dq

r2 + x2

dq = σ dA = Cr (2π r dr) and

V = C(2πke )

r2 dr

r2 + x20

∫ = C(πke ) R R2 + x2 + x2 ln

R + R2 + x2

25.72 dU = V dq where the potential V = ke qr .

The element of charge in a shell is dq = ρ (volume element) or dq = ρ (4π r 2 dr) and thecharge q in a sphere of radius r is

q = 4πρ r2 dr = ρ 4πr3

∫

Substituting this into the expression for dU, we have

dU = keq

dq = keρ

4πr3

ρ(4πr2 dr) = ke

16π2

ρ2r4 dr

U = dU∫ = ke

16π2

ρ2 r4 dr

∫ = ke16π2

ρ2R5

But the total charge, Q = ρ 43 πR3. Therefore,

U = 3

5keQ

Chapter 25 Solutions 83

*25.73 (a) From Problem 62, E = ∆V

ln ra rb( )1r

We require just outside the central wire

5.50 × 106 Vm

= 50.0 × 103 V

ln0.850 m

1rb

or 110 m-1( )rbln

0.850 mrb

= 1

We solve by homing in on the required value

rb (m) 0.0100 0.00100 0.00150 0.00145 0.00143 0.00142

110 m-1( )rb ln

0.850 mrb

4.89 0.740 1.05 1.017 1.005 0.999

Thus, to three significant figures, rb = 1.42 mm

(b) At ra , E = 50.0 kV

ln 0.850 m 0.00142 m( )1

0.850 m

= 9.20 kV m

Chapter 26 Solutions

*26.1 (a) Q = C (∆V) = (4.00 × 10–6 F)(12.0 V) = 4.80 × 10–5 C = 48.0 µC

(b) Q = C (∆V) = (4.00 × 10–6 F)(1.50 V) = 6.00 × 10–6 C = 6.00 µC

26.2 (a) C =

Q∆V

=10.0 × 10−6 C

10.0 V= 1.00 × 10−6 F = 1.00 µF

(b) ∆V =

=100 × 10−6 C1.00 × 10−6 F

= 100 V

26.3 E = keq

r2 : q = (4.90 × 104 N / C)(0.210 m)2

8.99 × 109 N ⋅ m2 / C2 = 0.240 µC

(a) σ = q

A= 0.240 × 10−6

4π(0.120)2 = 1.33 µC/m2

(b) C = 4πe0r = 4π(8.85 × 10−12 )(0.120) = 13.3 pF

26.4 (a) C = 4π e0 R

R = C

4π e0 = ke C = (8.99 × 109 N · m2/C 2)(1.00 × 10–12 F) = 8.99 mm

(b) C = 4π e0 R = 4π (8.85 × 10–12 C 2)(2.00 × 10–3 m)

N · m2 = 0.222 pF

26.5 (a)

Q2= R1

Q1 + Q2 = 1 + R1

Q2 = 3.50Q2 = 7.00 µC

Q2 = 2.00 µC Q1= 5.00 µC

(b)

V1 = V2 = Q1

C1= Q2

C2= 5.00 µC

8.99 × 109 m F( )−1(0.500 m)

= 8.99 × 104 V = 89.9 kV

Chapter 26 Solutions 85

*26.6 C = κ e0 A

d = (1.00)(8.85 × 10–12 C 2)(1.00 × 103 m)2

N · m2(800 m) = 11.1 nF

The potential between ground and cloud is

∆V = Ed = (3.00 × 106 N/C)(800 m) = 2.40 × 109 V

Q = C (∆V) = (11.1 × 10-9 C/V)(2.40 × 109 V) = 26.6 C

26.7 (a) ∆V = Ed

E = 20.0 V

1.80 × 10–3 m = 11.1 kV/m

(b) E = σ

σ = (1.11 × 104 N/C)(8.85 × 10–12 C 2/N · m2) = 98.3 nC/m2

8.85 × 10−12 C2 / N ⋅ m2( ) 7.60 cm2( ) 1.00 m /100 cm( )2

1.80 × 10−3 m= 3.74 pF

(d) ∆V = QC Q = (20.0 V)(3.74 × 10–12 F) = 74.7 pC

26.8 C = κe0 A

d= 60.0 × 10−15 F

d = κe0 A

1( ) 8.85 × 10−12( ) 21.0 × 10−12( )60.0 × 10−15

d = 3.10 × 10−9 m = 3.10 nm

26.9 Q =

e0 Ad

∆V( )

= σ =e0 ∆V( )

d =e0 ∆V( )

σ=

8.85 × 10−12 C 2 N ⋅ m2( ) 150 V( )30.0 × 10−9 C cm2( ) 1.00 × 104 cm2 m2( ) = 4.42 µm

86 Chapter 26 Solutions

26.10 With θ = π, the plates are out of mesh and the overlap area is zero.With θ = 0, the overlap area is that of a semi-circle, πR2 2 . Byproportion, the effective area of a single sheet of charge is

π − θ( )R2 2.

When there are two plates in each comb, the number of adjoiningsheets of positive and negative charge is 3, as shown in the sketch.When there are N plates on each comb, the number of parallelcapacitors is 2N − 1 and the total capacitance is

++++

+ -

+++

---- +

+++

----

C = 2N − 1( )e0 Aeffective

distance= 2N − 1( )e0 π − θ( )R2 2

d 2=

2N − 1( ) ∈ 0 π − θ( )R2

26.11 (a) C = l

2ke ln

= 50.0

2(8.99 × 109) ln

7.27

2.58

= 2.68 nF

(b) Method 1: ∆V = 2ke λ ln

λ = q / l = 8.10 × 10–6 C

50.0 m = 1.62 × 10–7 C/m

∆V = 2(8.99 × 10 9)(1.62 × 10–7) ln

7.27

2.58 = 3.02 kV

Method 2: ∆V = QC =

8.10 × 10–6

2.68 × 10–9 = 3.02 kV

26.12 Let the radii be b and a with b = 2a. Put charge Q on the inner conductor and – Q on the outer.Electric field exists only in the volume between them. The potential of the inner sphere is

Va = keQ a ; that of the outer is Vb = keQ b . Then

Va − Vb =

ke Qa

− ke Qb

4πe0

b − aab

and

C =

QVa − Vb

=4πe0abb − a

Here C =

4πe0 2a2

a= 8πe0 a

a =

C8πe0

The intervening volume is Volume = 4

3πb 3 − 4

3πa3 = 7 4

3πa3( ) = 7 4

3π( ) C 3

83 π3e03 =

7C 3

384π2e03

Volume =

7(20.0 × 10−6 C2 / N ⋅m)3

384π2(8.85 × 10−12 C2 / N ⋅m2 )3 = 2.13 × 1016 m3

The outer sphere is 360 km in diameter.

Chapter 26 Solutions 87

26.13 ΣFy = 0: T cosθ − mg = 0 ΣFx = 0: T sinθ − Eq = 0

Dividing, tanθ =

Eqmg

, so E = mg

qtanθ

∆V = Ed =

mgd tanθq

= (350 × 10−6 kg)(9.80 m / s2 )(4.00 × 10−2 m) tan 15.0°30.0 × 10−9 C

= 1.23 kV

26.14 ΣFy = 0: T cosθ − mg = 0 ΣFx = 0: T sinθ − Eq = 0

Dividing, tanθ =

Eqmg

, so E

mgq

= tanθ and ∆V = Ed =

mgd tanθq

26.15 (a) C = ab

ke(b − a)= (0.0700)(0.140)

(8.99 × 109)(0.140 − 0.0700)= 15.6 pF

(b) C

=∆

∆V = QC

= 4.00 × 10−6 C15.6 × 10−12 F

= 256 kV

Goal Solution An air-filled spherical capacitor is constructed with inner and outer shell radii of 7.00 and 14.0 cm,respectively. (a) Calculate the capacitance of the device. (b) What potential difference between thespheres results in a charge of 4.00 µ C on the capacitor?

G : Since the separation between the inner and outer shells is much larger than a typical electroniccapacitor with d ~ 0.1 mm and capacitance in the microfarad range, we might expect the capacitance ofthis spherical configuration to be on the order of picofarads, (based on a factor of about 700 timeslarger spacing between the conductors). The potential difference should be sufficiently low to preventsparking through the air that separates the shells.

O : The capacitance can be found from the equation for spherical shells, and the voltage can be foundfrom Q = C∆V .

A : (a) For a spherical capacitor with inner radius a and outer radius b,

C = ab

k(b − a)

= (0.0700 m)(0.140 m)

8.99 × 109 N ⋅ m2 C2( )(0.140 − 0.0700) m = 1.56 × 10−11 F = 15.6 pF

(b) ∆V = Q

C= (4.00 × 10−6 C)

1.56 × 10-11 F= 2.56 × 105 V = 256 kV

L : The capacitance agrees with our prediction, but the voltage seems rather high. We can check thisvoltage by approximating the configuration as the electric field between two charged parallel platesseparated by d = 7.00 cm, so

E ~

∆Vd

= 2.56 × 105 V0.0700 m

= 3.66 × 106 V / m

This electric field barely exceeds the dielectric breakdown strength of air 3 × 106 V / m( ) , so it may not

even be possible to place 4.00 µ C of charge on this capacitor!

88 Chapter 26 Solutions

26.16 C = 4πe0 R = 4π 8.85 × 10−12 C N ⋅ m2( ) 6.37 × 106 m( ) = 7.08 × 10−4 F

*26.17 (a) Capacitors in parallel add. Thus, the equivalent capacitor has a value of

Ceq = C1 + C2 = 5.00 µF + 12.0 µF = 17.0 µF

(b) The potential difference across each branch is the same and equal to the voltage of the battery.

∆V = 9.00 V

*26.18 (a) In series capacitors add as

1C eq

= 1

C 1 +

1C 2

= 1

5.00 µF + 1

12.0 µF and C eq = 3.53 µF

Q eq = C eq (∆V) = (3.53 µF)(9.00 V) = 31.8 µC

Each of the series capacitors has this same charge on it. So Q 1 = Q 2 = 31.8 µC

(b) The voltage across each is

∆V1 = Q 1C 1

= 31.8 µC5.00 µF = 6.35 V and ∆V2 =

Q 2C 2

= 31.8 µC12.0 µF = 2.65 V

26.19 C p = C 1 + C 21

C s =

1C 1

+ 1

C 2

Substitute C 2 = C p – C 11

C s =

1C 1

+ 1

C p – C 1 =

C p – C 1 + C 1C 1(C p – C 1)

Simplifying, C 21 – C 1C p + C pC s = 0

C1 =

Cp ± Cp2 − 4CpCs

2= 1

2Cp ± 1

4Cp

2 − CpCs

We choose arbitrarily the + sign. (This choice can be arbitrary, since with the case of theminus sign, we would get the same two answers with their names interchanged.)

C1 = 1

2Cp + 1

4Cp

2 − CpCs = 12

9.00 pF( ) + 14

9.00 pF( )2 − 9.00 pF( ) 2.00 pF( ) = 6.00 pF

C2 = Cp − C1 = 1

2Cp − 1

4Cp

2 − CpCs = 12 (9.00 pF) – 1.50 pF = 3.00 pF

Chapter 26 Solutions 89

26.20 Cp = C1 + C2 and

1Cs

= 1C1

+ 1C2

Substitute C2 = Cp − C1:

1Cs

= 1C1

+ 1Cp − C1

=Cp − C1 + C1

C1 Cp − C1

Simplifying, C12 − C1Cp + CpCs = 0

and C1 =

Cp ± Cp2 − 4CpCs

2Cp + 1

4Cp

2 − CpCs

where the positive sign was arbitrarily chosen (choosing the negative sign gives the samevalues for the capacitances, with the names reversed). Then, from C2 = Cp − C1

C2 =

2Cp − 1

4Cp

2 − CpCs

26.21 (a)1Cs

= 1

15.0 + 1

3.00 Cs = 2.50 µF

Cp = 2.50 + 6.00 = 8.50 µF

Ceq = 1

8.50 µF+ 1

20.0 µF

−1

= 5.96 µF

(b) Q = ∆V( )C = (15.0 V)(5.96 µF) = 89.5 µC on 20.0 µF

∆V = Q

C =

89.5 µC20.0 µF = 4.47 V

15.0 – 4.47 = 10.53 V

Q = ∆V( )C= (10.53)(6.00 µF) = 63.2 µC on 6.00 µF

89.5 – 63.2 = 26.3 µC on 15.0 µF and 3.00 µF

26.22 The circuit reduces first according to the rule forcapacitors in series, as shown in the figure, thenaccording to the rule for capacitors in parallel,shown below.

Ceq = C 1 + 1

2+ 1

= 11

6C = 1.83C

⇒

90 Chapter 26 Solutions

26.23 C = Q

∆V so 6.00 × 10–6 =

Q20.0 and Q = 120 µC

Q1 = 120 µC – Q2 and ∆V = QC

120 – Q 2C 1

= Q 2C 2

or120 – Q2

6.00 = Q 23.00

(3.00)(120 – Q2) = (6.00)Q2

Q 2 = 3609.00 = 40.0 µC

Q1 = 120 µC – 40.0 µC = 80.0 µC

*26.24 (a) In series , to reduce the effective capacitance:

132.0 µF

= 134.8 µF

+ 1Cs

Cs = 1

2.51× 10−3 µ F= 398 µF

(b) In parallel , to increase the total capacitance:

29.8 µF + Cp = 32.0 µF

Cp = 2.20 µF

26.25 With switch closed, distance d' = 0.500d and capacitance

′C =e0 A

′d=

2e0Ad

= 2C

(a) Q = ′C (∆V) = 2C(∆V) = 2(2.00 × 10−6 F)(100 V) = 400 µC

(b) The force stretching out one spring is

F =

Q 2

2e0 A=

4C 2(∆V)2

2e0 A =

2C 2 (∆V)2

(e0 A / d)d=

2C(∆V)2

One spring stretches by distance x = d /4, so

k =

=2C(∆V)2

d4d

8C(∆V)2

d 2 =

8(2.00 × 10−6 F)(100 V)2

(8.00 × 10−3 m)2 = 2.50 kN/m

Chapter 26 Solutions 91

26.26 Positive charge on A will induce equal negative charges on B, D,and F, and equal positive charges on C and E. The nesting spheresform three capacitors in series. From Example 26.3,

CAB =

abke(b − a)

=R(2R)

keR=

2Rke

CCD =

3R( ) 4R( )keR

=12Rke

CEF =

5R( ) 6R( )keR

=30Rke

Ceq =

1ke / 2R + ke /12R + ke / 30R

60R37 ke

26.27 nC = = 100

n/C

nC = 100C

n so n2 = 100 and n = 10

Goal Solution A group of identical capacitors is connected first in series and then in parallel. The combined capacitancein parallel is 100 times larger than for the series connection. How many capacitors are in the group?

G : Since capacitors in parallel add and ones in series add as inverses, 2 capacitors in parallel would have acapacitance 4 times greater than if they were in series, and 3 capacitors would give a ratio Cp /Cs = 9, so

maybe n = Cp /Cs = 100 = 10 .

O : The ratio reasoning above seems like an efficient way to solve this problem, but we should check theanswer with a more careful analysis based on the general relationships for series and parallelcombinations of capacitors.

A : Call C the capacitance of one capacitor and n the number of capacitors. The equivalent capacitance forn capacitors in parallel is

Cp = C1 + C2 + . . . + Cn = nC

The relationship for n capacitors in series is

1Cs

= 1C1

+ 1C2

+ . . . + 1Cn

= nC

Therefore

Cs= nC

C / n= n2 or

n =

Cs= 100 = 10

L : Our prediction appears to be correct. A qualitative reason that Cp /Cs = n2 is because the amount ofcharge that can be stored on the capacitors increases according to the area of the plates for a parallelcombination, but the total charge remains the same for a series combination.

92 Chapter 26 Solutions

26.28 Cs = 1

5.00+ 1

10.0

−1

= 3.33 µF

Cp1 = 2(3.33) + 2.00 = 8.66 µF

Cp2 = 2(10.0) = 20.0 µF

Ceq = 1

8.66+ 1

20.0

−1

= 6.04 µF

26.29 Qeq = Ceq ∆V( ) = 6.04 × 10−6 F( ) 60.0 V( ) = 3.62 × 10−4 C

Qp1 = Qeq, so ∆Vp1 =

Qeq

Cp1= 3.62 × 10−4 C

8.66 × 10−6 F= 41.8 V

Q3 = C3 ∆Vp1( ) = 2.00 × 10−6 F( ) 41.8 V( ) = 83.6 µC

26.30 Cs = 1

5.00+ 1

7.00

−1

= 2.92 µF

Cp = 2.92 + 4.00 + 6.00 = 12.9 µF

*26.31 (a) U = 12 C (∆V)2 =

12 (3.00 µF)(12.0 V) 2 = 216 µ J

(b) U = 12 C (∆V)2 =

12 (3.00 µF)(6.00 V) 2 = 54.0 µ J

*26.32 U = 12 C (∆V)2

The circuit diagram is shown at the right.

(a) Cp = C1 + C2 = 25.0 µF + 5.00 µF = 30.0 µF

U = 12 (30.0 × 10–6)(100) 2 = 0.150 J

(b) Cs =

C1 +

1C2

–1

25.0 µF + 1

5.00 µF –1

= 4.17 µF

U = 12 C (∆V)2

∆V = 2U

C= 0.150( ) 2( )

4.17 × 10−6 = 268 V

Chapter 26 Solutions 93

*26.33 Use U = 12

Q 2

C and C = e0 A

If d2 = 2d1, C2 = 12 C1. Therefore, the stored energy doubles .

26.34 u = U

V= 1

2e0E2

1.00 × 10−7

V= 1

2(8.85 × 10−12 )(3000)2

V = 2.51× 10−3 m3 = 2.51× 10−3 m3( ) 1000 L

= 2.51 L

26.35 W = U = F dx∫ so F =

dUdx

= ddx

= d

dxQ2x2e0A

Q 2

2 e0 A

26.36 Plate a experiences force – kx i from the spring and force QE i due to the electric field created byplate b according to E=σ / 2e0 = Q / 2 Ae0. Then,

kx = Q 2

2A e0 x =

2Ae0 k

where A is the area of one plate.

26.37 The energy transferred is W Q V= =12 ( )∆

12 (50.0 C)(1.00 × 108 V) = 2.50 × 109 J and 1% of this (or

W' = 2.50 × 107 J) is absorbed by the tree. If m is the amount of water boiled away, then

W' = m(4186 J/kg °C)(100 °C – 30.0 °C) + m(2.26 × 106 J/kg) = 2.50 × 107 J

giving m = 9.79 kg

94 Chapter 26 Solutions

26.38 U = 1

2C ∆V( )2 where

C = 4πe0R = R

ke and

∆V = keQ

R− 0 = keQ

U = 1

2Rke

keQR

keQ2

26.39

keQ2

2R= mc2

R =

ke e2

2mc2 =(8.99 × 109 N ⋅ m2 / C)(1.60 × 10−19 C)2

2(9.11× 10−31 kg)(3.00 × 108 m / s)2 = 1.40 fm

*26.40 C = κ ∈ 0 A

d = 4.90(8.85 × 10–12 F/m)(5.00 × 10–4 m2)

2.00 × 10–3 m = 1.08 × 10–11 F = 10.8 pF

*26.41 (a) C = κ ∈ 0 A

d = 2.10(8.85 × 10–12 F/m)(1.75 × 10–4 m2)

4.00 × 10–5 m = 8.13 × 10–11 F = 81.3 pF

(b) ∆Vmax = Emax d = (60.0 × 106 V/m)(4.00 × 10–5 m) = 2.40 kV

*26.42 Qmax = C (∆Vmax), but ∆Vmax = Emax d

Also, C = κ ∈ 0A

Thus, Qmax = κ ∈ 0A

d (Emax d) = κ ∈ 0 AEmax

(a) With air between the plates, κ = 1.00 and Emax = 3.00 × 106 V/m. Therefore,

Qmax = κ ∈ 0 AEmax = (8.85 × 10–12 F/m)(5.00 × 10–4 m2)(3.00 × 106 V/m) = 13.3 nC

(b) With polystyrene between the plates, κ = 2.56 and Emax = 24.0 × 106 V/m.

Qmax = κ ∈ 0 AEmax = 2.56(8.85 × 10–12 F/m)(5.00 × 10–4 m2)(24.0 × 106 V/m) = 272 nC

26.43 C = κ ∈ 0 A

dor

Chapter 26 Solutions 95

= 1.04 m

96 Chapter 26 Solutions

*26.44 Consider two sheets of aluminum foil, each 40 cm by 100 cm, with one sheet ofplastic between them. Suppose the plastic has κ ≅ 3, Emax ~ 107 V/m and thickness1 mil = 2.54 cm/1000. Then,

C = κ ∈ 0 A

d ~ 3(8.85 × 10–12 C2/N · m2)(0.4 m2)

2.54 × 10–5 m ~ 10–6 F

∆Vmax = Emaxd ~ 107 V

2.54 × 10−5 m( ) ~ 10 2 V

*26.45 (a) With air between the plates, we find C0 = Q

∆V =

48.0 µC12.0 V = 4.00 µF

(b) When Teflon is inserted, the charge remains the same (48.0 µC) because the plates are isolated.However, the capacitance, and hence the voltage, changes. The new capacitance is

C' = κ C 0 = 2.10(4.00 µF) = 8.40 µF

48.0 µC8.40 µF = 5.71 V

and the charge is 48.0 µC

26.46 Originally, C =∈ 0 A/ d = Q/(∆V)i

(a) The charge is the same before and after immersion, with value Q =∈ 0 A(∆V)i / d .

Q = (8.85 × 10−12 C2)(25.0 × 10−4 m2 )(250 V)

N ⋅ m2 (1.50 × 10−2 m) = 369 pC

(b) Finally, Cf = κ ∈ 0 A / d = Q /(∆V)f

Cf =

80.0(8.85 × 10−12 C2)(25.0 × 10−4 m2)N ⋅ m2 (1.50 × 10−2 m)

= 118 pF

(∆V) f =

Qdκ∈ 0 A

=∈ 0 A(∆V)i d

κ∈ 0 Ad=

(∆V)i

κ= 250 V

80.0 = 3.12 V

2 C(∆V)i2 =

∈ 0 A(∆V)i2

Finally, U f = 1

2 Cf (∆V) f2 =

κ∈ 0 A(∆V) i2

2d κ 2 =∈ 0 A(∆V) i

2dκ

So, ∆U = U f −U = − ∈ 0 A(∆V)i

2 (κ − 1)2d κ

∆U =

(− 8.85 × 10−12 C 2)(25.0 × 10−4 m2 )(250 V)2 (79.0)N ⋅ m2 2(1.50 × 10−2 m)80

= – 45.5 nJ

Chapter 26 Solutions 97

26.47

κ1abke(b − a)

κ 2bcke(c − b)

=ke(b − a)

κ1ab+

ke (c − b)κ 2bc

C =1

ke (b − a)κ1 ab

+ke (c − b)

κ 2bc

= κ1κ 2 abcke κ 2 (bc − ac) + ke κ1 (ac − ab)

4π κ1κ 2 abc∈ 0

κ 2bc − κ1ab + (κ1 − κ 2 )ac

26.48 (a) C = κ C0 = κ∈ 0 A

d= (173)(8.85 × 10−12 )(1.00 × 10−4 m2 )

0.100 × 10−3 m= 1.53 nF

(b) The battery delivers the free charge

Q = C (∆V) = (1.53 × 10-9 F)(12.0 V) = 18.4 nC

σ = Q

A= 18.4 × 10−9 C

1.00 × 10−4 m2 = 1.84 × 10-4 C/m2

The surface density of polarization charge is

σ σ

κσp = −

= −

1173

1.83 × 10-4 C/m2

(d) We have E = E0/κ and E0 = ∆V/d ; hence,

E = ∆V

κd= 12.0 V

(173)(1.00 × 10−4 m)= 694 V/m

26.49 The given combination of capacitors is equivalent to the circuitdiagram shown to the right.

Put charge Q on point A . Then,

Q = (40.0 µF)∆VAB = (10.0 µF)∆VBC = (40.0 µF)∆VCD

So, ∆VBC = 4 ∆VAB = 4 ∆VCD , and the center capacitor will break down first, at ∆VBC = 15.0 V. When this occurs,

∆ ∆ ∆V V VAB CD BC= = ( ) =14 3 75. V

and VAD = VAB + VBC + VCD = 3.75 V + 15.0 V + 3.75 V = 22.5 V

98 Chapter 26 Solutions

*26.50 (a) The displacement from negative to positive charge is

2 1 20 1 10 1 40 1 30a i j i j= − +( ) − −( ). . . .mm mm = −2.60i + 2.40 j( ) × 10−3 m

The electric dipole moment is

p = 2aq = 3.50 × 10−9 C( ) −2.60i + 2.40 j( ) × 10−3 m = −9.10i + 8.40 j( ) × 10−12 C ⋅ m

(b) τ = p × E = −9.10i + 8.40 j( ) × 10−12 C ⋅ m[ ] × 7.80i − 4.90 j( ) × 103 N C[ ] τ = +44.6k − 65.5k( ) × 10−9 N ⋅ m = −2.09 × 10−8 N ⋅ mk

(c) U = −p ⋅ E = − −9.10i + 8.40 j( ) × 10−12 C ⋅ m[ ] ⋅ 7.80i − 4.90 j( ) × 103 N C[ ] U = 71.0 + 41.2( ) × 10−9 J = 112 nJ

(d) p = 9.10( )2 + 8.40( )2 × 10−12 C ⋅ m = 12.4 × 10−12 C ⋅ m

E = 7.80( )2 + 4.90( )2 × 103 N C = 9.21× 103 N C

Umax = p E = 114 nJ, Umin = −114 nJ

Umax −Umin = 228 nJ

*26.51 (a) Let x represent the coordinate of the negative charge.Then x + 2acosθ is the coordinate of the positive charge.The force on the negative charge is F− = −qE x( )i . Theforce on the positive charge is

F+ = +qE x + 2acosθ( )i ≅ qE x( )i + q

dEdx

2acosθ( )iE

p F+

F-

The force on the dipole is altogether F = F− + F+ = q

dEdx

2acosθ( )i = p

dEdx

cosθ i

(b) The balloon creates field along the x − axis of

keqx2 i.

Thus,

dEdx

= −2( )keqx3

At x = 16.0 cm ,

dEdx

=−2( ) 8.99 × 109( ) 2.00 × 10−6( )

0.160( )3 = −8.78

MNC ⋅ m

F = 6.30 × 10−9 C ⋅ m( ) −8.78 × 106

NC ⋅ m

cos0˚ i = – 55.3 i mN

Chapter 26 Solutions 99

26.52 2πr E = qin

∈ 0so

E = λ

2πr∈ 0

∆V = − E ⋅ dr = λ

2πr ∈ 0r1

r2∫r1

r2∫ dr = λ2π ∈ 0

lnr1

λ max

2π∈ 0= Emaxrinner

∆V = 1.20 × 106 V

(0.100 × 10−3 m) ln

25.00.200

∆Vmax = 579 V

*26.53 (a) Consider a gaussian surface in the form of a cylindrical pillbox with ends of area ′A << Aparallel to the sheet. The side wall of the cylinder passes no flux of electric field since thissurface is everywhere parallel to the field. Gauss’s law becomes

E ′A + E ′A = Q

∈ A′A , so

E = Q

2 ∈ A directed away from the positive sheet.

(b) In the space between the sheets, each creates field Q 2 ∈ A away from the positive and towardthe negative sheet. Together, they create a field of

E = Q

∈ A

(c) Assume that the field is in the positive x − direction. Then, the potential of the positive platerelative to the negative plate is

∆V = − E ⋅ ds

−plate

+plate∫ = − Q

∈ Ai ⋅ −i dx( )

−plate

+plate∫ =

+ Qd

∈ A

(d) Capacitance is defined by: C = Q

∆V= Q

Qd ∈ A=

∈ Ad

= κ∈ 0 Ad

26.54 (a) C =

3.00 + 1

6.00 –1

2.00 + 1

4.00 –1

= 3.33 µF

Therefore, Q3 = Q6 = 180 µC

Qdf = Cdf ∆Vdf( ) = 1.33 µF( ) 90.0 V( ) = 120 µC

100 Chapter 26 Solutions

(b) ∆V3 = Q 3C 3

= 180 µC3.00 µF = 60.0 V

∆V6 = Q 6C 6

= 180 µC6.00 µF = 30.0 V

∆V2 = Q 2C 2

= 120 µC2.00 µF = 60.0 V

∆V4 = Q 4C 4

= 120 µC4.00 µF = 30.0 V

(d) UT = 12 C eq (∆V)2 =

12 (3.33 × 10–6)(90.0 V) 2 = 13.4 mJ

*26.55 The electric field due to the charge on the positive wire is perpendicular to the wire, radial,and of magnitude

E+ = λ

2π ∈ 0 r

The potential difference between wires due to the presence of this charge is

∆V1 = − E ⋅ dr

−wire

+wire∫ = − λ

2π ∈ 0

drrD−d

d∫ = λ

2π ∈ 0ln

D − dd

The presence of the linear charge density −λ on the negative wire makes an identicalcontribution to the potential difference between the wires. Therefore, the total potentialdifference is

∆V = 2 ∆V1( ) = λ

π∈ 0ln

D − dd

and the capacitance of this system of two wires, each of length , is

The capacitance per unit length is:

Chapter 26 Solutions 101

26.56 (a) We use Equation 26.11 to find the potential energy. As we will see, the potential difference∆V changes as the dielectric is withdrawn. The initial and final energies are

Ui = 12

Q 2

Ci and Uf =

Q 2

C f

But the initial capacitance (with the dielectric) is Ci = κ C f. Therefore, Uf = 12 κ

Q 2

Since the work done by the external force in removing the dielectric equals the change i npotential energy, we have

W = Uf – Ui = 12 κ

Q 2

Ci –

Q 2

Ci =

Q 2

C i (κ – 1)

To express this relation in terms of potential difference ∆Vi , we substitute Q = Ci (∆Vi), andevaluate:

W = 12 C i (∆Vi)2(κ – 1) =

12 (2.00 × 10–9 F)(100 V) 2(5.00 – 1.00) = 4.00 × 10–5 J

The positive result confirms that the final energy of the capacitor is greater than the initialenergy. The extra energy comes from the work done on the system by the external force thatpulled out the dielectric.

(b) The final potential difference across the capacitor is ∆Vf = QC f

Substituting C f = Ci

κ and Q = Ci (∆Vi) gives ∆Vf = κ ∆Vi = (5.00)(100 V) = 500 V

Even though the capacitor is isolated and its charge remains constant, the potential differenceacross the plates does increase in this case.

26.57 κ = 3.00, Emax = 2.00 × 108 V/m = ∆Vmax /d

For C = κ e0 A

d = 0.250 × 10–6 F,

A = Cd

κ e0=

C ∆Vmax( )κ e0Emax

= (0.250 × 10–6)(4000)

(3.00)(8.85 × 10–12)(2.00 × 108) = 0.188 m2

26.58 (a) C 1 = κ1 e0 A/2

d ; C 2 = κ2 e0 A/2

d/2 ; C 3 = κ3 e0 A/2

d/2

C 2 +

1C 3

–1

= C 2C 3

C 2 + C 3 =

e0 Ad

κ2κ3

κ2 + κ3

C = C 1 +

C 2 +

1C 3

–1

= e0 Ad

κ1

2 + κ2κ3

κ2 + κ3

102 Chapter 26 Solutions

(b) Using the given values we find: Ctotal = 1.76 × 10–12 F = 1.76 pF

Chapter 26 Solutions 103

26.59 The system may be considered to be two capacitors in series:

C1 = e0 At1

and C2 = e0 At2

1C =

1C1

+ 1

C2 =

t1 + t2

e0 A

C = e0 At1 + t2

= e0 As – d

Goal Solution A conducting slab of a thickness d and area A is inserted into the space between the plates of a parallel-plate capacitor with spacing s and surface area A, as shown in Figure P26.59. The slab is not necessarilyhalfway between the capacitor plates. What is the capacitance of the system?

G : It is difficult to predict an exact relationship for the capacitance of this system, but we can reason that C should increase if the distance between the slab and plates were decreased (until they touched andformed a short circuit). So maybe C ∝ 1/ s − d( ). Moving the metal slab does not change the amountof charge the system can store, so the capacitance should therefore be independent of the slab position.The slab must have zero net charge, with each face of the plate holding the same magnitude of chargeas the outside plates, regardless of where the slab is between the plates.

O : If the capacitor is charged with + Q on the top plate and −Q on the bottom plate, then free chargeswill move across the conducting slab to neutralize the electric field inside it, with the top face of theslab carrying charge −Q and the bottom face carrying charge + Q . Then the capacitor and slabcombination is electrically equivalent to two capacitors in series. (We are neglecting the slightfringing effect of the electric field near the edges of the capacitor.) Call x the upper gap, so that s − d − x is the distance between the lower two surfaces.

A : For the upper capacitor, C1 = ∈ 0 A x

and the lower has C2 = ∈ 0 A

s − d − x

So the combination has

C = 11

C1+ 1

= 1x

∈ 0 A+ s − d − x

∈ 0 A = ∈ 0 A

s − d

L : The equivalent capacitance is inversely proportional to s − d( ) as expected, and is also proportional to A . This result is the same as for the special case in Example 26.9 when the slab is just halfwaybetween the plates; the only critical factor is the thickness of the slab relative to the plate spacing.

104 Chapter 26 Solutions

26.60 (a) Put charge Q on the sphere of radius a and – Q on the other sphere. Relative to V = 0 atinfinity,

the potential at the surface of a is Va = keQ

a – keQd

and the potential of b is Vb = –keQ

b + keQd

The difference in potential is Va – Vb = keQ

a + keQ

b – keQd –

keQd

and C = Q

Va – Vb =

4π∈ 0

1 a( ) + 1 b( ) − 2 d( )

(b) As d → ∞, 1/d becomes negligible compared to 1/a. Then,

C = 4π ∈ 0

1 a + 1 b and

1C =

14π ∈ 0 a

+ 1

4π ∈ 0 b

as for two spheres in series.

26.61 Note that the potential difference between the plates is held constant at ∆Vi by the battery.

Ci = q0

∆Vi and

Cf =

∆Vi= q0 + q

∆Vi

But Cf = κ Ci , so

q0 + q∆Vi

= κ q0

∆Vi

Thus, κ = q0 + q

q0 or κ =

1 + q

26.62 (a)

(b) U C V= =12 ( )∆ 2

dx=

to the left

Chapter 26 Solutions 105

(d) F = (2000)2(8.85 × 10−12 ) 0.0500( )(4.50 − 1)

2(2.00 × 10−3 )= 1.55 × 10−3 N

106 Chapter 26 Solutions

*26.63 The portion of the capacitor nearly filled by metal has capacitance κ ∈ 0 ( x) d → ∞ and storedenergy Q

2 2C → 0 . The unfilled portion has capacitance ∈ 0 ( −x ) d. The charge on thisportion is Q = ( −x)Q0 / .

(a) The stored energy is

(b) F = − dU

dx=

F = to the right

(c)

(d)

26.64 Gasoline:

126000Btugal

1054J

Btu

1.00 gal3.786 × 10-3 m3

1.00 m3

670 kg

= 5.25 × 107

Jkg

Battery:

12.0 J C( ) 100 C s( ) 3600 s( )16.0 kg

= 2.70 × 105 J kg

Capacitor:

12 0.100 F( ) 12.0 V( )2

0.100 kg= 72.0 J kg

Gasoline has 194 times the specific energy content of the battery and 727 000 times that of the capacitor

Chapter 26 Solutions 107

26.65 Call the unknown capacitance Cu

Q = Cu(∆Vi ) = (Cu + C)(∆Vf )

Cu =

C(∆Vf )

(∆Vi ) − (∆Vf ) =

(10.0 µF)(30.0 V)(100 V − 30.0 V)

= 4.29 µF

Goal Solution An isolated capacitor of unknown capacitance has been charged to a potential difference of 100 V. Whenthe charged capacitor is then connected in parallel to an uncharged 10.0-µ F capacitor, the voltage acrossthe combination is 30.0 V. Calculate the unknown capacitance.

G : The voltage of the combination will be reduced according to the size of the added capacitance.(Example: If the unknown capacitance were C = 10.0 µF, then ∆V1 = 50.0 V because the charge is nowdistributed evenly between the two capacitors.) Since the final voltage is less than half the original,we might guess that the unknown capacitor is about 5.00 µF.

O : We can use the relationships for capacitors in parallel to find the unknown capacitance, along withthe requirement that the charge on the unknown capacitor must be the same as the total charge onthe two capacitors in parallel.

A : We name our ignorance and call the unknown capacitance Cu. The charge originally deposited oneach plate, + on one, − on the other, is

Q = Cu∆V = Cu 100 V( )

Now in the new connection this same conserved charge redistributes itself between the two capacitorsaccording to Q = Q1 + Q2 .

Q1 = Cu 30.0 V( ) and Q2 = 10.0 µF( ) 30.0 V( ) = 300 µC

We can eliminate Q and Q1 by substitution:

Cu 100 V( ) = Cu 30.0 V( ) + 300 µC so Cu = 300 µC

70.0 V= 4.29 µF

L : The calculated capacitance is close to what we expected, so our result seems reasonable. In this andother capacitance combination problems, it is important not to confuse the charge and voltage of thesystem with those of the individual components, especially if they have different values. Carefulattention must be given to the subscripts to avoid this confusion. It is also important to not confusethe variable “ C ” for capacitance with the unit of charge, “ C” for coulombs.

108 Chapter 26 Solutions

26.66 Put five 6.00 pF capacitors in series.

The potential difference across any one of the capacitors willbe:

∆V = ∆Vmax

5= 1000 V

5= 200 V

and the equivalent capacitance is:

1Ceq

= 51

6.00 pF

or Ceq = 6.00 pF

5= 1.20 pF

26.67 When ∆Vdb = 0, ∆Vbc = ∆Vdc , and

C2= Q3

Also, ∆Vba = ∆Vda or

From these equations we have C

C23

31=

However, from the properties of capacitors in series, we have Q 1 = Q 2 and Q 3 = Q 4

Therefore, C2 = C3

C1 = 9.0012.0

(4.00 µF) = 3.00 µF

26.68 Let C = the capacitance of an individual capacitor, and Cs represent the equivalent capacitanceof the group in series. While being charged in parallel, each capacitor receives charge

Q = C ∆Vchg = (5.00 × 10-4 F)(800 V) = 0.400 C

While being discharged in series,

∆Vdisch = Q

Cs= Q

C /10= 0.400 C

5.00 × 10−5 F= 8.00 kV

or 10 times the original voltage.

26.69 (a) C0 = ∈ 0 A

d= Q0

∆V0

When the dielectric is inserted at constant voltage,

Chapter 26 Solutions 109

C = κ C0 = Q

∆V0; U0 = C0(∆V0 )2

U = C(∆V0 )2

2= κ C0(∆V0 )2

2 and

UU0

= κ

The extra energy comes from (part of the) electrical work done by the battery in separating theextra charge.

(b) Q0 = C0 ∆V0 and Q = C ∆V0 = κ C0 ∆V0 so Q /Q 0 = κ

110 Chapter 26 Solutions

26.70 (a) A slice of width (dx) at coordinate x in 0 ≤ x ≤ L has thickness x d / L filled with dielectric κ2,and d – x d / L is filled with the material having constant κ1. This slice has a capacitance givenby

1dC

= 1κ 2 e0 (dx)W

xd / L

+ 1κ1e0 (dx)Wd − xd / L

= xd

κ 2 e0 W L(dx)+ dL − xd

κ1e0 W L(dx)= κ1xd + κ 2dL − κ 2xd

κ1κ 2 e0WL dx( )

dC = κ1 κ 2 e0 W L(dx)

κ 2 dL + (κ1 − κ 2 )xd

The whole capacitor is all the slices in parallel:

C = dC∫ = κ1 κ 2 e0W L(dx)

κ 2Ld + (κ1 − κ 2 )xdx=0

L∫

= κ1κ 2 e0 W L

(κ1 − κ 2 )dκ 2Ld + (κ1 − κ 2 )xd( )−1 (κ1 − κ 2 )d(dx)

x=0

∫

C =

κ1 κ 2 e0 W L(κ1 − κ 2 )d

ln κ 2Ld + (κ1 − κ 2 )xd[ ]0

L =

κ1 κ 2 e0W L(κ1 − κ 2 )d

lnκ1Ld − ln κ 2Ld[ ] =

κ1κ 2 e0 W L(κ1 − κ 2 )d

lnκ1

κ 2

(b) To take the limit κ κ1 2→ , write κ κ1 2= (1 + ) and let 0.x x → Then

C = κ 2

2 (1 + x)e0 W L(κ 2 + κ 2x − κ 2 )d

ln (1 + x)

Use the expansion of ln(1 + x) from Appendix B.5.

C =

κ 22 (1 + x) ∈ 0 W L

κ 2xd(x − 1

2 x 2 + 13 x 3 . . . )

κ 2(1 + x) ∈ 0 W Ld

(1 − 12 x + . . . )

limx → 0

C = κ 2 ∈ 0 W Ld

κ∈ 0 Ad

26.71 The vertical orientation sets up two capacitors in parallel, with equivalent capacitance

Cp =

∈ 0 A 2( )d

+κ∈ 0 A 2( )

d= κ + 1

∈ 0 Ad

where A is the area of either plate and d is the separation of the plates. The horizontalorientation produces two capacitors in series. If f is the fraction of the horizontal capacitorfilled with dielectric, the equivalent capacitance is

1Cs

= fdκ∈ 0 A

+1 − f( )d∈ 0 A

=f + κ 1 − f( )

d∈ 0 A

, or Cs = κ

f + κ 1 − f( )

∈ 0 Ad

Requiring that Cp = Cs gives

κ + 12

= κf + κ 1 − f( ) , or κ κ κ+( ) + −[ ] =1 1 2f f( )

For κ = 2.00, this yields 3 00 2 00 1 00 4 00. . ( . ) .−[ ] =f , with the solution f = 2 / 3 .

Chapter 26 Solutions 111

26.72 Initially (capacitors charged in parallel),

q1 = C1(∆V) = (6.00 µF)(250 V) = 1500 µC

q2 = C2(∆V) = (2.00 µF)(250 V) = 500 µC

After reconnection (positive plate to negative plate),

′qtotal = q1 – q2 = 1000 µC and ∆ ′V = ′qtotal

Ctotal= 1000 µC

8.00 µF= 125 V

Therefore,

′q1 = C1(∆ ′V ) = (6.00 µF)(125 V) = 750 µC

′q2 = C2(∆ ′V ) = (2.00 µF)(125 V) = 250 µC

26.73 Emax occurs at the inner conductor's surface.

Emax = 2keλ

a from Equation 24.7.

∆V = 2keλ ln

from Example 26.2

Emax = ∆V

a ln(b / a)

∆Vmax = Emaxa ln

= (18.0 × 106 V / m)(0.800 × 10−3 m) ln

3.000.800

= 19.0 kV

26.74 E = 2κλ

a ; ∆V = 2κλ ln

∆Vmax = Emaxa ln

dVmax

da= Emax ln

+ a

1b / a

− ba2

= 0

= 1 or

= e1 so a = be

112 Chapter 26 Solutions

26.75 Assume a potential difference across a and b, and notice that the potential difference across the 8.00 µF capacitor must be zero by symmetry. Then the equivalent capacitance can bedetermined from the following circuit:

⇒

Cab = 3.00 µF

26.76 By symmetry, the potential difference across 3C is zero, so the circuit reduces to

Ceq = (2C)(4C)

2C + 4C= 8

6C =

43 C

⇒

Chapter 27 Solutions

27.1 I = ∆Q

∆t ∆Q = I ∆t = (30.0 × 10–6 A)(40.0 s) = 1.20 × 10–3 C

N = Qe =

1.20 × 10–3 C1.60 × 10–19 C/electron

= 7.50 × 1015 electrons

*27.2 The atomic weight of silver = 107.9, and the volume V is

V = (area)(thickness) = (700 × 10-4 m2)(0.133 × 10-3 m) = 9.31 × 10-6 m3

The mass of silver deposited is mAg = ρV = 10.5 × 103 kg / m3( ) 9.31× 10−6 m3( ) = 9.78 × 10−2 kg.

and the number of silver atoms deposited is

N = (9.78 × 10-2 kg) 6.02 × 1026 atoms

107.9 kg = 5.45 × 1023

I = VR =

12.0 V1.80 Ω

= 6.67 A = 6.67 C/s

∆t = ∆QI =

NeI =

(5.45 × 1023)(1.60 × 10-19 C)6.67 C/s = 1.31 × 104 s = 3.64 h

27.3 Q(t) = ∫

0 Idt = I0τ (1 – e–t/τ )

(a) Q(τ) = I0τ (1 – e–1) = (0.632)I0τ

(b) Q(10τ ) = I0τ (1 – e–10) = (0.99995)I0τ

27.4 (a) Using

kee2

r2 = mv2

r , we get:

v = kee

mr= 2.19 × 106 m / s .

(b) The time for the electron to revolve around the proton once is:

t = 2πr

v= 2π(5.29 × 10−11 m)

(2.19 × 106 m / s)= 1.52 × 10−16 s

The total charge flow in this time is 1.60 × 10−19 C, so the current is

112 Chapter 27 Solutions

I = 1.60 × 10−19 C

1.52 × 10−16 s= 1.05 × 10−3 A = 1.05 mA

Chapter 27 Solutions 113

27.5 ω =

2πT

where T is the period.

I =

qω2π

=(8.00 × 10−9 C)(100π rad / s)

2π= 4.00 × 10−7 A = 400 nA

27.6 The period of revolution for the sphere is T = 2π

ω, and the average current represented by this

revolving charge is I = q

qω2π

27.7 q = 4t 3 + 5t + 6 A = (2.00 cm2 )

1.00 m100 cm

= 2.00 × 10−4 m2

(a) I 1.00 s( ) =

dqdt t=1.00 s

= 12t2 + 5( )t=1.00 s

= 17.0 A

(b) J =

=17.0 A

2.00 × 10−4 m2 = 85.0 kA / m2

27.8 I =

dqdt

q = dq∫ = I dt∫ = (100 A) sin

1/240 s∫ (120πt / s)dt

q =

−100 C120π

cos(π/ 2) − cos 0[ ] =+100 C120π

= 0.265 C

27.9 (a) J = I

A= 5.00 A

π(4.00 × 10−3 m)2 = 99.5 kA/m2

(b) J2 = 1

4J1 ;

IA2

= 14

IA1

A1 = 1

4A2 so

π(4.00 × 10−3 )2 = 1

4πr2

r2 = 2(4.00 × 10−3 ) = 8.00 × 10−3 m = 8.00 mm

114 Chapter 27 Solutions

27.10 (a) The speed of each deuteron is given by K = 12 m v 2

(2.00 × 106)(1.60 × 10–19 J) = 12 (2 × 1.67 × 10–27 kg) v2 and v = 1.38 × 107 m/s

The time between deuterons passing a stationary point is t in I = q /t

10.0 × 10–6 C/s = 1.60 × 10–19 C/t or t = 1.60 × 10–14 s

So the distance between them is vt = 1.38 × 107 m / s( ) 1.60 × 10−14 s( ) = 2.21 × 10–7 m

(b) One nucleus will put its nearest neighbor at potential

V = keqr =

(8.99 × 109 N · m2/C 2)(1.60 × 10-19 C)2 .21 × 10–7 m

= 6.49 × 10–3 V

This is very small compared to the 2 MV accelerating potential, so repulsion within the beamis a small effect.

27.11 (a)

J = IA

= 8.00 × 10−6 A

π 1.00 × 10−3 m( )2 = 2.55 A / m2

(b) From J = nevd , we have

n = J

evd= 2.55 A / m2

1.60 × 10-19 C( ) 3.00 × 108 m / s( ) = 5.31× 1010 m−3

I= NAe

6.02 × 1023( ) 1.60 × 10−19 C( )8.00 × 10−6 A

= 1.20 × 1010 s

(This is about 381 years!)

*27.12 We use I = nqAvd where n is the number of charge carriers per unit volume, and isidentical to the number of atoms per unit volume). We assume a contribution of 1 freeelectron per atom in the relationship above. For aluminum, which has a molecular weight of27, we know that Avogadro's number of atoms, NA, has a mass of 27.0 g. Thus, the mass peratom is

27.0 gNA

= 27.0 g

6.02 × 1023 = 4.49 × 10–23 g/atom

Thus, n = density of aluminum

mass per atom = 2.70 g/cm3

4.49 × 10–23 g/atom

n = 6.02 × 1022 atomscm3 = 6.02 × 1028

atomsm 3

Therefore, vd = I

n q A = 5.00 A

(6.02 × 1028 m–3)(1.60 × 10–19 C)(4.00 × 10–6 m2) = 1.30 × 10–4 m/s

or, vd = 0.130 mm/s

Chapter 27 Solutions 115

*27.13 I = ∆V

R= 120 V

240 Ω= 0.500 A = 500 mA

27.14 (a) Applying its definition, we find the resistance of the rod,

R = ∆V

I= 15.0 V

4.00 × 10−3 A= 3750 Ω = 3.75 kΩ

(b) The length of the rod is determined from Equation 27.11: R = ρ / A . Solving for andsubstituting numerical values for R, A, and the values of ρ given for carbon in Table 27.1, weobtain

= RA

ρ = (3.75 × 103 Ω)(5.00 × 10−6 m2 )

(3.50 × 10−5 Ω ⋅ m)= 536 m

27.15 ∆V = IR and R = ρl

A: A = 0.600 mm2

1.00 m

1000 mm 2 = 6.00 × 10–7 m2

∆V = Iρl

I = ∆VA

ρl =

(0.900 V)(6.00 × 10–7 m2)(5.60 × 10–8 Ω · m)(1.50 m)

I = 6.43 A

27.16 J = I

πr2 = σE = 3.00 A

π (0.0120 m)2 = σ (120 N/C)

σ = 55.3(Ω · m)-1 ρ = 1σ

= 0.0181 Ω · m

27.17 (a) Given M = ρdV = ρdAl where ρd ≡ mass density , we obtain: A = M

ρdl

Taking ρr ≡ resistivity,

R =ρrl

ρrl

Mρdl

= ρrρdl2

Thus, l = MR

ρrρd=

1.00 × 10−3( )(0.500)

(1.70 × 10−8 )(8.92 × 103)= 1.82 m

116 Chapter 27 Solutions

(b) V = M

ρd, or

πr2l = M

ρd

Thus, r = M

πρdl= 1.00 × 10−3

π(8.92 × 103)(1.82)= 1.40 × 10−4 m

The diameter is twice this distance: diameter = 280 µm

*27.18 (a) Suppose the rubber is 10 cm long and 1 mm in diameter.

R = ρlA

= 4ρlπd2 ~

4 1013 Ω ⋅ m( ) 10-1 m( )π 10−3 m( )2 = ~1018 Ω

(b) R = 4ρl

πd2 ~ 4(1.7 × 10–8 Ω · m)(10–3 m)

π (2 × 10–2 m)2 ~ 10–7 Ω

10 2 V1018 Ω

~ 10–16 A

I ~ 10 2 V10–7 Ω

~ 10 9 A

27.19 The distance between opposite faces of the cube is l = 90.0 g

10.5 g cm3

1 3

= 2.05 cm

(a) R = ρl

A= ρll2 = ρ

l = 1.59 × 10−8 Ω ⋅ m

2.05 × 10-2 m= 7.77 × 10−7 Ω = 777 nΩ

(b) I = ∆V

R= 1.00 × 10−5 V

7.77 × 10−7 Ω= 12.9 A

n = 10.5 g cm3

107.87 g mol6.02 × 1023

electronsmol

n = 5.86 × 1022

electronscm3

1.00 × 106 cm3

1.00 m3

= 5.86 × 1028 m3

I = nqvA and

v = InqA

= 12.9 C s

5.86 × 1028 m3( ) 1.60 × 10−19 C( ) 0.0205 m( )2 = 3.28 µm/s

Chapter 27 Solutions 117

27.20 Originally, R = ρl

Finally, Rf = ρ(l/ 3)

3A= ρl

9A=

27.21 The total volume of material present does not change, only its shape. Thus,

Aflf = Af 1.25li( ) = Aili giving Af = Ai 1.25

The final resistance is then: Rf =

ρlf

Af=

ρ 1.25li( )Ai 1.25

= 1.56ρli

= 1.56R

27.22

ρAll

π rAl( )2 = ρCul

π rCu( )2

rAlrCu

= ρAlρCu

= 2.82 × 10−8

1.70 × 10−8 = 1.29

27.23 J = σE so σ = J

E = 6.00 × 10−13 A / m2

100 V / m= 6.00 × 10-15 (Ω · m)-1

27.24 R =

ρ1l1

A1+

ρ2l2

A2= (ρ1l1 + ρ2l2 )/ d2

R = (4.00 × 10−3 Ω ⋅ m)(0.250 m) + (6.00 × 10−3 Ω ⋅ m)(0.400 m)

(3.00 × 10−3 m)2 = 378 Ω

27.25 ρ = m

nq2τ so τ =

ρnq2 = 9.11 × 10–31

(1.70 × 10–8)(8.49 × 1028)(1.60 × 1019)2 = 2.47 × 10–14 s

vd = qEm τ so 7.84 × 10–4 =

(1.60 × 10–19)E(2.47 × 10–14)9.11 × 10–31

Therefore E = 0.181 V/m

118 Chapter 27 Solutions

Goal Solution If the drift velocity of free electrons in a copper wire is 7.84 × 10−4 m / s, what is the electric field in theconductor?

G : For electrostatic cases, we learned that the electric field inside a conductor is always zero. On theother hand, if there is a current, a non-zero electric field must be maintained by a battery or othersource to make the charges flow. Therefore, we might expect the electric field to be small, butdefinitely not zero.

O : The drift velocity of the electrons can be used to find the current density, which can be used withOhm’s law to find the electric field inside the conductor.

A : We first need the electron density in copper, which from Example 27.1 is n = 8.49 × 1028 e- / m3. Thecurrent density in this wire is then

J = nqvd = (8.49 × 1028 e− / m3)(1.60 × 10−19 C / e- )(7.84 × 10−4 m / s) = 1.06 × 107 A / m2

Ohm’s law can be stated as J = σE = E /ρ where ρ = 1.7 × 10−8 Ω ⋅ m for copper, so then

E = ρJ = (1.70 × 10−8 Ω ⋅ m)(1.06 × 107 A / m2) = 0.181 V / m

L : This electric field is certainly smaller than typical static values outside charged objects. The directionof the electric field should be along the length of the conductor, otherwise the electrons would beforced to leave the wire! The reality is that excess charges arrange themselves on the surface of thewire to create an electric field that “steers” the free electrons to flow along the length of the wire fromlow to high potential (opposite the direction of a positive test charge). It is also interesting to notethat when the electric field is being established it travels at the speed of light; but the drift velocity ofthe electrons is literally at a “snail’s pace”!

27.26 (a) n is unaffected

(b) J = I

A∝ I so it doubles

(d) τ = mσnq2 is unchanged as long as σ does not change due to heating.

27.27 From Equation 27.17,

τ = me

nq2ρ= 9.11× 10−31

8.49 × 1028( ) 1.60 × 10−19( )21.70 × 10−8( )

= 2.47 × 10−14 s

l = vτ = 8.60 × 105 m / s( ) 2.47 × 10−14 s( ) = 2.12 × 10−8 m = 21.2 nm

Chapter 27 Solutions 119

27.28 At the low temperature TC we write RC = ∆VIC

= R0 1 + α (TC − T0 )[ ] where T0 = 20.0°C

At the high temperature Th , Rh = ∆VIh

= ∆V1 A = R0 1 + α (Th − T0 )[ ]

Then(∆V)/(1.00 A)

(∆V)/IC =

1 + (3.90 × 10–3)(38.0)1 + (3.90 × 10–3)(–108)

and IC = (1.00 A)(1.15/0.579) = 1.98 A

*27.29 R = R0 1 + α (∆T)[ ] gives 140 Ω = (19.0 Ω) 1 + (4.50 × 10−3/°C)∆T[ ]

Solving, ∆T = 1.42 × 103 °C = T – 20.0 °C

And, the final temperature is T = 1.44 × 103 °C

27.30 R = Rc + Rn = Rc 1 + α c(T − T0 )[ ] + Rn 1 + α n(T − T0 )[ ]0 = Rcα c (T – T0) + Rnα n (T – T0) so Rc = –Rn

α nα c

R = –Rn α nα c

+ Rn

Rn = R(1 – α n /α c)–1 Rc = R(1 – α c /α n)–1

Rn = 10.0 k Ω

1 – (0.400 × 10–3/C°)

(– 0.500 × 10–3/C°)

–1

Rn = 5.56 k Ω and Rc = 4.44 k Ω

27.31 (a) ρ = ρ0 1+ α (T − T0 )[ ] = (2.82 × 10−8 Ω ⋅ m) 1+ 3.90 × 10−3(30.0°)[ ] = 3.15 × 10-8 Ω · m

(b) J = E

ρ= 0.200 V / m

3.15 × 10−8 Ω ⋅ m= 6.35 × 106 A / m2

4J = π(1.00 × 10−4 m)2

4(6.35 × 106 A / m2 ) = 49.9 mA

(d)

n = 6.02 × 1023 electrons

26.98 g2.70 × 106 g / m3

= 6.02 × 1028 electrons/ m3

vd = J

ne= (6.35 × 106 A / m2 )

(6.02 × 1028 electrons/ m3)(1.60 × 10−19 C)= 659 µm/s

(e) ∆V = E = (0.200 V / m)(2.00 m) = 0.400 V

120 Chapter 27 Solutions

*27.32 For aluminum, αE = 3.90 × 10–3/°C (Table 27.1) α = 24.0 × 10–6/°C (Table 19.2)

R = ρl

ρ0 1 + αE∆T( ) 1 + α∆T( )A 1 + α∆T( )2 = R0

(1 + αE ∆T)(1 + α ∆T)

= (1.234 Ω) (1.39)

(1.0024) = 1.71 Ω

27.33 R = R0[1 + α ∆T]

R – R0 = R0α ∆T

R – R0R0

= α ∆T = (5.00 × 10-3)25.0 = 0.125

27.34 Assuming linear change of resistance with temperature, R = R0(1 + α ∆T)

R 77 K= 1.00 Ω( ) 1 + 3.92 × 10−3( ) −216°C( )[ ] = 0.153 Ω

27.35 ρ = ρ0 1 + α ∆T( ) or ∆TW = 1

αW

ρW

ρ0W− 1

Require that ρW = 4ρ0Cu so that

∆TW = 14.50 × 10−3 /°C

4 1.70 × 10−8( )

5.60 × 10−8 − 1

= 47.6 °C

Therefore, TW = 47.6 °C + T0 = 67.6°C

27.36 α = 1

∆R∆T

= 1

2R0 − R0

T − T0 = 1

T − T0

so, T = 1

+ T0 and

T = 1

0.400 × 10−3 C°−1

+20.0 °C so T = 2.52 × 103 °C

*27.37 I = P

∆V =

600 W120 V = 5.00 A

and R = ∆VI =

120 V5.00 A = 24.0 Ω

Chapter 27 Solutions 121

27.38 P = 0.800(1500 hp)(746 W/hp) = 8.95 × 105 W

P = I (∆V)

8.95 × 105 = I(2000)

I = 448 A

27.39 The heat that must be added to the water is

Q = mc ∆T = (1.50 kg)(4186 J/kg°C)(40.0°C) = 2.51 × 105 J

Thus, the power supplied by the heater is

P = Wt =

Qt =

2.51 × 105 J600 s = 419 W

and the resistance is R = ∆V( )2

P =

(110 V)2

419 W = 28.9 Ω

27.40 The heat that must be added to the water is Q = mc(T2 – T1)

Thus, the power supplied by the heat is P = W

∆t= Q

∆t=

mc T2 − T1( )t

and the resistance is R = ∆V( )2

∆V( )2 tmc T2 − T1( )

27.41

P0= (∆V)2 / R

(∆V0 )2 / R= ∆V

∆V0

= 140120

= 1.361

∆% = P −P0

100%( ) = P

P0− 1

100%( ) = (1.361 − 1)100 = 36.1%

122 Chapter 27 Solutions

Goal Solution Suppose that a voltage surge produces 140 V for a moment. By what percentage does the power output ofa 120-V, 100-W light bulb increase? (Assume that its resistance does not change.)

G : The voltage increases by about 20%, but since , the power will increase as the square ofthe voltage:

or a 36.1% increase.

O : We have already found an answer to this problem by reasoning in terms of ratios, but we can alsocalculate the power explicitly for the bulb and compare with the original power by using Ohm’s lawand the equation for electrical power. To find the power, we must first find the resistance of the bulb,which should remain relatively constant during the power surge (we can check the validity of thisassumption later).

A : From , we find that

The final current is, I f =

∆Vf

R= 140 V

144 Ω= 0.972 A

The power during the surge is

So the percentage increase is

136 W − 100 W100 W

= 0.361 = 36.1%

L : Our result tells us that this 100 - W light bulb momentarily acts like a 136 - W light bulb, whichexplains why it would suddenly get brighter. Some electronic devices (like computers) are sensitiveto voltage surges like this, which is the reason that surge protectors are recommended to protectthese devices from being damaged.

In solving this problem, we assumed that the resistance of the bulb did not change during thevoltage surge, but we should check this assumption. Let us assume that the filament is made oftungsten and that its resistance will change linearly with temperature according to equation 27.21.Let us further assume that the increased voltage lasts for a time long enough so that the filamentcomes to a new equilibrium temperature. The temperature change can be estimated from the powersurge according to Stefan’s law (equation 20.18), assuming that all the power loss is due to radiation.By this law, so that a 36% change in power should correspond to only about a 8% increase i ntemperature. A typical operating temperature of a white light bulb is about 3000 °C, so

∆T ≈ 0.08 3273 °C( ) = 260 °C. Then the increased resistance would be roughly

R = R0 1 + α T − T0( )( ) = 144 Ω( ) 1 + 4.5 × 10−3 260( )( ) ≅ 310 Ω

It appears that the resistance could change double from 144 Ω. On the other hand, if the voltagesurge lasts only a very short time, the 136 W we calculated originally accurately describes theconversion of electrical into internal energy in the filament.

Chapter 27 Solutions 123

27.42 P = I (∆V) = (∆V)2

R = 500 W

R = (110 V)2

(500 W) = 24.2 Ω

(a) R = ρ

Al so l

= RA

ρ= (24.2 Ω)π(2.50 × 10−4 m)2

1.50 × 10−6 Ω ⋅ m = 3.17 m

(b) R = R0 [1 + α ∆T] = 24.2 Ω 1 + (0.400 × 10−3 )(1180)[ ] = 35.6 Ω

P = (∆V)2

R= (110)2

35.6= 340 W

27.43 R = ρl

A =

(1.50 × 10–6 Ω · m)25.0 mπ (0.200 × 10–3 m)2 = 298 Ω

∆V = IR = (0.500 A)(298 Ω) = 149 V

(a) E = ∆V

l =

149 V25.0 m = 5.97 V/m

(b) P = (∆V)I = (149 V)(0.500 A) = 74.6 W

I = ∆VR =

(149 V)(337 Ω)

= 0.443 A

P = (∆V)I = (149 V)(0.443 A) = 66.1 W

27.44 (a) ∆U = q (∆V) = It (∆V) = (55.0 A · h)(12.0 V)

1 C

1 A · s

1 J

1 V · C

1 W · s

1 J = 660 W · h = 0.660 kWh

(b) Cost = 0.660 kWh

$0.0600

1 kWh = 3.96¢

27.45 P = I (∆V) ∆V = IR

P = (∆V)2

R= (10.0)2

120= 0.833 W

124 Chapter 27 Solutions

27.46 The total clock power is 270 × 106 clocks( ) 2.50

J sclock

3600 s1 h

= 2.43 × 1012 J h

From e = Wout

Qin, the power input to the generating plants must be:

Qin

t= Wout t

e= 2.43 × 1012 J h

0.250= 9.72 × 1012 J h

and the rate of coal consumption is

Rate = 9.72 × 1012 J h( ) 1.00 kg coal

33.0 × 106 J

= 2.95 × 105 kg coal

h =

295

metric tonh

27.47 P = I ∆V( ) = 1.70 A( ) 110 V( ) = 187 W

Energy used in a 24-hour day = (0.187 kW)(24.0 h) = 4.49 kWh

∴ cost = 4.49 kWh

$0.0600

k W h = $0.269 = 26.9¢

27.48 P = I (∆V) = (2.00 A)(120 V) = 240 W

∆U = (0.500 kg)(4186 J/kg°C)(77.0°C) = 161 kJ

= 1.61 × 105 J

240 W = 672 s

27.49 At operating temperature,

(a) P = I (∆V) = (1.53 A)(120 V) = 184 W

(b) Use the change in resistance to find the final operating temperature of the toaster.

R = R0(1 + α ∆T)

1201.53 =

1201.80

1 + (0.400 × 10−3 )∆T[ ]

∆T = 441°C

T = 20.0°C + 441°C = 461°C

Chapter 27 Solutions 125

Goal Solution A certain toaster has a heating element made of Nichrome resistance wire. When the toaster is firstconnected to a 120-V source of potential difference (and the wire is at a temperature of 20.0 °C), the initialcurrent is 1.80 A. However, the current begins to decrease as the resistive element warms up. When thetoaster has reached its final operating temperature, the current has dropped to 1.53 A. (a) Find the powerthe toaster consumes when it is at its operating temperature. (b) What is the final temperature of theheating element?

G : Most toasters are rated at about 1000 W (usually stamped on the bottom of the unit), so we mightexpect this one to have a similar power rating. The temperature of the heating element should be hotenough to toast bread but low enough that the nickel-chromium alloy element does not melt. (Themelting point of nickel is 1455 °C, and chromium melts at 1907 °C.)

O : The power can be calculated directly by multiplying the current and the voltage. The temperature canbe found from the linear conductivity equation for Nichrome, with α = 0.4 × 10−3 °C-1 from Table27.1.

A : (a) P = ∆V( )I = 120 V( ) 1.53 A( ) = 184 W

(b) The resistance at 20.0 °C is R0 = ∆V

I= 120 V

1.80 A= 66.7 Ω

At operating temperature, R = 120 V

1.53 A= 78.4 Ω

Neglecting thermal expansion, R = ρl

ρ0 1 + α T − T0( )( )lA

= R0 1 + α T − T0( )( )

T = T0 + R R0 − 1

α= 20.0 °C + 78.4 Ω 66.7 Ω − 1

0.4 × 10−3 °C-1 = 461 °C

L : Although this toaster appears to use significantly less power than most, the temperature seems highenough to toast a piece of bread in a reasonable amount of time. In fact, the temperature of a typical1000-W toaster would only be slightly higher because Stefan’s radiation law (Eq. 20.18) tells us that(assuming all power is lost through radiation) , so that the temperature might be about 700 °C.In either case, the operating temperature is well below the melting point of the heating element.

27.50 P = (10.0 W / ft2 )(10.0 ft)(15.0 ft) = 1.50 kW

Energy = P t = (1.50 kW)(24.0 h) = 36.0 kWh

Cost = (36.0 kWh)($0.0800 / kWh) = $2.88

*27.51 Consider a 400-W blow dryer used for ten minutes daily for a year. The energy converted is

P t = 400 J s( ) 600 s d( ) 365 d( ) ≅ 9 × 107 J

1 kWh3.6 × 106 J

≅ 20 kWh

We suppose that electrical energy costs on the order of ten cents per kilowatt-hour. Then thecost of using the dryer for a year is on the order of

Cost ≅ 20 kWh( ) $0.100 kWh( ) = $2 ~$1

126 Chapter 27 Solutions

*27.52 (a) I = ∆VR so P = (∆V)I =

(∆V)2

= (120 V)2

25.0 W = 576 Ω and = (120 V)2

100 W = 144 Ω

(b) I = = 25.0 W120 V = 0.208 A =

Qt =

1.00 Ct

t = 1.00 C

0.208 A = 4.80 s The charge has lower potential energy .

t = 1.00 J

25.0 W = 0.0400 s The energy changes from electrical to heat and light .

(d) ∆U = P t = (25.0 J/s)(86400 s/d)(30.0 d) = 64.8 × 106 J

The energy company sells energy .

Cost = 64.8 × 106 J

$0.0700

kWh

1000

W · s

3600 s = $1.26

Cost per joule = $0.0700k W h

k W h

3.60 × 106 J = $1.94 × 10–8/J

*27.53 We find the drift velocity from I = nqvd A = nqvd π r 2

vd = I

n qπ r 2 =

1000 A8.00 × 1028 m–3 (1.60 × 10–19 C)π (10–2 m)2 = 2.49 × 10–4 m/s

v = xt t =

xv =

200 × 103 m2.49 × 10–4 m/s

= 8.04 × 108 s = 25.5 yr

*27.54 The resistance of one wire is

0.500 Ω

m i (100 mi) = 50.0 Ω

The whole wire is at nominal 700 kV away from ground potential, but the potential differencebetween its two ends is

IR = (1000 A)(50.0 Ω) = 50.0 kV

Then it radiates as heat power P = (∆V)I = (50.0 × 103 V)(1000 A) = 50.0 MW

Chapter 27 Solutions 127

27.55 We begin with the differential equation α = 1ρ

dρd T

(a) Separating variables, ⌡⌠

ρ0 dρρ

= ⌡⌠

T0 α d T

ρ0 = α (T – T0) and ρ = ρ0eα(T – T0)

(b) From the series expansion e x ≅ 1 + x, (x << 1),

ρ ≅ ρ0 1 + α (T − T0 )[ ]

*27.56 Consider a 1.00-m length of cable. The potential difference between its ends is

= 6.67 mV

The resistance is R = ∆V

I = 6.67 × 10–3 V

300 A = 22.2 µ Ω

Then gives 1.56 cm

27.57

(m) R(Ω) ρ (Ω · m)0.540 10.4 1.41 × 10–6

1.028 21.1 1.50 × 10–6

1.543 31.8 1.50 × 10–6

ρ– = 1.47 × 10–6 Ω · m (in agreement with tabulated value)

1.50 × 10–6 Ω · m (Table 27.1)

128 Chapter 27 Solutions

27.58 2 wires → = 100 m

R = 0.108 Ω300 m (100 m) = 0.0360 Ω

(a) (∆V)home = (∆V)line – IR = 120 – (110)(0.0360) = 116 V

(b) P = I (∆V) = (110 A)(116 V) = 12.8 kW

*27.59 (a) E = − dV

dxi = − 0 − 4.00( )V

0.500 − 0( ) m = 8 00. i V m

(b)

R = ρlA

=4.00 × 10−8 Ω ⋅ m( ) 0.500 m( )

π 1.00 × 10−4 m( )2 = 0.637 Ω

R= 4.00 V

0.637 Ω= 6.28 A

(d)

J = IA

i = 6.28 A

π 1.00 × 10−4 m( )2 = 2.00 × 108 i A m2 = 200i MA m2

(e) ρJ = 4.00 × 10−8 Ω ⋅ m( ) 2.00 × 108 i A m2( ) = 8.00i V m = E

*27.60 (a) E = – dV(x)

dx i = VL i

(b) R = ρl

A =

4ρL

π d 2

Vπ d 2

4ρL

(d) J = I

A i = V

ρL i

(e) ρ J = VL i = E

Chapter 27 Solutions 129

27.61 R = R0 1 + α (T − T0 )[ ] so T = T0 + 1

αRR0

− 1

= T0 + 1

αI0

I− 1

In this case, I = I010 , so

T = T0 + 1

α(9) = 20° + 9

0.00450/°C = 2020 °C

27.62 R = ∆V

I = 12.0

I = 6.00

(I – 3.00) thus 12.0I – 36.0 = 6.00I and I = 6.00 A

Therefore, R = 12.0 V6.00 A = 2.00 Ω

27.63 (a) = I ∆V( ) so 667 A

(b) and d = vt = 20.0 m / s( ) 2.50 × 103 s( ) = 50.0 km

27.64 (a) We begin with R = ρl

ρ0 1 + α (T − T0 )[ ] l0 1 + ′α (T − T0 )[ ] A0 1 + 2 ′α (T − T0 )[ ]

which reduces to R =

R0 1 + α (T − T0 )[ ] 1 + ′α (T − T0 )[ ] 1 + 2 ′α (T − T0 )[ ]

(b) For copper: ρ0 = 1.70 × 10−8 Ω ⋅ m, α = 3.90 × 10−3 °C −1, and ′α = 17.0 × 10−6 °C−1

R0 = ρ0l0

A0= (1.70 × 10−8 )(2.00)

π(0.100 × 10−3 )2 = 1.08 Ω

The simple formula for R gives:

R = 1.08 Ω( ) 1 + (3.90 × 10−3 °C−1)(100°C − 20.0°C)[ ] = 1.420 Ω

while the more complicated formula gives:

R =1.08 Ω( ) 1+ (3.90 × 10−3 °C−1)(80.0°C)[ ] 1+ (17.0 × 10−6 °C−1)(80.0°C)[ ]

1 + 2(17.0 × 10−6 °C−1)(80.0°C)[ ] = 1.418 Ω

130 Chapter 27 Solutions

27.65 Let α be the temperature coefficient at 20.0°C, and α′ be the temperature coefficient at 0 °C.Then ρ = ρ0 1 + α T − 20.0°C( )[ ] , and ρ = ′ρ 1 + ′α T − 0°C( )[ ] must both give the correct resistivityat any temperature T. That is, we must have:

ρ0 1 + α (T − 20.0°C)[ ] = ρ′ 1 + α ′ T − 0°C( )[ ] (1)

Setting T = 0 in equation (1) yields: ′ρ = ρ0 1 − α (20.0°C)[ ] ,

and setting T = 20.0°C in equation (1) gives: ρ0 = ′ρ 1 + ′α (20.0°C)[ ]Put ′ρ from the first of these results into the second to obtain:

ρ0 = ρ0 1 − α (20.0°C)[ ] 1 + α ′ (20.0°C)[ ]Therefore

1 + α ′ 20.0°C( ) = 1

1 − α 20.0°C( )which simplifies to

α ′= α

1 − α (20.0 °C)[ ]

From this, the temperature coefficient, based on a reference temperature of 0°C, may becomputed for any material. For example, using this, Table 27.1 becomes at 0°C :

Material Temp Coefficients at 0°CSilver 4.1 × 10− 3/˚CCopper 4.2 × 10− 3/˚CGold 3.6 × 10− 3/˚CAluminum 4.2 × 10− 3/˚CTungsten 4.9 × 10− 3/˚CIron 5.6 × 10− 3/˚CPlatinum 4.25 × 10− 3/˚CLead 4.2 × 10− 3/˚CNichrome 0.4 × 10− 3/˚CCarbon −0.5 × 10− 3/˚CGermanium −24 × 10− 3/˚CSilicon −30 × 10− 3/˚C

27.66 (a) R = ρl

ρL

π rb2 − ra

2( )

(b)

R =3.50 × 105 Ω ⋅ m( ) 0.0400 m( )

π 0.0120 m( )2 − 0.00500 m( )2[ ] = 3.74 × 107 Ω = 37.4 MΩ

2πLdrrra

rb∫ =

ρ2πL

lnrb

(d) R =

3.50 × 105 Ω ⋅ m( )2π 0.0400 m( ) ln

1.200.500

= 1.22 × 106 Ω = 1.22 MΩ

Chapter 27 Solutions 131

27.67 Each speaker receives 60.0 W of power. Using = I 2 R, we then have

I = =

60.0 W4.00 Ω

= 3.87 A

The system is not adequately protected since the fuse should be set to melt at 3.87 A, or less.

27.68 ∆V = –E · l or dV = –E · dx

∆V = –IR = –E · l

I = dq

dt= E ⋅l

R= A

ρlE ⋅l = A

ρE = −σA

dVdx

= σA

dVdx

Current flows in the direction of decreasing voltage. Energy flows as heat in the direction ofdecreasing temperature.

27.69 R = ρ dx

A= ρ dx

wy∫∫ where y = y1 + y2 − y1

R = ρw

y1 + y2 − y1

= ρLw(y2 − y1)

ln y1 + y2 − y1

∫0

R =

ρLw(y2 − y1)

lny2

27.70 From the geometry of the longitudinal section of the resistor shown in thefigure, we see that

(b − r)y

= (b − a)h

From this, the radius at a distance y from the base is r = (a − b)

+ b

For a disk-shaped element of volume dR = ρ dy

πr 2 : R = ρ

πdy

(a − b)(y / h) + b[ ]20

∫ .

Using the integral formula

du(au + b)2 = − 1

a(au + b)∫ , R = ρ

πhab

132 Chapter 27 Solutions

27.71 I = I0 exp e ∆V / kBT( ) − 1[ ] and R = ∆V

with I0 = 1.00 × 10−9 A, e = 1.60 × 10−19 C, and kB = 1.38 × 10−23 J K .

The following includes a partial table of calculated valuesand a graph for each of the specified temperatures.

(i) For T = 280 K :

∆V V( ) I A( ) R Ω( )0.400 0.0156 25.60.440 0.0818 5.380.480 0.429 1.120.520 2.25 0.2320.560 11.8 0.04760.600 61.6 0.0097

(ii) For T = 300 K :

∆V V( ) I A( ) R Ω( )0.400 0.005 77.30.440 0.024 18.10.480 0.114 4.220.520 0.534 0.9730.560 2.51 0.2230.600 11.8 0.051

(iii) For T = 320 K :

∆V V( ) I A( ) R Ω( )0.400 0.0020 2030.440 0.0084 52.50.480 0.0357 13.40.520 0.152 3.420.560 0.648 0.8640.600 2.76 0.217

Chapter 28 Solutions

28.1 (a) P = ∆V( )2

R becomes 20.0 W =

(11.6 V)2

R so R = 6.73 Ω

(b) ∆V = IR so 11.6 V = I (6.73 Ω) and I = 1.72 A

ε = IR + Ir so 15.0 V = 11.6 V + (1.72 A)r

r = 1.97 Ω Figure for Goal

Solution

Goal SolutionA battery has an emf of 15.0 V. The terminal voltage of the battery is 11.6 V when it is delivering 20.0 Wof power to an external load resistor R. (a) What is the value of R? (b) What is the internal resistance ofthe battery?

G : The internal resistance of a battery usually is less than 1 Ω, with physically larger batteries having lessresistance due to the larger anode and cathode areas. The voltage of this battery drops significantly(23%), when the load resistance is added, so a sizable amount of current must be drawn from thebattery. If we assume that the internal resistance is about 1 Ω, then the current must be about 3 A togive the 3.4 V drop across the battery’s internal resistance. If this is true, then the load resistancemust be about R ≈ 12 V / 3 A = 4 Ω .

O : We can find R exactly by using Joule’s law for the power delivered to the load resistor when thevoltage is 11.6 V. Then we can find the internal resistance of the battery by summing the electricpotential differences around the circuit.

A : (a) Combining Joule's law, P = ∆VI , and the definition of resistance, ∆V = IR , gives

R = ∆V2

P= 11.6 V( )2

20.0 W= 6.73 Ω

(b) The electromotive force of the battery must equal the voltage drops across the resistances: ε = IR + Ir , where I = ∆V R .

r = ε − IR

I= ε − ∆V( )R

∆V= (15.0 V − 11.6 V)(6.73 Ω)

11.6 V= 1.97 Ω

L : The resistance of the battery is larger than 1 Ω, but it is reasonable for an old battery or for a batteryconsisting of several small electric cells in series. The load resistance agrees reasonably well with ourprediction, despite the fact that the battery’s internal resistance was about twice as large as weassumed. Note that in our initial guess we did not consider the power of the load resistance;however, there is not sufficient information to accurately solve this problem without this data.

Chapter 28 Solutions 135

28.2 (a) ∆Vterm = IR

becomes 10.0 V = I (5.60 Ω)

so I = 1.79 A

(b) ∆Vterm = ε – Ir

becomes 10.0 V = ε – (1.79 A)(0.200 Ω)

so ε = 10.4 V

28.3 The total resistance is R = 3.00 V

0.600 A = 5.00 Ω

(a) Rlamp = R – rbatteries = 5.00 Ω – 0.408 Ω = 4.59 Ω

(b)

Pbatteries

Ptotal =

(0.408 Ω)I 2

(5.00 Ω)I 2 = 0.0816 = 8.16%

28.4 (a) Here ε = I(R + r), so I = ε

R + r= 12.6 V

(5.00 Ω + 0.0800 Ω)= 2.48 A

Then, ∆V = IR = 2.48 A( ) 5.00 Ω( ) = 12.4 V

(b) Let I1 and I2 be the currents flowing through the battery and the headlights, respectively.

Then, I1 = I2 + 35.0 A , and ε − I1r − I2R = 0

so ε = (I2 + 35.0 A)(0.0800 Ω) + I2(5.00 Ω) = 12.6 V

giving I2 = 1.93 A

Thus, ∆V2 = (1.93 A)(5.00 Ω) = 9.65 V

28.5 ∆V = I 1R1 = (2.00 A)R 1 and ∆V = I 2(R 1 + R 2) = (1.60 A)(R 1 + 3.00 Ω)

Therefore, (2.00 A)R 1 = (1.60 A)(R 1 + 3.00 Ω) or R 1 = 12.0 Ω

136 Chapter 28 Solutions

28.6 (a) Rp = 1

1 7.00 Ω( ) + 1 10.0 Ω( ) = 4.12 Ω

Rs = R1 + R2 + R3 = 4.00 + 4.12 + 9.00 = 17.1 Ω

(b) ∆V = IR

34.0 V = I (17.1 Ω)

I = 1.99 A for 4.00 Ω, 9.00 Ω resistors

Applying ∆V = IR, (1.99 A)(4.12 Ω) = 8.18 V

8.18 V = I (7.00 Ω) so I = 1.17 A for 7.00 Ω resistor

8.18 V = I (10.0 Ω) so I = 0.818 A for 10.0 Ω resistor

*28.7 If all 3 resistors are placed in parallel ,

1R =

1500 +

2250 =

5500 and R = 100 Ω

*28.8 For the bulb in use as intended,

I = P

∆V= 75.0 W

120 V= 0.625 A and R =

∆VI =

120 V0.625 A = 192 Ω

Now, presuming the bulb resistance is unchanged,

I = 120 V

193.6 Ω = 0.620 A

Across the bulb is ∆V = IR = 192 Ω(0.620 A) = 119 V

so its power is P = (∆V)I = 119 V(0.620 A) = 73.8 W

Chapter 28 Solutions 137

28.9 If we turn the given diagram on its side, we find that it is thesame as Figure (a). The 20.0-Ω and 5.00-Ω resistors are in series, sothe first reduction is as shown in (b). In addition, since the 10.0-Ω,5.00-Ω, and 25.0-Ω resistors are then in parallel, we can solve fortheir equivalent resistance as:

Req = 11

10.0 Ω+ 1

5.00 Ω+ 1

25.0 Ω( ) = 2.94 Ω

This is shown in Figure (c), which in turn reduces to the circuitshown in (d).

Next, we work backwards through the diagrams applyingI = ∆V/R and ∆V = IR. The 12.94-Ω resistor is connected across25.0-V, so the current through the battery in every diagram is

I = ∆VR =

25.0 V12.94 Ω

= 1.93 A

In Figure (c), this 1.93 A goes through the 2.94-Ω equivalentresistor to give a potential difference of:

∆V = IR = (1.93 A)(2.94 Ω) = 5.68 V

From Figure (b), we see that this potential difference is the sameacross Vab, the 10-Ω resistor, and the 5.00-Ω resistor.

(b) Therefore, Vab = 5.68 V

(a) Since the current through the 20.0-Ω resistor is also the currentthrough the 25.0-Ω line ab,

I = VabR ab

= 5.68 V25.0 Ω

= 0.227 A = 227 mA

(a)

(b)

28.10 120 V = IReq = I

ρlA1

+ ρlA2

+ ρlA3

+ ρlA4

, or

Iρl = 120 V( )1

A1+ 1

A2+ 1

A3+ 1

∆V2 = IρlA2

= 120 V( )

A21

A1+ 1

A2+ 1

A3+ 1

= 29.5 V

138 Chapter 28 Solutions

28.11 (a) Since all the current flowing in the circuit must pass through theseries 100-Ω resistor, P = RI2

Pmax = RImax2 so

Imax = P

R= 25.0 W

100 Ω= 0.500 A

R eq = 100 Ω +

100 + 1

100 –1

Ω = 150 Ω

∆Vmax = R eq I max = 75.0 V

(b) P = (∆V)I = (75.0 V)(0.500 A) = 37.5 W total power

P 1 = 25.0 W P 2 = P 3 = RI 2 = (100 Ω)(0.250 A)2 = 6.25 W

28.12 Using 2.00-Ω, 3.00-Ω, 4.00-Ω resistors, there are 7 series, 4 parallel, and 6 mixed combinations:

Series Parallel Mixed The resistors may be arranged in patterns:2.00 Ω 6.00 Ω3.00 Ω 7.00 Ω4.00 Ω 9.00 Ω5.00 Ω

0.923 Ω1.56 Ω1.20 Ω 2.00 Ω1.33 Ω 2.22 Ω1.71 Ω 3.71 Ω

4.33 Ω5.20 Ω

28.13 The potential difference is the same across either combination.

∆V = IR = 3I

11R + 1

500( ) so R

+ 1500

= 3

1 + R

500 = 3 and R = 1000 Ω = 1.00 kΩ

28.14 If the switch is open, I = ε /( ′R + R) and P = ε 2 ′R /( ′R + R)2

If the switch is closed, I = ε /(R + ′R / 2) and P = ε 2 ( ′R / 2)/(R + ′R / 2)2

Then,

ε 2 ′R( ′R + R)2 =

ε 2 ′R2(R + ′R / 2)2

2R2 + 2R ′R + ′R 2 / 2 = ′R 2 + 2R ′R + R2

The condition becomes R2 = ′R 2 / 2 so ′R = 2 R = 2 (1.00 ) =Ω 1.41 Ω

Chapter 28 Solutions 139

28.15 Rp = 1

3.00+ 1

1.00

−1

= 0.750 Ω

Rs = 2.00 + 0.750 + 4.00( ) Ω = 6.75 Ω

Ibattery = ∆V

Rs= 18.0 V

6.75 Ω= 2.67 A

P = I2R: P 2 = 2.67 A( )2 2.00 Ω( )

P 2 = 14.2 W in 2.00 Ω

P 4 = 2.67 A( )2 4.00 Ω( ) = 28.4 W in 4.00 Ω

∆V2 = 2.67 A( ) 2.00 Ω( ) = 5.33 V, ∆V4 = 2.67 A( ) 4.00 Ω( ) = 10.67 V

∆Vp = 18.0 V − ∆V2 − ∆V4 = 2.00 V = ∆V3 = ∆V1( )

P 3 =

∆V3( )2

R3=

2.00 V( )2

3.00 Ω= 1.33 W in 3.00 Ω

P 1 =

∆V1( )2

R1=

2.00 V( )2

1.00 Ω= 4.00 W in 1.00 Ω

28.16 Denoting the two resistors as x and y,

x + y = 690, and 1

150 = 1x +

1150 =

1x +

1690 – x =

(690 – x) + xx(690 – x)

x 2 – 690x + 103,500 = 0

x = 690 ± (690)2 – 414,000

x = 470 Ω y = 220 Ω

140 Chapter 28 Solutions

28.17 (a) ∆V = IR : 33.0 V = I1 11.0 Ω( ) 33.0 V = I2 22.0 Ω( )

I1 = 3.00 A I2 = 1.50 A

P = I2R: P 1 = 3.00 A( )2 11.0 Ω( ) P 2 = 1.50 A( )2 22.0 Ω( )

P 1 = 99.0 W P 2 = 49.5 W

The 11.0-Ω resistor uses more power.

(b) P 1 + P 2 = 148 W P = I ∆V( ) = 4.50( ) 33.0( ) = 148 W

∆V = IR : 33.0 V = I 33.0 Ω( ), so I = 1.00 A

P = I2R: P 1 = 1.00 A( )2 11.0 Ω( ) P 2 = 1.00 A( )2 22.0 Ω( )

P 1 = 11.0 W P 2 = 22.0 W

The 22.0-Ω resistor uses more power.

(d) P 1 + P 2 = I2 R1 + R2( ) = 1.00 A( )2 33.0 Ω( ) = 33.0 W

P = I ∆V( ) = 1.00 A( ) 33.0 V( ) = 33.0 W

(e) The parallel configuration uses more power.

28.18 +15.0 – (7.00)I1 – (2.00)(5.00) = 0

5.00 = 7.00I1 so I1 = 0.714 A

I3 = I1 + I2 = 2.00 A

0.714 + I2 = 2.00 so I2 = 1.29 A

+ε – 2.00(1.29) – (5.00)(2.00) = 0 ε = 12.6 V

Chapter 28 Solutions 141

28.19 We name the currents I1, I2 , and I3 as shown.

From Kirchhoff's current rule, I3 = I1 + I2

Applying Kirchhoff's voltage rule to the loop containing I2 and I3 ,

12.0 V – (4.00)I3 – (6.00)I2 – 4.00 V = 0

8.00 = (4.00)I3 + (6.00)I2

Applying Kirchhoff's voltage rule to the loop containing I1 and I2 ,

– (6.00)I2 – 4.00 V + (8.00)I1 = 0

(8.00)I1 = 4.00 + (6.00)I2

Solving the above linear systems, I1 = 846 mA, I2 = 462 mA, I3 = 1.31 A

All currents flow in the directions indicated by the arrows in the circuit diagram.

*28.20 The solution figure is shown to the right.

*28.21 We use the results of Problem 19.

(a) By the 4.00-V battery: ∆U = (∆V)It = 4.00 V(– 0.462 A)120 s = – 222 J

By the 12.0-V battery: 12.0 V (1.31 A) 120 s = 1.88 kJ

(b) By the 8.00 Ω resistor: I 2 Rt = (0.846 A)2(8.00 Ω ) 120 s = 687 J

By the 5.00 Ω resistor: (0.462 A)2(5.00 Ω ) 120 s = 128 J

By the 1.00 Ω resistor: (0.462 A)2(1.00 Ω ) 120 s = 25.6 J

By the 3.00 Ω resistor: (1.31 A)2(3.00 Ω ) 120 s = 616 J

By the 1.00 Ω r e s i s t o r : (1.31 A)2(1.00 Ω ) 120 s = 205 J

687 J + 128 J + 25.6 J + 616 J + 205 J = 1.66 kJ from electrical to heat.

142 Chapter 28 Solutions

28.22 We name the currents I1, I2 , and I3 as shown.

[1] 70.0 – 60.0 – I2 (3.00 kΩ) – I1 (2.00 kΩ) = 0

[2] 80.0 – I3 (4.00 kΩ) – 60.0 – I2 (3.00 kΩ) = 0

[3] I2 = I1 + I3

(a) Substituting for I2 and solving the resulting simultaneousequations yields

I1 = 0.385 mA (through R1)

I3 = 2.69 mA (through R3)

I2 = 3.08 mA (through R2)

(b) ∆Vcf = – 60.0 V – (3.08 mA)(3.00 kΩ) = – 69.2 V

Point c is at higher potential.

28.23 Label the currents in the branches as shown in the first figure.Reduce the circuit by combining the two parallel resistors asshown in the second figure.

Apply Kirchhoff’s loop rule to both loops in Figure (b) toobtain:

2.71R( )I1 + 1.71R( )I2 = 250 and 1.71R( )I1 + 3.71R( )I2 = 500

With R = 1000 Ω , simultaneous solution of these equationsyields:

I1 = 10.0 mA and I2 = 130.0 mA

From Figure (b), Vc − Va = I1 + I2( ) 1.71R( ) = 240 V

Thus, from Figure (a), I4 = Vc − Va

4R= 240 V

4000 Ω= 60.0 mA

(a)

(b)

Finally, applying Kirchhoff’s point rule at point a in Figure (a) gives:

I = I4 − I1 = 60.0 mA − 10.0 mA = +50.0 mA,

or I = 50.0 mA flowing from point a to point e .

Chapter 28 Solutions 143

28.24 Name the currents as shown in the figure to the right. Thenw + x + z = y. Loop equations are

– 200w – 40.0 + 80.0x = 0

– 80.0x + 40.0 + 360 – 20.0y = 0

+ 360 – 20.0y – 70.0z + 80.0 = 0

Eliminate y by substitution.

x w

x w z

w x z

= 2.50 + 0.500400 100 20.0 20.0 = 0440 20.0 20.0 90.0 = 0

− − −− − −

Eliminate x :

350 270 20.0 = 0430 70.0 90.0 = 0

− −− −

w z

Eliminate z = 17.5 – 13.5w to obtain 430 70.0 1575 + 1215 = 0− −w w

w = 70.0/70.0 = 1.00 A upward in 200 Ω

Now z = 4.00 A upward in 70.0 Ω

x = 3.00 A upward in 80.0 Ω

y = 8.00 A downward in 20.0 Ω

and for the 200 Ω, ∆V = IR = (1.00 A)(200 Ω) = 200 V

28.25 Using Kirchhoff’s rules,

12.0 − 0.0100( )I1 − 0.0600( )I3 = 0

10.0 + 1.00( )I2 − 0.0600( )I3 = 0

and I1 = I2 + I3

12.0 − 0.0100( )I2 − 0.0700( )I3 = 0

10.0 + 1.00( )I2 − 0.0600( )I3 = 0

Solving simultaneously, I2 = 0.283 A downward in thedead battery,

and I3 = 171 A downward in the starter.

144 Chapter 28 Solutions

28.26 Vab = 1.00( )I1 + 1.00( ) I1 − I2( )

Vab = 1.00( )I1 + 1.00( )I2 + 5.00( ) I − I1 + I2( )

Vab = 3.00( ) I − I1( ) + 5.00( ) I − I1 + I2( )

Let I = 1.00 A, I1 = x , and I2 = y

Then, the three equations become:

Vab = 2.00x − y , or y = 2.00x − Vab

Vab = −4.00x + 6.00 y + 5.00

and Vab = 8.00 − 8.00x + 5.00 y

Substituting the first into the last two gives:

7.00Vab = 8.00x + 5.00 and 6.00Vab = 2.00x + 8.00

Solving these simultaneously yields Vab = 27

17 V

Then, Rab = Vab

I= 27 17 V

1.00 A or

Rab = 27

17 Ω

28.27 We name the currents I1, I2 , and I3 as shown.

(a) I1 = I2 + I3

Counterclockwise around the top loop,

12.0 V – (2.00 Ω)I3 – (4.00 Ω)I1 = 0

Traversing the bottom loop,

8.00 V – (6.00 Ω)I2 + (2.00 Ω)I3 = 0

I1 = 3.00 – 12 I3 I2 =

43 +

13 I3 and I3 = 909 mA

(b) Va – (0.909 A)(2.00 Ω) = Vb

Vb – Va = –1.82 V

Chapter 28 Solutions 145

28.28 We apply Kirchhoff's rules to the second diagram.

50.0 – 2.00I1 – 2.00I2 = 0 (1)

20.0 – 2.00I3 + 2.00I2 = 0 (2)

I1 = I2 + I3 (3)

Substitute (3) into (1), and solve for I1, I2, and I3

I1 = 20.0 A; I2 = 5.00 A; I3 = 15.0 A

Then apply P = I 2R to each resistor:

(2.00 Ω)1 : P = I12(2.00 Ω) = (20.0 A)2 (2.00 Ω) = 800 W

(4.00 Ω) : P =

5.002 A

2 (4.00 Ω) = 25.0 W

(Half of I2 goes through each)

(2.00 Ω)3 : P = I32(2.00 Ω) = (15.0 A)2(2.00 Ω) = 450 W

28.29 (a) RC = (1.00 × 106 Ω)(5.00 × 10–6 F) = 5.00 s

(b) Q = Cε = (5.00 × 10–6 C)(30.0 V) = 150 µC

Re−t/RC = 30.0

1.00 × 106 exp−10.0

(1.00 × 106)(5.00 × 10−6 )

= 4.06 µA

28.30 (a) I(t) = –I0e–t/RC

I0 = Q

RC = 5.10 × 10–6 C

(1300 Ω)(2.00 × 10–9 F) = 1.96 A

I(t) = – (1.96 A) exp

–9.00 × 10–6 s

(1300 Ω)(2.00 × 10–9 F) = – 61.6 mA

(b) q(t) = Qe–t/RC = (5.10 µC) exp

– 8.00 × 10–6 s

(1300 Ω)(2.00 × 10–9 F) = 0.235 µC

146 Chapter 28 Solutions

28.31 U = 1

2C ∆V( )2 and ∆V = Q C

Therefore, U = Q2 2C and when the charge decreases to half its original value, the stored

energy is one-quarter its original value: U f = 1

4U0

28.32 (a) τ = RC = (1.50 × 105 Ω)(10.0 × 10–6 F) = 1.50 s

(b) τ = (1.00 × 105 Ω)(10.0 × 10–6 F) = 1.00 s

50.0 × 103 Ω = 200 µA

The 100 kΩ carries current of magnitude I = I0e–t/RC =

10.0 V

100 × 103 Ω e–t/ 1.00 s

So the switch carries downward current 200 µA + (100 µA)e–t / 1.00 s

28.33 (a) Call the potential at the left junction VL and at the right VR. After a"long" time, the capacitor is fully charged.

VL = 8.00 V because of voltage divider: IL = 10.0 V5.00 Ω

= 2.00 A

VL = 10.0 V – (2.00 A)(1.00 Ω) = 8.00 V

Likewise, VR =

2.00 Ω

2.00 Ω + 8.00 Ω 10.0 V = 2.00 V

or IR = 10.0 V10.0 Ω

= 1.00 A

VR = (10.0 V) – (8.00 Ω)(1.00 A) = 2.00 V

Therefore, ∆V = VL – VR = 8.00 – 2.00 = 6.00 V

(b) Redraw the circuit R = 1

1/ 9.00 Ω( ) + 1/ 6.00 Ω( ) = 3.60 Ω

RC = 3.60 × 10–6 s

and e–t/RC = 1

10 so t = RC ln 10 = 8.29 µs

Chapter 28 Solutions 147

28.34 (a) τ = RC = 4.00 × 106 Ω( ) 3.00 × 10−6 F( ) = 12.0 s

(b) I = ε

Re− t/RC

= 12.0

4.00 × 106 e− t/12.0 s

q = Cε 1 − e− t/RC[ ]

= 3.00 × 10−6 12.0( ) 1 − e− t/12.0[ )

q = 36.0 µC 1 − e− t/12.0[ ] I = 3.00 µAe− t/12.0

28.35 ∆V0 = Q

Then, if q(t) = Qe− t/RC ∆V(t) = ∆V0e− t/RC

∆V t( )∆V0

= e− t/RC

Therefore

= exp − 4.00

R 3.60 × 10−6( )

ln12

= − 4.00

R 3.60 × 10−6( )

R = 1.60 MΩ

28.36 ∆V0 = Q

Then, if q(t) = Q e−t RC ∆V(t) = ∆V0( )e−t RC

and

∆V(t)∆V0( ) = e−t RC

When ∆V(t) = 1

2∆V0( ) , then

e−t RC = 1

− t

RC= ln

= −ln 2

Thus, R = t

C ln 2( )

148 Chapter 28 Solutions

28.37 q(t) = Q 1 − e−t/RC[ ] so

q(t)Q

= 1 − e−t/RC

0.600 = 1 − e−0.900/RC or e−0.900/RC = 1 − 0.600 = 0.400

− =0 9000 400

.ln( . )

RCthus

RC = − =0 900

0 400.

ln( . ) 0.982 s

28.38 Applying Kirchhoff’s loop rule, −Ig 75.0 Ω( ) + I − Ig( )Rp = 0

Therefore, if I = 1.00 A when Ig = 1.50 mA ,

Rp =Ig 75.0 Ω( )

I − Ig( ) =1.50 × 10−3 A( ) 75.0 Ω( )1.00 A − 1.50 × 10−3 A

= 0.113 Ω

28.39 Series Resistor → Voltmeter

∆V = IR: 25.0 = 1.50 × 10-3(Rs + 75.0)

Solving, Rs = 16.6 kΩ Figure for GoalSolution

Goal Solution The galvanometer described in the preceding problem can be used to measure voltages. In this case alarge resistor is wired in series with the galvanometer in a way similar to that shown in Figure P28.24bThis arrangement, in effect, limits the current that flows through the galvanometer when large voltagesare applied. Most of the potential drop occurs across the resistor placed in series. Calculate the value ofthe resistor that enables the galvanometer to measure an applied voltage of 25.0 V at full-scale deflection.

G : The problem states that the value of the resistor must be “large” in order to limit the current throughthe galvanometer, so we should expect a resistance of kΩ to MΩ.

O : The unknown resistance can be found by applying the definition of resistance to the portion of thecircuit shown in Figure 28.24b.

A : ∆Vab = 25.0 V; From Problem 38, I = 1.50 mA and Rg = 75.0 Ω . For the two resistors in series,

Req = Rs + Rg so the definition of resistance gives us: ∆Vab = I(Rs + Rg )

Therefore, Rs = ∆Vab

I− Rg = 25.0 V

1.50 × 10−3 A− 75.0 Ω = 16.6 kΩ

L : The resistance is relatively large, as expected. It is important to note that some caution would benecessary if this arrangement were used to measure the voltage across a circuit with a comparableresistance. For example, if the circuit resistance was 17 kΩ, the voltmeter in this problem wouldcause a measurement inaccuracy of about 50%, because the meter would divert about half the currentthat normally would go through the resistor being measured. Problems 46 and 59 address a similarconcern about measurement error when using electrical meters.

Chapter 28 Solutions 149

28.40 We will use the values required for the 1.00-V voltmeter to obtain the internal resistance of thegalvanometer. ∆V = Ig (R + rg)

Solve for rg : rg = ∆VIg

– R = 1.00 V

1.00 × 10-3 A – 900 Ω = 100 Ω

We then obtain the series resistance required for the 50.0-V voltmeter:

R = VIg

– rg = 50.0 V

1.00 × 10-3 A – 100 Ω = 49.9 kΩ

28.41 ∆V = Igrg = I − Ig( )Rp , or

Rp =Igrg

I − Ig( ) =Ig 60.0 Ω( )

I − Ig( )Therefore, to have I = 0.100 A = 100 mA when Ig = 0.500 mA:

Rp = 0.500 mA( ) 60.0 Ω( )

99.5 mA= 0.302 Ω

Figure for GoalSolution

Goal Solution Assume that a galvanometer has an internal resistance of 60.0 Ω and requires a current of0.500 mA to produce full-scale deflection. What resistance must be connected in parallel with thegalvanometer if the combination is to serve as an ammeter that has a full-scale deflection for a current of0.100 A?

G : An ammeter reads the flow of current in a portion of a circuit; therefore it must have a low resistanceso that it does not significantly alter the current that would exist without the meter. Therefore, theresistance required is probably less than 1 Ω.

O : From the values given for a full-scale reading, we can find the voltage across and the current throughthe shunt (parallel) resistor, and the resistance value can then be found from the definition ofresistance.

A : The voltage across the galvanometer must be the same as the voltage across the shunt resistor i nparallel, so when the ammeter reads full scale,

∆V = 0.500 mA( ) 60.0 Ω( ) = 30.0 mV

Through the shunt resistor, I = 100 mA − 0.500 mA = 99.5 mA

Therefore, R = ∆V

I= 30.0 mV

99.5 mA= 0.302 Ω

L : The shunt resistance is less than 1 Ω as expected. It is important to note that some caution would benecessary if this meter were used in a circuit that had a low resistance. For example, if the circuitresistance was 3 Ω, adding the ammeter to the circuit would reduce the current by about 10%, so thecurrent displayed by the meter would be lower than without the meter. Problems 46 and 59 address asimilar concern about measurement error when using electrical meters.

150 Chapter 28 Solutions

28.42 Rx =

R2R3

R1=

R2R3

2.50R2=

1000 Ω2.50

= 400 Ω

28.43 Using Kirchhoff’s rules with Rg << 1,

− 21.0 Ω( )I1 + 14.0 Ω( )I2 = 0, so I1 = 2

3I2

70.0 − 21.0I1 − 7.00 I1 + Ig( ) = 0 , and

70.0 − 14.0I2 − 7.00 I2 − Ig( ) = 0

The last two equations simplify to

10.0 − 4.00 2

3I2( ) = Ig , and 10.0 − 3.00I2 = −Ig

Solving simultaneously yields: Ig = 0.588 A

21.0 Ω

14.0 Ω

7.00 Ω

I2 - Ig

I1 + Ig

70.0 V

28.44 R = ρL

A and

Ri = ρLi

But, V = AL = AiLi , so R = ρL2

V and

Ri = ρLi

Therefore, R =

ρ Li + ∆L( )2

ρLi 1 + ∆L Li( )[ ]2

V= Ri 1 + α[ ]2 where

α ≡ ∆L

This may be written as: R = Ri (1 + 2α + α 2)

28.45

εx

Rs=

εs

Rs; εx =

εsRx

Rs=

48.0 Ω36.0 Ω

(1.0186 V) = 1.36 V

Chapter 28 Solutions 151

*28.46 (a) In Figure (a), the emf sees anequivalent resistance of 200.00 Ω.

I = 6.000 0 V

200.00 Ω= 0.030 000 A

6.0000 V

20.000 Ω

180.00 Ω 180.00 Ω 180.00 Ω

AV AV

(a) (b) (c)

20.000 Ω 20.000 Ω

The terminal potential difference is ∆V = IR = 0.030 000 A( ) 180.00 Ω( ) = 5.400 0 V

(b) In Figure (b), Req = 1

180.00 Ω+ 1

20 000 Ω

−1

= 178.39 Ω

The equivalent resistance across the emf is 178.39 Ω + 0.500 00 Ω + 20.000 Ω = 198.89 Ω

The ammeter reads I = ε

R= 6.000 0 V

198.89 Ω= 0.030 167 A

and the voltmeter reads ∆V = IR = 0.030 167 A( ) 178.39 Ω( ) = 5.381 6 V

1180.50 Ω

+ 120 000 Ω

−1

= 178.89 Ω

Therefore, the emf sends current through Rtot = 178.89 Ω + 20.000 Ω = 198.89 Ω

The current through the battery is I = 6.000 0 V

198.89 Ω= 0.030 168 A

but not all of this goes through the ammeter.

The voltmeter reads ∆V = IR = 0.030 168 A( ) 178.89 Ω( ) = 5.396 6 V

The ammeter measures current I = ∆V

R= 5.396 6 V

180.50 Ω= 0.029 898 A

The connection shown in Figure (c) is better than that shown in Figure (b) for accurate readings.

28.47 (a) P = I(∆V) So for the Heater, I = P

∆V= 1500 W

120 V= 12.5 A

For the Toaster, I = 750 W120 W = 6.25 A

And for the Grill, I = 1000 W120 V = 8.33 A (Grill)

(b) 12.5 + 6.25 + 8.33 = 27.1 A The current draw is greater than 25.0 amps, so this would not besufficient.

152 Chapter 28 Solutions

28.48 (a) P = I2R = I2 ρl

= (1.00 A)2(1.70 × 10−8 Ω ⋅ m)(16.0 ft)(0.3048 m / ft)

π(0.512 × 10−3 m)2 = 0.101 W

(b) P = I 2 R = 100(0.101 Ω) = 10.1 W

28.49 IAl2 RAl = ICu

2 RCu so IAl = RCu

RAlICu = ρCu

ρAlICu = 1.70

2.8220.0( ) = 0.776(20.0) = 15.5 A

*28.50 (a) Suppose that the insulation between either of your fingers and the conductor adjacent is achunk of rubber with contact area 4 mm2 and thickness 1 mm. Its resistance is

R = ρl

A≅

1013 Ω ⋅ m( ) 10−3 m( )4 × 10−6 m2 ≅ 2 × 1015 Ω

The current will be driven by 120 V through total resistance (series)

2 × 1015 Ω + 104 Ω + 2 × 1015 Ω ≅ 5 × 1015 Ω

It is: I = ∆V

120 V5 × 1015 Ω

~ 10−14 A

(b) The resistors form a voltage divider, with the center of your hand at potential Vh 2 , where Vhis the potential of the "hot" wire. The potential difference between your finger and thumb is

∆V = IR ~ 10−14 A( ) 104 Ω( ) ~ 10−10 V . So the points where the rubber meets your fingers are atpotentials of

2+ 10−10 V and

2− 10−10 V

*28.51 The set of four batteries boosts the electric potential of each bit of charge that goes through themby 4 × 1.50 V = 6.00 V. The chemical energy they store is

∆U = q∆V = (240 C)(6.00 J/C) = 1440 J

The radio draws current I = ∆VR =

6.00 V200 Ω

= 0.0300 A

So, its power is P = (∆V)I = (6.00 V)(0.0300 A) = 0.180 W = 0.180 J/s

Then for the time the energy lasts, we have P = E t : t = E

P= 1440 J

0.180 J / s= 8.00 × 103 s

We could also compute this from I = Q/t: t = QI =

240 C0.0300 A = 8.00 × 103 s = 2.22 h

Chapter 28 Solutions 153

*28.52 I = ε

R + r, so

P = I2R = ε 2R

R + r( )2 or R + r( )2 = ε 2

Let x ≡ ε 2

P, then R + r( )2 = xR or R

2 + 2r − x( )R − r2 = 0

With r = 1.20 Ω , this becomes R2 + 2.40 − x( )R − 1.44 = 0,

which has solutions of R =

− 2.40 − x( ) ± 2.40 − x( )2 − 5.762

(a) With ε = 9 20. V and P = 12.8 W , x = 6.61: R =

+4.21 ± 4.21( )2 − 5.762

= 3.84 Ω or

0.375 Ω

(b) For ε = 9 20. V and P = 21.2 W, x ≡ ε 2

P= 3.99

R =

+1.59 ± 1.59( )2 − 5.762

= 1.59 ± −3.222

The equation for the load resistance yields a complex number, so there is no resistance that willextract 21.2 W from this battery. The maximum power output occurs when R = r = 1.20 Ω , andthat maximum is: Pmax = ε 2 4r = 17.6 W

28.53 Using Kirchhoff’s loop rule for the closed loop, +12.0 − 2.00 I − 4.00 I = 0, so I = 2.00 A

Vb − Va = +4.00 V − 2.00 A( ) 4.00 Ω( ) − 0( ) 10.0 Ω( ) = −4.00 V

Thus, ∆Vab = 4.00 V and point a is at the higher potential .

28.54 The potential difference across the capacitor ∆V t( ) = ∆Vmax 1 −e−t RC[ ]

Using 1 Farad = 1 s Ω , 4.00 V = 10.0 V( ) 1 −e

− 3.00 s( ) R 10.0 × 10−6 s Ω( )

Therefore, 0.400 = 1.00 −e− 3.00 × 105 Ω( ) R

or e− 3.00 × 105 Ω( ) R

= 0.600

Taking the natural logarithm of both sides, − 3.00 × 105 Ω

R= ln 0.600( )

and R = − 3.00 × 105 Ω

ln 0.600( ) = + 5.87 × 105 Ω = 587 kΩ

154 Chapter 28 Solutions

28.55 Let the two resistances be x and y.

Then, Rs = x + y = Ps

I2 = 225 W

5.00 A( )2 = 9.00 Ω y = 9.00 Ω – x

and Rp = xy

x + y=Pp

I2 = 50.0 W

5.00 A( )2 = 2.00 Ω

x y

x 9.00 Ω − x( )x + 9.00 Ω − x( ) = 2.00 Ω x

2 − 9.00x + 18.0 = 0

Factoring the second equation, x − 6.00( ) x − 3.00( ) = 0

so x = 6.00 Ω or x = 3.00 Ω

Then, y = 9.00 Ω − x gives y = 3.00 Ω or y = 6.00 Ω

The two resistances are found to be 6.00 Ω and 3.00 Ω .

28.56 Let the two resistances be x and y.

Then, Rs = x + y = Ps

I2 and Rp = xy

x + y=Pp

I2 .

From the first equation, y = Ps

I2 − x , and the second

becomes

x Ps I2 − x( )x + Ps I2 − x( ) =

I2 or x2 − Ps

x +

PsPp

I 4 = 0.

Using the quadratic formula, x =

Ps ± Ps2 − 4PsPp

2I2 .

Then, y = Ps

I2 − x gives y =

Ps > Ps2 − 4PsPp

2I2 .

The two resistances are

Ps + Ps2 − 4PsPp

2I2 and

Ps − Ps2 − 4PsPp

2I2

Chapter 28 Solutions 155

28.57 The current in the simple loop circuit will be I = ε

R + r

(a) ∆Vter = ε – Ir =

εRR + r

and ∆Vter → ε as R → ∞

(b) I = ε

R + r and I →

εr

as R → 0

(R + r)2

dPdR

= ε 2R(−2)(R + r)−3 + ε 2(R + r)−2

= −2ε 2R

(R + r)3 + ε 2

(R + r)2 = 0

Then 2R = R + r and R = r

Figure for GoalSolution

Goal Solution A battery has an emf ε and internal resistance r. A variable resistor R is connected across the terminals ofthe battery. Determine the value of R such that (a) the potential difference across the terminals is amaximum, (b) the current in the circuit is a maximum, (c) the power delivered to the resistor is amaximum.

G : If we consider the limiting cases, we can imagine that the potential across the battery will be amaximum when R = ∞ (open circuit), the current will be a maximum when R = 0 (short circuit), andthe power will be a maximum when R is somewhere between these two extremes, perhaps whenR = r.

O : We can use the definition of resistance to find the voltage and current as functions of R, and thepower equation can be differentiated with respect to R.

A : (a) The battery has a voltage ∆Vterminal = ε − Ir = εR

R + ror as R → ∞ , ∆Vterminal → ε

(b) The circuit's current is I = ε

R + ror as R → 0,

I → ε

(R + r)2

To maximize the power P as a function of R , we differentiate with respect to R , and require that dP/dR= 0

dPdR

= ε 2R(−2)(R + r)−3 + ε 2(R + r)−2

= −2ε 2R

(R + r)3 + ε 2

(R + r)2 = 0

Then 2R = R + r and R = r

L : The results agree with our predictions. Making load resistance equal to the source resistance tomaximize power transfer is called impedance matching.

156 Chapter 28 Solutions

28.58 (a) ε − I(ΣR) − (ε1 + ε2 ) = 0

40.0 V − (4.00 A) (2.00 + 0.300 + 0.300 + R)Ω[ ] − (6.00 + 6.00) V = 0; so R = 4.40 Ω

(b) Inside the supply, P = I2R = 4.00 A( )2 2.00 Ω( ) = 32.0 W

Inside both batteries together, P = I2R = 4.00 A( )2 0.600 Ω( ) = 9.60 W

For the limiting resistor, P = 4.00 A( )2 4.40 Ω( ) = 70.4 W

28.59 Let Rm = measured value, R = actual value,

IR = current through the resistor R

I = current measured by the ammeter.

(a) When using circuit (a), IRR = ∆V = 20 000(I – IR) or R = 20 000

IIR

− 1

But since I = ∆VRm

and IR = ∆VR , we have

IIR =

RRm

(a)

(b)

Figure for Goalsolution

and R = 20 000 (R – Rm)

Rm (1)

When R > Rm, we require (R – Rm)

R ≤ 0.0500

Therefore, Rm ≥ R (1 – 0.0500) and from (1) we find R ≤ 1050 Ω

(b) When using circuit (b), IRR = ∆V – IR (0.5 Ω).

But since IR = ∆VRm

, Rm = (0.500 + R) (2)

When Rm > R, we require (Rm – R)

R ≤ 0.0500

From (2) we find R ≥ 10.0 Ω

Chapter 28 Solutions 157

Goal Solution The value of a resistor R is to be determined using the ammeter-voltmeter setup shown in Figure P28.59.The ammeter has a resistance of 0.500 Ω, and the voltmeter has a resistance of 20000 Ω . Within whatrange of actual values of R will the measured values be correct to within 5.00% if the measurement ismade using (a) the circuit shown in Figure P28.59a (b) the circuit shown in Figure P28.59b?

G : An ideal ammeter has zero resistance, and an ideal voltmeter has infinite resistance, so that addingthe meter does not alter the current or voltage of the existing circuit. For the non-ideal meters in thisproblem, a low values of R will give a large voltage measurement error in circuit (b), while a largevalue of R will give significant current measurement error in circuit (a). We could hope that thesemeters yield accurate measurements in either circuit for typical resistance values of 1 Ω to 1 MΩ .

O : The definition of resistance can be applied to each circuit to find the minimum and maximum currentand voltage allowed within the 5.00% tolerance range.

A : (a) In Figure P28.59a, at least a little current goes through the voltmeter, so less current flows throughthe resistor than the ammeter reports, and the resistance computed by dividing the voltage by theinflated ammeter reading will be too small. Thus, we require that ∆V/ I = 0.950R where I is thecurrent through the ammeter. Call IR the current through the resistor; then I − IR is the current i nthe voltmeter. Since the resistor and the voltmeter are in parallel, the voltage across the meter equalsthe voltage across the resistor. Applying the definition of resistance:

∆V = IRR = I − IR( ) 20000 Ω( ) so I = IR(R + 20000 Ω)

20000 Ω

Our requirement is

IRRIR(R + 20000 Ω)

20000 Ω

≥ 0.95R

Solving, 20000 Ω ≥ 0.95(R + 20000 Ω) = 0.95R + 19000 Ω

and R ≤ 1000 Ω

0.95 or R ≤ 1.05 kΩ

(b) If R is too small, the resistance of an ammeter in series will significantly reduce the current thatwould otherwise flow through R . In Figure 28.59b, the voltmeter reading is I 0.500 Ω( ) + IR , at least alittle larger than the voltage across the resistor. So the resistance computed by dividing the inflatedvoltmeter reading by the ammeter reading will be too large.

We require

≤ 1.05R so that

I (0.500 Ω) + IRI

≤ 1.05R

Thus, 0.500 Ω ≤ 0.0500R and R ≥ 10.0 Ω

L : The range of R values seems correct since the ammeter’s resistance should be less than 5% of thesmallest R value ( 0.500 Ω ≤ 0.05R means that R should be greater than 10 Ω), and R should be lessthan 5% of the voltmeter’s internal resistance ( R ≤ 0.05 × 20 kΩ = 1 kΩ ). Only for the restricted rangebetween 10 ohms and 1000 ohms can we indifferently use either of the connections (a) and (b) for areasonably accurate resistance measurement. For low values of the resistance R , circuit (a) must beused. Only circuit (b) can accurately measure a large value of R .

158 Chapter 28 Solutions

28.60 The battery supplies energy at a changing rate

dEdt

= P = E I = E E

Re−1/RC

Then the total energy put out by the battery is

dE∫ = ε 2

Rt=0

∞∫ exp − t

dE∫ = ε 2

R(−RC)

∞∫ exp − t

− dt

= −ε 2C exp − t

∞= −ε 2C[0 − 1] = ε 2C

The heating power of the resistor is

dEdt

= P = ∆VRI = I2R = Rε 2

R2 exp − 2tRC

So the total heat is

dE∫ = ε 2

∞∫ exp − 2t

dE∫ = ε 2

R− RC

exp

∞∫ − 2t

− 2dt

= −ε 2C

2exp − 2t

∞= −ε 2C

2[0 − 1] = ε 2C

The energy finally stored in the capacitor is U = 12 C (∆V)2 =

12 C ε 2. Thus, energy is conserved:

ε 2C = 12 ε 2C +

12 ε 2C and resistor and capacitor share equally in the energy from the battery.

28.61 (a) q = C ∆V( ) 1 − e−t RC[ ]

q = 1.00 × 10−6 F( ) 10.0 V( ) = 9.93 µC

(b) I = dq

dt= ∆V

e−t RC

I = 10.0 V

2.00 × 106 Ω

e−5.00 = 3.37 × 10−8 A = 33.7 nA

(c)

dUdt

= ddt

= q

Cdqdt

= qC

dUdt

= 9.93 × 10−6 C1.00 × 10−6 C V

3.37 × 10−8 A( ) = 3.34 × 10−7 W = 334 nW

(d) Pbattery = IE = 3.37 × 10−8 A( ) 10.0 V( ) = 3.37 × 10−7 W = 337 nW

Chapter 28 Solutions 159

28.62 Start at the point when the voltage has just reached

V and the switch has just closed. The voltage is 23

and is decaying towards 0 V with a time constant RBC.

VC(t) = 2

e−t/RBC

We want to know when VC(t) will reach 13

V .

Therefore,

V = 2

e−t/RBC or

e−t/RBC = 1

or t1 = RBC ln 2

After the switch opens, the voltage is 13

V , increasing toward V with time constant RA + RB( )C :

V C (t) = V –

e−t/(RA +RB )C

When VC (t) = 23

V ,

V = V − 23

Ve−t/(RA +RB )C or e−t/(RA +RB )C = 1

so t2 = (RA + RB)C ln 2 and T = t1 + t2 = (RA + 2RB)C ln 2

28.63 (a) First determine the resistance of each light bulb: P = (∆V)2 R

R = (∆V)2

P= (120 V)2

60.0 W= 240 Ω

We obtain the equivalent resistance Req of the network of light bulbsby applying Equations 28.6 and 28.7:

Req = R1 + 1

1/ R2( ) + 1/ R3( ) = 240 Ω + 120 Ω = 360 Ω

The total power dissipated in the 360 Ω is P = (∆V)2

Req= (120 V)2

360 Ω= 40.0 W

(b) The current through the network is given by P = I2Req : I = P

Req= 40.0 W

360 Ω= 1

The potential difference across R1 is ∆V1 = IR1 = 1

(240 Ω) = 80.0 V

The potential difference ∆V23 across the parallel combination of R2 and R3 is

∆V23 = IR23 = 1

11/ 240 Ω( ) + 1/ 240 Ω( )

= 40.0 V

160 Chapter 28 Solutions

28.64 ∆V = IR

(a) 20.0 V = (1.00 × 10-3 A)(R1 + 60.0 Ω)

R1 = 1.994 × 104 Ω = 19.94 kΩ

(b) 50.0 V = (1.00 × 10-3 A)(R2 + R1 + 60.0 Ω) R2 = 30.0 kΩ

28.65 Consider the circuit diagram shown, realizing that

Ig = 1.00 mA. For the 25.0 mA scale:

24.0 mA( ) R1 + R2 + R3( ) = 1.00 mA( ) 25.0 Ω( )

or R1 + R2 + R3 = 25.0

24.0

Ω (1)

For the 50.0 mA scale: 49.0 mA( ) R1 + R2( ) = 1.00 mA( ) 25.0 Ω + R3( )

or 49.0 R1 + R2( ) = 25.0 Ω + R3 (2)

For the 100 mA scale: 99.0 mA( )R1 = 1.00 mA( ) 25.0 Ω + R2 + R3( )

or 99.0R1 = 25.0 Ω + R2 + R3 (3)

Solving (1), (2), and (3) simultaneously yields

R1 = 0.260 Ω, R2 = 0.261 Ω, R3 = 0.521 Ω

28.66 Ammeter: Igr = 0.500 A − Ig( ) 0.220 Ω( )

or Ig r + 0.220 Ω( ) = 0.110 V (1)

Voltmeter: 2.00 V = Ig r + 2500 Ω( ) (2)

Solve (1) and (2) simultaneously to find:

Ig = 0.756 mA and r = 145 Ω

Chapter 28 Solutions 161

28.67 (a) After steady-state conditions have been reached, there is no DC current through the capacitor.

Thus, for R3: IR3= 0 (steady-state)

For the other two resistors, the steady-state current is simply determined by the 9.00-V emfacross the 12-k Ω and 15-k Ω resistors in series:

For R1 and R2: I(R1 +R2 ) = ε

R1 + R2= 9.00 V

(12.0 kΩ + 15.0 kΩ)= 333 µA (steady-state)

(b) After the transient currents have ceased, the potential difference across C is the same as thepotential difference across R2(= IR2) because there is no voltage drop across R3. Therefore, thecharge Q on C is

Q = C (∆V)R2 = C (IR2) = (10.0 µF)(333 µA)(15.0 k Ω) = 50.0 µC

(c) When the switch is opened, the branch containing R1 is no longer part of the circuit. Thecapacitor discharges through (R2 + R3) with a time constant of (R2 + R3)C = (15.0 k Ω + 3.00k Ω)(10.0 µF) = 0.180 s. The initial current Ii in this discharge circuit is determined by the initialpotential difference across the capacitor applied to (R2 + R3) in series:

Ii = (∆V)C

(R2 + R3 )= IR2

(R2 + R3 )= (333 µA)(15.0 kΩ)

(15.0 kΩ + 3.00 kΩ)= 278 µA

Thus, when the switch is opened, the current through R2changes instantaneously from 333 µA (downward) to278 µA (downward) as shown in the graph. Thereafter, itdecays according to

IR2= Iie

−t/(R2 +R3 )C = (278 µA)e−t/(0.180 s) (for t > 0)

(d) The charge q on the capacitor decays from Q i to Q i/5according to

q = Qie−t/(R2 +R3 )C

5 = Qie

(−t/0.180 s)

5 = et/0.180 s

ln 5 = t

180 ms

t = (0.180 s)(ln 5) = 290 ms

(a)

162 Chapter 28 Solutions

28.68 ∆V = ε e−t RC so ln

ε∆V

= 1

A plot of ln

ε∆V

versus t should be a straight line with slope =

1RC

Using the given data values: t s( ) ∆V (V) ln ε ∆V( )0 6.19 04.87 5.55 0.10911.1 4.93 0.22819.4 4.34 0.35530.8 3.72 0.50946.6 3.09 0.69567.3 2.47 0.919102.2 1.83 1.219

(a) A least-square fit to this datayields the graph to the right.

Σxi = 282 , Σxi2 = 1.86 × 104 , Σxiyi = 244, Σyi = 4.03 , N = 8

Slope =N Σxiyi( ) − Σxi( ) Σyi( )

N Σxi2( ) − Σxi( )2 = 0.0118

Intercept =Σxi

2( ) Σyi( ) − Σxi( ) Σxiyi( )N Σxi

2( ) − Σxi( )2 = 0.0882

The equation of the best fit line is: ln

ε∆V

= 0.0118( )t + 0.0882

(b) Thus, the time constant is τ = = = =RC

1 10 0118slope .

84.7 s

and the capacitance is C = τ

R= 84.7 s

10.0 × 106 Ω= 8.47 µF

Chapter 28 Solutions 163

28.69r

3 junctions atthe same potential

another set of3 junctions at the same potential

a b

i/3

i/3 i/3

i/3

i/6

28.70 (a) For the first measurement, the equivalent circuit is as shownin Figure 1.

Rab = R1 = Ry + Ry = 2Ry

so Ry = 1

2R1 (1)

For the second measurement, the equivalent circuit is shownin Figure 2.

Thus, Rac = R2 = 1

2Ry + Rx (2)

a bc

RyRyRx

Figure 1

a c

RxRyRy

Figure 2

Substitute (1) into (2) to obtain: R2 = 1

212

+ Rx , or

Rx = R2 − 1

4R1

(b) If R1 = 13.0 Ω and R2 = 6.00 Ω , then Rx = 2.75 Ω

The antenna is inadequately grounded since this exceeds the limit of 2.00 Ω.

28.71 Since the total current passes through R3, that resistor will dissipatethe most power. When that resistor is operating at its power limit of32.0 W, the current through it is

Itotal

2 = P

R= 32.0 W

2.00 Ω= 16.0 A2 , or Itotal = 4.00 A

Half of this total current (2.00 A) flows through each of the other tworesistors, so the power dissipated in each of them is:

R1 = R2 = R3 = 2.00 Ω

P = 1

2Itotal( )2

R = (2.00 A)2(2.00 Ω) = 8.00 W

Thus, the total power dissipated in the entire circuit is:

164 Chapter 28 Solutions

Ptotal = 32.0 W + 8.00 W + 8.00 W = 48.0 W 28.72 The total resistance between points b and c is:

R = ( )( )

2 00 3 002 00 3 00

1 20. .. .

. k k k k

kΩ ΩΩ Ω

The total capacitance between points d and e is:

C = 2.00 µF + 3.00 µF = 5.00 µF

The potential difference between point d and e in this series RCcircuit at any time is:

∆V = ε 1 − e−t RC[ ] = 120.0 V( ) 1 − e−1000t 6[ ]

20 V

.00 kΩ

1 = 2.00 µF

2 = 3.00 µF

Therefore, the charge on each capacitor between points d and e is:

q1 = C1 ∆V( ) = 2.00 µF( ) 120.0 V( ) 1 − e−1000t 6[ ] =

240 µC( ) 1 − e−1000t 6[ ]

and q2 = C2 ∆V( ) = 3.00 µF( ) 120.0 V( ) 1 − e−1000t 6[ ] =

360 µC( ) 1 − e−1000t 6[ ]

*28.73 (a) Req = 3R I = ε

3R Pseries = ε I = ε 2 3R

(b) Req = 1

1/ R( ) + 1/ R( ) + 1/ R( ) = R3

I = 3εR Pparallel = ε I =

3ε 2

Chapter 29 Solutions

29.1 (a) up

(b) out of the page, since thecharge is negative.

(d) into the page

29.2 At the equator, the Earth's magnetic field is horizontally north. Because an electron hasnegative charge, F = q v × B is opposite in direction to v × B. Figures are drawn looking down.

(a) Down × North = East, so the force is directed West

(b) North × North = sin 0° = 0: Zero deflection

(d) Southeast × North = Up, so the force is Down (a) (c) (d)

29.3 FB = q v × B; FB (–j) = –e v i × B

Therefore, B = B (–k) which indicates the negative z direction

*29.4 (a) FB = qvB sin θ = (1.60 × 10–19 C)(3.00 × 106 m/s)(3.00 × 10–1 T) sin 37.0°

FB = 8.67 × 10–14 N

(b) a = Fm =

8.67 × 10–14 N1.67 × 10–27 kg

= 5.19 × 1013 m/s2

29.5 F = ma = (1.67 × 10–27 kg)(2.00 × 1013 m/s2) = 3.34 × 10–14 N = qvB sin 90°

B = F

qv = 3.34 × 10–14 N

(1.60 × 10–19 C)(1.00 × 107 m/s) = 2.09 × 10–2 T

The right-hand rule shows that B must be in the –y direction to yield aforce in the +x direction when v is in the z direction.

2 Chapter 29 Solutions

*29.6 First find the speed of the electron: ∆K = 12 mv2 = e(∆V) = ∆U

v = 2e ∆V( )m

=2 1.60 × 10−19 C( ) 2400 J / C( )

9.11× 10-31 kg( ) = 2.90 × 107 m / s

(a) FB, max = qvB = (1.60 × 10–19 C)(2.90 × 107 m/s)(1.70 T) = 7.90 × 10–12 N

(b) FB, min = 0 occurs when v is either parallel to or anti-parallel to B

29.7 Gravitational force: Fg = mg = (9.11 × 10–31 kg)(9.80 m/s2) = 8.93 × 10–30 N down

Electric force: Fe = qE = (–1.60 × 10–19 C)100 N/C down = 1.60 × 10–17 N up

Magnetic force: FB = qv × B = −1.60 × 10−19 C( ) 6.00 × 106

× 50.0 × 10−6 N ⋅ s

C ⋅ m N

FB = – 4.80 × 10–17 N up = 4.80 × 10–17 N down

29.8 We suppose the magnetic force is small compared to gravity. Then its horizontal velocitycomponent stays nearly constant. We call it v i.

From vy2 = vyi

2 + 2ay y − yi( ) , the vertical component at impact is − 2gh j. Then,

FB = qv × B = Q vi − 2gh j( ) × Bk = QvB −j( ) − Q 2gh Bi

FB = QvB vertical + Q 2gh B horizontal

FB = 5.00 × 10–6 C(20.0 m/s)(0.0100 T) j + 5.00 × 10–6 C 2(9.80 m/s2)(20.0 m) (0.0100 T) i

FB = (1.00 × 10–6 N) vertical + (0.990 × 10–6 N) horizontal

29.9 FB = qvB sin θ so 8.20 × 10-13 N = (1.60 × 10-19 C)(4.00 × 106 m/s)(1.70 T) sin θ

sin θ = 0.754 and θ = sin-1(0.754) = 48.9° or 131°

Chapter 29 Solutions 3

29.10 qE = −1.60 × 10−19 C( ) 20.0 N / C( )k = −3.20 × 10−18 N( )kΣF = qE + qv × B = ma

−3.20 × 10−18 N( )k – 1.60 × 10-19 C (1.20 × 104 m/s i) × B = (9.11 × 10-31)(2.00 × 1012 m/s2)k

– (3.20 × 10-18 N)k – (1.92 × 10-15 C · m/s)i × B = (1.82 × 10-18 N)k

(1.92 × 10-15 C · m/s)i × B = – (5.02 × 10-18 N)k

The magnetic field may have any x-component . Bz = 0 and By = –2.62 mT

29.11 FB = qv × B

v × B =i j k

+2 −4 +1+1 +2 −3

= 12 − 2( )i + 1 + 6( )j + 4 + 4( )k = 10i + 7 j + 8k

v × B = 102 + 72 + 82 = 14.6 T ⋅ m / s

FB = q v × B = 1.60 × 10−19 C( ) 14.6 T ⋅ m s( ) = 2.34 × 10−18 N

29.12

FB = qv × B = −1.60 × 10−19( )i j k

0 3.70 × 105 01.40 2.10 0

FB = −1.60 × 10−19 C( ) 0 − 0( )i + 0 − 0( )j + 0 − 1.40 T( ) 3.70 × 105 m s( )( ) k[ ] =

8.29 × 10−14 k( ) N

29.13 FB = ILB sin θ with FB = Fg = mg

mg = ILB sin θ som L g = IB sin θ

I = 2.00 A and

= 0.500 g / cm( ) 100 cm / m1000 g / kg

= 5.00 × 10−2 kg / m

Thus (5.00 × 10–2)(9.80) = (2.00)B sin 90.0°

B = 0.245 Tesla with the direction given by right-hand rule: eastward

4 Chapter 29 Solutions

Goal Solution A wire having a mass per unit length of 0.500 g/cm carries a 2.00-A current horizontally to the south.What are the direction and magnitude of the minimum magnetic field needed to lift this wire verticallyupward?

G : Since I = 2.00 A south, B must be to the east to make F upward according to theright-hand rule for currents in a magnetic field.

The magnitude of B should be significantly greater than the earth’s magnetic field (~ 50 µT), since wedo not typically see wires levitating when current flows through them.

O : The force on a current-carrying wire in a magnetic field is FB = I1× B , from which we can find B.

A : With I to the south and B to the east, the force on the wire is simply FB = IlBsin90°, which mustoppose the weight of the wire, mg. So,

B = FB

Il= mg

Il= g

Iml

= 9.80 m / s2

2.00 A

0.500

gcm

102 cm / m103 g / kg

= 0.245 T

L : The required magnetic field is about 5000 times stronger than the earth’s magnetic field. Thus it wasreasonable to ignore the earth’s magnetic field in this problem. In other situations the earth’s fieldcan have a significant effect.

29.14 FB = I L × B = (2.40 A)(0.750 m)i × (1.60 T)k = (–2.88 j) N

29.15 (a) FB = ILB sin θ = (5.00 A)(2.80 m)(0.390 T) sin 60.0° = 4.73 N

(b) FB = (5.00 A)(2.80 m)(0.390 T) sin 90.0° = 5.46 N

29.16

L= mg

L= I L × B

I = mgBL =

(0.0400 kg/m)(9.80 m/s2)3.60 T = 0.109 A

The direction of I in the bar is to the right .

Chapter 29 Solutions 5

29.17 The magnetic and gravitational forces must balance. Therefore, it is necessary to have FB =BIL = mg, or I = (mg/BL) = (λ g/B) [λ is the mass per unit length of the wire].

Thus, I = (1.00 × 10-3 kg/m)(9.80 m/s2)

(5.00 × 10-5 T) = 196 A (if B = 50.0 µT)

The required direction of the current is eastward , since East × North = Up.

29.18 For each segment, I = 5.00 A and B = 0.0200 N / A ⋅ m j

Segment L FB = I L × B( )

d a

−0.400 m j

0.400 m k

– 0.400 m i + 0.400 m j

0.400 m i – 0.400 m k

(40.0 mN)(– i)

(40.0 mN)(– k)

(40.0 mN)(k + i)

29.19 The rod feels force FB = I d × B( ) = Id k( ) × B −j( ) = IdB i( )

The work-energy theorem is Ktrans + Krot( )i + ∆E = Ktrans + Krot( ) f

0 + 0 + Fscosθ = 12 mv2 + 1

2 Iω2

IdBLcos 0˚ = 1

2 mv2 + 12

12 mR2( ) v

and IdBL = 34 mv2

v =

4IdBL3m

4(48.0 A)(0.120 m)(0.240 T)(0.450 m)3(0.720 kg)

= 1.07 m/s

29.20 The rod feels force FB = I d × B( ) = Id k( ) × B −j( ) = IdB i( )

The work-energy theorem is Ktrans + Krot( )i + ∆E = Ktrans + Krot( ) f

0 + 0 + Fscosθ = 12 mv2 + 1

2 Iω2

IdBLcos 0˚ = 1

2 mv2 + 12

12 mR2( ) v

and v =

4IdBL3m

6 Chapter 29 Solutions

29.21 The magnetic force on each bit of ring is I ds × B = I ds B radiallyinward and upward, at angle θ above the radial line. The radiallyinward components tend to squeeze the ring but all cancel out asforces. The upward components I ds B sin θ all add to

I 2π rB sin θ up .

*29.22 Take the x-axis east, the y-axis up, and the z-axis south. The field is

B = 52.0 µT( ) cos 60.0˚ −k( ) + 52.0 µT( ) sin 60.0˚ −j( )The current then has equivalent length: ′L = 1.40 m −k( ) + 0.850 m j( )

FB = I ′L × B = 0.0350 A( ) 0.850 j − 1.40k( )m × −45.0 j − 26.0k( )10−6 T

FB = 3.50 × 10−8 N −22.1i − 63.0i( ) = 2.98 × 10−6 N −i( ) = 2.98 µN west

29.23 (a) 2π r = 2.00 m so r = 0.318 m

µ = IA = (17.0 × 10-3 A) π(0.318)2 m2[ ] = 5.41 mA · m2

(b) τ = µ × B so τ = (5.41 × 10-3 A · m2)(0.800 T) = 4.33 mN · m

*29.24 τ = µB sin θ so 4.60 × 10-3 N · m = µ(0.250) sin 90.0°

µ = 1.84 × 10–2 A · m2 = 18.4 mA · m2

29.25 τ = NBAI sin θ

τ = 100 0.800 T( ) 0.400 × 0.300 m2( ) 1.20 A( )sin60°

τ = 9.98 N · m

Note that θ is the angle between the magneticmoment and the B field. The loop will rotate soas to align the magnetic moment with the B field.Looking down along the y-axis, the loop willrotate in a clockwise direction.

(a) (b)

Chapter 29 Solutions 7

29.26 (a) Let θ represent the unknown angle; L , the total length of the wire; and d , the length of oneside of the square coil. Then, use the right-hand rule to find

µ = NAI = L

d2I at angle θ with the horizontal.

At equilibrium, Στ = (µ × B) – (r × mg) = 0

ILBd4

sin(90.0°− θ) − mgd

sinθ = 0 and

mgd2

sinθ = ILBd

cosθ

θ = tan−1 ILB

2mg

= tan−1 (3.40 A)(4.00 m)(0.0100 T)2(0.100 kg)(9.80 m / s2)

= 3.97°

(b) τm = ILBd

cosθ = 1

4(3.40 A)(4.00 m)(0.0100 T)(0.100 m) cos 3.97° = 3.39 mN · m

29.27 From τ = µ × B = IA × B, the magnitude of the torque is IAB sin 90.0°

(a) Each side of the triangle is 40.0 cm/3.

Its altitude is 13.3 2 – 6.67 2 cm = 11.5 cm and its area is

A = 12 (11.5 cm)(13.3 cm) = 7.70 × 10-3 m2

Then τ = (20.0 A)(7.70 × 10-3 m2)(0.520 N · s/C · m) = 80.1 mN · m

(b) Each side of the square is 10.0 cm and its area is 100 cm2 = 10-2 m2.

τ = (20.0 A)(10-2 m2)(0.520 T) = 0.104 N· m

A = π r 2 = 1.27 × 10-2 m2

τ = (20.0 A)(1.27 × 10-2 m2)(0.520) = 0.132 N · m

(d) The circular loop experiences the largest torque.

*29.28 Choose U = 0 when the dipole moment is at θ = 90.0° to the field. The field exerts torque ofmagnitude µBsinθ on the dipole, tending to turn the dipole moment in the direction ofdecreasing θ . Its energy is given by

U − 0 = µBsinθ dθ

90.0°

θ∫ = µB −cosθ( ) 90.0°

θ = −µBcosθ + 0 or U = – µ · B

8 Chapter 29 Solutions

*29.29 (a) The field exerts torque on the needle tending to align it with the field, so the minimumenergy orientation of the needle is:

pointing north at 48.0 below the horizontal°

where its energy is Umin = −µBcos0˚ = − 9.70 × 10−3 A ⋅ m2( ) 55.0 × 10−6 T( ) = −5.34 × 10−7 J

It has maximum energy when pointing in the opposite direction,

south at 48.0 above the horizontal°

where its energy is Umax = −µB cos 180˚ = + 9.70 × 10−3 A ⋅ m2( ) 55.0 × 10−6 T( ) = + 5.34 × 10−7 J

(b) Umin +W = Umax: W = Umax −Umin = + 5.34 × 10−7 J − −5.34 × 10−7 J( ) = 1.07 µJ

29.30 (a) τ = µ × B, so τ = µ × B = µB sin θ = NIAB sin θ

τmax = NIABsin90.0˚ = 1 5.00 A( ) π 0.0500 m( )2[ ] 3.00 × 10−3 T( ) = 118 µN · m

(b) U = −µ ⋅B, so −µB ≤ U ≤ +µB

Since µB = NIA( )B = 1 5.00 A( ) π 0.0500 m( )2[ ] 3.00 × 10−3 T( ) = 118 µJ,

the range of the potential energy is: −118 µJ ≤ U ≤ +118 µJ

29.31 (a) B = 50.0 × 10-6 T; v = 6.20 × 106 m/s

Direction is given by the right-hand-rule: southward

FB = qvB sin θ

FB = (1.60 × 10-19 C)(6.20 × 106 m/s)(50.0 × 10-6 T) sin 90.0° = 4.96 × 10-17 N

(b) F = mv2

r so r = mv2

F = (1.67 × 10-27 kg)(6.20 × 106 m/s)2

4.96 × 10-17 N = 1.29 km

29.32 (a)12 m v2 = q(∆V)

12 (3.20 × 10-26 kg) v2 = (1.60 × 10-19 C)(833 V) v = 91.3 km/s

The magnetic force provides the centripetal force: qvB sin θ = mv2

Chapter 29 Solutions 9

r = m v

qB sin 90.0° = (3.20 × 10-26 kg)(9.13 × 104 m/s)

(1.60 × 10-19 C)(0.920 N · s/C · m) = 1.98 cm

10 Chapter 29 Solutions

29.33 For each electron, q vB sin 90.0° = mv2

r and v = eBrm

The electrons have no internal structure to absorb energy, so the collision must be perfectlyelastic:

K = 12 mv1i

2 + 0 = 12 mv1f

2 + 12 mv2 f

K = 1

2 me2B2R1

+ 1

2 me2B2R2

e2B2

2mR1

2 + R22( )

K = e(1.60 × 10−19 C)(0.0440 N ⋅ s / C ⋅ m)2

2(9.11× 10−31 kg)(0.0100 m)2 + (0.0240 m)2[ ] = 115 keV

29.34 We begin with qvB = mv2

R , so v = qRBm

The time to complete one revolution is

T = 2πRv

= 2πRqRBm

= 2πmqB

Solving for B, B = 2πmqT = 6.56 × 10-2 T

29.35 q ∆V( ) = 12 mv2 or

v = 2q ∆V( )

Also, qvB = mv2

rso

r = mv

qB= m

qB2q ∆V( )

m= 2m ∆V( )

qB2

Therefore, rp

2 =2mp ∆V( )

eB2

2 = 2md ∆V( )qdB2 =

2 2mp( ) ∆V( )eB2 = 2

2mp ∆V( )eB2

= 2rp

and rα

2 = 2mα ∆V( )qα B2 =

2 4mp( ) ∆V( )2e( )B2 = 2

2mp ∆V( )eB2

= 2rp

The conclusion is: rα = rd = 2 rp

Chapter 29 Solutions 11

Goal Solution 29.35 A proton (charge +e , mass mp ), a deuteron (charge +e , mass 2mp), and an alpha particle, (charge

+ 2e , mass 4mp) are accelerated through a common potential difference ∆V . The particles enter auniform magnetic field B with a velocity in a direction perpendicular to B. The proton moves in acircular path of radius rp . Determine the values of the radii of the circular orbits for the deuteron rd andthe alpha particle rα in terms of rp .

G : In general, particles with greater speed, more mass, and less charge will have larger radii as theymove in a circular path due to a constant magnetic force. Since the effects of mass and charge haveopposite influences on the path radius, it is somewhat difficult to predict which particle will have thelarger radius. However, since the mass and charge ratios of the three particles are all similar i nmagnitude within a factor of four, we should expect that the radii also fall within a similar range.

O : The radius of each particle’s path can be found by applying Newton’s second law, where the forcecausing the centripetal acceleration is the magnetic force: F = qv × B . The speed of the particles can befound from the kinetic energy resulting from the change in electric potential given.

A : An electric field changes the speed of each particle according to K +U( )i = K +U( ) f . Therefore,

assuming that the particles start from rest, we can write q∆V = 1

2mv2.

The magnetic field changes their direction as described by ΣF = ma : qvBsin 90°= mv2

thus r = mv

qB= m

qB2q∆V

m= 1

B2m∆V

For the protons, rp = 1

2mp∆V

For the deuterons, rd = 1

2(2mp )∆V

e= 2rp

For the alpha particles, rα = 1

2(4mp )∆V

2e= 2rp

L : Somewhat surprisingly, the radii of the deuterons and alpha particles are the same and are only 41%greater than for the protons.

29.36 (a) We begin with qvB = mv2

R, or qRB = mv . But, L = mvR = qR2B.

Therefore,

R = LqB

= 4.00 × 10−25 J ⋅ s

1.60 × 10−19 C( ) 1.00 × 10−3 T( ) = 0.0500 m = 5.00 cm

(b) Thus,

v = LmR

= 4.00 × 10−25 J ⋅ s

9.11× 10−31 kg( ) 0.0500 m( )= 8.78 × 106 m s

12 Chapter 29 Solutions

29.37 ω = qB

m= (1.60 × 10−19 C)(5.20 T)

1.67 × 10−27 kg= 4.98 × 108 rad/s

29.3812 mv2 = q(∆V) so

v = 2q ∆V( )

r = m vqB so

r =

m 2q(∆V)/ mqB

r2 = m

q⋅ 2(∆V)

B2 and

′r( )2 = ′m′q

⋅ 2(∆V)B2

m = qB2r2

2(∆V)and

′m( ) =

′q( )B2 ′r( )2

2(∆V)so

′mm

= ′qq

⋅ ′r( )2

r2 = 2ee

2RR

= 8

29.39 E = 12 mv2 = e(∆V) and evBsin90°= mv2 R

B = mv

eR= m

eR2e(∆V)

m= 1

R2m(∆V)

B = 1

5.80 × 1010 m2(1.67 × 10−27 kg)(10.0 × 106 V)

1.60 × 10−19 C= 7.88 × 10-12 T

29.40 r = mv

qBso

m = rqB

7.94 × 10−3 m( ) 1.60 × 10−19 C( ) 1.80 T( )4.60 × 105 m s

m = 4.97 × 10−27 kg

1 u1.66 × 10−27 kg

= 2.99 u

The particle is singly ionized: either a tritium ion, 13H+ , or a helium ion, 2

3He+ .

29.41 FB = Fe so qvB = qE where v = 2K / m . K is kinetic energy of the electrons.

E = vB = 2Km

B =2 750( ) 1.60 × 10−19( )

9.11× 10−31

1/2

0.0150( ) = 244 kV/m

Chapter 29 Solutions 13

29.42 K = 12 mv2 = q(∆V) so

v = 2q ∆V( )

FB = qv × B = mv2

r r = mv

qB= m

q2q(∆V)/ m

B= 1

B2m(∆V)

(a) r238 = 2(238 × 1.66 × 10−27 )2000

1.60 × 10−191

1.20

= 8.28 × 10−2 m = 8.28 cm

(b) r235 = 8.23 cm

r238

r235= m238

m235= 238.05

235.04= 1.0064

The ratios of the orbit radius for different ions are independent of ∆V and B.

29.43 In the velocity selector: v = E

B= 2500 V m

0.0350 T= 7.14 × 104 m s

In the deflection chamber:

r = mvqB

=2.18 × 10−26 kg( ) 7.14 × 104 m s( )

1.60 × 10−19 C( ) 0.0350 T( )= 0.278 m

29.44 K = 12 mv 2:

34.0 × 106 eV( ) 1.60 × 10−19 J / eV( ) = 1

21.67 × 10−27 kg( )v2

v = 8.07 × 107 m/s r = m vqB =

(1.67 × 10-27 kg)(8.07 × 107 m/s)

(1.60 × 10-19 C)(5.20 T) = 0.162 m

29.45 (a) FB = qvB = mv2

ω = vR =

qBRm R =

qBm =

(1.60 × 10-19 C)(0.450 T)

1.67 × 10-27 kg = 4.31 × 107 rad/s

(b) v = qBRm =

(1.60 × 10-19 C)(0.450 T)(1.20 m)

1.67 × 10-27 kg = 5.17 × 107 m/s

29.46 FB = qvB = mv2

B = m vqr =

4.80 × 10–16 kg · m/s(1.60 × 10–19 C)(1000 m)

= 3.00 T

14 Chapter 29 Solutions

29.47 θ = tan-1 25.010.0 = 68.2° and R =

1.00 cmsin 68.2° = 1.08 cm

Ignoring relativistic correction, the kinetic energy of the electronsis

12 mv2 = q(∆V) so

v = 2q ∆V( )

m= 1.33 × 108 m / s

From the centripetal force m v 2

R = qvB, we find the magnetic

field

B = mv

q R= (9.11× 10−31 kg)(1.33 × 108 m / s)

(1.60 × 10−19 C)(1.08 × 10−2 m)= 70.1 mT

29.48 (a) RH ≡ 1

nqso

n = 1qRH

= 1

1.60 × 10−19 C( ) 0.840 × 10−10 m3 C( ) = 7.44 × 1028 m−3

(b) ∆VH = IB

nqt

B =

nqt ∆VH( )I

=7.44 × 1028 m−3( ) 1.60 × 10−19 C( ) 0.200 × 10−3 m( ) 15.0 × 10−6 V( )

20.0 A= 1.79 T

29.49

1nq

=t ∆VH( )

IB= (35.0 × 10−6 V)(0.400 × 10-2 m)

(21.0 A)(1.80 T)= 3.70 × 10-9 m3/C

29.50 Since ∆VH = IB

nqt , and given that I = 50.0 A, B = 1.30 T, and t = 0.330 mm, the number of

charge carriers per unit volume is

n = IB

e(∆VH)t = 1.28 × 1029 m-3

The number density of atoms we compute from the density:

n0 = 8.92 g

cm31 mole63.5 g

6.02 × 1023 atomsmole

106 cm3

1 m3

= 8.46 × 1028 atom / m3

So the number of conduction electrons per atom is

nn0

= 1.28 × 1029

8.46 × 1028 = 1.52

Chapter 29 Solutions 15

29.51 B =

nqt ∆VH( )I

=8.48 × 1028 m−3( ) 1.60 × 10−19 C( ) 5.00 × 10−3 m( ) 5.10 × 10−12 V( )

8.00 A

B = 4.32 × 10−5 T = 43.2 µT

Goal Solution In an experiment designed to measure the Earth's magnetic field using the Hall effect, a copper bar 0.500cm thick is positioned along an east-west direction. If a current of 8.00 A in the conductor results in a Hallvoltage of 5.10 pV, what is the magnitude of the Earth's magnetic field? (Assume thatn = 8.48 × 1028 electrons/m3 and that the plane of the bar is rotated to be perpendicular to the direction ofB.)

G : The Earth’s magnetic field is about 50 µT (see Table 29.1), so we should expect a result of that order ofmagnitude.

O : The magnetic field can be found from the Hall effect voltage:

∆VH = IB

nqt or

B = nqt∆VH

A : From the Hall voltage,

B =

8.48 × 1028 e- m3( ) 1.60 × 10−19 C e-( ) 0.00500 m( ) 5.10 × 10-12 V( )8.00 A = 4.32 × 10−5 T = 43.2 µT

L : The calculated magnetic field is slightly less than we expected but is reasonable considering that theEarth’s local magnetic field varies in both magnitude and direction.

29.52 (a) ∆VH = IB

nqtso

nqtI

= B∆VH

= 0.0800 T0.700 × 10−6 V

= 1.14 × 105 TV

Then, the unknown field is B = nqt

∆VH( )

B = 1.14 × 105 T V( ) 0.330 × 10−6 V( ) = 0.0377 T = 37.7 mT

(b)

nqtI

= 1.14 × 105 TV

so n = 1.14 × 105

Iqt

n = 1.14 × 105 TV

0.120 A

1.60 × 10−19 C( ) 2.00 × 10−3 m( ) = 4.29 × 1025 m−3

16 Chapter 29 Solutions

29.53 q vB sin 90° = mv2

r ∴ ω = vr =

eBm =

θt

(a) The time it takes the electron to complete π radians is

t = θω

= θme B =

(π rad)(9.11 × 10–31 kg)(1.60 × 10–19 C)(0.100 N · s/C · m)

= 1.79 × 10–10 s

(b) Since v = q Br

m ,

Ke = 1

2 mv2 = q2B2r2

2m= e(1.60 × 10−19 C)(0.100 N ⋅ s / Cm)2(2.00 × 10−2 m)2

2(9.11× 10−31 kg) = 351 keV

29.54 ∑ Fy = 0: +n – mg = 0

∑ Fx = 0: –µkn + IBd sin 90.0° = 0

B = µkm g

Id = 0.100(0.200 kg)(9.80 m/s2)

(10.0 A)(0.500 m) = 39.2 mT

29.55 (a) The electric current experiences a magnetic force .

I(h × B) in the direction of L.

(b) The sodium, consisting of ions and electrons, flows along the pipetransporting no net charge. But inside the section of length L,electrons drift upward to constitute downward electric currentJ × (area) = JLw.

The current then feels a magnetic force I h × B = JLwhB sin 90°

This force along the pipe axis will make the fluid move, exertingpressure

Farea =

JLwhBhw = JLB

29.56 The magnetic force on each proton, FB = qv × B = qvB sin 90°

downward perpendicular to velocity, supplies centripetal force, guiding itinto a circular path of radius r, with

qvB = mv2

r and r = m vqB

We compute this radius by first finding the proton's speed: K = 12 m v 2

v = 2K

2 5.00 × 106 eV( ) 1.60 × 10−19 J / eV( )1.67 × 10−27 kg

= 3.10 × 107 m / s

Chapter 29 Solutions 17

Now, r = mvqB =

(1.67 × 10–27 kg)(3.10 × 107 m/s)(C · m)(1.60 × 10–19 C)(0.0500 N · s)

= 6.46 m

(b) From the figure, observe that sin ∝ = 1.00 m

r = 1 m

6.46 m α = 8.90°

(a) The magnitude of the proton momentum stays constant, and its final y component is

– (1.67 × 10–27 kg)(3.10 × 107 m/s) sin(8.90°) = – 8.00 × 10–21 kg · m/s

*29.57 (a) If B = Bx i + By j + Bz k , FB = qv × B = e vi i( ) × Bx i + By j + Bz k( ) = 0 + eviBy k − eviBz j

Since the force actually experienced is FB = Fi j, observe that

Bx could have any value , By = 0 , and Bz = −Fi evi

(b) If v = −vi i , then FB = qv × B = e −vi i( ) × Bx i + 0 j −Fi evi k( ) = –Fi j

Reversing either the velocity or the sign of the charge reverses the force.

29.58 A key to solving this problem is that reducing the normal forcewill reduce the friction force: FB = BIL or B = FB IL

When the wire is just able to move, ΣFy = n + FB cosθ − mg = 0

so n = mg − FB cosθ

and f = µ mg − FB cosθ( )Also, ΣFx = FB sinθ − f = 0

so FB sinθ = f : FB sinθ = µ mg − FB cosθ( ) and FB = µmg

sinθ + µ cosθ

We minimize B by minimizing FB:

dFB

dθ= µmg( ) cosθ − µ sinθ

sinθ + µ cosθ( )2 = 0 ⇒ µ sinθ = cosθ

Thus, θ = tan−1 1

= tan−1 5.00( ) = 78.7° for the smallest field, and

B = FB

IL= µg

m L( )sinθ + µ cosθ

Bmin =0.200( ) 9.80 m s2( )

1.50 A

0.100 kg msin 78.7° + 0.200( ) cos 78.7°

= 0.128 T

Bmin = 0.128 T pointing north at an angle of 78.7˚ below the horizontal

18 Chapter 29 Solutions

29.59 (a) The net force is the Lorentz force given by F = qE + qv × B = q E + v × B( )

F = 3.20 × 10−19( ) 4 i − 1j − 2k( ) + 2i + 3 j − 1k( ) × 2i + 4 j + 1k( )[ ] N

Carrying out the indicated operations, we find: F = 3.52i − 1.60 j( ) × 10−18 N

(b)

θ = cos−1 Fx

= cos−1 3.52

3.52( )2 + 1.60( )2

= 24.4°

29.60 r = mv

qB= (1.67 × 10−27 )(1.50 × 108)

(1.60 × 10−19)(5.00 × 10−5) m = 3.13 × 104 m = 31.3 km

No, the proton will not hit the Earth .

29.61 Let ∆x1 be the elongation due to the weight of the wire andlet ∆x2 be the additional elongation of the springs when themagnetic field is turned on. Then Fmagnetic = 2k ∆x2 where kis the force constant of the spring and can be determinedfrom k = mg/2∆x1. (The factor 2 is included in the twoprevious equations since there are 2 springs in parallel.)Combining these two equations, we find

Fmagnetic = 2

mg2∆x1

∆x2 = mg ∆x2

∆x1; but FB = I L × B = ILB

Therefore, where I = 24.0 V

12.0 Ω= 2.00 A,

B = mg ∆x2

IL∆x1 = (0.0100)(9.80)(3.00 × 10−3 )

(2.00)(0.0500)(5.00 × 10−3 )= 0.588 T

*29.62 Suppose the input power is 120 W = 120 V( )I : I ~ 1 A = 100 A

Suppose ω = 2000

revmin

1 min60 s

2π rad1 rev

~200

rads

and the output power is 20 W = τω = τ 200

rads

τ ~10−1 N ⋅ m

Suppose the area is about 3 cm 4 cm( ) × ( ), or A~10−3 m2

From Table 29.1, suppose that the field is B~10−1 T

Then, the number of turns in the coil may be found from τ ≅ NIAB:

0.1 N ⋅ m ~ N 1

10−3 m2( ) 10−1

N ⋅ sCm

giving N ~103

Chapter 29 Solutions 19

29.63 Call the length of the rod L and the tension in each wire alone T 2. Then, at equilibrium:

ΣFx = T sinθ − ILB sin 90.00 = 0 or T sinθ = ILB

ΣFy = T cosθ − mg = 0 , or T cosθ = mg

Therefore, tanθ = ILB

mg= IB

m L( )gor

B =

m L( )gI

tanθ

B =

0.0100 kg m( ) 9.80 m s2( )5.00 A

tan 45.0˚( ) = 19.6 mT

29.64 Call the length of the rod L and the tension in each wire alone T 2. Then, at equilibrium:

ΣFx = T sinθ − ILBsin90.00 = 0 or T sinθ = ILB

ΣFy = T cosθ − mg = 0 , or T cosθ = mg

tanθ = ILB

mg= IB

m L( )gor

B =

m L( )gI

tanθ =

µgI

tanθ

29.65 ΣF = ma or qvB sin 90.0° = mv2

∴ the angular frequency for each ion is vr = ω =

qBm = 2π f and

∆f = f12 − f14 = qB

2π1

m12− 1

m14

= (1.60 × 10−19 C)(2.40 T)2π(1.66 × 10−27 kg / u)

112.0 u

− 114.0 u

∆f = f12 – f14 = 4.38 × 105 s-1 = 438 kHz

29.66 Let vx and v ⊥ be the components of the velocity of the positronparallel to and perpendicular to the direction of the magneticfield.

(a) The pitch of trajectory is the distance moved along x by thepositron during each period, T (see Equation 29.15).

p = vxT = (vcos 85.0°)

2πmBq

p = (5.00 × 106 )(cos 85.0°)(2π)(9.11× 10−31)

(0.150)(1.60 × 10−19)= 1.04 × 10-4 m

(b) From Equation 29.13, r = mv⊥Bq =

mv sin 85.0°Bq

r = (9.11× 10−31)(5.00 × 106 )(sin 85.0°)

(0.150)(1.60 × 10−19)= 1.89 × 10-4 m

20 Chapter 29 Solutions

29.67 τ = IAB where the effective current due to the orbiting electrons is I = ∆q

∆t =

and the period of the motion is T = 2πR

The electron's speed in its orbit is found by requiring

keq2

R2 = mv2

R or

v = q

Substituting this expression for v into the equation for T, we find T = 2π mR3

q2ke

T = 2π (9.11× 10−31)(5.29 × 10−11)3

(1.60 × 10−19)2(8.99 × 109)= 1.52 × 10−16 s

Therefore, τ = q

AB = 1.60 × 10−19

1.52 × 10−16 π(5.29 × 10−11)2(0.400) = 3.70 × 10-24 N · m

Goal Solution Consider an electron orbiting a proton and maintained in a fixed circular path of radiusR = 5.29 × 10-11 m by the Coulomb force. Treating the orbiting charge as a current loop, calculate theresulting torque when the system is in a magnetic field of 0.400 T directed perpendicular to the magneticmoment of the electron.

G : Since the mass of the electron is very small (~10-30 kg), we should expect that the torque on theorbiting charge will be very small as well, perhaps ~10-30 N⋅m.

O : The torque on a current loop that is perpendicular to a magnetic field can be found from

τ = IAB sinθ . The magnetic field is given, θ = 90°, the area of the loop can be found from the radiusof the circular path, and the current can be found from the centripetal acceleration that results fromthe Coulomb force that attracts the electron to proton.

A : The area of the loop is A = πr2 = π 5.29 × 10−11 m( )2

= 8.79 × 10−21 m2 .

If v is the speed of the electron, then the period of its circular motion will be T = 2πR v, and theeffective current due to the orbiting electron is I = ∆Q / ∆t = e T . Applying Newton’s second law withthe Coulomb force acting as the central force gives

ΣF = keq

R2 = mv2

R so that

v = q

mR and

T = 2π mR3

q2ke

T = 2π (9.10 × 10−31 kg)(5.29 × 10-11 m)3

1.60 × 10-19 C( )28.99 × 109 N ⋅ m2 C2( )

= 1.52 × 10−16 s

The torque is τ = q

AB: τ

= 1.60 × 1019 C

1.52 × 10-16 s(π)(5.29 × 10−11 m)2(0.400 T) = 3.70 × 10−24 N ⋅ m

L : The torque is certainly small, but a million times larger than we guessed. This torque will cause theatom to precess with a frequency proportional to the applied magnetic field. A similar process on thenuclear, rather than the atomic, level leads to nuclear magnetic resonance (NMR), which is used formagnetic resonance imaging (MRI) scans employed for medical diagnostic testing (see Section 44.2).

Chapter 29 Solutions 21

29.68 Use the equation for cyclotron frequency ω = qB

m or

m = qB

ω= qB

2πf

m = (1.60 × 10−19 C)(5.00 × 10−2 T)

(2π)(5.00 rev /1.50 × 10−3 s)= 3.82 × 10-25 kg

29.69 (a) K = 12 mv2 = 6.00 MeV

= 6.00 × 106 eV( ) 1.60 × 10−19

JeV

K = 9.60 × 10-13 J

v =

2 9.60 × 10−13 J( )1.67 × 10−27 kg

= 3.39 × 107 m s

x x x x x

45˚R

45.0˚

45˚

Bin = 1.00 T

θ’

FB = qvB = mv2

R so

R = mv

=1.67 × 10−27 kg( ) 3.39 × 107 m s( )

1.60 × 10−19 C( ) 1.00 T( )= 0.354 m

Then, from the diagram, x = 2R sin 45.0˚ = 2 0.354 m( )sin 45.0˚ = 0.501 m

(b) From the diagram, observe that θ' = 45.0° .

29.70 (a) See graph to the right. The Hallvoltage is directly proportional to themagnetic field. A least-square fit tothe data gives the equation of the bestfitting line as:

∆VH = 1.00 × 10−4 V T( )B

(b) Comparing the equation of the linewhich fits the data best to

∆VH = I

nqt

0.00

20.00

40.00

60.00

80.00

100.00

120.00

0.00 0.20 0.40 0.60 0.80 1.00 1.20

B (T)

∆V H (µV)

observe that:

Inqt

= 1.00 × 10−4 V T , or

t = I

nq 1.00 × 10−4 V T( )Then, if I = 0.200 A, q = 1.60 × 10−19 C, and n = 1.00 × 1026 m−3, the thickness of the sample is

t = 0.200 A

1.00 × 1026 m−3( ) 1.60 × 10−19 C( ) 1.00 × 10−4 V T( ) = 1.25 × 10−4 m = 0.125 mm

22 Chapter 29 Solutions

*29.71 (a) The magnetic force acting on ions in the blood stream willdeflect positive charges toward point A and negativecharges toward point B. This separation of chargesproduces an electric field directed from A toward B. Atequilibrium, the electric force caused by this field mustbalance the magnetic force,

so qvB = qE = q

∆Vd

v = ∆VBd

= 160 × 10−6 V

0.040 0 T( ) 3.00 × 10−3 m( ) = 1.33 m/s

(b) No . Negative ions moving in the direction of v would be deflected toward point B, givingA a higher potential than B. Positive ions moving in the direction of v would be deflectedtoward A, again giving A a higher potential than B. Therefore, the sign of the potentialdifference does not depend on whether the ions in the blood are positively or negativelycharged.

*29.72 When in the field, the particles follow a circularpath according to qvB = mv2 r , so the radius ofthe path is: r = mv / qB

(a) When r = h = mv

qB, that is, when

v = qBh

m, the

particle will cross the band of field. It will movein a full semicircle of radius h , leaving the field at

2h, 0, 0( ) with velocity v f = −v j .

(b) When v < qBh

m, the particle will move in a smaller semicircle of radius

r = mv

qB< h. It will

leave the field at 2r, 0, 0( ) with velocity v f = −v j .

m, the particle moves in a circular arc of radius

r = mv

qB> h, centered at r, 0, 0( ).

The arc subtends an angle given by θ = sin−1 h r( ). It will leave the field at the point with

coordinates r 1 − cosθ( ), h, 0[ ] with velocity v f = vsinθ i + vcosθ j .

Chapter 30 Solutions

30.1 B = µ0I2R =

µ0q(v/2π R)2R = 12.5 T

*30.2 We use the Biot-Savart law. For bits of wire along the straight-line sections, ds is at 0° or 180°to ~, so ds × ~= 0. Thus, only the curved section of wire contributes to B at P. Hence, ds is

tangent to the arc and ~ is radially inward; so ds × ~= ds l sin 90° = ds ⊗ . All pointsalong the curve are the same distance r = 0.600 m from the field point, so

B = ⌡⌠all current

dB = ⌡⌠

µ0

4π I ds × ~

r 2 =

µ0

4π

Ir 2

⌡⌠

ds =

µ0

4π

Ir 2

where s is the arclength of the curved wire,

s = rθ = (0.600 m)30.0°

2π

360° = 0.314 m

Then, B =

10–7

T · mA

(3.00 A)(0.600 m)2 (0.314 m)

B = 261 nT into the page

30.3 (a) B = 4µ0I

4π a

cos π4 – cos

3π4 where a =

is the distance from any side to the center.

B = 4.00 × 10–6

0.200

2 + 2

2 = 2 2 × 10–5 T = 28.3 µT into the paper

(b) For a single circular turn with 4 l = 2π R,

B = µ0I2R =

µ0π I4 l =

(4π 2 × 10–7)(10.0)4(0.400) = 24.7 µT into the paper

Figure for GoalSolution

Chapter 30 Solutions 191

Goal Solution (a) A conductor in the shape of a square of edge length l= 0.400 m carries a currentI = 10.0 A (Fig. P30.3). Calculate the magnitude and direction of the magnetic field at the center of thesquare. (b) If this conductor is formed into a single circular turn and carries the same current, what is thevalue of the magnetic field at the center?

G : As shown in the diagram above, the magnetic field at the center is directed into the page from theclockwise current. If we consider the sides of the square to be sections of four infinite wires, then wecould expect the magnetic field at the center of the square to be a little less than four times thestrength of the field at a point l/2 away from an infinite wire with current I.

B < 4µ0I2πa

= 44π× 10−7 T ⋅ m / A( ) 10.0 A( )

2π 0.200 m( )

= 40.0 µT

Forming the wire into a circle should not significantly change the magnetic field at the center sincethe average distance of the wire from the center will not be much different.

O : Each side of the square is simply a section of a thin, straight conductor, so the solution derived fromthe Biot-Savart law in Example 30.1 can be applied to part (a) of this problem. For part (b), the Biot-Savart law can also be used to derive the equation for the magnetic field at the center of a circularcurrent loop as shown in Example 30.3.

A : (a) We use Equation 30.4 for the field created by each side of the square. Each side contributes a fieldaway from you at the center, so together they produce a magnetic field:

B = 4µ0I

4πacos

π4

− cos3π4

4 4π× 10−6 T ⋅ m / A( ) 10.0 A( )4π 0.200 m( )

+ 22

so at the center of the square, B = 2.00 2 × 10−5 T = 28.3 µT perpendicularly into the page

(b) As in the first part of the problem, the direction of the magnetic field will be into the page. Thenew radius is found from the length of wire: 4 = 2πR, so R = 2 /π = 0.255 m. Equation 30.8 givesthe magnetic field at the center of a circular current loop:

B = µ0I

2R= (4π× 10−7 T ⋅ m / A) 10.0 A( )

2(0.255 m)= 2.47 × 10−5 T = 24.7 µT

Caution! If you use your calculator, it may not understand the keystrokes: Toget the right answer, you may need to use .

L : The magnetic field in part (a) is less than 40µT as we predicted. Also, the magnetic fields from thesquare and circular loops are similar in magnitude, with the field from the circular loop being about15% less than from the square loop.

Quick tip: A simple way to use your right hand to find the magnetic field due to a currentloop is to curl the fingers of your right hand in the direction of the current. Your extendedthumb will then point in the direction of the magnetic field within the loop or solenoid.

192 Chapter 30 Solutions

30.4 B = µ0I

2πr= 4π× 10−7 (1.00 A)

2π(1.00 m)= 2.00 × 10-7 T

30.5 For leg 1, ds × ~= 0, so there is no contribution tothe field from this segment. For leg 2, the wire isonly semi-infinite; thus,

B = 1

2µ0I2πx

= µ0I

4π x into the paper

30.6 B = µ0I

2RR = µ0I

2B= 20.0π× 10−7

2.00 × 10−5 = 31.4 cm

30.7 We can think of the total magnetic field as the superposition of the field due to the longstraight wire (having magnitude µ0I 2πR and directed into the page) and the field due to thecircular loop (having magnitude µ0I 2R and directed into the page). The resultant magneticfield is:

B = 1 +

1π

µ0I2R

= 1 + 1π

4π × 10−7 T ⋅ m / A( ) 7.00 A( )2 0.100 m( ) = 5.80 × 10−5 T

or B = 58.0 µT directed into the page( )

30.8 We can think of the total magnetic field as the superposition of the field due to the longstraight wire (having magnitude µ0I 2πR and directed into the page) and the field due to thecircular loop (having magnitude µ0I 2R and directed into the page). The resultant magneticfield is:

B = 1 +

1π

µ0I2R

directed into the page( )

30.9 For the straight sections ds × ~= 0. The quarter circle makes one-fourth the field of a fullloop:

B =

µ0I2R

=µ0I8R

into the paper B =

(4π× 10−7 T ⋅ m / A)(5.00 A)8(0.0300 m)

= 26.2 µT into the paper

Chapter 30 Solutions 193

30.10 Along the axis of a circular loop of radius R,

B = µ0IR2

2 x2 + R2( )3 2

BB0

= 1

x R( )2 + 1

3 2

where B0 ≡ µ0I 2R.

B Along Axis of Circular Loop

0.00

0.20

0.40

0.60

0.80

1.00

0.00 1.00 2.00 3.00 4.00 5.00

x/R

B/B 0

x R B B00.00 1.001.00 0.3542.00 0.08943.00 0.03164.00 0.01435.00 0.00754

30.11 dB = µ0I

4πd1×~

B = µ0I

4π

16 2πa

a2 −16 2πb

B =

µ0I12

− 1b

directed out of the paper

30.12 Apply Equation 30.4 three times:

B =µ 0I

4πacos 0 − d

d2 + a2

toward you

µ0I4πd

d2 + a2+

d2 + a2

away from you

µ0I4πa

− d

d2 + a2− cos 180°

toward you

B =

µ 0I a2 + d2 − d a2 + d2

2πad a2 + d2 away from you

194 Chapter 30 Solutions

30.13 The picture requires L = 2R

B =

µ0I2 R

+µ0I

4πR(cos 90.0° − cos 135°)

µ0I4πR

(cos 45.0° − cos 135°)

µ0I4πR

(cos 45.0° − cos 90.0°) into the page

B =

µ0IR

π 2

= 0.475

µ0IR

(into the page)

30.14 Label the wires 1, 2, and 3 as shown in Figure (a) and let themagnetic field created by the currents in these wires be

B1, B2 , and B3 respectively.

(a) At Point A : B1 = B2 = µ0I

2π a 2( ) and B3 = µ0I

2π 3a( ) .

The directions of these fields are shown in Figure (b).Observe that the horizontal components of B1 and B2cancel while their vertical components both add to B3. Figure (a)

Therefore, the net field at point A is:

BA = B1 cos 45.0˚ + B2 cos 45.0˚ + B3 = µ0I

2πa22

cos 45.0˚ + 13

BA =4π× 10−7 T ⋅ m A( ) 2.00 A( )

2π 1.00 × 10−2 m( )22

cos 45˚ + 13

= 53.3 µT

(b) At point B : B1 and B2 cancel, leaving BB = B3 = µ0I

2π 2a( ) .

BB =4π× 10−7 T ⋅ m A( ) 2.00 A( )

2π 2( ) 1.00 × 10−2 m( ) = 20.0 µT

Figure (b)

Figure (c)

2π a 2( ) and B3 = µ0I

2πa with the directions shown in Figure (c). Again,

the horizontal components of B1 and B2 cancel. The vertical components both oppose B3giving

BC = 2µ0I

2π a 2( ) cos 45.0˚

− µ0I2πa

= µ0I2πa

2 cos 45.0˚2

− 1

= 0

Chapter 30 Solutions 195

30.15 Take the x-direction to the right and the y-direction up in the plane ofthe paper. Current 1 creates at P a field

B1 = µ0I

2πa=

2.00 × 10−7 T ⋅ m( ) 3.00 A( )A 0.0500 m( )

B1 = 12.0 µT downward and leftward, at angle 67.4° below the –x axis.

Current 2 contributes

B2 =

2.00 × 10−7 T ⋅ m( ) 3.00 A( )A 0.120 m( ) clockwise perpendicular to 12.0 cm

B2 = 5.00 µT to the right and down, at angle –22.6°

Then, B = B1 + B2 = 12.0 µT( ) −i cos 67.4°−j sin 67.4°( ) + 5.00 µT( ) i cos 22.6°−j sin 22.6°( )

B = −11.1 µT( )j − 1.92 µT( )j = (–13.0 µT)j

*30.16 Let both wires carry current in the x direction, the first aty = 0 and the second at y = 10.0 cm.

(a) B = µ0I

2πrk =

4π× 10−7 T ⋅ m A( ) 5.00 A( )2π 0.100 m( ) k

B = 1.00 × 10−5 T out of the page

(b) FB = I2L × B = 8.00 A( ) 1.00 m( )i × 1.00 × 10−5 T( )k[ ] = 8.00 × 10−5 N( ) − j( )

FB = 8.00 × 10−5 N toward the first wire

2πr−k( ) =

4π× 10−7 T ⋅ m A( ) 8.00 A( )2π 0.100 m( ) −k( ) = (1.60 × 10−5 T) −k( )

B = 1.60 × 10−5 T into the page

(d) FB = I1L × B = 5.00 A( ) 1.00 m( )i × 1.60 × 10−5 T( ) −k( )[ ] = 8.00 × 10−5 N( ) + j( )

FB = 8.00 × 10−5 N toward the second wire

30.17 By symmetry, we note that the magnetic forces on the top andbottom segments of the rectangle cancel. The net force on thevertical segments of the rectangle is (using Equation 30.12)

FB = µ0 I1I2 l

2π

c + a – 1c i

Substituting given values FB = –2.70 × 10–5 i N = – 27.0 µN i

196 Chapter 30 Solutions

Goal Solution In Figure P30.17, the current in the long, straight wire is I 1 = 5.00 A and the wire lies in the plane of therectangular loop, which carries 10.0 A. The dimensions are c = 0.100 m, a = 0.150 m, and l = 0.450 m. Findthe magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire.

G : Even though there are forces in opposite directions on the loop, we must remember that themagnetic field is stronger near the wire than it is farther away. By symmetry the forces exerted onsides 2 and 4 (the horizontal segments of length a) are equal and opposite, and therefore cancel. Themagnetic field in the plane of the loop is directed into the page to the right of I1. By the right-handrule, F = I1× B is directed toward the left for side 1 of the loop and a smaller force is directed towardthe right for side 3. Therefore, we should expect the net force to be to the left, possibly in the µNrange for the currents and distances given.

O : The magnetic force between two parallel wires can be found from Equation 30.11, which can beapplied to sides 1 and 3 of the loop to find the net force resulting from these opposing force vectors.

A : F = F1 + F2 = µ0I1I2l

2π1

c + a− 1

i = µ0I1I2l

2π−a

c c + a( )

F =

4π× 10−7 N / A2( ) 5.00 A( )(10.0 A)(0.450 m)

2π−0.150 m

(0.100 m)(0.250 m)

F i= − × −( .2 70 10 5 ) N or F = × −2 70 10 5. N toward the left

L : The net force is to the left and in the µN range as we expected. The symbolic representation of the netforce on the loop shows that the net force would be zero if either current disappeared, if eitherdimension of the loop became very small ( a → 0 or l → 0), or if the magnetic field were uniform( c → ∞) .

30.18 The separation between the wires is

a = 2(6.00 cm) sin 8.00° = 1.67 cm.

(a) Because the wires repel, the currents are in

opposite directions .

(b) Because the magnetic force acts horizontally,

FBFg

= µ0I 2 l

2π a mg = tan 8.00°

I 2 = mg 2π a l µ0

tan 8.00° so I = 67.8 A

Chapter 30 Solutions 197

30.19 Each wire is distant from P by (0.200 m) cos 45.0° = 0.141 m

Each wire produces a field at P of equal magnitude:

BA = µ0I

2π a =

(2.00 × 10–7 T · m)(5.00 A)A(0.141 m) = 7.07 µT

Carrying currents into the page, A produces at P a field of7.07 µT to the left and down at –135°, while B creates afield to the right and down at – 45°. Carrying currentstoward you, C produces a field downward and to the rightat – 45°, while D 's contribution is downward and to theleft. The total field is then

4 (7.07 µT) sin 45.0° = 20.0 µT toward the page's bottom

30.20 Let the current I flow to the right. It creates a field B = µ0I 2πd at the proton's location.And we have a balance between the weight of the proton and the magnetic force

mg(− j) + qv(− i) × µ0I

2πd(k) = 0 at a distance d from the wire

d =

qvµ0I2πmg

(1.60 × 10−19 C)(2.30 × 104 m / s)(4π × 10−7 T ⋅ m / A)(1.20 × 10−6 A)2π(1.67 × 10−27 kg) (9.80 m / s2 )

= 5.40 cm

30.21 From Ampère's law, the magnetic field at point a is given by Ba = µ0Ia 2πra , where Ia is thenet current flowing through the area of the circle of radius ra . In this case, Ia = 1.00 A out ofthe page (the current in the inner conductor), so

Ba =

4π× 10−7 T ⋅ m / A( )(1.00 A)

2π(1.00 × 10−3 m)= 200 µT toward top of page

Similarly at point b : Bb = µ0 Ib

2πrb , where Ib is the net current flowing through the area of the

circle having radius rb .

Taking out of the page as positive, Ib = 1.00 A − 3.00 A = −2.00 A , or Ib = 2.00 A into the page.Therefore,

Bb = (4π× 10−7 T ⋅ m / A)(2.00 A)

2π(3.00 × 10−3 m)= 133 µT toward bottom of page

198 Chapter 30 Solutions

*30.22 (a) In B = µ0I

2π r , the field will be one-tenth as large at a ten-times larger distance: 400 cm

(b) B = µ0I

2π r1 k +

µ0I

2π r2 (–k) so B =

4π × 10–7 T · m (2.00 A)2π A

0.3985 m – 1

0.4015 m = 7.50 nT

B = µ0I

2π

r – d – 1

r + d = µ0I

2π

2dr 2 – d 2

7.50 × 10–10 T =

2.00 × 10–7

T · mA (2.00 A)

(3.00 × 10–3 m)r 2 – 2.25 × 10–6 m2 so r = 1.26 m

The field of the two-conductor cord is weak to start with and falls off rapidly with distance.

(d) The cable creates zero field at exterior points, since a loop in Ampère's law encloses zerototal current. Shall we sell coaxial-cable power cords to people who worry about biologicaldamage from weak magnetic fields?

30.23 (a) Binner = µ0NI

2π r = 3.60 T

(b) Bouter = µ0NI

2π r = 1.94 T

*30.24 (a) B = µ0I

2π a2 r for r ≤ a so B = µ0(2.50 A)

2π (0.0250 m)2 (0.0125 m) = 10.0 µT

(b) r = µ0I

2π B =

µ0(2.50 A)2π (10.0 × 10–6 T)

= 0.0500 m = 2.50 cm beyond the conductor's surface

30.25 (a) One wire feels force due to the field of the other ninety-nine.

Within the bundle, B = µ0I

2πR2

r = 3.17 × 10−3 T .

The force, acting inward, is FB = I lB, and the force per unit lengthis

l = 6.34 × 10–3 N/m inward

(b) B ∝ r, so B is greatest at the outside of the bundle. Since each wirecarries the same current, F is greatest at the outer surface .

Figures for GoalSolution

Chapter 30 Solutions 199

Goal Solution A packed bundle of 100 long, straight, insulated wires forms a cylinder of radius R = 0.500 cm. (a) If eachwire carries 2.00 A, what are the magnitude and direction of the magnetic force per unit length acting on awire located 0.200 cm from the center of the bundle? (b) Would a wire on the outer edge of the bundleexperience a force greater or less than the value calculated in part (a)?

G : The force on one wire comes from its interaction with the magnetic field created by the other ninety-nine wires. According to Ampere’s law, at a distance r from the center, only the wires enclosedwithin a radius r contribute to this net magnetic field; the other wires outside the radius producemagnetic field vectors in opposite directions that cancel out at r . Therefore, the magnetic field (andalso the force on a given wire at radius r ) will be greater for larger radii within the bundle, and willdecrease for distances beyond the radius of the bundle, as shown in the graph to the right. Applying F = I1× B, the magnetic force on a single wire will be directed toward the center of the bundle, so thatall the wires tend to attract each other.

O : Using Ampere’s law, we can find the magnetic field at any radius, so that the magnetic force F = I1× Bon a single wire can then be calculated.

A : (a) Ampere’s law is used to derive Equation 30.15, which we can use to find the magnetic field atr = 0.200 cm from the center of the cable:

B = µoIor

2πR2 =4π× 10−7 T ⋅ m / A( ) 99( ) 2.00 A( ) 0.200 × 10−2 m( )

2π (0.500 × 10−2 m)2 = 3.17 × 10−3 T

This field points tangent to a circle of radius 0.200 cm and exerts a force F = I1× B toward the center ofthe bundle, on the single hundredth wire:

F l= IBsinθ = 2.00 A( ) 3.17 × 10−3 T( ) sin 90°( ) = 6.34 mN / m

(b) As is shown above in Figure 30.12 from the text, the magnetic field increases linearly as afunction of r until it reaches a maximum at the outer surface of the cable. Therefore, the force on asingle wire at the outer radius r = 5.00 cm would be greater than at r = 2.00 cm by a factor of 5/2.

L : We did not estimate the expected magnitude of the force, but 200 amperes is a lot of current. It wouldbe interesting to see if the magnetic force that pulls together the individual wires in the bundle isenough to hold them against their own weight: If we assume that the insulation accounts for abouthalf the volume of the bundle, then a single copper wire in this bundle would have a cross sectionalarea of about

1 2( ) 0.01( )π 0.500 cm( )2 = 4 × 10−7 m2

with a weight per unit length of ρ gA = 8920 kg / m3( ) 9.8 N / kg( ) 4 × 10−7 m2( ) = 0.03 N / m

Therefore, the outer wires experience an inward magnetic force that is about half the magnitude oftheir own weight. If placed on a table, this bundle of wires would form a loosely held moundwithout the outer sheathing to hold them together.

30.26 From

B ⋅ d1 = µ0I ,∫ I = 2π rB

µ0 =

(2π)(1.00 × 10-3)(0.100)4π × 10–7 = 500 A

200 Chapter 30 Solutions

30.27 Use Ampère’s law,

B ⋅ ds = µ0I∫ . For current density J , thisbecomes

B ⋅ ds = µ0 J∫∫ ⋅ dA

(a) For r1 < R , this gives

2πr1 = µ0 br( ) 2πr dr( )

r1∫ and

B =

µ0br12

3 for r1 < R or inside the cylinder( )

(b) When r2 > R , Ampère’s law yields

2πr2( )B = µ0 br( ) 2πr dr( )

R∫ = 2π µ0bR3 3 ,

or B =

µ0bR3

3r2 for r2 > R or outside the cylinder( )

30.28 (a) See Figure (a) to the right.

(b) At a point on the z axis, the contribution from each wire has

magnitude B = µ0I

2π a2 + z2 and is perpendicular to the line from

this point to the wire as shown in Figure (b). Combining fields,the vertical components cancel while the horizontalcomponents add, yielding

By = 2µ0I

2π a2 + z2sinθ

= µ0I

π a2 + z2

a2 + z2

= µ0I z

π a2 + z2( )The condition for a maximum is:

dBy

dz= −µ0I z 2z( )

π a2 + z2( )2 + µ0I

π a2 + z2( ) = 0, or

µ0Iπ

a2 − z2( )a2 + z2( )2 = 0

Thus, along the z axis, the field is a maximum at d = a .

(Currents are intothe paper)

Figure (a)

Figure (b)

Chapter 30 Solutions 201

30.29 B = µ0 N l I so I =

Bµ0n = 31.8 mA

30.30 (a) I = 10.0

(4π × 10–7)(2000) = 3.98 kA

(b)FB

l = IB = 39.8 kN/m radially outward

This is the force the windings will have to resist when the magnetic field in the solenoid is10.0 T.

30.31 The resistance of the wire is Re = ρl

πr2 , so it carries current I = ε

Re= επr2

ρl.

If there is a single layer of windings, the number of turns per length is the reciprocal of thewire diameter: n = 1/ 2r .

So, B = nµ0I = µ0ε π r 2

ρ l 2r =

µ0ε π r2ρ l

= (4π × 10–7 T · m/A)(20.0 V)π (2.00 × 10–3 m)

2(1.70 × 10–8 Ω · m)(10.0 m) = 464 mT

*30.32 The field produced by the solenoid in its interior isgiven by

B = µ0nI −i( ) = 4π× 10−7

T ⋅ mA

30.010-2 m

15.0 A( ) −i( )

B = − 5.65 × 10−2 T( )i

The force exerted on side AB of the square currentloop is

FB( )AB = IL × B = 0.200 A( ) 2.00 × 10−2 m( )j × 5.65 × 10−2 T( ) −i( )[ ]

FB( )AB = 2.26 × 10−4 N( )k

Similarly, each side of the square loop experiences a force, lying in theplane of the loop, of 226 µN directed away from the center . From the

above result, it is seen that the net torque exerted on the square loop bythe field of the solenoid should be zero. More formally, the magneticdipole moment of the square loop is given by

µ = IA = 0.200 A( ) 2.00 × 10−2 m( )2

−i( ) = −80.0 µA ⋅ m2 i

The torque exerted on the loop is then τ = µ × B = −80.0 µA ⋅ m2 i( ) × −5.65 × 10−2 T i( ) = 0

202 Chapter 30 Solutions

30.33 (a) ΦB = B ⋅ dA∫ = B ⋅ A = 5i + 4 j + 3k( )T ⋅ 2.50 × 10−2 m( )2

ΦB = 3.13 × 10−3 T ⋅ m2 = 3.13 × 10−3 Wb = 3.13 mWb

(b) ΦB( )total = B ⋅ dA∫ = 0 for any closed surface (Gauss’s law for magnetism)

30.34 (a) ΦB = B ⋅ A = BA where A is the cross-sectional area of the solenoid.

ΦB = µ0NI

πr2( ) = 7.40 µWb

(b) ΦB = B ⋅ A = BA = µ0NI

π r2

2 − r12( )[ ]

ΦB =4π× 10−7 T ⋅ m A( ) 300( ) 12.0 A( )

0.300 m( )

π 8.00( )2 − 4.00( )2[ ] 10−3 m( )2

= 2.27 µWb

30.35 (a) ΦB( )flat = B ⋅ A = BπR2 cos 180 − θ( )= –Bπ R 2 cos θ

(b) The net flux out of the closed surface is zero: ΦB( )flat + ΦB( )curved = 0

ΦB( )curved = Bπ R 2 cos θ

30.36

dΦE

dt= d

dt(EA) = dQ / dt

e0= Ie0

(a)

dEdt

= Ie0A

= 7.19 × 1011 V/m · s

(b)

B ⋅ ds = e0µ0ΦE

dt∫ so 2πrB = e0µ0

ddt

Qe0A

⋅ πr2

B = µ0Ir

2A= µ0(0.200)(5.00 × 10−2 )

2π(0.100)2 = 2.00 × 10-7 T

30.37 (a)

dΦE

dt=

dQ / dte0

=Ie0

=(0.100 A)

8.85 × 10−12 C2 / N ⋅ m2 = 11.3 × 109 V ⋅ m / s

(b) Id = e0

dΦE

dt= I = 0.100 A

Chapter 30 Solutions 203

30.38 (a) I = ev

2π r

µ = IA =

2π r π r 2 = 9.27 × 10–24 A · m2

The Bohr model predicts the correct magnetic moment.However, the "planetary model" is seriously deficient in otherregards.

(b) Because the electron is (–), its [conventional] current isclockwise, as seen from above, and µ points downward .

30.39 Assuming a uniform B inside the toroid is equivalent

to assuming r << R, then B0 ≅ µ0 N I

2π R and a tightly

wound solenoid.

B0 = µ0

(630)(3.00)2π(0.200)

= 0.00189 T

With the steel, B = κmB0 = (1 + χ)B0 = (101)(0.00189 T) B = 0.191 T

30.40 B = µnI = µ N

2πr

I so

I =2πr( )BµN

= 2π 0.100 m( ) 1.30 T( )5000 4π× 10−7 Wb A ⋅ m( ) 470( )

= 277 mA

30.41 ΦB = µnIA

B = µnI = (750 × 4π× 10−7 )

5002π(0.200)

(0.500) = 0.188 T

A = 8.00 × 10-4 m2 and ΦB = (0.188 T)(8.00 × 10-4 m2) = 1.50 × 10-4 T · m2) = 150 µT · m2

30.42 The period is T = 2π /ω. The spinning constitutes a current I =

=Qω2π

µ = IA =

Qω2π

πR2

QωR2

2 in the direction of ω

µ =

(6.00 × 10−6 C)(4.00 / s)(0.0200 m)2

2= 4.80 × 10-9 A · m2

204 Chapter 30 Solutions

30.43 B = µ0(H + M) so H =

Bµ0

− M = 2.62 × 106 A/m

30.44 B = µ0 H + M( )

If µ0M = 2.00 T , then the magnetization of the iron is M = 2.00 T

µ0.

But M = xnµB where µB is the Bohr magneton, n is the number of atoms per unit volume,and x is the number of electrons that contribute per atom. Thus,

x = MnµB

= 2.00 TnµBµ0

= 2.00 T

8.50 × 1028 m−3( ) 9.27 × 10−24 N ⋅ m T( ) 4π× 10−7 T ⋅ m A( ) = 2.02

*30.45 (a) Comparing Equations 30.29 and 30.30, we see that the applied field is described by B0 = µ0H.

Then Eq. 30.35 becomes M = C

T= C

Tµ0H , and the definition of susceptibility (Eq. 30.32) is

χ = M

H= C

Tµ0

(b) C = χT

µ0=

2.70 × 10−4( ) 300 K( )4π× 10−7 T ⋅ m A

= 6.45 × 104

K ⋅ AT ⋅ m

30.46 (a) Bh = Bcoil = µ0NI

2R= (4π× 10−7 )(5.00)(0.600)

0.300= 12.6 µT

(b) Bh = Bsinφ → B = Bh

sinφ= 12.6 µT

sin 13.0°= 56.0 µT

30.47 (a) Number of unpaired electrons = 8.00 × 1022 A · m2

9.27 × 10–24 A · m2 = 8.63 × 1045

Each iron atom has two unpaired electrons, so the number of iron atoms required is

8.63 × 1045( ) .

(b) Mass = (4.31 × 1045 atoms)(7900 kg/m3)

8.50 × 1028 atoms/m3 = 4.01 × 1020 kg

Chapter 30 Solutions 205

Goal Solution The magnetic moment of the Earth is approximately 8.00 × 1022 A·m2. (a) If this were caused by thecomplete magnetization of a huge iron deposit, how many unpaired electrons would this correspond to?(b) At two unpaired electrons per iron atom, how many kilograms of iron would this correspond to?(Iron has a density of 7 900 kg/m3, and approximately 8.50 × 1028 atoms/m3.)

G : We know that most of the Earth is not iron, so if the situation described provides an accurate model,then the iron deposit must certainly be less than the mass of the Earth ( MEarth = 5.98 × 1024

kg ). Onemole of iron has a mass of 55.8 g and contributes 2(6.02 × 10 23 ) unpaired electrons, so we shouldexpect the total unpaired electrons to be less than 1050.

O : The Bohr magneton µB is the measured value for the magnetic moment of a single unpairedelectron. Therefore, we can find the number of unpaired electrons by dividing the magnetic momentof the Earth by µB . We can then use the density of iron to find the mass of the iron atoms that eachcontribute two electrons.

A : (a) µB = 9.27 × 10−24 J

N ⋅ mJ

1 TN ⋅ s C ⋅ m

1 AC / s

= 9.27 × 10−24 A ⋅ m2

The number of unpaired electrons is N = 8.00 × 1022 A ⋅ m2

9.27 × 10−24 A ⋅ m2 = 8.63 × 1045 e-

(b) Each iron atom has two unpaired electrons, so the number of iron atoms required is

12 N = 1

2 (8.63 × 1045) = 4.31× 1045 iron atoms.

Thus, MFe =

4.31× 1045 atoms( ) 7900 kg / m3( )8.50 × 1028 atoms / m3 = 4.01× 1020 kg

L : The calculated answers seem reasonable based on the limits we expected. From the data in thisproblem, the iron deposit required to produce the magnetic moment would only be about 1/15 000the mass of the Earth and would form a sphere 500 km in diameter. Although this is certainly a largeamount of iron, it is much smaller than the inner core of the Earth, which is estimated to have adiameter of about 3000 km.

30.48 B = µ0I

2π R = 2.00 × 10–5 T = 20.0 µT

30.49 B = µ0IR 2

2(R 2 + R 2)3/2 so I = 2.00 × 109 A flowing west

30.50 (a) BC = µ0I

2π (0.270) –

µ0(10.0)2π (0.0900)

= 0 so I = 30.0 A

(b) BA = 4µ0(10.0)2π (0.0900)

= 88.9 µT out of paper

206 Chapter 30 Solutions

*30.51 Suppose you have two 100-W headlights running from a 12-V battery, with the whole200 W12 V = 17 A current going through the switch 60 cm from the compass. Suppose the

dashboard contains little iron, so µ ≅ µ0. Model the current as straight. Then,

B = µ0I

2π r =

(4π × 10–7)172π (0.6)

~ 10– 5 T

If the local geomagnetic field is 5 × 10–5 T, this is ~10–1 times as large, enough to affect thecompass noticeably.

30.52 A ring of radius r and width dr has area dA = 2π r dr. The current inside radius r is

I = 2πJ r dr

r∫ = 2πJ0 r dr − 2π J0 R2( )0

r∫ r3 dr

r∫ = 2πJ0 r2 2 − 2π J0 R2( ) r4 4( )

(a) Ampère's law says B 2πr( ) = µ0I = µ0π J0 r2 − r4 2R2( ) ,

or B = µ0J0R

− 1

4rR

for r ≤ R

and B 2πr( ) = µ0Itotal = µ0 πJ0R2 − πJ0R2 2[ ] = µ0 πJ0R2 2

or B = µ0J0R2

4r= µ0J0R

4 r R( ) for r ≥ R

(b)

0.000

0.050

0.100

0.150

0.200

0.250

0.300

0 2 4 6

r/R

B /µ 0J 0R

dBdr

= µ0J0

2− 3

µ0J0r2

4R2 = 0

This gives the position of the maximum as r = 2 / 3 R .

Here B = µ0J0R

1 2

− 14

3 2

= 0.272µ0J0R

Chapter 30 Solutions 207

30.53 Consider a longitudinal filament of the strip of widthdr as shown in the sketch. The contribution to thefield at point P due to the current dI in the element dris

dB = µ0dI

2π r where dI = I dr w( )

B = ∫ d B = ⌡⌠b

b + w

µ0I dr

2π w r k =

µ0I

2π w ln

1 +

wb k

30.54 We find the total number of turns: B = µ0NI

N = B l

µ0I = (0.0300 T)(0.100 m)A

(4π × 10–7 T · m)(1.00 A) = 2.39 × 103

Each layer contains (10.0 cm/0.0500 cm) = 200 closely wound turns

so she needs (2.39 × 103/200) = 12 layers .

The inner diameter of the innermost layer is 10.0 mm. The outer diameter of the outermostlayer is 10.0 mm + 2 × 12 × 0.500 mm = 22.0 mm. The average diameter is 16.0 mm, so thetotal length of wire is

(2.39 × 103)π (16.0 × 10–3 m) = 120 m

30.55 On the axis of a current loop, the magnetic field is given by B = µ0IR 2

2(x 2 + R 2)3/2

where in this case I = q

(2π /ω) . The magnetic field is directed away from the center, with a

strength of

B = µ0ωR 2q

4π (x 2 + R 2)3/2

= µ0(20.0)(0.100)2(10.0 × 10−6 )

4π (0.0500)2 + (0.100)2[ ]3/2 = 1.43 × 10–10 T

30.56 On the axis of a current loop, the magnetic field is given by B = µ0IR 2

2(x 2 + R 2)3/2

where in this case I = q

(2π /ω) . Therefore, B =

µ0ωR 2q

4π (x 2 + R 2)3/2

when x = R2 , then

B = µ0ωqR2

4π 54 R2( )3/2 =

µ0qω

2.5 5 π R

208 Chapter 30 Solutions

30.57 (a) Use Equation 30.7 twice: Bx = µ0IR 2

2(x 2 + R 2)3/2

B = Bx1 + Bx2 = µ0IR 2

(x 2 + R 2)3/2 + 1

((R – x)2 + R 2)3/2

B = µ0IR 2

(x 2 + R 2)3/2 + 1

(2R 2 + x 2 – 2xR)3/2

(b)

dBdx

= µ0IR2

2− 3

22x( ) x2 + R2( )−5 2

− 32

2R2 + x2 − 2xR( )−5 22x − 2R( )

Substituting x = R

2 and cancelling terms,

dBdx

= 0

d2Bdx2 = − 3µ0IR2

2(x2+ R2 )−5 2− 5x2(x2+ R2 )−7 2+ (2R2 + x2− 2xR)−5 2− 5(x − R)2(2R2 + x2− 2xR)−7 2[ ]

Again substituting x = R

2 and cancelling terms,

d2Bdx2 = 0

30.58 "Helmholtz pair" → separation distance = radius

B = 2µ0IR 2

2 R / 2( )2 + R 2[ ]3/2 = µ0IR 2

+ 1

3/2

R 3= µ0I

1.40R for 1 turn

For N turns in each coil, B = µ0NI

1.40R=

4π× 10−7( )100 10.0( )1.40 0.500( ) = 1.80 × 10- 3 T

Chapter 30 Solutions 209

30.59 Model the two wires as straight parallel wires (!)

(a) FB = µ0I 2L

2π a (Equation 30.12)

FB = (4π × 10–7)(140)22π(0.100)

2π (1.00 × 10–3) = 2.46 N upward

(b) aloop = 2.46 N – mloop g

m loop = 107 m/s2 upward

*30.60 (a) In dB = µ0

4πr2 Ids × ~, the moving charge constitutes a bit of current as in I = nqvA. For a

positive charge the direction of ds is the direction of v , so dB = µ0

4πr2 nqA ds( )v × ~. Next, A ds( )

is the volume occupied by the moving charge, and nA ds( ) = 1 for just one charge. Then,

B = µ0

4πr2 qv × ~

(b)

B =4π× 10−7 T ⋅ m A( ) 1.60 × 10−19 C( ) 2.00 × 107 m s( )

4π 1.00 × 10−3( )2 sin 90.0˚ = 3.20 × 10−13 T

FB = 1.02 × 10−24 N directed away from the first proton

(d)

Fe = qE = keq1q2

r2 =8.99 × 109 N ⋅ m2 C2( ) 1.60 × 10−19 C( )2

1.00 × 10−3( )2

Fe = 2.30 × 10−22 N directed away from the first proton

Both forces act together. The electrical force is stronger by two orders of magnitude. It isproductive to think about how it would look to an observer in a reference frame movingalong with one proton or the other.

*30.61 (a) B = µ0I

2πr=

4π× 10−7 T ⋅ m A( ) 24.0 A( )2π 0.0175 m( ) = 2.74 × 10−4 T

(b) At point C, conductor AB produces a field 12 2.74 × 10−4 T( ) − j( ), conductor DE produces a

field of 12 2.74 × 10−4 T( ) − j( ), BD produces no field, and AE produces negligible field. The

total field at C is 2.74 × 10−4 T − j( ) .

210 Chapter 30 Solutions

1.15 × 10−3 N( )i

(d) a = ΣF

1.15 × 10−3 N( )i

3.00 × 10−3 kg=

0.384

ms2

(e) The bar is already so far from AE that it moves through nearly constant magnetic field. Theforce acting on the bar is constant, and therefore the bar’s acceleration is constant .

(f) vf

2 = vi2 + 2ax = 0 + 2 0.384 m s2( ) 1.30 m( ), so v f = 0 999. m s( )i

Chapter 30 Solutions 211

30.62 At equilibrium,

l= µ0IAIB

2πa= mg

l or

IB =

2πa m l( )gµ0IA

IB =2π 0.0250 m( ) 0.0100 kg m( ) 9.80 m s2( )

4π× 10−7 T ⋅ m A( ) 150 A( )= 81.7 A

30.63 (a) The magnetic field due to an infinite sheet of charge (or the magneticfield at points near a large sheet of charge) is given by B = µ0Js 2 .The current density Js = I l and in this case the equivalent currentof the moving charged belt is

I = dq

dt= d

dt(σlx) = σlv ; v =

dxdt

Therefore, Js = σ v and B = µ0σ v

(b) If the sheet is positively charged and moving in the direction shown,the magnetic field is out of the page, parallel to the roller axes.

30.64 C =

TMB

=(4.00 K)(10.0%)(8.00 × 1027 atoms / m3)(5.00)(9.27 × 10−24 J / T2)

5.00 T=

2.97 × 104 K ⋅ J

T2 ⋅ m3

30.65 At equilibrium, Στ = + µ ×B − mg

cos 5.00˚

= 0,

or µB sin 5.00˚ = mgL

2cos 5.00˚

Therefore, B = mgL

2µ tan 5.00˚=

0.0394 kg( ) 9.80 m s2( ) 0.100 m( )2 7.65 J T( )tan 5.00˚

B = 28.8 mT

30.66 The central wire creates field B = µ0I1 2πR counterclockwise. The curved portions of theloop feels no force since 1 × B = 0 there. The straight portions both feel I1 × B forces to theright, amounting to

FB = I2 2L

µ0 I1

2πR=

µ0 I1 I2 LπR

to the right

212 Chapter 30 Solutions

30.67 When the conductor is in the rectangular shape shown in figure(a), the segments carrying current straight toward or away frompoint P1 do not contribute to the magnetic field at P1. Each of theother four setions of length l makes an equal contribution to thetotal field into the page at P1. To find the contribution of thehorizontal section of current in the upper right, we use

B = µ0I

4πa(cos θ1 – cos θ2 ) with a =l, θ1 = 90°, and θ2 = 135°

So B1 = 4µ0I

4πl0 –

= µ0I2 πl

When the conductor is in the shape of a circular arc, the magnitude or the field at the center is

given by Equation 30.6, B = µ0I

4πRθ . From the geometry in this case, we find

R = 4l

π and θ = π.

Therefore, B2 = µ0Iπ

4π(4l/ π)= µ0Iπ

16l ; so that

B2= 8 2

π2

30.68 I = 2πrB

µ0=

2π 9.00 × 103( ) 1.50 × 10−8( )4π× 10−7 = 675 A

Flow of positive current is downward or negative charge flows upward .

30.69 By symmetry of the arrangement, the magnitude of the netmagnetic field at point P is B = 8B0x where B0 is the contributionto the field due to current in an edge length equal to L/2. Inorder to calculate B0, we use the Biot-Savart law and consider theplane of the square to be the yz-plane with point P on the x-axis.The contribution to the magnetic field at point P due to a currentelement of length dz and located a distance z along the axis isgiven by Equation 30.3.

B0 = µ0I

4πd1×~

r2∫

From the figure we see that

r = x2 + (L2 / 4) + z2 and d1×~ = dzsinθ = dz

L2 / 4 + x2

L2 / 4 + x2 + z2

By symmetry all components of the field B at P cancel except the components along x(perpendicular to the plane of the square); and

Chapter 30 Solutions 213

B0x = B0 cosφ where cosφ = L / 2

L2 / 4 + x2.

Therefore,

B0x = µ0I4π

sinθ cosφ dzr2

L/2

∫ and B = 8B0x .

Using the expressions given above for sin θ cos φ, and r, we find

B = µ0IL2

2π x2 + L2

x2 + L2

30.70 (a) From Equation 30.10, the magnetic field produced by one loop at the center of the second loop

is given by B = µ0IR2

2x3 =µ0I πR2( )

2πx3 = µ0µ2πx3 where the magnetic moment of either loop is

µ = I πR2( ). Therefore,

Fx = µ dB

dx= µ µ0µ

2π

3x4

3µ0 πR2 I( )2

2πx4 =

3π2

µ0I2R4

(b)

Fx = 3π2

µ0I2R4

x4 = 3π2

4π× 10−7 T ⋅ m A( ) 10.0 A( )2 5.00 × 10−3 m( )4

5.00 × 10−2 m( )4 = 5.92 × 10−8 N

30.71 There is no contribution from the straight portion ofthe wire since ds×~= 0. For the field of the spiral,

dB = µ0I

(4π)(ds×~)

B = µ0I4π

ds sinθ ~

r2θ=0

2π

∫

= µ0I4π

2 dr( ) sin3π4

1r2

θ=0

2π

∫

B = µ0I4π

r−2 dr = −θ=0

2π

∫ µ0I4π

r−1( )θ=0

2π

Substitute r = eθ: B = −µ0I

4πe−θ[ ]0

2π= −µ0I

4πe−2π − e0[ ] =

µ0I4π

1 − e−2π( ) (out of the page)

214 Chapter 30 Solutions

30.72 (a) B = B0 + µ0M

M = B − B0

µ0and

M =

B − B0

µ0

Assuming that B and B0 are parallel, thisbecomes M = B − B0( ) µ0

The magnetization curve gives a plot of Mversus B0.

(b) The second graph is a plot of the relativepermeability B B0( ) as a function of the appliedfield B0.

30.73 Consider the sphere as being built up of little rings of radius r , centered on the rotation axis. The contribution to thefield from each ring is

dB =µ0 r2 dI

2 x2 + r2( )3 2 where dI = dQ

t= ω dQ

2π

dQ = ρ dV = ρ 2πr dr( ) dx( )

dB = µ0ρωr3 dr dx

2 x2 + r2( )3 2 where

ρ = Q43 πR3( )

B = µ0ρω2r=0

R2 −x2

∫x=−R

+R∫

r3 dr dx

x2 + r2( )3 2

Let v = r2 + x2 , dv = 2r dr , and r2 = v − x2

B = µ0ρω

2v=x2

∫x=−R

+R∫

v − x2( )dv

2v3 2 dx = µ0ρω4

v−1 2 dvv=x2

∫ − x2 v−3 2 dvv=x2

∫

x=−R

R∫

B = µ0ρω

42v1 2

+ 2x2( )v−1 2x2

x=−R

R∫ = µ0ρω

42 R − x( ) + 2x2 1

R− 1

dxx=−R

R∫

B = µ0ρω

R− 4 x + 2R

−R

R∫ = 2µ0ρω

R− 4x + 2R

R∫

B = 2µ0ρω

42R3

3R− 4R2

2+ 2R2

µ0ρωR2

Chapter 30 Solutions 215

30.74 Consider the sphere as being built up of little rings of radius r ,centered on the rotation axis. The current associated with eachrotating ring of charge is

dI = dQ

t= ω

2πρ 2πr dr( ) dx( )[ ]

The magnetic moment contributed by this ring is

dµ = A dI( ) = πr2 ω

2πρ 2πr dr( ) dx( )[ ] = πωρr3 dr dx

µ = πωρ r3 dr

r=0

R2 −x2

∫

x=−R

+R∫ dx

= πωρ

R2 − x2

4x=−R

+R∫ dx = πωρ

R2 − x2( )2

4x=−R

+R∫ dx

µ = πωρ

4R4 − 2R2x2 + x4( )x=−R

+R∫ dx = πωρ

4R4 2R( ) − 2R2 2R2

+ 2R5

µ = πωρ

4R5 2 − 4

3+ 2

= πωρR5

41615

4πωρR5

15 up

30.75 Note that the current I exists in the conductor witha current density J = I A , where

A = π a2 − a2 4 − a2 4[ ] = πa2 2

Therefore, J = 2I πa2 .

To find the field at either point P1 or P2 , find Bswhich would exist if the conductor were solid,using Ampère’s law. Next, find B1 and B2 thatwould be due to the conductors of radius a 2 thatcould occupy the void where the holes exist. Thenuse the superposition principle and subtract thefield that would be due to the part of the conductorwhere the holes exist from the field of the solidconductor.

θ θ

-B1

-B2

r2 + a 2( )2

a/2

− ′B1 − ′B2

(a) At point P1, Bs =

µ0J πa2( )2πr

, B1 =

µ0Jπ a 2( )2

2π r − a 2( ) , and B2 =

µ0Jπ a 2( )2

2π r + a 2( ) .

B = Bs − B1 − B2 = µ0Jπa2

2π1r

− 14 r − a 2( ) − 1

4 r + a 2( )

B = µ0 2I( )2π

4r2 − a2 − 2r2

4r r2 − a2 4( )

µ0Iπr

2r2 − a2

4r2 − a2

directed to the left

216 Chapter 30 Solutions

(b) At point P2 , Bs =

µ0J πa2( )2πr

and

′B1 = ′B2 =µ0Jπ a 2( )2

2π r2 + a 2( )2.

The horizontal components of ′B1 and ′B2 cancel while their vertical components add.

B = Bs − ′B1 cosθ − ′B2 cosθ =µ0J πa2( )

2πr− 2

µ0Jπa2 4

2π r2 + a2 4

r2 + a2 4

B =µ0J πa2( )

2πr1 − r2

2 r 2 + a2 4( )

= µ0 2I( )2πr

1 − 2r2

4r2 + a2

µ0Iπr

2r2 + a2

4r2 + a2

directed toward thetop of the page

Chapter 31 Solutions

31.1 ε = ∆ΦB

∆t= ∆ NBA( )

∆t= 500 mV

31.2 ε = ∆ΦB

∆t= ∆ B ⋅ A( )

∆t= 1.60 mV and

Iloop = ε

R= 1.60 mV

2.00 Ω= 0.800 mA

31.3 ε = –N ∆BA cos θ

∆t = –NB π r 2

cos θf – cos θi

∆t

= −25.0 50.0 × 10−6 T( )π 0.500m( )2 cos 180°−cos 0

0.200 s

E = + 9.82 mV

31.4 (a) ε = – dΦBdt = –A

dBdt =

ABmax

τ e–t/τ

(b) ε = (0.160 m2)(0.350 T)

2.00 s e– 4.00/2.00 = 3.79 mV

31.5 ε = N dΦBdt =

∆(NBA)∆t

= 3.20 kV so I = εR

= 160 A

Chapter 31 Solutions 219

Goal Solution A strong electromagnet produces a uniform field of 1.60 T over a cross-sectional area of 0.200 m2. A coilhaving 200 turns and a total resistance of 20.0 Ω is placed around the electromagnet. The current in theelectromagnet is then smoothly decreased until it reaches zero in 20.0 ms. What is the current inducedin the coil?

G : A strong magnetic field turned off in a short time ( 20.0 ms) will produce a large emf, maybe on theorder of 1 kV . With only 20.0 Ω of resistance in the coil, the induced current produced by this emfwill probably be larger than 10 A but less than 1000 A.

O : According to Faraday’s law, if the magnetic field is reduced uniformly, then a constant emf will beproduced. The definition of resistance can be applied to find the induced current from the emf.

A : Noting unit conversions from F = qv × B

and U = qV , the induced voltage is

ε = −N

d(B ⋅ A)dt

= −N0 − BiAcosθ

∆t

=+200 1.60 T( ) 0.200 m2( ) cos 0°( )

20.0 × 10−3 s1 N ⋅ s / C ⋅ m

1 V ⋅ CN ⋅ m

= 3200 V

I =

εR

= 3200 V20.0 Ω

= 160 A

L : This is a large current, as we expected. The positive sign is indicative that the induced electric field isin the positive direction around the loop (as defined by the area vector for the loop).

31.6 ε = –N dΦBdt = –

N(BA – 0)∆t

∆t = NBA

ε = NB(πr 2)ε =

500(0.200)π (5.00 × 10-2)2

10.0 × 103 = 7.85 × 10–5 s

31.7 ε = d(BA)

dt = 0.500 µ0nA dIdt = 0.480 × 10–3 V

(a) Iring = εR

= 4.80 × 10-4

3.00 × 10-4 = 1.60 A

(b) Bring = µ0I

2rring = 20.1 µT

Bring points upward

31.8 ε = d(BA)

dt = 0.500 µ0nA dIdt = 0.500 µ0nπ r 22

∆I

∆t

220 Chapter 31 Solutions

(a) Iring = εR

= µ0nπ r 22

2R ∆I

∆t

(b) B = µ0I2r1

= µ2

0nπ r22

4r1R ∆I

∆t

31.9 (a) dΦB = B ⋅ dA = µ0I

2πxLdx :

ΦB = µ0IL

2πx=h

h+w∫ dx

µ0IL2π

lnh + w

(b) ε = − dΦB

dt= − d

dtµ0IL2π

lnh + w

= − µ0L2π

lnh + w

dIdt

ε = −

4π× 10−7 T ⋅ m A( ) 1.00 m( )2π

ln1.00 + 10.0

1.00

10.0

= −4.80 µV

The long wire produces magnetic flux into the page through the rectangle (first figure, above).As it increases, the rectangle wants to produce its own magnetic field out of the page, which itdoes by carrying counterclockwise current (second figure, above).

Chapter 31 Solutions 221

31. 10 ΦB = (µ0nI)Asolenoid

ε = −N

dΦB

dt= −Nµ0n πrsolenoid

2( ) dIdt

= −Nµ0n πrsolenoid2( ) 600 A s( ) cos(120t)

ε = −15.0 4π× 10−7 T ⋅ m A( ) 1.00 × 103 m( )π 0.0200 m( )2 600 A s( ) cos(120t)

E = −14.2 cos(120t) mV

31.11 For a counterclockwise trip around the left-hand loop,with B = At

ddt

At(2a2 )cos 0°[ ] − I1(5R) − IPQR = 0

and for the right-hand loop,

ddt

At a2[ ] + IPQ R − I2(3R) = 0

where IPQ = I1 − I2 is the upward current in QP

Thus, 2Aa2 − 5R(IPQ + I2 ) − IPQR = 0

and Aa2 + IPQR = I2(3R)

2Aa2 − 6RIPQ − 53 (Aa2 + IPQR) = 0

IPQ =

Aa2

23R upward, and since R = (0.1 0 /m)(0.65 m) = 0.0650 0 0Ω Ω

IPQ =

(1.00 × 10−3 T / s)(0.650 m)2

23(0.0650 Ω) = 283 µA upward

31.12 ε = ∆ΦB

∆t= N

dBdt

A = N 0.0100 + 0.0800t( )A

At t = 5.00 s, ε = 30.0(0.410 T) π 0.0400 m( )2[ ] = 61.8 mV

222 Chapter 31 Solutions

31.13 B = µ0nI = µ0n 30.0 A( ) 1 − e−1.60t( )

ΦB = BdA∫ = µ0n 30.0 A( ) 1 − e−1.60t( ) dA∫

ΦB = µ0n 30.0 A( ) 1 − e−1.60t( )πR2

ε = − N

dΦB

dt = − Nµ0n 30.0 A( )πR2 1.60( )e− 1.60t

ε = −(250)(4π× 10−7 N A2)(400 m−1)(30.0 A) π(0.0600 m)2[ ]1.60 s−1 e−1.60t

ε = 68.2 mV( )e−1.60t counterclockwise

31.14 B = µ0nI = µ0nImax(1 − e−α t )

ΦB = BdA∫ = µ0nImax(1 − e−α t ) dA∫

ΦB = µ0nImax(1 − e−α t )πR2

ε = − N

dΦB

dt= − Nµ0nImaxπR2α e−α t

ε = Nµ0nImaxπR2α e−α t counterclockwise

31.15 ε = d

dt(NBl2 cos θ) = Nl2 ∆B cos θ

∆t

l= ε ∆t

N ∆B cosθ = (80.0 × 10−3 V)(0.400 s)

(50)(600 × 10−6 T − 200 × 10−6 T)cos(30.0°)= 1.36 m

Length = 4 lN = 4(1.36 m)(50) = 272 m

Chapter 31 Solutions 223

Goal Solution A coil formed by wrapping 50.0 turns of wire in the shape of a square is positioned in a magnetic field sothat the normal to the plane of the coil makes an angle of 30.0° with the direction of the field. When themagnetic field is increased uniformly from 200 µ T to 600 µ T in 0.400 s, an emf of80.0 mV is induced in the coil. What is the total length of the wire?

G : If we assume that this square coil is some reasonable size between 1 cm and 1 m across, then the totallength of wire would be between 2 m and 200 m.

O : The changing magnetic field will produce an emf in the coil according to Faraday’s law of induction.The constant area of the coil can be found from the change in flux required to produce the emf.

A : By Faraday’s law, ε = −N

dΦB

dt = −N

ddt

(BAcosθ) = −NAcosθ dBdt

For magnitudes, ε = NA cosθ

∆B∆t

and the area is

A =ε

N cosθ ∆B∆t

= 80.0 × 10−3 V

50(cos 30.0°)600 × 10−6 T − 200 × 10−6 T

0.400 s

= 1.85 m2

Each side of the coil has length d = A , so the total length of the wire is

L = N 4d( ) = 4N A = (4)(50) 1.85 m2 = 272 m

L : The total length of wire is slightly longer than we predicted. With d = 1.36 m , a normal person couldeasily step through this large coil! As a bit of foreshadowing to a future chapter on AC circuits, aneven bigger coil with more turns could be hidden in the ground below high-power transmission linesso that a significant amount of power could be “stolen” from the electric utility. There is a story ofone man who did this and was arrested when investigators finally found the reason for a large powerloss in the transmission lines!

31.16 The average induced emf is given by ε = – N

∆ΦB

∆t

Here N = 1, and ∆ΦB = B(Asquare − Acircle )

with Acircle = πr2 = π(0.500 m)2 = 0.785 m2

Also, the circumference of the circle is 2π r = 2π (0.500 m) = 3.14 m

Thus, each side of the square has a length L =

3.14 m4

= 0.785 m,

and Asquare = L2 = 0.617 m2

So ∆ΦB = (0.400 T)(0.617 m2 − 0.785 m2 ) = − 0.0672 T ⋅ m2

The average induced emf is therefore: ε = –

− 0.0672 T ⋅ m2

0.100 s = 0.672 V

224 Chapter 31 Solutions

31.17 In a toroid, all the flux is confined to the inside ofthe toroid.

B = µ0NI

2πr= 500 µ0I

2πr

ΦB = BdA∫ = 500 µ0Imax

2πsinωt

dzdrr∫

ΦB = 500 µ0Imax

2πasinωt ln

b + RR

ε = ′N

dΦB

dt = 20

500 µ0Imax

2π

ωa ln

b + RR

cosωt

ε = 104

2π4π× 10−7 N

50.0 A( ) 377

rads

0.0200 m( )ln 3.00 + 4.00( ) cm

4.00 cm

cosωt = (0.422 V) cos ω t

31.18 The field inside the solenoid is: B = µ0nI = µ0

Thus, through the single-turn loop ΦB = BAsolenoid = µ0

πr2( )I

and the induced emf in the loop is ε = − ∆ΦB

∆t= −µ0

πr2( ) ∆I

∆t

− µ0Nπr2

I2 – I1

∆t

31.19 ε = −N

dΦB

dt IR = −N

dΦB

Idt = − N

RdΦB

Idt∫ = − N

RdΦB∫

Q = − N

R∆ΦB = − N

RA Bf − Bi( )

Q = − 200

5.00 Ω

(100 × 10−4 m2)(−1.10 − 1.10) T = 0.880 C

31.20 I = ε

R= Blv

v = 1.00 m/s

Chapter 31 Solutions 225

31.21 (a) FB = I 1× B = IlB . When I = E/ R and ε = Blv , we get

FB = Blv

R(lB) = B2l2v

R= (2.50)2(1.20)2(2.00)

6.00= 3.00 N

The applied force is 3.00 N to the right

(b) P = I2R = B2l2v2

R= 6.00 W or P = Fv = 6.00 W

*31.22 FB = IlB and E = Blv

I = E

R= Blv

R so

B = IR

(a) FB = I2lR

lv and

I = FBv

R= 0.500 A

(b) I 2R = 2.00 W

31.23 The downward component of B, perpendicular to v, is (50.0 × 10–6 T) sin 58.0° = 4.24 × 10–5 T

E = Blv = 4.24 × 10−5 T( ) 60.0 m( ) 300 m / s( ) = 0.763 V

The left wing tip is positive relative to the right.

31.24 ε = –N ddt BA cos θ = –NB cos θ

∆A

∆t

ε = –1(0.100 T) cos 0° (3.00 m × 3.00 m sin 60.0°) – (3.00 m)2

0.100 s = 1.21 V

I = 1.21 V10.0 Ω

= 0.121 A

The flux is into the page and decreasing. The loop makes itsown magnetic field into the page by carrying clockwise current.

31.25 ω = (2.00 rev/s)(2π rad/rev) = (4.00)π rad/s

E = 1

2Bωl2 = 2.83 mV

226 Chapter 31 Solutions

31.26 (a) Bext = Bext i and Bext decreases; therefore, the inducedfield is B0 = B0 i (to the right). Therefore, the currentis to the right in the resistor.

(b) Bext = Bext (–i) increases; therefore, the induced fieldB0 = B0 (+ i) is to the right, and the current isto the right in the resistor.

(c) Bext = Bext (–k) into the paper and Bext decreases;therefore, the induced field is B0 = B0 (–k) into thepaper. Therefore, the current is to the right in theresistor.

(d) By the Lorentz force law, FB = q (v × B). Therefore, apositive charge will move to the top of the bar if B isinto the paper .

(a) (b)

31.27 (a) The force on the side of the coil entering the field(consisting of N wires) is

F = N ILB( ) = N IwB( )

The induced emf in the coil is

ε = N

dΦB

dt= N

d Bwx( )dt

= NBwv,

so the current is I = ε

R= NBwv

R counterclockwise.

The force on the leading side of the coil is then:

F = N

NBwvR

wB =

N2B2w2vR

to the left

(b) Once the coil is entirely inside the field, ΦB = NBA = constant , so ε = 0, I = 0, and F = 0.

(c) As the coil starts to leave the field, the flux decreases at the rate Bwv , so the magnitude of thecurrent is the same as in part (a), but now the current flows clockwise. Thus, the force exertedon the trailing side of the coil is:

Chapter 31 Solutions 227

F =

N2B2w2vR

to the left again

31.28 (a) Motional emf ε = Bwv appears in the conducting water. Itsresistance, if the plates are submerged, is

ρLA =

ρwab

Kirchhoff's loop theorem says Bwv – IR – Iρwab = 0

I = B w v

R + ρwab

= abvB

ρ + abRw

(b) Isc = (100 m)(5.00 m)(3.00 m/s)(50.0 × 10– 6 T)

100 Ω · m = 0.750 mA

31.29 Look in the direction of ba. The bar magnet creates a field into the page, and the fieldincreases. The loop will create a field out of the page by carrying a counterclockwise current.Therefore, current must flow from b to a through the resistor. Hence, Va – Vb will be

negative .

31.30 E = 1

2Bωl2 = 0.259 mV

31.31 Name the currents as shown in the diagram:

Left loop: + Bdv2 − I2R2 − I1R1 = 0

Right loop: + Bdv3 − I3R3 + I1R1 = 0

At the junction: I2 = I1 + I3

Then, Bdv2 − I1R2 − I3R2 − I1R1 = 0

I3 =

Bdv3

R3+

I1R1

So, Bdv2 − I1 (R1 + R2 ) −

Bdv3 R2

R3−

I1R1R2

R3= 0

228 Chapter 31 Solutions

I1 = Bd

v2R3 − v3R2

R1R2 + R1R3 + R2R3

upward

I1 = (0.0100 T)(0.100 m)

(4.00 m / s)(15.0 Ω) − (2.00 m / s)(10.0 Ω)(5.00 Ω)(10.0 Ω) + (5.00 Ω)(15.0 Ω) + (10.0 Ω)(15.0 Ω)

= 145 µA

upward

31.32 (a)d Bdt = 6.00t 2 – 8.00t ε =

dΦBdt

At t = 2.00 s, E = π R2(dB/dt)

2π r2 =

8.00π (0.0250)2

2π (0.0500)

F = qE = 8.00 × 10–21 N clockwise for electron

(b) When 6.00t2 – 8.00t = 0, t = 1.33 s

31.33d Bdt = 0.0600t ε =

dΦBdt

At t = 3.00 s, E = π r21

d B

2π r1 dt = 1.80 × 10–3 N/C perpendicular to r1 and counterclockwise

*31.34 ε = dΦB

dt= πr2 dB

= E ⋅ d1∫

E(2πR) = πr2 dB

dt, or

E = πr2

2πR

dBdt

B = µ0nI

dBdt

= µ0ndIdt

I = 3.00 e0.200t

dIdt

= 0.600 e0.200t

At t = 10.0 s, E = πr2

2πRµ0n( )(0.600 e0.200t )

becomes E = (0.0200 m)2

2(0.0500 m)(4π× 10−7 N / A2 )(1000 turns / m)(0.600)e2.00= 2.23 × 10−5 N / C

Chapter 31 Solutions 229

31.35 (a)

E ⋅ d1 = dΦB

dt∫

2πrE = (πr2 )

dBdt

so E = (9.87 mV/m) cos (100 π t)

(b) The E field is always opposite to increasing B. ∴ clockwise

230 Chapter 31 Solutions

31.36 For the alternator, ω = 3000 revmin

2π rad

1 rev

1 min

60 s = 314 rad/s

ε = −N

dΦB

dt= −250

ddt

(2.50 × 10−4 T ⋅ m2)cos(314 t / s)[ ] = +250(2.50 × 10–4 T · m2)(314/s) sin(314t)

(a) ε = (19.6 V) sin(314t)

(b) ε max = 19.6 V

31.37 (a) εmax = NABω = (1000)(0.100)(0.200)(120π) = 7.54 kV

(b) ε(t) = –NBAω · sin ωt = –NBAω sin θ

ε is maximal when sin θ = 1, or θ = ± π2 ,

so the plane of coil is parallel to B

31.38 Let θ represent the angle through which the coil turns, starting from θ = 0 at an instant whenthe horizontal component of the Earth's field is perpendicular to the area. Then,

ε = − N

ddt

BA cos θ = − NBAddt

cos ωt = + NBAω sin ωt

Here sin ω t oscillates between +1 and –1, so the spinning coil generates an alternating voltagewith amplitude

εmax = NBAω = NBA2πf = 100(2.00 × 10−5 T)(0.200 m)2(1500)

2π rad60.0 s

= 12.6 mV

31.39 B = µ0nI = 4π× 10−7 T ⋅ m A( ) 200 m−1( ) 15.0 A( ) = 3.77 × 10−3 T

For the small coil, ΦB = NB ⋅ A = NBA cosωt = NB πr2( )cosωt

Thus, ε = − dΦB

dt= NBπr2ωsinωt

ε = 30.0( ) 3.77 × 10−3 T( )π 0.0800 m( )2 4.00π s−1( )sin 4.00πt( )= 28 6 4 00. sin . mV( ) ( )πt

Chapter 31 Solutions 231

31.40 As the magnet rotates, the fluxthrough the coil variessinusoidally in time with ΦB = 0at t = 0. Choosing the flux aspositive when the field passesfrom left to right through thearea of the coil, the flux at anytime may be written as

ΦB = −Φmax sinωt so theinduced emf is given by

ε = − dΦB

dt= ωΦmax cosωt .

- 1

-0 .5

0.5

0 0.5 1 1.5 2

t/T = (ω t / 2 π )

I / I max

The current in the coil is then I = ε

R= ωΦmax

Rcosωt = Imax cosωt

31.41 (a) F = NI lB

τmax = 2Fr = NI lwB = 0.640 N · m

(b) P = τω = (0.640 N · m)(120π rad/s)

Pmax = 241 W (about 13 hp)

31.42 (a) εmax = BAω = B 1

2 πR2( )ω

εmax = 1.30 T( ) π

20.250 m( )2 4.00π

rads

εmax = 1.60 V

(b) ε = ε

2π0

2π∫ dθ = BAω

2πsinθ dθ =

2π∫ 0

(d) See Figure 1 at the right.

(e) See Figure 2 at the right.

Figure 2

Figure 1

31.43 (a) ΦB = BA cos θ = BA cos ωt = (0.800 T)(0.0100 m2) cos 2π (60.0)t = (8.00 mT · m2) cos (377t)

232 Chapter 31 Solutions

(b) ε = – dΦBdt = (3.02 V) sin (377t)

(d) P = I 2R = (9.10 W) sin2 (377t)

(e) P = Fv = τ ω so τ = P

ω= (24.1 mN · m) sin2 (377t)

31.44 At terminal speed, the upward magnetic force exerted onthe lower edge of the loop must equal the weight of theloop. That is,

Mg = FB = IwB = ε

= Bwvt

wB = B2w2vt

Thus,

B = MgR

w2vt=

0.150 kg( ) 9.80 m s2( ) 0.750 Ω( )1.00 m( )2 2.00 m s( )

= 0.742 T

31.45 See the figure above with Problem 31.44.

(a) At terminal speed, Mg = FB = IwB = ε

wB = Bwvt

wB = B2w2vt

or vt = MgR

B2w2

(b) The emf is directly proportional to vt, but the current is inversely proportional to R. A large Rmeans a small current at a given speed, so the loop must travel faster to get Fm = mg .

(c) At given speed, the current is directly proportional to the magnetic field. But the force isproportional to the product of the current and the field. For a small B, the speed mustincrease to compensate for both the small B and also the current, so vt ∝ B2 .

*31.46 The current in the magnet creates an upward magnetic field, so the N and S poles on the

solenoid core are shown correctly. On the rail in front of the brake, the upward flux of B

increases as the coil approaches, so a current is induced here to create a downward magneticfield. This is clockwise current, so the S pole on the rail is shown correctly. On the rail

behind the brake, the upward magnetic flux is decreasing. The induced current in the rail willproduce upward magnetic field by being counterclockwise as the picture correctly shows.

Chapter 31 Solutions 233

31.47 F = ma = qE + qv × B

a = e

m[E + v × B] where

v × B =i j k

200 0 00.200 0.300 0.400

= −200(0.400)j + 200(0.300)k

a = 1.60 × 10−19

1.67 × 10−27 [50.0j − 80.0j + 60.0k] = 9.58 × 107[−30.0j + 60.0k]

a = 2.87 × 109[−j + 2k] m s2 = (−2.87 × 109 j + 5.75 × 109 k) m s2

31.48 F = ma = qE + qv × B so a = −e

m[E + v × B] where

v × B =i j k

10.0 0 00 0 0.400

= −4.00j

a =

−1.60 × 10−19( )9.11× 10−31 [2.50i + 5.00 j − 4.00 j] = −1.76 × 1011( )[2.50i + 1.00 j]

a = −4.39 × 1011 i − 1.76 × 1011 j( ) m s2

*31.49 ε = −N

ddt

BAcosθ( ) = −N πr2( ) cos0˚dBdt

ε = − 30.0( )π 2.70 × 10−3 m( )2

1( ) ddt

50.0 mT + 3.20 mT( ) sin 2π 523t s( )[ ]

ε = − 30.0( )π 2.70 × 10−3 m( )2

3.20 × 10−3 T( ) 2π( ) 523 s( ) cos 2π 523t s( )

ε = − 7.22 × 10−3 V( ) cos 2π 523t s( )

*31.50 (a) Doubling the number of turns.Amplitude doubles: period unchanged

(b) Doubling the angular velocity.doubles the amplitude: cuts the period in half

(c) Doubling the angular velocity while reducing thenumber of turns to one half the original value.Amplitude unchanged: cuts the period in half

234 Chapter 31 Solutions

*31.51 ε = −N

∆∆t

BAcosθ( ) = −N πr2( )cos0˚∆B∆t

= −1 0.00500 m2( ) 1( ) 1.50 T − 5.00 T20.0 × 10−3 s

= 0.875 V

(a) I = ε

R= 0.875 V

0.0200 Ω= 43.8 A

(b) P = EI = 0.875 V( ) 43.8 A( ) = 38.3 W

31.52 In the loop on the left, the induced emf is

ε = dΦB

dt= A

dBdt = π 0.100 m( )2 100 T s( ) = π V

and it attempts to produce a counterclockwisecurrent in this loop.

In the loop on the right, the induced emf is

ε = dΦB

dt= π 0.150 m( )2 100 T s( ) = 2.25π V

and it attempts to produce a clockwise current. Assume that I1 flows down through the6.00-Ω resistor, I2 flows down through the 5.00-Ω resistor, and that I3 flows up through the3.00-Ω resistor.

From Kirchhoff’s point rule: I3 = I1 + I2 (1)

Using the loop rule on the left loop: 6.00 I1 + 3.00I3 = π (2)

Using the loop rule on the right loop: 5.00 I2 + 3.00I3 = 2.25π (3)

Solving these three equations simultaneously,

I1 = 0.0623 A , I2 = 0.860 A , and I3 = 0.923 A

*31.53 The emf induced between the ends of the moving bar is

ε = Blv = 2.50 T( ) 0.350 m( ) 8.00 m s( ) = 7.00 V

The left-hand loop contains decreasing flux away from you, so the induced current in it willbe clockwise, to produce its own field directed away from you. Let I1 represent the current

flowing upward through the 2.00-Ω resistor. The right-hand loop will carry counterclockwisecurrent. Let I3 be the upward current in the 5.00-Ω resistor.

Chapter 31 Solutions 235

(a) Kirchhoff’s loop rule then gives: +7.00 V − I1 2.00 Ω( ) = 0 I1 = 3.50 A

and +7.00 V − I3 5.00 Ω( ) = 0 I3 = 1.40 A

(b) The total power dissipated in the resistors of the circuit is

P = EI1 + EI3 = E I1 + I3( ) = 7.00 V( ) 3.50 A + 1.40 A( ) = 34.3 W

I2 = 3.50 A + 1.40 A = 4.90 A. The magnetic field exerts a force of

Fm = IlB = 4.90 A( ) 0.350 m( ) 2.50 T( ) = 4.29 N directed toward the right on this conductor. A n

outside agent must then exert a force of 4.29 N to the left to keep the bar moving.

Method 2 : The agent moving the bar must supply the power according to P = F ⋅ v = Fvcos0˚ . The force required is then:

F = P

v= 34.3 W

8.00 m s= 4.29 N

*31.54 Suppose we wrap twenty turns of wire into a flat compact circular coil of diameter 3 cm.Suppose we use a bar magnet to produce field 10−3 T through the coil in one direction alongits axis. Suppose we then flip the magnet to reverse the flux in 10−1 s. The average inducedemf is then

ε = −N

∆ΦB

∆t= −N

∆ BAcosθ[ ]∆t

= −NB πr2( ) cos180˚ − cos0˚∆t

ε = − 20( ) 10−3 T( )π 0.0150 m( )2 −2

10−1 s

~10−4 V

31.55 I = ε + ε Induced

R and ε Induced = – ddt (BA)

F = m d vdt = IBd

d vdt =

IBdm =

B dm R (ε + ε Induced) =

Bdm R (ε – Bvd)

To solve the differential equation, let u = (ε – Bvd),d udt = –Bd

dvdt .

– 1

Bd dudt =

Bdm R u so

u∫ du

u= −

t=0

t∫ (Bd)2

mRdt

Integrating from t = 0 to t = t, ln uu0

= – (Bd)2

m R t or uu0

e−B2d2t/mR

Since v = 0 when t = 0, u0 = ε and u = ε – Bvd

236 Chapter 31 Solutions

ε – Bvd = ε e−B2d2t/mR and v = εBd

(1 − e−B2d2t/mR )

Chapter 31 Solutions 237

31.56 (a) For maximum induced emf, with positive charge at the top of the antenna,

F+ = q+ (v × B), so the auto must move east

(b) ε = Blv = (5.00 × 10–5 T)(1.20 m)

65.0 × 103 m

3600 s cos 65.0° = 4.58 × 10– 4 V

31.57 I = ε

R= B

R∆A∆t

so q = I ∆t = (15.0 µT)(0.200 m)2

0.500 Ω= 1.20 µC

Goal Solution The plane of a square loop of wire with edge length a = 0.200 m is perpendicular to the Earth's magneticfield at a point where B = 15.0 µT, as shown in Figure P31.57. The total resistance of the loop and thewires connecting it to the galvanometer is 0.500 Ω. If the loop is suddenly collapsed by horizontal forcesas shown, what total charge passes through the galvanometer?

G : For the situation described, the maximum current is probably less than 1 mA. So if the loop is closedin 0.1 s, then the total charge would be

Q = I∆t = 1 mA( ) 0.1 s( ) = 100 µC

O : We do not know how quickly the loop is collapsed, but we can find the total charge by integrating thechange in magnetic flux due to the change in area of the loop ( a

2 → 0).

A : Q = Idt∫ =

ε dt

R∫ = 1

R− dΦB

dt∫ = − 1

RdΦB∫

= − 1R

d(BA) = − BR∫ dA

A1 =a2

A2 =0

∫

Q = − BR

A A1 =a2

A2 =0

= Ba2

R = (15.0 × 10−6 T)(0.200 m)2

0.500 Ω= 1.20 × 10−6 C

L : The total charge is less than the maximum charge we predicted, so the answer seems reasonable. It isinteresting that this charge can be calculated without knowing either the current or the time tocollapse the loop. Note: We ignored the internal resistance of the galvanometer. D’Arsonvalgalvanometers typically have an internal resistance of 50 to 100 Ω, significantly more than theresistance of the wires given in the problem. A proper solution that includes RG would reduce thetotal charge by about 2 orders of magnitude ( Q ~ 0.01 µC).

238 Chapter 31 Solutions

*31.58 (a) I = dqdt =

εR

where E = −N

dΦB

dt so

dq∫ = NR

dΦBΦ1

Φ2

∫

and the charge through the circuit will be Q = NR (Φ2 – Φ1)

(b) Q = N

RBAcos0 − BAcos

π2

= BANR

so B = RQNA =

(200 Ω)(5.00 × 10– 4 C)(100)(40.0 × 10– 4 m2)

= 0.250 T

31.59 (a) ε = B lv = 0.360 V I = εR

= 0.900 A

(b) FB = IlB = 0.108 N

(c) Since the magnetic flux B · A is in effect decreasing, theinduced current flow through R is from b to a. Point b is athigher potential.

(d) No . Magnetic flux will increase through a loop to the left of ab. Here counterclockwisecurrent will flow to produce upward magnetic field. The in R is still from b to a.

31.60 ε = Blv at a distance r from wire

ε = µ0I

2πr

31.61 (a) At time t , the flux through the loop is ΦB = BAcosθ = a + bt( ) πr2( )cos0˚ = π a + bt( )r2

At t = 0, ΦB = πar2

(b) ε = − dΦB

dt= −πr2 d a + bt( )

dt= −πbr2

− πbr2

(d) P = ε I = − πbr2

−πbr2( ) =

π2b2r4

Chapter 31 Solutions 239

31.62 ε = − d

dt(NBA) = −1

dBdt

πa2 = πa2K

(a) Q = C ε = C π a2K (b) B into the paper is decreasing; therefore, current will attempt to counteract this. Positive

charge will go to upper plate .

(c) The changing magnetic field through the enclosed area induces an electric field ,surrounding the B-field, and this pushes on charges in the wire.

31.63 The flux through the coil is ΦB = B ⋅ A = BAcosθ = BAcosωt . The induced emf is

ε = −N

dΦB

dt= −NBA

d cosωt( )dt

= NBAωsinωt.

(a) εmax = NBAω = 60.0 1.00 T( ) 0.100 × 0.200 m2( ) 30.0 rad s( ) = 36.0 V

(b)

dΦB

dt= ε

N, thus

dΦB

dt max= εmax

N= 36.0 V

60.0= 0.600 V = 0.600 Wb/s

(d) The torque on the coil at any time is τ = µ × B = NIA × B = NAB( )I sinωt = εmax

εR

sinωt

When ε = εmax , sin .ωt = 1 00 and τ = εmax

ωR= 36.0 V( )2

30.0 rad s( ) 10.0 Ω( ) = 4.32 N · m

31.64 (a) We use ε = – N

∆ΦB

∆t , with N = 1.

Taking a = 5.00 × 10- 3 m to be the radius of the washer, and h = 0.500 m,

∆ΦB = B2A − B1A = A(B2 − B1) = πa2 µ0I

2π(h + a)− µ0I

2πa

a2µ0I2

1 +

1h a a

−

−µ0ahI2(h + a)

The time for the washer to drop a distance h (from rest) is: ∆t =

2hg

Therefore, ε =

µ0ahI2(h + a)∆t

µ0ahI2(h + a)

g2h

=µ0aI

2(h + a)gh2

and ε =

T m/A) m) Am m)

m/s m)( ( . ( . )( . .

( . )( .4 10 5 00 10 10 02 0 500 0 00500

9 80 0 5002

7 3 2π × ⋅ ×+

− −

= 97.4 nV

(b) Since the magnetic flux going through the washer (into the plane of the paper) is decreasingin time, a current will form in the washer so as to oppose that decrease. Therefore, thecurrent will flow in a clockwise direction .

240 Chapter 31 Solutions

31.65 ε = −N

dΦB

dt= −N

ddt

BAcosθ( )

ε = −NBcosθ ∆A

∆t

= −200 50.0 × 10−6 T( ) cos 62.0˚( ) 39.0 × 10−4 m2

1.80 s

= –10.2 µV

31.66 Find an expression for the flux through a rectangular area"swept out" by the bar in time t. The magnetic field at a distancex from wire is

B = µ0I

2πx and ΦB =

BdA.∫ Therefore,

ΦB = µ0Ivt

2πdxx

r+l

∫ where vt is the distance the bar has moved in time t.

Then, ε = dΦB

dt=

µ0Iv2π

ln 1 + lr

31.67 The magnetic field at a distance x from a long wire is B = µ0I

2πx. Find an expression for the

flux through the loop.

dΦB = µ0I

2πx(ldx) so

ΦB = µ0Il

2πdxx

r+w

∫ = µ0Il

2πln 1 + w

Therefore, ε = − dΦB

dt= µ0Ilv

2πrw

(r + w) and

I = ε

µ0Ilv2πRr

w(r + w)

31.68 As the wire falls through the magnetic field, a motional emf ε = Blv is induced in it. Thus, acounterclockwise induced current of I = ε R = Blv R flows in the circuit. The falling wireis carrying a current toward the left through the magnetic field. Therefore, it experiences anupward magnetic force given by FB = IlB = B2l2v R . The wire will have attained terminalspeed when the magnitude of this magnetic force equals the weight of the wire.

Thus,

B2l2vt

R= mg , or the terminal speed is vt =

mgRB2l2

31.69 ΦB = (6.00t 3 – 18.0t 2) T · m2 and ε = – dΦBdt = –18.0t 2 + 36.0t

Maximum ε occurs when dεdt

= – 36.0t + 36.0 = 0, which gives t = 1.00 s.

Therefore, the maximum current (at t = 1.00 s) is I = εR

= (–18.0 + 36.0)V

3.00 Ω = 6.00 A

Chapter 31 Solutions 241

31.70 For the suspended mass, M: ΣF = Mg – T = Ma

For the sliding bar, m: ΣF = T – I lB = ma, where I = ε

R= Blv

Mg − B2l2v

R= (m + M)a or

a = dv

dt= Mg

m + M− B2l2v

R(M + m)

dv(α − βv)

= dt0

t∫0

v∫ where

α = Mg

M + m and

β = B2l2

R(M + m)

Therefore, the velocity varies with time as v = α

β(1 − e−β t ) =

MgRB2l2 1 − e−B2l2t/R(M+m)

*31.71 (a) ε = –N dΦBdt = –NA

dBdt = –NA

ddt (µ0nI)

where A = area of coil, N = number of turns in coil, and n = number of turns per unit lengthin solenoid. Therefore,

ε = Nµ0An ddt 4 120sin( )πt[ ] = Nµ0An(480π) cos (120π t)

ε = 40(4π × 10–7 ) π( . )0 0500 2m[ ] (2.00 × 103)(480π) cos(120π t) = (1.19 V) cos(120π t)

(b) I = ∆VR and P = ∆VI =

(1.19 V)2 cos2(120π t)(8.00 Ω)

From cos2 θ = 12 +

12 cos 2θ, the average value of cos2θ is

12 , so

P = 1

21.19 V( )2

8.00 Ω( ) = 88.5 mW

31.72 The induced emf is ε = Blv where B = µ0I

2πy,

v = vi + gt = 9.80 m s2( )t , and

y = yi − 1

2 gt2 = 0.800 m − 4.90 m s2( )t2.

ε =4π× 10−7 T ⋅ m A( ) 200 A( )

2π 0.800 m − 4.90 m s2( )t2[ ] 0.300 m( ) 9.80 m s2( )t =

1.18 × 10−4( )t0.800 − 4.90t2[ ] V

At t = 0.300 s ,

ε =1.18 × 10−4( ) 0.300( )

0.800 − 4.90 0.300( )2[ ] V = 98.3 µV

242 Chapter 31 Solutions

31.73 The magnetic field produced by the current in the straight wire isperpendicular to the plane of the coil at all points within the coil.The magnitude of the field is B = µ0I 2πr . Thus, the flux linkage is

NΦB = µ0NIL

2πdrrh

h+w∫ = µ0NImaxL

2πln

h + wh

sin(ωt + φ)

Finally, the induced emf is ε = −µ0NImaxLω

2πln 1 + w

cos(ωt + φ)

ε = −

4π× 10−7( )(100)(50.0)(0.200 m)(200π s−1)

2πln 1 + 5.00 cm

5.00 cm

cos(ωt + φ)

ε = − 87.1 mV( ) cos(200πt + φ)

The term sin(ωt + φ)in the expression for the current in the straight wire does not changeappreciably when ωt changes by 0.100 rad or less. Thus, the current does not changeappreciably during a time interval

t < 0.100

(200π s−1)= 1.60 × 10−4 s .

We define a critical length, ct = (3.00 × 108 m / s)(1.60 × 10 −4 s) = 4.80 × 104 m equal to thedistance to which field changes could be propagated during an interval of 1.60 × 10 −4 s. Thislength is so much larger than any dimension of the coilor its distance from the wire that,although we consider the straight wire to be infinitely long, we can also safely ignore the fieldpropagation effects in the vicinity of the coil. Moreover, the phase angle can be considered tobe constant along the wire in the vicinity of the coil.

If the frequency ω were much larger, say, 200π× 105 s−1, the corresponding critical lengthwould be only 48.0 cm. In this situation propagation effects would be important and theabove expression for ε would require modification. As a "rule of thumb" we can considerfield propagation effects for circuits of laboratory size to be negligible for frequencies, f = ω 2π,that are less than about 106 Hz.

31.74 ΦB = BA cos θ dΦBdt = –ωBA sinθ ;

I ∝ – sin θ

τ ∝ IB sin θ ∝ – sin2 θ

31.75 The area of the tent that is effective in intercepting magnetic field lines is the areaperpendicular to the direction of the magnetic field. This is the same as the base of the tent.In the initial configuration, this is

A1 = L(2L cos θ) = 2(1.50 m)2 cos 60.0˚ = 2.25 m2

After the tent is flattened, A2 = L(2L) = 2L2 = 2(1.50 m)2 = 4.50 m2

The average induced emf is: ε = − ∆ΦB

∆t= − B ∆A( )

∆t= −

0.300 T( ) 4.50 − 2.25( ) m2

0.100 s= – 6.75 V

Chapter 32 Solutions

*32.1 ε = L ∆I

∆t = (3.00 × 10– 3 H)

1.50 A − 0.200 A0.200 s

= 1.95 × 10–2 V = 19.5 mV

32.2 Treating the telephone cord as a solenoid, we have:

L =

µ0N2Al

(4π× 10−7 T ⋅ m / A)(70.0)2(π)(6.50 × 10−3 m)2

0.600 m = 1.36 Hµ

32.3 ε = + L

∆I∆t

= (2.00 H)

0.500 A0.0100 s

= 100 V

32.4 L = µ0n2Al so n =

Lµ0Al

= 7.80 × 103 turns/m

32.5 L =

N ΦB

I→ ΦB =

L IN

= 240 nT · m2 (through each turn)

32.6 ε = L

dIdt

where L = µ0N2A

Thus, ε = µ0N2A

dIdt

=4π× 10−7 T ⋅ m A( ) 300( )2 π× 10−4 m2( )

0.150 m10.0 A s( ) = 2.37 mV

32.7 ε back = –ε = L dIdt = L

ddt (Imax sin ω t) = Lω Imax cos ω t = (10.0 × 10-3)(120π )(5.00) cos ω t

ε back = (6.00π ) cos (120π t) = (18.8 V) cos (377t)

Chapter 32 Solutions 243

*32.8 From ε = L

∆I∆t

, we have L =

ε∆Ι∆t

= 24.0 × 10– 3 V

10.0 A/s = 2.40 × 10– 3 H

From L = NΦB

I , we have ΦB = LIN =

(2.40 × 10– 3 H)(4.00 A)500 = 19.2 µT · m2

32.9 L = µ0N 2A

l = µ0(420)2(3.00 × 10– 4)

0.160 = 4.16 × 10– 4 H

ε = –L dIdt →

dIdt =

−εL

= –175 × 10– 6 V4.16 × 10– 4 H

= – 0.421 A/s

32.10 The induced emf is ε = −L

dIdt

, where the self-inductance of a solenoid is given by L = µ0N2A

Thus,

dIdt

= − εL

= − εl

µ0N2A

32.11 ε = L dIdt = (90.0 × 10-3)

ddt (t

2 – 6t) V

(a) At t = 1.00 s, ε = 360 mV

(b) At t = 4.00 s, ε = 180 mV

32.12 (a) B = µ0nI = µ0

450

0.120 (0.0400 mA) = 188 µT

(b) ΦB = BA = 3.33 × 10-8 T · m2

I = 0.375 mH

244 Chapter 32 Solutions

(d) B and ΦB are proportional to current; L is independent of current

Chapter 32 Solutions 245

32.13 (a) L = µ0N 2A

l = µ0(120)2π ( 5.00 × 10–3)2

0.0900 = 15.8 µH

(b) ′ΦB = µmµ0

ΦB → L = µmN 2A

l = 800(1.58 × 10– 5 H) = 12.6 mH

32.14 L = NΦB

I = NBA

I ≈ N A

I · µ0NI

2π R =

µ0N 2A2π R

32.15 ε = ε0e−k t = − L

dIdt

dI = − ε0

Le−k t dt

If we require I → 0 as t → ∞, the solution is I =

ε0

kLe−k t =

dqdt

Q = I dt∫ =

ε0

kLe−k t

∞∫ dt = − ε0

k2L Q =

ε0

k2L

32.16 I = ε

R(1 − e−Rt/L )

0.900 ε

R= ε

R1 − e−R(3.00 s)/2.50 H[ ]

exp − R(3.00 s)

2.50 H

= 0.100

R = 2.50 H

3.00 sln 10.0 = 1.92 Ω

246 Chapter 32 Solutions

32.17 τ = LR = 0.200 s:

IImax

= 1 – e–t/τ

(a) 0.500 = 1 – e–t/0.200 → t = τ ln 2.00 = 0.139 s

(b) 0.900 = 1 – e–t/0.200 → t = τ ln 10.0 = 0.461 s

Figure for GoalSolution

Goal Solution A 12.0-V battery is about to be connected to a series circuit containing a 10.0-Ω resistor and a 2.00-Hinductor. How long will it take the current to reach (a) 50.0% and (b) 90.0% of its final value?

G : The time constant for this circuit is τ = L R = 0.2 s, which means that in 0.2 s, the current will reach1/e = 63% of its final value, as shown in the graph to the right. We can see from this graph that thetime to reach 50% of Imax should be slightly less than the time constant, perhaps about 0.15 s, and thetime to reach 0.9Imax should be about 2.5τ = 0.5 s.

O : The exact times can be found from the equation that describes the rising current in the above graphand gives the current as a function of time for a known emf, resistance, and time constant. We settime t = 0 to be the moment the circuit is first connected.

A : At time t , I t( ) = ε(1 − e−t/τ )

where, after a long time, Imax = ε(1 − e−∞)

R= ε

At I t( ) = 0.500Imax , 0.500( ) ε

R= ε(1 − e−t/0.200 s )

Rso 0.500 = 1 − e−t/0.200 s

Isolating the constants on the right, ln e−t/2.00 s( ) = ln 0.500( )

and solving for t , − t

0.200 s= −0.693 or t = 0.139 s

(b) Similarly, to reach 90% of Imax , 0.900 = 1 − e−t/τ and t = −τ ln 1 − 0.900( )

Thus, t = − 0.200 s( ) ln 0.100( ) = 0.461 s

L : The calculated times agree reasonably well with our predictions. We must be careful to avoidconfusing the equation for the rising current with the similar equation for the falling current.Checking our answers against predictions is a safe way to prevent such mistakes.

Chapter 32 Solutions 247

32.18 Taking τ = L R, I = I0e−t/τ :

dIdt

= I0e−t/τ − 1τ

IR + L

dIdt

= 0 will be true if I0Re−t/τ + L I0 e−t/τ( ) − 1

= 0

Because τ = L R , we have agreement with 0 = 0

*32.19 (a) τ = L R = 2.00 × 10– 3 s = 2.00 ms

(b) I = Imax 1 − e−t/τ( ) = 6.00 V

4.00 Ω

1 − e−0.250/2.00( ) = 0.176 A

= 6.00 V4.00 Ω

= 1.50 A

(d) 0.800 = 1 – e–t/2.00 ms → t = – (2.00 ms) ln(0.200) = 3.22 ms

*32.20 I = εR

(1 – e–t/τ ) = 1209.00 (1 – e–1.80/7.00) = 3.02 A

∆VR = IR = (3.02)(9.00) = 27.2 V

∆VL = ε – ∆VR = 120 – 27.2 = 92.8 V

32.21 (a) ∆VR = IR = (8.00 Ω)(2.00 A) = 16.0 V and

∆VL = ε − ∆VR = 36.0 V − 16.0 V = 20.0 V

Therefore,

∆VR

∆VL= 16.0 V

20.0 V= 0.800

(b) ∆VR = IR = (4.50 A)(8.00 Ω) = 36.0 V

∆VL = ε − ∆VR = 0

Figure for GoalSolution

248 Chapter 32 Solutions

Goal Solution For the RL circuit shown in Figure P32.19, let L = 3.00 H, R = 8.00 Ω, and ε = 36.0 V. (a) Calculate the ratioof the potential difference across the resistor to that across the inductor when I = 2.00 A. (b) Calculate thevoltage across the inductor when I = 4.50 A.

G : The voltage across the resistor is proportional to the current, ∆ IRVR = , while the voltage across theinductor is proportional to the rate of change in the current, εL = −LdI dt . When the switch is firstclosed, the voltage across the inductor will be large as it opposes the sudden change in current. As thecurrent approaches its steady state value, the voltage across the resistor increases and the inductor’semf decreases. The maximum current will be ε /R = 4.50 A, so when I = 2.00 A, the resistor andinductor will share similar voltages at this mid-range current, but when I = 4.50 A, the entire circuitvoltage will be across the resistor, and the voltage across the inductor will be zero.

O : We can use the definition of resistance to calculate the voltage across the resistor for each current.We will find the voltage across the inductor by using Kirchhoff's loop rule.

A : (a) When I = 2.00 A, the voltage across the resistor is ∆VR = IR = 2.00 A( ) 8.00 Ω( ) = 16.0 V

Kirchhoff's loop rule tells us that the sum of the changes in potential around the loop must be zero:

ε − ∆VR − εL = 36.0 V − 16.0 V − εL = 0 so εL = 20.0 V and

∆VR

εL= 16.0 V

20.0 V= 0.800

(b) Similarly, for I = 4.50 A, ∆VR = IR = 4.50 A( ) 8.00 Ω( ) = 36.0 V

ε − ∆VR − εL = 36.0 V − 36.0 V − εL = 0 so εL = 0

L : We see that when I = 2.00 A, ∆VR < εL , but they are similar in magnitude as expected. Also aspredicted, the voltage across the inductor goes to zero when the current reaches its maximum value.A worthwhile exercise would be to consider the ratio of these voltages for several different times afterthe switch is reopened.

*32.22 After a long time, 12.0 V = (0.200 A)R Thus, R = 60.0 Ω. Now, τ = LR gives

L = τ R = (5.00 × 10– 4 s)(60.0 V/A) = 30.0 mH

32.23 I = Imax 1 − e−t/τ( ) :

dIdt

= −Imax e−t/τ( ) − 1τ

τ = LR =

15.0 H30.0 Ω

= 0.500 s :dIdt =

RL Imax e–t/τ and Imax =

εR

(a) t = 0: dIdt =

RL Imax e0 =

εL

= 100 V15.0 H = 6.67 A/s

(b) t = 1.50 s: dIdt =

εL

e–t/τ = (6.67 A/s)e– 1.50/(0.500) = (6.67 A/s)e–3.00 = 0.332 A/s

Chapter 32 Solutions 249

32.24 I = Imax 1 − e−t/τ( )

0.980 = 1 − e−3.00×10− 3 /τ

0.0200 = e−3.00×10− 3 /τ

τ = − 3.00 × 10−3

ln(0.0200)= 7.67 × 10−4 s

τ = L R, so L = τ R = (7.67 × 10−4 )(10.0) = 7.67 mH

32.25 Name the currents as shown. By Kirchhoff’s laws:

I1 = I2 + I3 (1)

+10.0 V − 4.00 I1 − 4.00 I2 = 0 (2)

+10.0 V − 4.00 I1 − 8.00 I3 − 1.00( ) dI3

dt= 0 (3)

From (1) and (2), +10.0 − 4.00 I1 − 4.00 I1 + 4.00 I3 = 0 and I1 = 0.500 I3 + 1.25 A

Then (3) becomes 10.0 V − 4.00 0.500 I3 + 1.25 A( ) − 8.00 I3 − 1.00( ) dI3

dt= 0

1.00 H( ) dI3 dt( ) + 10.0 Ω( ) I3 = 5.00 V

We solve the differential equation using Equations 32.6 and 32.7:

I3 t( ) = 5.00 V

10.0 Ω1 − e− 10.0 Ω( )t 1.00 H[ ] =

0.500 A( ) 1 − e−10t/s[ ]

I1 = 1.25 + 0.500 I3 = 1.50 A − 0.250 A( )e−10t/s

32.26 (a) Using τ = RC = L

R, we get

R = L

C= 3.00 H

3.00 × 10−6 F= 1.00 × 103 Ω = 1.00 kΩ

(b) τ = RC = 1.00 × 103 Ω( ) 3.00 × 10−6 F( ) = 3.00 × 10−3 s = 3.00 ms

250 Chapter 32 Solutions

32.27 For t ≤ 0, the current in the inductor is zero. At t = 0, it starts togrow from zero toward 10.0 A with time constant

τ = L R = 10.0 mH( ) 100 Ω( ) = 1.00 × 10−4 s .

For 0 ≤ t ≤ 200 µs, I = Imax 1 − e

−t/τ

10.00 A( ) 1 − e−10000t/s( )

At t = 200 µs, I = 10.00 A( ) 1 − e−2.00( ) = 8.65 A

Thereafter, it decays exponentially as I = I0e− ′t τ , so for t ≥ 200 µs,

I = 8.65 A( )e−10000 t−200 µs( ) s = 8.65 A( )e−10000t s +2.00 = 8.65e2.00 A( )e−10000t s = 63.9 A( )e−10000t s

32.28 (a) I = εR

= 12.0 V12.0 Ω

= 1.00 A

(b) Initial current is 1.00 A, : ∆V12 = (1.00 A)(12.00 Ω) = 12.0 V

∆V1200 = (1.00 A)(1200 Ω) = 1.20 kV

∆VL = 1.21 kV

= – Imax RL e–Rt/L and –L

dIdt = ∆VL = Imax Re–Rt/L

Solving 12.0 V = (1212 V)e–1212t/2.00 so 9.90 × 10– 3 = e– 606t

Thus, t = 7.62 ms

32.29 τ = L

R= 0.140

4.90= 28.6 ms;

Imax = ε

R= 6.00 V

4.90 Ω= 1.22 A

(a) I = Imax 1 − e−t/τ( ) so

0.220 = 1.22 1 − e−t/τ( )

e−t/τ = 0.820 t = −τ ln(0.820) = 5.66 ms

(b) I = Imax 1 − e

− 10.00.0286

= (1.22 A) 1 − e−350( ) = 1.22 A

Chapter 32 Solutions 251

32.30 (a) For a series connection, both inductors carry equal currents at every instant, so dI/dt is thesame for both. The voltage across the pair is

L eq

dIdt

= L1dIdt

+ L 2dIdt

so L eq = L1 + L 2

(b) L eq

dIdt

= L1dI1

dt= L 2

dI2

dt= ∆VL where I = I1 + I2 and

dIdt

=dI1

dt+

dI2

Thus,

∆VL

L eq=

∆VL

L1+

∆VL

L 2 and

1L eq

L1+

1L 2

dIdt

+ R eq I = L1dIdt

+ IR1 + L 2dIdt

+ IR 2

Now I and dI/dt are separate quantities under our control, so functional equality requires both

L eq = L1 + L 2 and R eq = R 1 + R 2

(d) ∆V = L eq

dIdt

+ R eqI = L1dI1

dt+ R1I1 = L 2

dI2

dt+ R 2I2 where I = I1 + I2 and

dIdt

=dI1

dt+

dI2

We may choose to keep the currents constant in time. Then,

1R eq

R 1+

1R 2

We may choose to make the current swing through 0. Then,

1L eq

L1+

1L 2

This equivalent coil with resistance will be equivalentto the pair of real inductors for all other currents as well.

32.31 L = N ΦB

I = 200(3.70 × 10– 4)

1.75 = 42.3 mH so U = 12 LI 2 =

12 (0.423 H)(1.75 A) 2 = 0.0648 J

32.32 (a) The magnetic energy density is given by

u = B 2

2µ0 =

(4.50 T)2

2(1.26 × 10– 6 T · m/A) = 8.06 × 106 J/m3

(b) The magnetic energy stored in the field equals u times the volume of the solenoid (thevolume in which B is non-zero).

U = uV = (8.06 × 106 J/m3) (0.260 m)π(0.0310 m)2[ ] = 6.32 kJ

252 Chapter 32 Solutions

32.33 L = µ0

N 2Al

= µ0(68.0)2 π(0.600 × 10−2 )2

0.0800= 8.21 µH

U = 12 LI 2 = 1

2 (8.21× 10−6 H)(0.770 A)2 = 2.44 µJ

32.34 (a) U = 12 LI 2 =

L ε

= Lε 2

8R2 = (0.800)(500)2

8(30.0)2 = 27.8 J

(b) I = εR

1 − e−(R/L)t[ ] so

ε2R

= εR

1 − e−(R/L)t[ ] → e−(R/L)t = 1

t = ln 2 so t = L

Rln 2 = 0.800

30.0ln 2 = 18.5 ms

32.35 u = ε 0 E 2

2 = 44.2 nJ/m3 u = B 2

2µ0 = 995 µ J/m3

*32.36 (a) U = 12 LI 2 =

12 (4.00 H)(0.500 A) 2 = 0.500 J

(b)d Udt = LI = (4.00 H)(1.00 A) = 4.00 J/s = 4.00 W

32.37 From Equation 32.7, I = ε

R1 − e− Rt L( )

(a) The maximum current, after a long time t , is I = ε

R= 2.00 A.

At that time, the inductor is fully energized and P = I(∆V) = (2.00 A)(10.0 V) = 20.0 W

(b) Plost = I 2R = (2.00 A)2(5.00 Ω) = 20.0 W

(d) U = LI 2

2= (10.0 H)(2.00 A)2

2= 20.0 J

Chapter 32 Solutions 253

32.38 We have u = e0

2and

u = B2

2µ0

Therefore e0

2= B2

2µ0so B

2 = e0µ0E2

B = E e0µ0 = 6.80 × 105 V / m

3.00 × 108 m / s= 2.27 × 10– 3 T

32.39 The total magnetic energy is the volume integral of the energy density, u = B2

2µ0

Because B changes with position, u is not constant. For B = B0 R / r( )2 , u = B0

2µ0

Next, we set up an expression for the magnetic energy in a spherical shell of radius r andthickness dr. Such a shell has a volume 4π r 2 dr, so the energy stored in it is

dU = u 4πr2dr( ) = 2πB0

2R4

µ0

drr2

We integrate this expression for r = R to r = ∞ to obtain the total magnetic energy outside thesphere. This gives

U = 2π B2

0 R 3

µ0 =

2π (5.00 × 10–5 T)2(6.00 × 106 m)3

(1.26 × 10– 6 T · m/A) = 2.70 × 1018 J

32.40 I1(t) = Imaxe−α t sinωt with Imax = 5.00 A, α = 0.0250 s−1, and ω = 377 rad s .

dI1

dt= Imaxe−α t −α sinωt + ωcosωt( )

At t = 0.800 s ,

dI1

dt= 5.00 A s( )e−0.0200 − 0.0250( )sin 0.800 377( )( ) + 377 cos 0.800 377( )( )[ ]

dI1

dt= 1.85 × 103 A s

Thus, ε2 = −M

dI1

dt:

M = −ε2

dI1 dt= +3.20 V

1.85 × 103 A s= 1.73 mH

254 Chapter 32 Solutions

32.41 ε2 = −M

dI1

dt= −(1.00 × 10−4 H)(1.00 × 104 A / s) cos(1000t)

ε2( )max = 1.00 V

32.42 M =

ε2

dI1 dt=

96.0 mV1.20 A / s

= 80.0 mH

32.43 (a) M = NBΦBA

IA= 700(90.0 × 10−6 )

3.50= 18.0 mH

(b) LA = ΦA

IA= 400(300 × 10−6 )

3.50= 34.3 mH

dIA

dt= −(18.0 mH)(0.500 A / s) = – 9.00 mV

32.44 M = N2Φ12

I1=

N2 B1A1( )I1

=N2 µ0n1I1( )A1[ ]

I1= N2 µ 0n1A1

M = (1.00) 4π× 10−7 T ⋅ m A( ) 70.0

0.0500 m

π 5.00 × 10−3 m( )2

= 138 nH

32.45 B at center of (larger) loop: B1 = µ0 I1

(a) M = Φ2

I1= B1A 2

I1= (µ0I1 / 2R)(πr2 )

I1=

µ π02

(b) M =

µ0 π(0.0200)2

2(0.200)= 3.95 nH

Chapter 32 Solutions 255

*32.46 Assume the long wire carries current I. Then the magnitude of the magnetic field itgenerates at distance x from the wire is B = µ0I 2πx, and this field passes perpendicularlythrough the plane of the loop. The flux through the loop is

ΦB = B ⋅ dA = BdA∫ = B ldx( )∫ =∫

µ0Il2π

dxx0.400 mm

1.70 mm∫ = µ0Il

2πln

1.700.400

The mutual inductance between the wire and the loop is then

M = N2Φ12

I1= N2µ0Il

2πIln

1.700.400

= N2µ0l

2π1.45( )

= 1(4π× 10−7 T ⋅ m A)(2.70 × 10−3 m)

2π1.45( )

M = 7.81× 10−10 H = 781 pH

32.47 With I = I1 + I2, the voltage across the pair is:

∆V = − L1

dI1

dt− M

dI2

dt= − L 2

dI2

dt− M

dI1

dt= − L eq

dIdt

So, − dI1

dt=

∆VL1

+ML1

dI2

and − L 2

dI2

dt+

M ∆V( )L1

+M2

dI2

dt= ∆V

(a) (b)

(−L1L2 + M2 )

dI2

dt= ∆V(L1 − M) [1]

By substitution, − dI2

dt=

∆VL 2

+ML 2

dI1

leads to (− L1L 2 + M 2 )

dI1

dt= ∆V (L2 − M) [2]

Adding [1] to [2], (− L1L 2 + M 2 )

dIdt

= ∆V (L1 + L 2 − 2M)

So, L eq = − ∆V

dI / dt=

L1L 2 − M 2

L1 + L 2 − 2M

32.48 At different times, UC( )max = UL( )max so

12 C ∆V( )2[ ]max

= 12 LI2( )max

Imax = C

L∆V( )max = 1.00 × 10−6 F

10.0 × 10−3 H40.0 V( ) = 0.400 A

256 Chapter 32 Solutions

32.49

12 C ∆V( )2[ ]max

= 12 LI2( )max

so ∆VC( )max = L

CImax = 20.0 × 10−3 H

0.500 × 10−6 F0.100 A( ) = 20.0 V

32.50 When the switch has been closed for a long time, battery, resistor,and coil carry constant current Imax = ε / R . When the switch isopened, current in battery and resistor drops to zero, but the coilcarries this same current for a moment as oscillations begin in theLC loop.

We interpret the problem to mean that the voltage amplitude ofthese oscillations is ∆V, in

12 C ∆V( )2 = 1

2 LImax2 .

Then, L = C ∆V( ) 2

Imax2 = C ∆V( )2 R2

ε 2 =

0.500 × 10−6 F( ) 150 V( )2 250 Ω( )2

50.0 V( )2 = 0.281 H

32.51 C =

1(2πf )2 L

(2π ⋅ 6.30 × 106 )2 (1.05 × 106 )= 608 pF

Goal Solution A fixed inductance L = 1.05 µ H is used in series with a variable capacitor in the tuning section of a radio.What capacitance tunes the circuit to the signal from a station broadcasting at 6.30 MHz?

G : It is difficult to predict a value for the capacitance without doing the calculations, but we might expecta typical value in the µF or pF range.

O : We want the resonance frequency of the circuit to match the broadcasting frequency, and for a simpleRLC circuit, the resonance frequency only depends on the magnitudes of the inductance andcapacitance.

A : The resonance frequency is f0 = 1

2π LC

Thus,

C = 1(2π f0 )2 L

= 1

(2π)(6.30 × 106 Hz)[ ]2(1.05 × 10−6 H)

= 608 pF

L : This is indeed a typical capacitance, so our calculation appears reasonable. However, you probablywould not hear any familiar music on this broadcast frequency. The frequency range for FM radiobroadcasting is 88.0 – 108.0 MHz, and AM radio is 535 – 1605 kHz. The 6.30 MHz frequency falls in theMaritime Mobile SSB Radiotelephone range, so you might hear a ship captain instead of Top 40tunes! This and other information about the radio frequency spectrum can be found on the NationalTelecommunications and Information Administration (NTIA) website, which at the time of thisprinting was at http://www.ntia.doc.gov/osmhome/allochrt.html

Chapter 32 Solutions 257

32.52 f =

12π LC

: L =1

(2πf )2C=

1(2π ⋅ 120)2(8.00 × 10−6 )

= 0.220 H

32.53 (a) f = 1

2π LC= 1

2π (0.0820 H)(17.0 × 10−6 F)= 135 Hz

(b) Q = Qmax cosωt = (180 µC) cos(847 × 0.00100) = 119 µC

dt= −ωQmax sinωt = −(847)(180) sin(0.847) = – 114 mA

32.54 (a) f =

12π LC

2π (0.100 H)(1.00 × 10−6 F)= 503 Hz

(b) Q = Cε = (1.00 × 10−6 F)(12.0 V) = 12.0 µC

2 LImax2

Imax = ε C

L = 12 V

1.00 × 10−6 F0.100 H

= 37.9 mA

(d) At all times U = 12 Cε 2 = 1

2 (1.00 × 10−6 F)(12.0 V)2 = 72.0 µ J

32.55

ω = 1LC

= 1

3.30 H( ) 840 × 10−12 F( )= 1.899 × 104 rad s

Q = Qmax cosωt, I = dQ

dt= −ωQmax sinωt

(a)

UC = Q2

2C=

105 × 10−6[ ] cos 1.899 × 104 rad s( ) 2.00 × 10−3 s( )[ ]( )2

2 840 × 10−12( ) = 6.03 J

(b) UL = 1

2 LI2 = 12 Lω2Qmax

2 sin2 ωt( ) =Qmax

2 sin2 ωt( )2C

UL =105 × 10−6 C( )2

sin2 1.899 × 104 rad s( ) 2.00 × 10−3 s( )[ ]2 840 × 10−12 F( ) = 0.529 J

258 Chapter 32 Solutions

32.56 (a)

ωd = 1LC

− R2L

= 1

2.20 × 10−3( ) 1.80 × 10−6( ) − 7.60

2 2.20 × 10−3( )

= 1.58 × 104 rad / s

Therefore, fd = ωd

2π= 2.51 kHz

(b) Rc = 4L

C= 69.9 Ω

32.57 (a) ω0 = 1

LC= 1

(0.500)(0.100 × 10−6 )= 4.47 krad/s

(b) ωd = 1

LC− R

= 4.36 krad/s

(c)

∆ωω0

= 2.53% lower

32.58 Choose to call positive current clockwise in Figure 32.19. It drains charge from the capacitoraccording to I = – dQ/dt. A clockwise trip around the circuit then gives

− IR − LdIdt

= 0

+dQdt

R + Lddt

dQdt

= 0, identical with Equation 32.29.

32.59 (a) Q = Qmaxe− Rt

2L cos ωdt so Imax ∝ e− Rt

0.500 = e− Rt

2L and

Rt2L

= −ln(0.500)

t = − 2L

Rln 0.500( ) =

0.693

2LR

(b) U0 ∝ Qmax2 and U = 0.500U0 so Q = 0.500 Qmax = 0.707Qmax

t = − 2L

Rln(0.707) =

0.347

2LR

(half as long)

Chapter 32 Solutions 259

32.60 With Q = Qmax at t = 0, the charge on the capacitor at any time is Q = Qmax cosωt where

ω = 1 LC . The energy stored in the capacitor at time t is then

U = Q2

2C= Qmax

2Ccos2 ωt = U0 cos2 ωt .

When U = 1

4U0 ,

cos ω t = 1

2and

ω t = 1

3π rad

Therefore,

tLC

= π3

LC= π2

The inductance is then: L =

9t 2

π2C

32.61 (a) εL = −L

dIdt

= − 1.00 mH( ) d 20.0t( )dt

= – 20.0 mV

(b) Q = I dt

t∫ = 20.0t( )dt

t∫ = 10.0t 2

∆VC = −Q

C= −10.0t 2

1.00 × 10−6 F=

− 10.0 MV s2( )t 2

Q 2

2C≥ 1

2LI 2, or

−10.0t 2( )2

2 1.00 × 10−6( ) ≥ 12

1.00 × 10−3( ) 20.0t( )2 ,

then 100t 4 ≥ 400 × 10−9( )t 2 . The earliest time this is true is at t = 4.00 × 10−9 s = 63.2 µs

32.62 (a) ε L = − L

dIdt

= − Lddt

(Kt) = –LK

(b) I =

dQdt

, so Q = I dt

t∫ = Kt dt

t∫ = 1

2 Kt 2

and ∆VC = − Q

− K t 2

2 LI 2 ,

12 C

K2 t 4

4C 2

= 1

2 L K 2t 2( )

Thus t = 2 LC

260 Chapter 32 Solutions

32.63

12C

+12

LI2 so I =

3Q 2

4CL

The flux through each turn of the coil is ΦB =

LIN

Q2 N

3LC

where N is the number of turns.

32.64 Equation 30.16: B = µ0NI

2πr

(a) ΦB = BdA∫ = µ0NI

2πrhdr

∫ = µ0NIh2π

drr

∫ = µ0NIh2π

lnba

L = NΦB

µ0N2h2π

lnba

(b) L = µ0(500)2(0.0100)

2πln

12.010.0

= 91.2 µH

2πAR

= µ0(500)2

2π2.00 × 10−4 m2

0.110

= 90.9 µH

*32.65 (a) At the center, B = Nµ0IR2

2(R2 + 02)3/2 = Nµ0I2R

So the coil creates flux through itself ΦB ≈ BAcosθ = Nµ0I

2RπR2 cos0°= π

2Nµ0IR

When the current it carries changes, εL = −N

dΦB

dt≈ − N

π2

Nµ0RdIdt

= − LdIdt

so L ≈ π

2N2µ0R

(b) 2π r ≈ 3(0.3 m), so r ≈ 0.14 m; L ≈ π2 12

4π × 10–7

T · mA 0.14 m = 2.8 × 10–7 H ~ 100 nH

(c)LR ≈

2.8 × 10–7 V · s/A270 V/A = 1.0 × 10– 9 s ~ 1 ns

Chapter 32 Solutions 261

32.66 (a) If unrolled, the wire forms the diagonal of a 0.100 m(10.0 cm) rectangle as shown. The length of this rectangleis

′L = 9.80 m( )2 − 0.100 m( )2 ′L

0.100 m9.80 m

The mean circumference of each turn is C r= ′2π , where

′r = 24.0 + 0.6442

mm is the mean

radius of each turn. The number of turns is then:

N = ′LC

=9.80 m( )2 − 0.100 m( )2

2π 24.0 + 0.6442

× 10−3 m

= 127

(b)

R = ρlA

=1.70 × 10−8 Ω ⋅ m( ) 10.0 m( )

π 0.322 × 10−3 m( )2 = 0.522 Ω

′l= 800µ0

′l′L

π ′r( )2

L =800 4π× 10−7( )

0.100 m9.80 m( )2 − 0.100 m( )2

π 24.0 + 0.644( ) × 10−3 m

π 24.0 + 0.6442

× 10−3 m

L = 7.68 × 10−2 H = 76.8 mH

32.67 From Ampere’s law, the magnetic field at distance r ≤ R is found as:

B 2πr( ) = µ0J πr 2( ) = µ0

IπR2

πr 2( ) , or

B = µ0Ir

2πR2

The magnetic energy per unit length within the wire is then

= B2

2µ02πr dr( )

R∫ =

µ0 I2

4πR4 r 3 dr0

R∫ =

µ0 I2

4πR4R4

µ0 I 2

16π

This is independent of the radius of the wire.

262 Chapter 32 Solutions

32.68 The primary circuit (containing the battery and solenoid) is anRL circuit with R = 14.0 Ω , and

L = µ0N2A

4π× 10−7( ) 12 500( )2 1.00 × 10−4( )0.0700

= 0.280 H

(a) The time for the current to reach 63.2% of the maximumvalue is the time constant of the circuit:

τ = L

R= 0.280 H

14.0 Ω= 0.0200 s = 20.0 ms

(b) The solenoid's average back emf is ε L = L

∆I∆t

= L

I f − 0

∆t

where I f = 0.632 Imax = 0.632

∆VR

= 0.632

60.0 V14.0 Ω

= 2.71 A

Thus, εL = 0.280 H( ) 2.71 A

0.0200 s

= 37.9 V

(c) The average rate of change of flux through each turn of the overwrapped concentric coil is thesame as that through a turn on the solenoid:

∆ΦB

∆t= µ0n ∆I( )A

∆t=

4π× 10−7 T ⋅ m A( ) 12500 0.0700 m( ) 2.71 A( ) 1.00 × 10−4 m2( )0.0200 s

= 3.04 mV

(d) The magnitude of the average induced emf in the coil is εL = N ∆ΦB ∆t( ) and magnitude ofthe average induced current is

I =

ε L

R= N

R∆ΦB

∆t

= 820

24.0 Ω3.04 × 10−3 V( ) = 0.104 A = 104 mA

32.69 Left-hand loop: E − (I + I2 )R 1 − I2R2 = 0

Outside loop: E − (I + I2 )R 1 − L

dIdt

= 0

Eliminating I 2 gives

′E − I ′R − LdIdt

= 0

This is of the same form as Equation 32.6, so its solution is of the same form as Equation 32.7:

I t( ) = ′E′R(1 − e− ′R t L )

But ′R = R1R2 / R1 + R2( ) and ′E = R2E/ R1 + R2( ), so

′E′R

= ER 2 /(R 1 + R2 )R 1R2 /(R1 + R2 )

= E

R 1

Thus I(t) =

R 1(1 − e− ′R t L )

Chapter 32 Solutions 263

32.70 When switch is closed, steady current I0 = 1.20 A. Whenthe switch is opened after being closed a long time, thecurrent in the right loop is

I = I0e−R2 t L

so e Rt L = I0

Iand

RtL

= lnI0

Therefore, L = R2 t

ln I0 I( ) = 1.00 Ω( ) 0.150 s( )ln 1.20 A 0.250 A( ) = 0.0956 H = 95.6 mH

32.71 (a) While steady-state conditions exist, a 9.00 mA flows clockwise around the right loop of thecircuit. Immediately after the switch is opened, a 9.00 mA current will flow around the outerloop of the circuit. Applying Kirchhoff’s loop rule to this loop gives:

+ε0 − 2.00 + 6.00( ) × 103 Ω[ ] 9.00 × 10−3 A( ) = 0

+ε0 = 72 0. V with end at the higher potentialb

(b)

I = Imaxe−Rt L with Imax = 9.00 mA, R = 8.00 kΩ , and L = 0.400 H

Thus, when the current has reached a value I = 2.00 mA, the elapsed time is:

t = L

Imax

= 0.400 H

8.00 × 103 Ω

9.002.00

= 7.52 × 10−5 s = 75.2 µs

264 Chapter 32 Solutions

32.72 (a) The instant after the switch is closed, the situation is as shown i nthe circuit diagram of Figure (a). The requested quantities are:

IL = 0, IC = ε0 R , IR = ε0 R

∆VL = ε0 , ∆VC = 0, ∆VR = ε0

(b) After the switch has been closed a long time, the steady-stateconditions shown in Figure (b) will exist. The currents andvoltages are:

IL = 0, IC = 0, IR = 0

∆VL = 0, ∆VC = ε0 , ∆VR = 0

IR = 0

+ -

ε0

Q = 0∆VC = 0

IR = ε0/R

+ -IL = 0 ∆VL = ε0

∆VR = ε0

Figure (a)

+ -

ε0

Q = Cε0

∆VC = ε0

+ -IL = 0 ∆VL = 0

∆VR = 0

Figure (b)

IC = ε0/R

32.73 When the switch is closed, asshown in Figure (a), the currentin the inductor is I :

12.0 – 7.50I – 10.0 = 0 → I = 0.267 A

When the switch is opened, theinitial current in the inductorremains at 0.267 A.

IR = ∆V: (0.267 A)R ≤ 80.0 V

R ≤ 300 Ω

(a) (b)

Goal Solution To prevent damage from arcing in an electric motor, a discharge resistor is sometimes placed in parallelwith the armature. If the motor is suddenly unplugged while running, this resistor limits the voltagethat appears across the armature coils. Consider a 12.0-V dc motor with an armature that has a resistanceof 7.50 Ω and an inductance of 450 mH. Assume that the back emf in the armature coils is 10.0 V whenthe motor is running at normal speed. (The equivalent circuit for the armature is shown in FigureP32.73.) Calculate the maximum resistance R that limits the voltage across the armature to 80.0 V whenthe motor is unplugged.

Chapter 32 Solutions 265

G : We should expect R to be significantly greater than the resistance of the armature coil, for otherwise alarge portion of the source current would be diverted through R and much of the total power wouldbe wasted on heating this discharge resistor.

O : When the motor is unplugged, the 10-V back emf will still exist for a short while because the motor’sinertia will tend to keep it spinning. Now the circuit is reduced to a simple series loop with an emf,inductor, and two resistors. The current that was flowing through the armature coil must now flowthrough the discharge resistor, which will create a voltage across R that we wish to limit to 80 V. Astime passes, the current will be reduced by the opposing back emf, and as the motor slows down, theback emf will be reduced to zero, and the current will stop.

A : The steady-state coil current when the switch is closed is found from applying Kirchhoff's loop rule tothe outer loop:

+ 12.0 V − I 7.50 Ω( ) − 10.0 V = 0

so I = 2.00 V

7.50 Ω= 0.267 A

We then require that ∆VR = 80.0 V = 0.267 A( )R

so R = ∆VR

I= 80.0 V

0.267 A= 300 Ω

L : As we expected, this discharge resistance is considerably greater than the coil’s resistance. Note thatwhile the motor is running, the discharge resistor turns P = (12 V)2 300 Ω = 0.48 W of power intoheat (or wastes 0.48 W). The source delivers power at the rate of about P = IV = 0.267 A + 12 V / 300 Ω( )[ ] 12 V( ) = 3.68 W, so the discharge resistor wastes about 13% of thetotal power. For a sense of perspective, this 4-W motor could lift a 40-N weight at a rate of 0.1 m/s.

32.74 (a) L1 = µ0N1

2Al1

=4π× 10−7 T ⋅ m A( ) 1000( )2 1.00 × 10−4 m2( )

0.500 m = 2.51× 10−4 H = 251 µH

(b) M = N2Φ2

I1= N2Φ1

I1= N2BA

I1=

N2 µ0 N1 l1( )I1[ ]A

I1= µ0N1N2A

M =

4π× 10−7 T ⋅ m A( ) 1000( ) 100( ) 1.00 × 10−4 m2( )0.500 m = 2.51× 10−5 H = 25.1 µH

dI2

dt, or

I1R1 = −M

dI2

dt and

I1 = dQ1

dt= − M

dI2

Q1 = − M

R1dI20

t f∫ = − MR1

I2 f − I2i( ) = − MR1

0 − I2i( ) = M I2i

Q1 =

2.51× 10−5 H( ) 1.00 A( )1000 Ω

= 2.51× 10−8 C = 25.1 nC

266 Chapter 32 Solutions

32.75 (a) It has a magnetic field, and it stores energy, so L = 2UI 2

is non-zero.

(b) Every field line goes through the rectangle between the conductors.

I= 1

IBda

y=a

w−a∫

L = 1

Ix dy

w−a∫ µ0I

2π y+ µ0I

2π w −y( )

= 2

Iµ0Ix2π y

dy =∫ 2µ0x2π

ln ya

w−a

Thus L = µ0x

πln

w − aa

32.76 For an RL circuit, I(t) = Imaxe− R

Lt:

I(t)Imax

= 1 − 10−9 = e− R

≅ 1 − RL

t = 10−9 so Rmax = (3.14 × 10−8 )(10−9)

(2.50 yr)(3.16 × 107 s / yr)= 3.97 × 10−25 Ω

(If the ring were of purest copper, of diameter 1 cm, and cross-sectional area 1 mm2, itsresistance would be at least 10– 6 Ω).

32.77 (a) UB = 12 LI 2 =

12 (50.0 H)(50.0 × 10 3 A) 2 = 6.25 × 1010 J

(b) Two adjacent turns are parallel wires carrying current in the same direction. Since the loopshave such large radius, a one-meter section can be regarded as straight.

Then one wire creates a field of B = µ0I

2π r

This causes a force on the next wire of F = IlB sin θ

giving F = Il

µ0I2πr

sin90°= µ0lI2

2πr

Solving for the force, F = (4π × 10–7 N/A2) (1.00 m)(50.0 × 10 3 A) 2

(2π)(0.250 m) = 2000 N

Chapter 32 Solutions 267

32.78 P = I ∆V( )

I = P

∆V= 1.00 × 109 W

200 × 103 V= 5.00 × 103 A

From Ampere’s law, B 2πr( ) = µ0Ienclosed or B = µ0Ienclosed

2πr

(a) At r = a = 0.0200 m, Ienclosed = 5.00 × 103 A and

B =

4π× 10−7 T ⋅ m A( ) 5.00 × 103 A( )2π 0.0200 m( ) = 0.0500 T = 50.0 mT

(b) At r = b = 0.0500 m, Ienclosed = I = 5.00 × 103 A and

B =

4π× 10−7 T ⋅ m A( ) 5.00 × 103 A( )2π 0.0500 m( ) = 0.0200 T = 20.0 mT

B r( )[ ]2 2πrldr( )2µ 0r=a

r=b∫ = µ0I2l

4πdrra

b∫ = µ0I2l

4πln

U =

4π× 10−7 T ⋅ m A( ) 5.00 × 103 A( )21000 × 103 m( )

4πln

5.00 cm2.00 cm

= 2.29 × 106 J = 2.29 MJ

(d) The magnetic field created by the inner conductor exerts a force of repulsion on the current i nthe outer sheath. The strength of this field, from part (b), is 20.0 mT. Consider a smallrectangular section of the outer cylinder of length l and width w . It carries a current of

5.00 × 103 A( ) w

2π 0.0500 m( )

and experiences an outward force F = IlBsinθ =

5.00 × 103 A( )w2π 0.0500 m( ) l 20.0 × 10−3 T( ) sin 90.0˚

The pressure on it is P = F

A= F

wl=

5.00 × 103 A( ) 20.0 × 10−3 T( )2π 0.0500 m( ) = 318 Pa

268 Chapter 32 Solutions

*32.79 (a) B = µ0NI

4π× 10−7 T ⋅ m A( ) 1400( ) 2.00 A( )1.20 m

= 2.93 × 10−3 T (upward)

(b)

u = B2

2µ0=

2.93 × 10−3 T( )2

2 4π× 10−7 T ⋅ m A( ) = 3.42 J

m31 N ⋅ m

1 J

= 3.42 N

m2 = 3.42 Pa

(d) The vertical component of the field of the solenoid exerts an inward force on thesuperconductor. The total horizontal force is zero. Over the top end of the solenoid, its fielddiverges and has a radially outward horizontal component. This component exerts upwardforce on the clockwise superconductor current. The total force on the core is upward . Youcan think of it as a force of repulsion between the solenoid with its north end pointing up,and the core, with its north end pointing down.

(e) F = PA = 3.42 Pa( ) π 1.10 × 10−2 m( )2

= 1.30 × 10−3 N

Note that we have not proven that energy density is pressure. In fact, it is not in some cases;see problem 12 in Chapter 21.

Chapter 33 Solutions

33.1 ∆v(t) = ∆Vmax sin(ωt) = 2 ∆Vrms sin(ωt) = 200 2 sin[2π(100t)] = (283 V) sin (628 t)

33.2 ∆Vrms = 170 V

2= 120 V

(a) P = (∆Vrms)2

R→ R = (120 V)2

75.0 W= 193 Ω

(b) R = (120 V)2

100 W= 144 Ω

33.3 Each meter reads the rms value.

∆Vrms = 100 V

2= 70.7 V

Irms = ∆Vrms

R= 70.7 V

24.0 Ω= 2.95 A

33.4 (a) ∆vR = ∆Vmax sinωt

∆vR = 0.250 ∆Vmax( ), so sinωt = 0.250, or ωt = sin−1 0.250( )The smallest angle for which this is true is ωt = 0 253. rad. Thus, if t = 0.0100 s ,

ω = 0.253 rad

0.0100 s= 25.3 rad/s

(b) The second time when ∆vR = 0.250 ∆Vmax( ), ωt = sin−1 0.250( ) again. For this occurrence,

ω πt = − =0 253 2 89. . rad rad (to understand why this is true, recall the identity sin π − θ( ) = sinθfrom trigonometry). Thus,

t = 2.89 rad

25.3 rad s= 0.114 s

33.5 iR = Imax sinωt becomes 0.600 = sin(ω 0.00700)

Thus, (0.00700)ω = sin–1(0.600) = 0.644

and ω = 91.9 rad/s = 2π f so f = 14.6 Hz

Chapter 33 Solutions 271

33.6 P = Irms ∆Vrms( ) and ∆Vrms = 120 V for each bulb (parallel circuit), so:

I1 = I2 = P1

∆Vrms= 150 W

120 V= 1.25 A , and

R1 = ∆Vrms

I1= 120 V

1.25 A= 96.0 Ω = R2

I3 = P3

∆Vrms= 100 W

120 V= 0.833 A , and

R3 = ∆Vrms

I3= 120 V

0.833 A= 144 Ω

33.7 ∆Vmax = 15.0 V and Rtotal = 8.20 Ω + 10.4 Ω = 18.6 Ω

Imax = ∆Vmax

Rtotal= 15.0 V

18.6 Ω= 0.806 A = 2 Irms

Pspeaker = Irms

2 Rspeaker = 0.806 A2

10.4 Ω( ) = 3.38 W

33.8 For Imax = 80.0 mA, Irms = 80.0 mA

2 = 56.6 mA

(XL)min = VrmsIrms

= 50.0 V

0.0566 A = 884 Ω

XL = 2π f L → L = XL2π f ≥

884 Ω2π (20.0) ≥ 7.03 H

33.9 (a) XL = ∆Vmax

Imax= 100

7.50= 13.3 Ω

L = XL

ω= 13.3

2π(50.0)= 0.0424 H = 42.4 mH

(b) XL = ∆Vmax

Imax= 100

2.50= 40.0 Ω

ω = XL

L= 40.0

42.4 × 10−3 = 942 rad/s

33.10 At 50.0 Hz,

XL = 2π 50.0 Hz( )L = 2π 50.0 Hz( )XL 60.0 Hz

2π 60.0 Hz( )

= 50.060.0

54.0 Ω( ) = 45.0 Ω

Imax = ∆Vmax

XL=

2 ∆Vrms( )XL

= 2 100 V( )45.0 Ω

= 3.14 A

272 Chapter 33 Solutions

33.11

iL t( ) = ∆Vmax

ωLsin ωt − π 2( ) =

80.0 V( )sin 65.0π ( ) 0.0155( ) − π 2[ ]65.0π rad s( ) 70.0 × 10−3 H( )

iL t( ) = 5.60 A( )sin 1.59 rad( ) = 5.60 A

33.12 ω = 2πf = 2π(60.0 / s) = 377 rad / s

XL = ωL = (377 / s)(0.0200 V ⋅ s / A) = 7.54 Ω

Irms =

∆Vrms

XL=

120 V7.54 Ω

= 15.9 A

Imax = 2 Irms = 2 (15.9 A) = 22.5 A

i(t) = Imax sinωt = (22.5 A)sin

2π 60.0( )s

⋅ 1 s180

= (22.5 A) sin 120° = 19.5 A

U = 1

2 Li 2 = 12 0.0200

V ⋅ sA

(19.5 A)2 = 3.80 J

33.13 L =

N ΦB

I where ΦB is the flux through each turn.

N ΦB,max = LIB, max = XL

ω∆VL, max( )

N ΦB,max =

2 ∆VL, rms( )2πf

= 120 V ⋅ s2 π(60.0)

T ⋅ C ⋅ mN ⋅ s

N ⋅ mJ

JV ⋅ C

= 0.450 T · m2

33.14 (a) XC = 1

2πfC:

12πf (22.0 × 10−6 )

< 175 Ω

12π(22.0 × 10−6 )(175)

< f f > 41.3 Hz

(b) XC ∝ 1

C, so X(44) = 1

2 X(22): XC < 87.5 Ω

33.15 Imax = 2 Irms =

2 ∆Vrms( )XC

= 2 ∆Vrms( )2πfC

(a) Imax = 2 (120 V)2π(60.0 / s)(2.20 × 10−6 C / V) = 141 mA

(b) Imax = 2 (240 V)2π(50.0 / s)(2.20 × 10−6 F) = 235 mA

Chapter 33 Solutions 273

33.16 Qmax = C ∆Vmax( ) = C 2 ∆Vrms( )[ ] = 2 C ∆Vrms( )

33.17 Imax = ∆Vmax( )ωC = (48.0 V)(2π)(90.0 s−1)(3.70 × 10−6 F) = 100 mA

33.18 XC =

1ωC

2π(60.0 / s)(1.00 × 10−3 C / V)= 2.65 Ω

vC(t) = ∆Vmax sin ωt , to be zero at t = 0

iC =

∆Vmax

XCsin(ωt + φ) =

2 (120 V)2.65 Ω

sin 2π 60 s-1

180 s-1 + 90.0°

= (64.0 A)sin(120°+ 90.0°) = – 32.0 A

33.19 (a) XL = ω L = 2π (50.0)(400 × 10- 3) = 126 Ω

XC = 1

ωC= 1

2π(50.0)(4.43 × 10−6 )= 719 Ω

Z = R2 + (XL − XC )2 = 5002 + (126 − 719)2 = 776 Ω

∆Vmax = Imax Z = (250 × 10- 3)(776) = 194 V

(b) φ = tan−1 XL − XC

= tan−1 126 − 719

500

= – 49.9° Thus, the Current leads the voltage.

33.20 ωL = 1

ωC→ ω = 1

LC= 1

(57.0 × 10−6 )(57.0 × 10−6 )= 1.75 × 104 rad / s

f = ω2π = 2.79 kHz

33.21 (a) XL = ω L = 2π (50.0 s-1)(250 × 10-3 H) = 78.5 Ω

(b) XC = 1

ωC= 2π(50.0 s−1)(2.00 × 10−6 F)[ ] −1

= 1.59 kΩ

(d) Imax = ∆Vmax

Z= 210 V

1.52 × 103 Ω= 138 mA

(e) φ = tan−1 XL − XC

= tan−1(−10.1) = – 84.3°

274 Chapter 33 Solutions

33.22 (a) Z = R2 + XL − XC( ) = 68.02 + 16.0 − 101( )2 = 109 Ω

XL = ωL = 100( ) 0.160( ) = 16.0 Ω

XC = 1ωC

= 1

100( ) 99.0 × 10−6( ) = 101 Ω

(b) Imax = ∆Vmax

Z= 40.0 V

109 Ω= 0.367 A

R= 16.0 − 101

68.0= −1.25:

φ = −0.896 rad = −51.3°

Imax = 0.367 A ω = 100 rad/s φ = – 0.896 rad = – 51.3°

33.23 XL = 2πf L = 2π 60.0( ) 0.460( ) = 173 Ω

XC = 12πfC

= 1

2π 60.0( ) 21.0 × 10−6( ) = 126 Ω

(a) tanφ = XL − XC

R= 173 Ω − 126 Ω

150 Ω= 0.314

φ = 0.304 rad = 17.4°

(b) Since XL > XC , φ is positive; so voltage leads the current .

33.24 XC =

12πfC

2π(60.0 Hz)(20.0 × 10−12 F)= 1.33 × 108 Ω

Z = (50.0 × 103 Ω)2 + (1.33 × 108 Ω)2 ≈ 1.33 × 108 Ω

Irms = ∆Vrms

Z = 5000 V

1.33 × 108 Ω = 3.77 × 10–5 A

∆Vrms( )body = IrmsRbody = (3.77 × 10−5 A)(50.0 × 10 3 Ω) = 1.88 V

Chapter 33 Solutions 275

33.25 XC = 1

ωC= 1

2π(50.0)(65.0 × 10−6 )= 49.0 Ω

XL = ωL = 2π(50.0)(185 × 10−3 ) = 58.1 Ω

Z = R2 + (XL − XC )2 = (40.0)2 + (58.1 − 49.0)2 = 41.0 Ω

Imax = ∆Vmax

Z= 150

41.0= 3.66 A

(a) ∆VR = Imax R = (3.66)(40) = 146 V

(b) ∆VL = Imax XL = (3.66)(58.1) = 212.5 = 212 V

(d) ∆VL – ∆VC = 212.5 – 179.1 = 33.4 V

33.26 R = 300 Ω

XL = ωL = 2π 500

π s−1

0.200 H( ) = 200 Ω

XC = 1

ωC= 2π 500

π s−1

11.0 × 10−6 F( )

−1

= 90.9 Ω

XL = 200 Ω

XC = 90.9 ΩR = 300 Ω

ZXL - XC = 109 Ω

Z = R2 + XL − XC( )2 = 319 Ω and φ = tan−1 XL − XC

= 20.0°

33.27 (a) XL = 2π 100 Hz( ) 20.5 H( ) = 1.29 × 104 Ω

Z = ∆Vrms

Irms= 200 V

4.00 A= 50.0 Ω

XL − XC( )2 = Z2 − R2 = 50.0 Ω( )2 − 35.0 Ω( )2

XL − XC = 1.29 × 104 Ω − 1

2π 100 Hz( )C = ±35.7 Ω C = 123 nF or 124 nF

(b) ∆VL,rms = Irms XL = 4.00 A( ) 1.29 × 104 Ω( ) = 51.5 kV

Notice that this is a very large voltage!

276 Chapter 33 Solutions

33.28 XL = ωL = (1000 / s)(0.0500 H)[ ] = 50.0 Ω

XC = 1/ωC = (1000 / s)(50.0 × 10−6 F)[ ]−1

= 20.0 Ω

Z = R2 + (XL − XC )2

Z = (40.0)2 + (50.0 − 20.0)2 = 50.0 Ω

(a) Irms = ∆Vrms( )/ Z = 100 V / 50.0 Ω

Irms = 2.00 A

φ = Arctan

XL − XC

φ = Arctan

30.0 Ω40.0 Ω

= 36.9°

(b) P = ∆Vrms( ) Irms cos φ = 100 V(2.00 A) cos 36.9° = 160 W

33.29 ω = 1000 rad/s, R = 400 Ω, C = 5.00 × 10– 6 F, L = 0.500 H

∆Vmax = 100 V, ωL = 500 Ω ,

1ωC

= 200 Ω

Z = R2 + ωL − 1

ωC

= 4002 + 3002 = 500 Ω

Imax = ∆Vmax

Z= 100

500= 0.200 A

The average power dissipated in the circuit is P = Irms

2 R = Imax2

P = (0.200 A)2

2(400 Ω) = 8.00 W

Chapter 33 Solutions 277

Goal Solution An ac voltage of the form ∆v = 100 V( )sin 1000 t( ) is applied to a series RLC circuit. If R = 400 Ω,C = 5.00 µ F, and L = 0.500 H, what is the average power delivered to the circuit?

G : Comparing ∆v = 100 V( )sin 1000 t( ) with ∆v = ∆Vmax sin ωt , we see that

∆Vmax = 100 V and ω = 1000 s-1

Only the resistor takes electric energy out of the circuit, but the capacitor and inductor will impede thecurrent flow and therefore reduce the voltage across the resistor. Because of this impedance, theaverage power dissipated by the resistor must be less than the maximum power from the source:

Pmax =

∆Vmax( )2

2R= 100 V( )2

2 400 Ω( ) = 12.5 W

O : The actual power dissipated by the resistor can be found from P = Irms2 R , where Irms = ∆Vrms / Z.

A : ∆Vrms = 100

2= 70.7 V

In order to calculate the impedance, we first need the capacitive and inductive reactances:

XC = 1

ωC= 1

(1000 s-1)(5.00 × 10−6 F)= 200 Ω and

XL = ωL = 1000 s-1( ) 0.500 H( ) = 500 Ω

Then, Z = R2 + (XL − XC )2 = (400 Ω)2 + (500 Ω − 200 Ω)2 = 500 Ω

Irms = ∆Vrms

Z= 70.7 V

500 Ω= 0.141 A and P = Irms

2 R = 0.141 A( )2 400 Ω( ) = 8.00 W

L : The power dissipated by the resistor is less than 12.5 W, so our answer appears to be reasonable. Aswith other RLC circuits, the power will be maximized at the resonance frequency where XL = XC sothat Z = R . Then the average power dissipated will simply be the 12.5 W we calculated first.

33.30 Z = R2 + XL − XC( )2 or XL − XC( ) = Z2 − R2

XL − XC( ) = 75.0 Ω( )2 − 45.0 Ω( )2 = 60.0 Ω

φ = tan−1 XL − XC

= tan−1 60.0 Ω

45.0 Ω

= 53.1°

Irms = ∆Vrms

Z= 210 V

75.0 Ω= 2.80 A

P = ∆Vrms( )Irms cosφ = 210 V( ) 2.80 A( )cos 53.1˚( ) = 353 W

278 Chapter 33 Solutions

33.31 (a) P = Irms(∆Vrms)cosφ = (9.00)(180) cos(– 37.0°) = 1.29 × 103 W

P = Irms2 R so 1.29 × 103 = (9.00)2 R and R = 16.0 Ω

(b) tanφ = XL − XC

Rbecomes

tan( . )− ° = −

37 016

X XL C : so XL – XC = – 12.0 Ω

*33.32 XL = ωL = 2π(60.0 / s)(0.0250 H) = 9.42 Ω

Z = R2 + (XL − XC )2 = 20.0( )2 + 9.42( )2 Ω = 22.1 Ω

(a) Irms = ∆Vrms

Z= 120 V

22.1 Ω= 5.43 A

(b) φ = tan−1 9.42 / 20.0( ) = 25.2° so power factor = cos φ = 0.905

12π(60.0 s−1)C

and C = 281 µF

(d) Pb = Pd or ∆Vrms( )b Irms( )b cos φb =

∆Vrms( )d2

∆Vrms( )d = R ∆Vrms( )b Irms( )b cos φb = 20.0 Ω( ) 120 V( ) 5.43 A( ) 0.905( ) = 109 V

33.33 Consider a two-wire transmission line:

Irms = P

∆Vrms= 100 × 106 W

50.0 × 103 V= 2.00 × 103 A

loss = 0.0100( )P = Irms2 R line = Irms

2 2R1( )

RL∆Vrms

Thus,

R1 =0.0100( )P

2 Imax2 =

0.0100( ) 100 × 106 W( )2 2.00 × 103 A( )2 = 0.125 Ω

But R1 = ρl

A or

A = πd2

4= ρl

Therefore d = 4ρl

πR1=

4 1.70 × 10−8 Ω ⋅ m( ) 100 × 103 m( )π 0.125 Ω( ) = 0.132 m = 132 mm

Chapter 33 Solutions 279

33.34 Consider a two-wire transmission line:

Irms = P

∆Vrmsand power loss =

Irms

2 R line = P

100

Thus,

∆Vrms

2R1( ) = P

100or

R1 =

∆Vrms( )2

200P

R1 = ρ d

∆Vrms( )2

200Por

A = π 2r( )2

4= 200ρP d

∆Vrms( )2

and the diameter is 2r =

800ρP d

π ∆Vrms( )2

RL∆Vrms

33.35 One-half the time, the left side of the generator is positive, thetop diode conducts, and the bottom diode switches off. Thepower supply sees resistance

12R

+ 12R

−1

= R and the power is

∆Vrms( )2

The other half of the time the right side of the generator ispositive, the upper diode is an open circuit, and the lower diodehas zero resistance. The equivalent resistance is then

RL∆Vrms

Req = R + 1

3R+ 1

−1

= 7R4

and P =

∆Vrms( )2

Req=

4 ∆Vrms( )2

The overall time average power is:

∆Vrms( )2 R[ ] + 4 ∆Vrms( )2 7R[ ]2

11 ∆Vrms( )2

14 R

33.36 At resonance,

12πfC

= 2πf L and

2π f( )2 L= C

The range of values for C is 46.5 pF to 419 pF

33.37 ω0 = 2π(99.7 × 106 ) = 6.26 × 108 rad / s = 1

C = 1

ω02L

= 1(6.26 × 108)2(1.40 × 10−6 )

= 1.82 pF

280 Chapter 33 Solutions

33.38 L = 20.0 mH, C = 1.00 × 10–7, R = 20.0 Ω, ∆Vmax = 100 V

(a) The resonant frequency for a series –RLC circuit is f = 1

2π1

LC= 3.56 kHz

(b) At resonance, Imax = ∆Vmax

R= 5.00 A

R= 22.4

(d) ∆VL,max = XLImax = ω0L Imax = 2.24 kV

33.39 The resonance frequency is ω0 = 1 LC . Thus, if ω = 2ω0,

XL = ωL = 2

L = 2LC

and XC = 1

ωC= LC

2C= 1

2LC

Z = R2 + XL − XC( )2 = R2 + 2.25 L C( ) so

Irms = ∆Vrms

Z= ∆Vrms

R2 + 2.25 L C( )

and the energy dissipated in one period is Q = P∆t :

Q =

∆Vrms( )2 R

R2 + 2.25 L C( )2πω

∆Vrms( )2 RC

R2C + 2.25Lπ LC( ) =

4π ∆Vrms( )2 RC LC

4R2C + 9.00L

With the values specified for this circuit, this gives:

Q =4π 50.0 V( )2 10.0 Ω( ) 100 × 10−6 F( ) 3 2

10.0 × 10−3 H( )1 2

4 10.0 Ω( )2 100 × 10−6 F( ) + 9.00 10.0 × 10−3 H( ) = 242 mJ

33.40 The resonance frequency is ω0 = 1 LC . Thus, if ω = 2ω0,

XL = ωL = 2

L = 2LC

and XC = 1

ωC= LC

2C= 1

2LC

Then Z = R2 + XL − XC( )2 = R2 + 2.25 L C( ) so

Irms = ∆Vrms

Z= ∆Vrms

R2 + 2.25 L C( )and the energy dissipated in one period is

Q = P∆t =

∆Vrms( )2 R

R2 + 2.25 L C( )2πω

∆Vrms( )2 RC

R2C + 2.25Lπ LC( ) =

4π ∆Vrms( )2 RC LC

4R2C + 9.00L

Chapter 33 Solutions 281

*33.41 For the circuit of problem 22,

ω0 = 1LC

= 1

160 × 10−3 H( ) 99.0 × 10−6 F( )= 251 rad s

Q = ω0L

251 rad s( ) 160 × 10−3 H( )68.0 Ω

= 0.591

For the circuit of problem 23, Q = ω0L

R= L

R LC= 1

RLC

= 1150 Ω

460 × 10−3 H21.0 × 10−6 F

= 0.987

The circuit of problem 23 has a sharper resonance.

33.42 (a) ∆V2,rms = 113 120 V( ) = 9.23 V

(b) ∆V1,rms I1,rms = ∆V2,rms I2,rms

(120 V)(0.350 A) = (9.23 V)I2,rms

I2,rms =

42.0 W9.23 V

= 4.55 A for a transformer with no energy loss

33.43 ∆Vout( )max = N2

N1∆Vin( )max = 2000

350

(170 V) = 971 V

∆Vout( )rms = (971 V)

2= 687 V

33.44 (a) ∆V2, rms( ) =

N1∆V1, rms( )

N2 =

(2200)(80)110

= 1600 windings

(b) I1,rms ∆V1,rms( ) = I2,rms ∆V2,rms( )

I1,rms =

(1.50)(2200)110

= 30.0 A

I1,rms =

(1.20)(2200)110(0.950)

= 25.3 A

282 Chapter 33 Solutions

33.45 The rms voltage across the transformer primary is

N2∆V2,rms( )

so the source voltage is ∆Vs, rms = I1,rms Rs +

N2∆V2,rms( )

The secondary current is

∆V2,rms( )RL

, so the primary current is

∆V2,rms( )RL

= I1,rms

Then ∆Vs, rms =

N2 ∆V2,rms( )Rs

N1RL+

N1 ∆V2,rms( )N2

and

Rs =N1RL

N2 ∆V2,rms( ) ∆Vs, rms −N1 ∆V2,rms( )

= 5(50.0 Ω)2(25.0 V)

80.0 V − 5(25.0 V)2

= 87.5 Ω

33.46 (a) ∆V2,rms = N2

N1∆V1,rms( )

N1=

∆V2,rms

∆V1,rms= 10.0 × 103 V

120 V= 83.3

(b) I2,rms ∆V2,rms( ) = 0.900 I1,rms ∆V1,rms( )

I2,rms 10.0 × 103 V( ) = 0.900

120 V24.0 Ω

120 V( ) I2,rms = 54.0 mA

∆V2,rms

I2,rms= 10.0 × 103 V

0.054 A= 185 kΩ

33.47 (a) R = (4.50 × 10−4 Ω / m)(6.44 × 105 m) = 290 Ω and Irms =

∆Vrms=

5.00 × 106 W5.00 × 105 V

= 10.0 A

Ploss = Irms2 R = (10.0 A)2(290 Ω) = 29.0 kW

(b)

Ploss

2.90 × 104

5.00 × 106 = 5.80 × 10−3

(c) It is impossible to transmit so much power at such low voltage. Maximum power transferoccurs when load resistance equals the line resistance of 290 Ω, and is

(4.50 × 103 V)2

2 ⋅ 2(290 Ω)= 17.5 kW, far below the required 5 000 kW

Chapter 33 Solutions 283

33.48 For the filter circuit,

∆Vout

∆Vin= XC

R2 + XC2

(a) At f = 600 Hz,

XC = 12πfC

= 1

2π 600 Hz( ) 8.00 × 10−9 F( ) = 3.32 × 104 Ω

and

∆Vout

∆Vin= 3.32 × 104 Ω

90.0 Ω( )2 + 3.32 × 104 Ω( )2≈ 1.00

(b) At f = 600 kHz,

XC = 12πfC

= 1

2π 600 × 103 Hz( ) 8.00 × 10−9 F( ) = 33.2 Ω

and

∆Vout

∆Vin= 33.2 Ω

90.0 Ω( )2 + 33.2 Ω( )2= 0.346

33.49 For this RC high-pass filter,

∆Vout

∆Vin= R

R2 + XC2

(a) When

∆Vout

∆Vin= 0.500,

then

0.500 Ω

0.500 Ω( )2 + XC2

= 0.500 or XC = 0.866 Ω

If this occurs at f = 300 Hz, the capacitance is

C = 1

2πf XC= 1

2π 300 Hz( ) 0.866 Ω( ) = 6.13 × 10−4 F = 613 µF

(b) With this capacitance and a frequency of 600 Hz,

XC = 1

2π 600 Hz( ) 6.13 × 10−4 F( ) = 0.433 Ω

∆Vout

∆Vin= R

R2 + XC2

= 0.500 Ω

0.500 Ω( )2 + 0.433 Ω( )2= 0.756

(a)

(b)

(c)

Figures for GoalSolution

284 Chapter 33 Solutions

Goal Solution The RC high-pass filter shown in Figure 33.22 has a resistance R = 0.500 Ω. (a) What capacitance gives anoutput signal that has one-half the amplitude of a 300-Hz input signal? (b) What is the gain ( ∆Vout / ∆Vin)for a 600-Hz signal?

G : It is difficult to estimate the capacitance required without actually calculating it, but we might expect atypical value in the µF to pF range. The nature of a high-pass filter is to yield a larger gain at higherfrequencies, so if this circuit is designed to have a gain of 0.5 at 300 Hz, then it should have a highergain at 600 Hz. We might guess it is near 1.0 based on Figure (b) above.

O : The output voltage of this circuit is taken across the resistor, but the input sees the impedance of theresistor and the capacitor. Therefore, the gain will be the ratio of the resistance to the impedance.

A :

∆Vout

∆Vin= R

R2 + 1 ωC( )2

(a) When ∆Vout / ∆Vin = 0.500

solving for C gives

C = 1

ωR∆Vin

∆Vout

− 1 = 1

(2π)(300 Hz)(0.500 Ω) (2.00)2 − 1= 613 µF

(b) At 600 Hz, we have ω = 2π rad( ) 600 s-1( )

∆Vout

∆Vin= 0.500 Ω

0.500 Ω( )2 + 11200π rad / s( ) 613 µF( )

2= 0.756

L : The capacitance value seems reasonable, but the gain is considerably less than we expected. Based onour calculation, we can modify the graph in Figure (b) to more transparently represent thecharacteristics of this high-pass filter, now shown in Figure (c). If this were an audio filter, it wouldreduce low frequency “humming” sounds while allowing high pitch sounds to pass through. A lowpass filter would be needed to reduce high frequency “static” noise.

33.50 ∆V1 = I r + R( )2 + XL2 , and ∆V2 = I R2 + XL

Thus, when ∆V1 = 2∆V2 r + R( )2 + XL

2 = 4 R2 + XL2( )

or 25.0 Ω( )2 + XL2 = 4 5.00 Ω( )2 + 4XL

∆V1 ∆V2

r = 20.0 Ω

R =5.00 Ω

L = 250 mH

which gives XL = 2πf 0.250 H( ) = 625 − 100

3 Ω and f = 8.42 Hz

Chapter 33 Solutions 285

*33.51

∆Vout

∆Vin= R

R2 + XL − XC( )2

(a) At 200 Hz:

= 8.00 Ω( )2

8.00 Ω( )2 + 400πL − 1400πC

At 4000 Hz: 8.00 Ω( )2 + 8000πL − 1

8000πC

= 4 8.00 Ω( )2

At the low frequency, XL − XC < 0 . This reduces to 400

1400

13 9ππ

− = − . Ω [1]

For the high frequency half-voltage point, 8000

18000

13 9ππ

− = + . Ω [2]

Solving Equations (1) and (2) simultaneously gives C = 54.6 µF and L = 580 µH

(b) When XL = XC,

∆Vout

∆Vin= ∆Vout

∆Vin

max

= 1.00

f0 = 12π LC

= 1

2π 5.80 × 10−4 H( ) 5.46 × 10−5 F( )= 894 Hz

(d) At 200 Hz,

∆Vout

∆Vin= R

Z= 1

2 and XC > XL ,

so the phasor diagram is as shown:

ZXL - XC

φ or φ∆Vout

∆Vin

φ = −cos−1 R

= −cos−1 1

so ∆Vout leads ∆Vin by 60.0°

At f0 , XL = XC so ∆Vout and ∆Vin have a phase difference of 0°

At 4000 Hz,

∆Vout

∆Vin= R

Z= 1

2and XL − XC > 0

Thus, φ = cos−1 1

= 60.0°

ZXL - XC φ

orφ ∆Vout

∆Vin

or ∆Vout lags ∆Vin by 60.0°

(e) At 200 Hz and at 4 kHz, P =

∆Vout,rms( )2

12 ∆Vin, rms( )2

12 ∆Vin,max( )2

R = 10.0 V( )2

8 8.00 Ω( ) = 1.56 W

At f0 , P =

∆Vout,rms( )2

∆Vin,rms( )2

12 ∆Vin,max( )2

R= 10.0 V( )2

2 8.00 Ω( ) = 6.25 W

(f) We take: Q =

ω0L

R= 2π f0 L

2π 894 Hz( ) 5.80 × 10−4 H( )8.00 Ω

= 0.408

286 Chapter 33 Solutions

33.52 For a high-pass filter,

∆Vout

∆Vin= R

R2 + 1ωC

∆Vout( )1∆Vin( )1

= R

R2 + 1ωC

2and

∆Vout( )2∆Vin( )2

= R

R2 + 1ωC

Now ∆Vin( )2 = ∆Vout( )1 so

∆Vout( )2∆Vin( )1

= R2

R2 + 1ωC

2 =

1 + 1ωRC

33.53 Rewrite the circuit in terms of impedance as shown in Fig. (b).

Find: ∆Vout = ZR

ZR + ZC∆Vab [1]

From Figure (c), ∆Vab =

ZC || ZR + ZC( )ZR + ZC || ZR + ZC( ) ∆Vin

So Eq. [1] becomes ∆Vout =

ZR ZC || ZR + ZC( )[ ]ZR + ZC( ) ZR + ZC|| ZR + ZC( )[ ] ∆Vin

∆Vout

∆Vin=

ZR1

ZC+ 1

ZR + ZC

−1

ZR + ZC( ) ZR + 1ZC

+ 1ZR + ZC

−1

∆Vout

∆Vin = ZRZC

ZC ZC + ZR( ) + ZR ZR + 2ZC( ) = ZR

3ZR + ZC + ZR( )2 ZC

Now, ZR = R and ZC = −j

ωC where j= −1

∆Vout

∆Vin= R

3R − 1ωC

j+ R2ωCj

where we used

= −j.

ZC∆Vin ∆Vout

bFigure (a)

∆Vout∆Vab

Figure (b)

ZCZC

∆Vin

∆Vab

bFigure (c)

− j

ω C

∆Vout

∆Vin= R

3R − 1ωC

− R2ωC

= R

3R( )2 + 1ωC

− R2ωC

= 1.00 × 103

3.00 × 103( )2+ 1592 − 628( )2

= 0.317

Chapter 33 Solutions 287

33.54 The equation for ∆v t( ) during the first period (using

y = mx + b ) is:

∆v t( ) =

2 ∆Vmax( )tT

− ∆Vmax

∆v( )2[ ]ave

= 1T

∆v t( )[ ]0

T∫

2dt

∆Vmax( )2

T2T

t − 1

T∫

∆v( )2[ ]ave

=∆Vmax( )2

TT2

2t T − 1[ ] 3

3t=0

t=T

=∆Vmax( )2

6+1( )3 − −1( )3[ ] =

∆Vmax( )2

∆Vrms = ∆v( )2[ ]ave

=∆Vmax( )2

∆Vmax

33.55 ω0 = 1

LC= 1

(0.0500 H)(5.00 × 10−6 F)= 2000 s−1

so the operating frequency of the circuit is ω = ω0

2= 1000 s−1

Using Equation 33.35,

P =∆Vrms( )2 Rω2

R2ω2 + L2 ω2 − ω02( )2

P = (400)2(8.00)(1000)2

(8.00)2(1000)2 + (0.0500)2 (1.00 − 4.00) × 106[ ]2 = 56.7 W

Q ≈ 12.5( )

Figure for GoalSolution

Goal Solution A series RLC circuit consists of an 8.00-Ω resistor, a 5.00-µ F capacitor, and a 50.0-mH inductor. A variablefrequency source applies an emf of 400 V (rms) across the combination. Determine the power deliveredto the circuit when the frequency is equal to one half the resonance frequency.

G : Maximum power is delivered at the resonance frequency, and the power delivered at otherfrequencies depends on the quality factor, Q. For the relatively small resistance in this circuit, wecould expect a high Q = ω0L R . So at half the resonant frequency, the power should be a small

fraction of the maximum power, Pav, max = ∆Vrms2 R = 400 V( )2 8 Ω = 20 kW.

O : We must first calculate the resonance frequency in order to find half this frequency. Then the powerdelivered by the source must equal the power taken out by the resistor. This power can be foundfrom Pav = Irms

2 R where Irms = ∆Vrms / Z.

288 Chapter 33 Solutions

A : The resonance frequency is

f0 = 12π LC

= 1

2π 0.0500 H( ) 5.00 × 10−6 F( )= 318 Hz

The operating frequency is f = f0 / 2 = 159 Hz. We can calculate the impedance at this frequency:

XL = 2π f L = 2π(159 Hz) 0.0500 H( ) = 50.0 Ω and

XC = 12π f C

= 1

2π(159 Hz) 5.00 × 10-6 F( ) = 200 Ω

Z = R2 + (XL − XC )2 = 8.002 + (50.0 − 200)2 Ω = 150 Ω

So, Irms = ∆Vrms

Z= 400 V

150 Ω= 2.66 A

The power delivered by the source is the power dissipated by the resistor:

Pav = Irms2R = (2.66 A)2 8.00 Ω( ) = 56.7 W

L : This power is only about 0.3% of the 20 kW peak power delivered at the resonance frequency. Thesignificant reduction in power for frequencies away from resonance is a consequence of the relativelyhigh Q -factor of about 12.5 for this circuit. A high Q is beneficial if, for example, you want to listento your favorite radio station that broadcasts at 101.5 MHz, and you do not want to receive the signalfrom another local station that broadcasts at 101.9 MHz.

33.56 The resistance of the circuit is R = ∆V

I = 12.0 V0.630 A = 19.0 Ω

The impedance of the circuit is Z = ∆Vrms

Irms= 24.0 V

0.570 A= 42.1 Ω

Z2 = R2 + ω2L2

L = 1

ωZ2 − R2 = 1

377(42.1)2 − (19.0)2 = 99.6 mH

33.57 (a) When ω L is very large, the bottom branch carries negligible current. Also, 1/ω C will benegligible compared to 200 Ω and 45 0 V/200 =. Ω 225 mA flows in the power supply and thetop branch.

(b) Now 1/ω C → ∞ and ω L → 0 so the generator and bottom branch carry 450 mA

Chapter 33 Solutions 289

33.58 (a) With both switches closed, the current goes only throughgenerator and resistor.

i(t) = ∆Vmax

Rcosωt

(b) P = 1

2∆Vmax( )2

R2 + ω2L2cos ωt + Arctan(ωL / R)[ ]

(d) For 0 = φ = Arctan

ω0L − 1ω0C

We require ω0L = 1

ω0 C, so

C = 1

ω02 L

(e) At this resonance frequency, Z = R

(f) U = 12 C ∆VC( )2 = 1

2 C I2XC2

Umax = 1

2 CImax2 XC

2 = 12 C

∆Vmax( )2

R21

ω02C2 =

∆Vmax( )2 L

2R2

(g) Umax = 12 LImax

2 = 12 L

∆Vmax( )2

(h) Now ω = 2ω0 = 2

So φ = Arctan

ωL − 1ωC

= Arctan 2

− 12

= Arctan

32R

(i) Now ωL = 1

ωC ω =

12LC

= ω0

290 Chapter 33 Solutions

33.59 (a) As shown in part (b), circuit (a) is a high-pass filter

and circuit (b) is a low-pass filter .

(b) For circuit (a),

∆Vout

∆Vin=

RL2 + XL

RL2 + XL − XC( )2

RL2 + ωL( )2

RL2 + ωL − 1 ωC( )2

As ω → 0,

∆Vout

∆Vin≈ ωRLC ≈ 0

As ω → ∞ ,

∆Vout

∆Vin≈ 1 (high-pass filter)

For circuit (b),

∆Vout

∆Vin= XC

RL2 + XL − XC( )2

1 ωC

RL2 + ωL − 1 ωC( )2

As ω → 0,

∆Vout

∆Vin≈ 1

As ω → ∞ ,

∆Vout

∆Vin≈ 1

ω2LC≈ 0 (low-pass filter)

∆Vout∆Vin

Circuit (b)

∆Vout∆Vin

Circuit (a)

33.60 (a) IR, rms = ∆Vrms

R= 100 V

80.0 Ω= 1.25 A

(b) The total current will lag the applied voltage as seen in thephasor diagram at the right.

IL, rms = ∆Vrms

XL= 100 V

2π 60.0 s−1( ) 0.200 H( )= 1.33 A

∆V

Thus, the phase angle is: φ = tan−1 IL, rms

IR, rms

= tan−1 1.33 A

1.25 A

= 46.7°

*33.61 Suppose each of the 20 000 people uses an average power of 500 W. (This means 12 kWh perday, or $36 per 30 days at 10¢ per kWh). Suppose the transmission line is at 20 kV. Then

Irms = P

∆Vrms=

20 000( ) 500 W( )20 000 V

~103 A

If the transmission line had been at 200 kV, the current would be only ~102 A .

Chapter 33 Solutions 291

33.62 L = 2.00 H, C = 10.0 × 10– 6 F, R = 10.0 Ω, ∆v(t) = (100 sin ω t)

(a) The resonant frequency ω0 produces the maximum current and thus the maximum powerdissipation in the resistor.

ω0 = 1

LC= 1

(2.00)(10.0 × 10−6 )= 224 rad/s

(b) P =

∆Vmax( )2

2R= (100)2

2(10.0)= 500 W

(c)

Irms = ∆Vrms

Z= ∆Vrms

R2 + ωL − 1ωC

2and

Irms( )max = ∆Vrms

Irms

2 R = 12

Irms2( )max

R or

∆Vrms( )2

Z2 R = 12

∆Vrms( )2

R2 R

This occurs where Z 2 = 2R 2: R2 + ωL − 1

ωC

= 2R2

ω4L2C 2 − 2Lω2C − R2ω2C 2 + 1 = 0 or L2C 2ω4 − (2LC + R2C 2 )ω2 + 1 = 0

(2.00)2(10.0 × 10−6 )2[ ]ω4 − 2(2.00)(10.0 × 10−6 ) + (10.0)2(10.0 × 10−6 )2[ ]ω2 + 1 = 0

Solving this quadratic equation, we find that ω2 = 51130, 48 894

ω1 = 48 894 = 221 rad/s and ω2 = 51130 = 226 rad/s

33.63 R = 200 Ω, L = 663 mH, C = 26.5 µF, ω = 377 s−1, ∆Vmax = 50.0 V

ωL = 250 Ω,

1ωC

= 100 Ω, Z = R2 + XL − XC( )2 = 250 Ω

(a) Imax = ∆Vmax

Z= 50.0 V

250 Ω= 0.200 A

φ = tan−1 XL − XC

= 36.8° (∆V leads I)

(b) ∆VR,max = Imax R = 40.0 V at φ = 0°

ωC= 20.0 V at φ = – 90.0° (I leads ∆V)

(d) ∆VL,max = ImaxωL = 50.0 V at φ = + 90.0° (∆V leads I)

292 Chapter 33 Solutions

*33.64 P = Irms

2 R = ∆Vrms

R , so 250 W = 120 V( )2

Z2 40.0 Ω( ): Z = R2 + ωL − 1 ωC( )2

250 = 120( )2 40.0( )

40.0( )2 + 2π f 0.185( ) − 1

2π f 65.0 × 10−6( )

2 and

250 = 576 000 f 2

1600 f 2 + 1.1624 f 2 − 2448.5( )2

1 = 2304 f 2

1600 f 2 + 1.3511 f 4 − 5692.3 f 2 + 5 995 300so 1.3511 f 4 − 6396.3 f 2 + 5995300 = 0

f 2 =

6396.3 ± 6396.3( )2 − 4 1.3511( ) 5 995 300( )2 1.3511( ) = 3446.5 or 1287.4

f = 58.7 Hz or 35.9 Hz

33.65 (a) From Equation 33.39,

N2= ∆V1

∆V2

Let output impedance Z1 = ∆V1

I1and the input impedance

Z2 = ∆V2

so that

N2= Z1I1

Z2I2But from Eq. 33.40,

I2= ∆V2

∆V1= N2

So, combining with the previous result we have

N2= Z1

(b)

N2= Z1

Z2= 8000

8.00= 31.6

33.66 IR = ∆Vrms

IL = ∆Vrms

ωL;

IC = ∆Vrms

(ωC)−1

(a) Irms = IR2 + (IC − IL )2 =

∆Vrms

1R2

+ ωC − 1

ωL

(b) tanφ = IC − IL

IR= ∆Vrms

1XC

− 1XL

1∆Vrms / R

tanφ = R

1XC

− 1XL

Chapter 33 Solutions 293

33.67 (a) Irms = ∆Vrms

1R2 + ωC − 1

ωL

∆Vrms → ∆Vrms( )max when ω

ωC

L= 1

f = 1

2π LC

f = 1

2π 200 × 10−3 H)(0.150 × 10−6 F)= 919 Hz

(b) IR = ∆Vrms

R= 120 V

80.0 Ω= 1.50 A

IL = ∆Vrms

ωL= 120 V

(374 s−1)(0.200 H)= 1.60 A

IC = ∆Vrms(ωC) = (120 V)(374 s−1)(0.150 × 10−6 F) = 6.73 mA

(d) φ = tan−1 IC − IL

= tan−1 0.00673 − 1.60

1.50

= – 46.7°

The current is lagging the voltage .

33.68 (a) tanφ = ∆VL

∆VR=

I ωL( )IR

= ωLR

Thus, R = ωL

tanφ=

200 s−1( ) 0.500 H( )tan 30.0°( ) = 173 Ω

(b)

∆Vout

∆Vin= ∆VR

∆Vin= cosφ

∆Vout = ∆Vin( )cosφ = 10.0 V( ) cos 30.0°= 8.66 V ∆VR = IR

∆VL = IXL

∆V = IZ

∆Vin

R ∆Vout

294 Chapter 33 Solutions

33.69 (a) XL = XC = 1884 Ω when f = 2000 Hz

L = XL

2π f= 1884 Ω

4000π rad s= 0.150 H and

C = 1

2π f( )XC= 1

4000π rad s( ) 1884 Ω( ) = 42.2 nF

XL = 2π f 0.150 H( ) XC = 1

2π f( ) 4.22 × 10−8 F( ) Z = 40.0 Ω( )2 + XL − XC( )2

f (Hz) XL (Ω) XC (Ω) Z (Ω)

300 283 12600 12300600 565 6280 5720800 754 4710 3960

1000 942 3770 28301500 1410 2510 11002000 1880 1880 403000 2830 1260 15704000 3770 942 28306000 5650 628 5020

10000 9420 377 9040

(b) Impedence, Ω

33.70 ω0 = 1

LC= 1.00 × 106 rad s

ω ω0 ωL Ω( )

1ωC

Ω( ) Z Ω( ) P = I2R W( )

For each angular frequency, wefind

Z = R2 + ωL − 1/ωC( )2

then I = 1.00 V( )/ Z

and P = I2 1.00 Ω( )

0.99900.99910.99930.99950.99970.99991.00001.00011.00031.00051.00071.00091.0010

999.0999.1999.3999.5999.7999.9

10001000.11000.31000.51000.71000.91001

1001.01000.91000.71000.51000.31000.11000.0999.9999.7999.5999.3999.1999.0

2.242.061.721.411.171.021.001.021.171.411.722.062.24

0.199840.235690.337680.499870.735240.961531.000000.961540.735350.500120.337990.236010.20016

The full width at half maximum is:

∆f = ∆ω

2π= 1.0005 − 0.9995( )ω0

2π

∆f = 1.00 × 10 3 s−1

2π= 159 Hz

while

R2πL

= 1.00 Ω2π 1.00 × 10−3 H( ) = 159 Hz

Chapter 33 Solutions 295

33.71

∆Vout

∆Vin= R

R2 + 1 ωC( )2= R

R2 + 1 2π f C( )2

(a)

∆Vout

∆Vin= 1

2 when

1ωC

= R 3

Hence, f = ω

2π= 1

2πRC 3= 1.84 kHz

∆Vin R

∆Vout

(b) Log Gain versus Log Frequency

- 4

- 3

- 2

- 1

0 1 2 3 4 5 6

Log f

Log∆V out/∆V in

Chapter 34 Solutions

34.1 Since the light from this star travels at 3.00 × 108 m/s, the last bit of light will hit the Earth i n

6.44 × 1018 m3.00 × 108 m / s

= 2.15 × 1010 s = 680 years. Therefore, it will disappear from the sky in the year

1999 + 680 = 2.68 × 10 3 A.D.

34.2 v = 1

kµ0e0= 1

1.78c = 0.750c = 2.25 × 108 m/s

34.3EB = c or

220B = 3.00 × 108; so B = 7.33 × 10–7 T = 733 nT

34.4EmaxBmax

= v is the generalized version of Equation 34.13.

Bmax = Emax

v= 7.60 × 10−3 V / m

(2 / 3)(3.00 × 108 m / s)N ⋅ mV ⋅ C

T ⋅ C ⋅ mN ⋅ s

= 3.80 × 10–11 T = 38.0 pT

34.5 (a) f λ = c or f (50.0 m) = 3.00 × 108 m/s so f = 6.00 × 106 Hz = 6.00 MHz

(b)EB = c or

22.0Bmax

= 3.00 × 108 so Bmax = (73.3 nT)(–k)

= 2π

50.0 = 0.126 m–1 and ω = 2π f = 2π (6.00 × 106 s–1) = 3.77 × 107

rad/s

B = Bmax cos(kx – ω t) = (73.3 nT) cos (0.126x – 3.77 × 107 t)(–k)

34.6 ω = 2π f = 6.00π × 109 s–1 = 1.88 × 1010 s-1

k = 2πλ

= ωc =

6.00π × 109 s–1

3.00 × 108 m/s = 20.0π = 62.8 m–1 Bmax =

Ec =

300 V/m3.00 × 108 m/s

= 1.00 µT

E = 300

cos 62.8x − 1.88 × 1010t( ) B = (1.00 µT) cos (62.8x – 1.88 × 1010 t)

Chapter 34 Solutions 299

34.7 (a) B = Ec =

100 V/m3.00 × 10 8 m/s

= 3.33 × 10–7 T = 0.333 µT

(b) λ = 2πk =

2π1.00 × 10 7 m–1 = 0.628 µm

λ =

3.00 × 108 m/s6.28 × 10–7 m

= 4.77 × 1014 Hz

34.8 E = Emax cos(kx – ω t)

∂E

∂ x = –Emax sin(kx – ω t)(k)

∂E

∂ t = –Emax sin(kx – ω t)(–ω)

∂ 2E

∂ x 2 = –Emax cos(kx – ω t)(k 2)

∂ 2E

∂ t 2 = –Emax cos(kx – ω t)(–ω)2

We must show:

∂E∂x2 = µ0e0

∂ 2E∂t2

That is, − k2( )Emax cos kx − ωt( ) = −µ0e0 −ω( )2Emax cos kx − ωt( )

But this is true, because

ω2 = 1fλ

= 1c2 = µ0e0

The proof for the wave of magnetic field is precisely similar.

*34.9 In the fundamental mode, there is a single loop in the standing wave between the plates.Therefore, the distance between the plates is equal to half a wavelength.

λ = 2L = 2(2.00 m) = 4.00 m

Thus, f = c

λ =

3.00 × 108 m/s4.00 m = 7.50 × 10 7 Hz = 75.0 MHz

*34.10 dA to A = 6 cm ± 5% = λ

λ = ±12 cm 5%

300 Chapter 34 Solutions

v = λ f = 0.12 m ± 5%( ) 2.45 × 109 s−1( ) = 2.9 × 108 m s ± 5%

34.11 S = I = UAt =

UcV = uc

EnergyUnit Volume = u =

Ic =

1000 W/m2

3.00 × 108 m/s = 3.33 µ J/m3

34.12 Sav =

4πr2 = 4.00 × 10 3 W

4π (4.00 × 1609 m)2 = 7.68 µW/m2

Emax = 2µ0cSav = 0.0761 V / m

∆Vmax = Emax · L = (76.1 mV/m)(0.650 m) = 49.5 mV (amplitude) or 35.0 mV (rms)

34.13 r = 5.00 mi( ) 1609 m / mi( ) = 8.04 × 103 m

S =

4πr 2 = 250 × 10 3 W

4π (8.04 × 10 3 m)2 = 307 µW/m2

Chapter 34 Solutions 301

Goal Solution What is the average magnitude of the Poynting vector 5.00 miles from a radio transmitter broadcastingisotropically with an average power of 250 kW?

G : As the distance from the source is increased, the power per unit area will decrease, so at a distance of5 miles from the source, the power per unit area will be a small fraction of the Poynting vector nearthe source.

O : The Poynting vector is the power per unit area, where A is the surface area of a sphere with a 5-mileradius.

A : The Poynting vector is

In meters, r = 5.00 mi( ) 1609 m / mi( ) = 8045 m

and the magnitude is S = 250 × 103 W

(4π)(8045)2 = 3.07 × 10−4 W / m2

L : The magnitude of the Poynting vector ten meters from the source is 199 W/m2, on the order of amillion times larger than it is 5 miles away! It is surprising to realize how little power is actuallyreceived by a radio (at the 5-mile distance, the signal would only be about 30 nW, assuming areceiving area of about 1 cm2).

34.14 I = 100 W

4π (1.00 m)2 = 7.96 W/m2

u = Ic = 2.65 × 10 – 8 J/m3 = 26.5 n J/m3

(a) uE = 12 u = 13.3 n J/m3

(b) uB = 12 u = 13.3 n J/m3

34.15 Power output = (power input)(efficiency)

Thus, Power input = power out

eff = 1.00 × 106 W

0.300 = 3.33 × 106 W

and A =

I =

3.33 × 10 6 W1.00 × 10 3 W/m2 = 3.33 × 10 3 m2

302 Chapter 34 Solutions

*34.16 I =

Bmax2 c2µ0

= P

4π r 2

Bmax =

4πr2

2µ0

10.0 × 103( ) 2( ) 4π× 10−7( )4π 5.00 × 103( )2

3.00 × 108( ) = 5.16 × 10–10 T

Since the magnetic field of the Earth is approximately 5 × 10 –5 T, the Earth's field is some100,000 times stronger.

34.17 (a) P = I 2R = 150 W; A = 2π rL = 2π (0.900 × 10–3 m)(0.0800 m) = 4.52 × 10 –4 m2

S =

A = 332 kW/m2 (points radially inward)

(b) B = µ0 I

2π r = µ0(1.00)

2π (0.900 × 10 –3) = 222 µT

E = ∆V

∆x =

IRL =

150 V0.0800 m = 1.88 kV/m

Note: S = EBµ0

= 332 kW/m2

Chapter 34 Solutions 303

34.18 (a) E · B = (80.0 i + 32.0 j – 64.0 k)(N/C) · (0.200 i + 0.0800 j + 0.290 k)µT

E · B= (16.0 + 2.56 – 18.56)N 2 · s/C 2 · m = 0

(b) S = 1

µ0 E × B=

(80.0 i + 32.0 j – 64.0 k)(N/C) × (0.200 i + 0.0800 j + 0.290 k)µT4π × 10 –7 T · m/A

S = (6.40 k – 23.2 j – 6.40 k + 9.28 i – 12.8 j + 5.12 i)10 – 6 W/m2

4π × 10–7

S = (11.5 i – 28.6 j) W/m2 = 30.9 W/m2 at – 68.2° from the +x axis

34.19 We call the current Irms and the intensity I . The power radiated at this frequency is

P = (0.0100)(∆Vrms)Irms = 0.0100(∆Vrms)2

R = 1.31 W

If it is isotropic, the intensity one meter away is

I =

A =

1.31 W4π (1.00 m)2 = 0.104 W/m2 = Sav =

c2µ0

B2max

Bmax = 2µ0I

2 4π× 10−7 T ⋅ m / A( ) 0.104 W / m2( )3.00 × 108 m / s

= 29.5 nΤ

*34.20 (a) efficiency = useful power output

total power input× 100% = 700 W

1400 W

× 100% = 50.0%

(b) Sav = P

A= 700 W

0.0683 m( ) 0.0381 m( ) = 2.69 × 105 W m2

Sav = 269 kW m2 toward the oven chamber

2µ0c

Emax = 2 4π× 10−7

T ⋅ mA

3.00 × 108

2.69 × 105

Wm2

= 1.42 × 104

= 14.2 kV m

304 Chapter 34 Solutions

34.21 (a) Bmax = Emax

c = 7.00 × 105 N/C3.00 × 108 m/s

= 2.33 mT

(b) I = E2

max2µ0c =

(7.00 × 10 5)2

2(4π × 10 –7 )(3.00 × 108) = 650 MW/m2

A: P = I A = (6.50 × 108 W/m2)

π4 (1.00 × 10 – 3 m) 2 = 510 W

34.22 Power = SA =

Emax2

2µ0c(4πr2 ) ; solving for r ,

r = Pµ0c

Emax2 2π

= (100 W)µ0c2π(15.0 V / m)2 = 5.16 m

34.23 (a) I =

( . )( .

10 0 100 800 10

××

=−

−W m)2π

4.97 kW/m2

(b) uav = I

c= 4.97 × 103 J / m2 ⋅ s

3.00 × 108 m / s= 16.6 µ J/m3

34.24 (a) E cB= = × × −( .3 00 10 08 m/s)(1.8 10 T) =6 540 V/m

(b) uav = B2

µ0= (1.80 × 10−6 )2

4π× 10−7 = 2.58 µ J/m3

(d) This is 77.3% of the flux in Example 34.5 . It may be cloudy, or the Sun may be setting.

34.25 For complete absorption, P = Sc =

25.03.00 × 108 = 83.3 nPa

*34.26 (a) P = Sav( ) A( ) = 6.00 W / m2( ) 40.0 × 10−4 m2( ) = 2.40 × 10−2 J / s

In one second, the total energy U impinging on the mirror is 2.40 × 10–2 J. The momentum ptransferred each second for total reflection is

p = 2Uc =

2(2.40 × 10–2 J)3.00 × 108 m/s

= 1.60 × 10 –10 kg · m

s (each second)

(b) F = dpdt =

1.60 × 10–10 kg · m/s1 s = 1.60 × 10–10 N

Chapter 34 Solutions 305

34.27 (a) The radiation pressure is

(2)(1340 W / m2)3.00 × 108 m / s2 = 8.93 × 10−6 N / m2

Multiplying by the total area, A = 6.00 × 105 m2 gives: F = 5.36 N

(b) The acceleration is: a = Fm =

5.36 N6000 kg = 8.93 × 10– 4 m/s2

t = 2da

=2 3.84 × 108 m( )

8.93 × 10−4 m / s2( ) = 9.27 × 105 s = 10.7 days

34.28 The pressure P upon the mirror is P = 2Sav

where A is the cross-sectional area of the beam and Sav = P

The force on the mirror is then F = PA = 2

A = 2P

Therefore, F = 2(100 × 10−3 )

(3 × 108)= 6.67 × 10–10 N

34.29 I = P

πr2 = Emax2

2µ0c

(a) Emax =

P 2µ0c( )πr2 = 1.90 kN/C

(b)15 × 10–3 J/s

3.00 × 108 m/s (1.00 m) = 50.0 pJ

5 × 10–11

3.00 × 108 = 1.67 × 10–19 kg · m/s

306 Chapter 34 Solutions

34.30 (a) If PS is the total power radiated by the Sun, and rE and rM are the radii of the orbits of theplanets Earth and Mars, then the intensities of the solar radiation at these planets are:

IE = PS

4πrE2 and

IM = PS

4πrM2

Thus, IM = IE

= 1340 W m2( ) 1.496 × 1011 m2.28 × 1011 m

= 577 W/m2

(b) Mars intercepts the power falling on its circular face:

PM = IM πRM

2( ) = 577 W m2( )π 3.37 × 106 m( )2= 2.06 × 1016 W

c= IM

and force F = PA = IM

cπRM

2( ) = PM

c= 2.06 × 1016 W

3.00 × 108 m s= 6.87 × 107 N

(d) The attractive gravitational force exerted on Mars by the Sun is

Fg = GMSMM

rM2 =

6.67 × 10−11 N ⋅ m2 kg2( ) 1.991× 1030 kg( ) 6.42 × 1023 kg( )2.28 × 1011 m( )2 = 1.64 × 1021 N

which is ~1013 times stronger than the repulsive force of (c).

34.31 (a) The total energy absorbed by the surface is

U = 1

2 I( )At = 12 750

Wm2

0.500 × 1.00 m2( ) 60.0 s( ) = 11.3 kJ

(b) The total energy incident on the surface in this time is 2U = 22.5 kJ, with U = 11.3 kJ beingabsorbed and U = 11.3 kJ being reflected. The total momentum transferred to the surface is

p = momentum from absorption( ) + momentum from reflection( )

p = U

+ 2U

= 3U

3 11.3 × 103 J( )3.00 × 108 m s

= 1.13 × 10−4 kg ⋅ m s

34.32 Sav = µ0Jmax

2 c8

or 570 = (4π× 10−7 )Jmax

2 (3.00 × 108)8

so Jmax = 3.48 A/m2

Chapter 34 Solutions 307

34.33 (a) P = SavA = µ0Jmax

2 c8

= 4π× 10−7(10.0)2(3.00 × 108)

(1.20 × 0.400) = 2.26 kW

(b) Sav = µ0Jmax

2 c8

= (4π× 10−7(10.0)2(3.00 × 108)8

= 4.71 kW/m2

*34.34 P = ∆V( )2

R or P ∝ ∆ V( )2

∆V = −( )Ey ⋅ ∆y = Ey ⋅lcosθ

∆V ∝ cosθ so P ∝ cos2 θ

(a) θ = 15.0°: P = Pmax cos2 15.0°( ) = 0.933Pmax= 93.3%

(b) θ = 45.0° : P = Pmax cos2 45.0°( ) = 0.500Pmax= 50.0%

θ l

eceivingntenna∆y

34.35 (a) Constructive interference occurs when d cos θ = nλ forsome integer n.

cosθ = n

λd

= nλ

λ / 2

= 2n n = 0, ± 1, ± 2, . . .

∴ strong signal @ θ = cos–1 0 = 90°, 270°

(b) Destructive interference occurs when

dcosθ = 2n + 1

λ : cos θ = 2n + 1

∴ weak signal @ θ = cos–1 (±1) = 0°, 180°

308 Chapter 34 Solutions

Goal Solution Two radio-transmitting antennas are separated by half the broadcast wavelength and are driven in phasewith each other. In which directions are (a) the strongest and (b) the weakest signals radiated?

G : The strength of the radiated signal will be a function of the location around the two antennas andwill depend on the interference of the waves.

O : A diagram helps to visualize this situation. The two antennas are driven in phase, which meansthat they both create maximum electric field strength at the same time, as shown in the diagram. Theradio EM waves travel radially outwards from the antennas, and the received signal will be the vectorsum of the two waves.

A : (a) Along the perpendicular bisector of the line joining the antennas, the distance is the same toboth transmitting antennas. The transmitters oscillate in phase, so along this line the two signals willbe received in phase, constructively interfering to produce a maximum signal strength that is twicethe amplitude of one transmitter.

(b) Along the extended line joining the sources, the wave from the more distant antenna musttravel one-half wavelength farther, so the waves are received 180° out of phase. They interferedestructively to produce the weakest signal with zero amplitude.

L : Radio stations may use an antenna array to direct the radiated signal toward a highly-populatedregion and reduce the signal strength delivered to a sparsely-populated area.

34.36 λ = c

f= 536 m so

h = λ

4= 134 m

λ = c

f= 188 m so

h = λ

4= 46.9 m

34.37 For the proton: ΣF = ma ⇒ qvB sin 90.0° = mv2 R

The period and frequency of the proton’s circular motion are therefore:

T = 2πRv

= 2πmqB

=2π 1.67 × 10−27 kg( )

1.60 × 10 −19 C( ) 0.350 T( )= 1.87 × 10−7 s f = 5.34 × 106 Hz.

The charge will radiate at this same frequency, with λ = c

f= 3.00 × 108 m s

5.34 × 106 Hz= 56.2 m

34.38 For the proton, ΣF = ma yields qvB sin 90.0° = mv2

The period of the proton’s circular motion is therefore: T = 2πR

v= 2πm

The frequency of the proton's motion is f = 1/T

The charge will radiate electromagnetic waves at this frequency, with λ = = =c

fcT

2πmcqB

Chapter 34 Solutions 309

*34.39 From the electromagnetic spectrum chart and accompanying text discussion, the followingidentifications are made:

Frequency

fWavelength,

λ = c fClassification

2 Hz = 2 × 100 Hz 150 Mm Radio

2 kHz = 2 × 103 Hz 150 km Radio

2 MHz = 2 × 106 Hz 150 m Radio

2 GHz = 2 × 109 Hz 15 cm Microwave

2 THz = 2 × 1012 Hz 150 µm Infrared

2 PHz = 2 × 1015 Hz 150 nm Ultraviolet

2 EHz = 2 × 1018 Hz 150 pm x-ray

2 ZHz = 2 × 1021 Hz 150 fm Gamma ray

2 YHz = 2 × 1024 Hz 150 am Gamma Ray

Wavelength,λ

Frequency

f c= λClassification

2 km = 2 × 103 m 1.5 × 105 Hz Radio

2 m = 2 × 100 m 1.5 × 108 Hz Radio

2 mm = 2 × 10−3 m 1.5 × 1011 Hz Microwave

2 µm = 2 × 10−6 m 1.5 × 1014 Hz Infrared

2 nm = 2 × 10−9 m 1.5 × 1017 Hz Ultraviolet/x-ray

2 pm = 2 × 10−12 m 1.5 × 1020 Hz x-ray/Gamma ray

2 fm = 2 × 10−15 m 1.5 × 1023 Hz Gamma ray

2 am = 2 × 10−18 m 1.5 × 1026 Hz Gamma ray

*34.40 (a) f = c

λ =

3 × 108 m/s1.7 m ~ 108 Hz radio wave

(b) 1000 pages, 500 sheets, is about 3 cm thick so one sheet is about 6 × 10 – 5 m thick

f = 3 × 108 m/s6 × 10–5 m

~ 1013 Hz infrared

*34.41 f = c

λ =

3.00 × 108 m/s5.50 × 10–7 m

= 5.45 × 1014 Hz

34.42 (a) λ = cf =

3.00 × 108 m/s1150 × 103/s

= 261 m so180 m261 m = 0.690 wavelengths

(b) λ = cf =

3.00 × 108 m/s98.1 × 106/s

= 3.06 m so180 m3.06 m = 58.9 wavelengths

310 Chapter 34 Solutions

34.43 (a) f λ = c gives (5.00 × 1019 Hz)λ = 3.00 × 108 m/s: λ = 6.00 × 10 –12 m = 6.00 pm

(b) f λ = c gives (4.00 × 109 Hz)λ = 3.00 × 108 m/s: λ = 0.075 m = 7.50 cm

*34.44 Time to reach object = 12 (total time of flight) =

12 (4.00 × 10 – 4 s) = 2.00 × 10 – 4 s

Thus, d = vt = (3.00 × 10 8 m/s)(2.00 × 10 – 4 s) = 6.00 × 10 4 m = 60.0 km

34.45 The time for the radio signal to travel 100 km is: tr =

100 × 103 m3.00 × 108 m / s

= 3.33 × 10 – 4 s

The sound wave to travel 3.00 m across the room in: ts = 3.00 m

343 m/s = 8.75 × 10– 3 s

Therefore, listeners 100 km away will receive the news before the people in the newsroomby a total time difference of

∆t = 8.75 × 10– 3 s – 3.33 × 10– 4 s = 8.41 × 10– 3 s

*34.46 The wavelength of an ELF wave of frequency 75.0 Hz is λ = c

f= 3.00 × 108 m s

75.0 Hz= 4.00 × 106 m

The length of a quarter-wavelength antenna would be L = 1.00 × 106 m = 1.00 × 103 km

or L = 1000 km( ) 0.621 mi

1.00 km

= 621 mi

Thus, while the project may be theoretically possible, it is not very practical.

34.47 (a) For the AM band, λ max = c

fmin= 3.00 × 108 m / s

540 × 103 Hz= 556 m

λ min = c

fmax= 3.00 × 108 m / s

1600 × 103 Hz= 187 m

(b) For the FM band, λ max = c

fmin= 3.00 × 108 m / s

88.0 × 106 Hz= 3.41 m

λ min = c

fmax= 3.00 × 108 m / s

108 × 106 Hz= 2.78 m

Chapter 34 Solutions 311

34.48 CH 4 : fmin = 66 MHz λ max = 4.55 m

fmax = 72 MHz λ min = 4.17 m

CH 6 : fmin = 82 MHz λ max = 3.66 m

fmax = 88 MHz λ min = 3.41 m

CH 8 : fmin = 180 MHz λ max = 1.67 m

fmax = 186 MHz λ min = 1.61 m

34.49 (a) P = SA = (1340 W/m2)4π (1.496 × 1011 m)2 = 3.77 × 1026 W

(b) S =

cBmax2

2µ0so Bmax =

2µ0Sc

2(4π× 10−7 N / A2)(1340 W / m2 )3.00 × 108 m / s

= 3.35 µT

S =

Emax2

2µ0cso

Emax = 2µ0cS = 2 4π× 10−7( ) 3.00 × 108( ) 1340( ) = 1.01 kV/m

*34.50 Suppose you cover a 1.7 m-by-0.3 m section of beach blanket. Suppose the elevation angle ofthe Sun is 60°. Then the target area you fill in the Sun's field of view is

(1.7 m )( 0.3 m )( cos 30° ) = 0.4 m2

Now I = P

A= E

At: E = IAt = 1340

Wm 2 (0.6)(0.5)(0.4 m2) 3600 s ~ 106 J

34.51 (a) ε = − dΦB

dt= − d

dt(BA cos θ) = −A

ddt

(Bmax cos ωt cos θ) = ABmax ω(sin ωt cos θ)

ε(t) = 2πf Bmax A sin 2πf t cos θ = 2π2r 2 f Bmax cos θ sin 2πf t

Thus, εmax = 2π2r 2 f Bmax cos θ , where θ is the angle between the magnetic field and the

normal to the loop.

(b) If E is vertical, then B is horizontal, so the plane of the loop should be vertical and the

plane should contain the line of sight to the transmitter .

312 Chapter 34 Solutions

34.52 (a) Fgrav = GMsm

R2 = GMs

ρ 4 / 3( )πr3

where Ms = mass of Sun, r = radius of particle and R = distance from Sun to particle.

Since Frad = Sπ r 2

c ,

Frad

Fgrav= 1

3SR2

4cGMsρ

∝ 1

(b) From the result found in part (a), when Fgrav = Frad, we have r = 3SR2

4cGMs ρ

r =3 214 W / m2( ) 3.75 × 1011 m( )2

4 6.67 × 10−11 N ⋅ m2 kg2( ) 1.991× 1030 kg( ) 1500 kg m3( ) 3.00 × 108 m s( ) = 3.78 × 10–7 m

34.53 (a) Bmax = Emax

c = 6.67 × 10–16 T

(b) Sav =

max2

02µ = 5.31 × 10–17 W/m2

(d) F = PA = Sav

A = 5.56 × 10–23 N (≈ weight of 3000 H atoms!)

*34.54 (a) The electric field between the plates is E = ∆V l, directeddownward in the figure. The magnetic field between the plate'sedges is B = µ0 i 2πr counterclockwise.

The Poynting vector is: S = 1

µ0

E × B =

∆V( )i2πrl

(radially outward)

(b) The lateral surface area surrounding the electric field volume is

A = 2πrl, so the power output is P = SA = ∆V( )i

2πrl

2πrl( ) = (∆V)i

- -- - -- -

-- -

--- - -

+ ++ + +

+ ++

+++ + ++ES

(c) As the capacitor charges, the polarity of the plates and hence the direction of the electric fieldis unchanged. Reversing the current reverses the direction of the magnetic field, andtherefore the Poynting vector.

The Poynting vector is now directed radially inward.

Chapter 34 Solutions 313

*34.55 (a) The magnetic field in the enclosed volume is directed upward,

with magnitude B = µ0ni and increasing at the rate

dBdt

= µ0ndidt

The changing magnetic field induces an electric field around anycircle of radius r , according to Faraday’s Law:

E 2πr( ) = −µ0n

didt

πr2( )

E = −µ0nr

2didt

or E = µ0nr

2didt

(clockwise)

Then, S =

1µ0

E × B = 1µ0

µ0nr2

didt

µ0ni( ) inward,

or the Poynting vector is S =

µ0n2r i2

didt

(radially inward)

(b) The power flowing into the volume is P = SAlat where Alat is the lateral area perpendicular to S . Therefore,

P = µ0n2r i

2didt

2πrl( ) =

µ0πn2r2li

didt

(c) Taking Across to be the cross-sectional area perpendicular to B, the induced voltage betweenthe ends of the inductor, which has N = nl turns, is

∆V = ε = N

dBdt

Across

= nl µ0n

didt

πr2( ) = µ0πn2r2l

didt

and it is observed that P = ∆V( )i

*34.56 (a) The power incident on the mirror is: P I= IA = 1340

Wm2

π 100 m( )2[ ] = 4.21× 107 W

The power reflected through the atmosphere is PR = 0.746 4.21× 107 W( ) = 3.14 × 107 W

(b)

S = PR

A= 3.14 × 107 W

π 4.00 × 103 m( )2 = 0.625 W/m2

PN = 0.746 1340 W / m2( )sin 7.00°= 122 W / m2

The radiation intensity received from the mirror is

0.625 W m2

122 W m2

100% = 0.513% of that from the noon Sun in January.

314 Chapter 34 Solutions

34.57 u = 12 e0Emax

2 (Equation 34.21)

Emax = 2u

e0 = 95.1 mV/m

*34.58 The area over which we model the antenna as radiating is the lateral surface of a cylinder,

A = 2πrl= 2π 4.00 × 10−2 m( ) 0.100 m( ) = 2.51× 10−2 m2

(a) The intensity is then: S = P

A= 0.600 W

2.51× 10−2 m2 = 23.9 W m2

(b) The standard is: 0.570

mWcm2 = 0.570

mWcm2

1.00 × 10−3 W1.00 mW

1.00 × 104 cm2

1.00 m2

= 5.70

Wm2

While it is on, the telephone is over the standard by

23.9 W m2

5.70 W m2 = 4.19 times

34.59 (a) Bmax = Emax

c = 175 V/m

3.00 × 108 m/s = 5.83 × 10–7 T

k = 2πλ

= 2π

(0.0150 m) = 419 rad/m

ω = kc = 1.26 × 1011 rad/s

Since S is along x, and E is along y, B must be in the z direction . (That is S ∝ E × B.)

(b) Sav = EmaxBmax

2µ0 = 40.6 W/m2

(d) a = ΣF

m= PA

2.71× 10−7 N / m2( ) 0.750 m2( )0.500 kg

= 4.06 × 10−7 m / s2

Chapter 34 Solutions 315

*34.60 (a) At steady-state, Pin = Pout and the power radiated out is Pout = eσAT4.

Thus, 0.900 1000

Wm2

A = 0.700( ) 5.67 × 10−8

Wm2 ⋅ K4

AT4

T = 900 W m2

0.700 5.67 × 10−8 W m2 ⋅ K4( )

1 4

= 388 K = 115°C

(b) The box of horizontal area A , presents projected area A sin .50 0° perpendicular to thesunlight. Then by the same reasoning,

0.900 1000

Wm2

A sin 50.0° = 0.700( ) 5.67 × 10−8

Wm2 ⋅ K4

AT4

T =900 W m2( )sin 50.0°

0.700 5.67 × 10−8 W m2 ⋅ K4( )

1 4

= 363 K = 90.0 °C

34.61 (a) P = FA =

F = IA

c= P

c= 100 J / s

3.00 × 108 m / s= 3.33 × 10−7 N = 110 kg( )a

a = 3.03 × 10–9 m/s2 and x = 12 at 2

t = 2x

a= 8.12 × 104 s = 22.6 h

(b) 0 = (107 kg)v – (3.00 kg)(12.0 m/s – v) = (107 kg)v – 36.0 kg · m/s + (3.00 kg)v

v = 36.0110 = 0.327 m/s

t = 30.6 s

316 Chapter 34 Solutions

Goal Solution An astronaut, stranded in space 10.0 m from his spacecraft and at rest relative to it, has a mass (includingequipment) of 110 kg. Since he has a 100-W light source that forms a directed beam, he decides to use thebeam as a photon rocket to propel himself continuously toward the spacecraft. (a) Calculate how long ittakes him to reach the spacecraft by this method. (b) Suppose, instead, he decides to throw the lightsource away in a direction opposite the spacecraft. If the mass of the light source has a mass of 3.00 kg and,after being thrown, moves at 12.0 m/s relative to the recoiling astronaut , how long does it take for theastronaut to reach the spacecraft?

G : Based on our everyday experience, the force exerted by photons is too small to feel, so it may take avery long time (maybe days!) for the astronaut to travel 10 m with his “photon rocket.” Using themomentum of the thrown light seems like a better solution, but it will still take a while (maybe a fewminutes) for the astronaut to reach the spacecraft because his mass is so much larger than the mass ofthe light source.

O : In part (a), the radiation pressure can be used to find the force that accelerates the astronaut towardthe spacecraft. In part (b), the principle of conservation of momentum can be applied to find the timerequired to travel the 10 m.

A : (a) Light exerts on the astronaut a pressure P = F A = S c , and a force of

F = SA

= 100 J / s

3.00 × 108 m / s= 3.33 × 10−7 N

By Newton’s 2nd law, a = F

m= 3.33 × 10−7 N

110 kg= 3.03 × 10−9 m / s2

This acceleration is constant, so the distance traveled is x = 12 at2 , and the amount of time it travels is

t = 2x

a= 2 10.0 m( )

3.03 × 10−9 m / s2 = 8.12 × 104 s = 22.6 h

(b) Because there are no external forces, the momentum of the astronaut before throwing the lightis the same as afterwards when the now 107-kg astronaut is moving at speed v towards the spacecraftand the light is moving away from the spacecraft at 12.0 m / s − v( ) . Thus, pi = p f gives

0 = 107 kg( )v − 3.00 kg( ) 12.0 m / s − v( )

0 = 107 kg( )v − 36.0 kg ⋅ m / s( ) + 3.00 kg( )v

v = 36.0

110= 0.327 m / s

t = x

v= 10.0 m

0.327 m / s= 30.6 s

L : Throwing the light away is certainly a more expedient way to reach the spacecraft, but there is notmuch chance of retrieving the lamp unless it has a very long cord. How long would the cord need tobe, and does its length depend on how hard the astronaut throws the lamp? (You should verify thatthe minimum cord length is 367 m, independent of the speed that the lamp is thrown.)

Chapter 34 Solutions 317

34.62 The 38.0% of the intensity S = 1340

Wm2 that is reflected exerts a pressure

P1 = 2Sr

c= 2 0.380( )S

The absorbed light exerts pressure P2 = Sa

c= 0.620( )S

Altogether the pressure at the subsolar point on Earth is

(a) Ptot = P1 + P2 = 1.38S

c = 1.38(1340 W/m2)

3.00 × 108 m/s = 6.16 × 10– 6 Pa

(b)

Ptot= 1.01× 105 N / m2

6.16 × 10−6 N / m2 = 1.64 × 1010 times smaller than atmospheric pressure

34.63 Think of light going up and being absorbed by the bead which presents a face area π r 2b .

The light pressure is P = Sc =

Ic .

(a) Fl = Iπ r 2b

c = mg = ρ 43 π r 3b g and

I = 4ρ gc

33m4πρ

1/3

= 8.32 × 10 7 W/m2

(b) P = IA = (8.32 × 10 7 W/m2)π (2.00 × 10–3 m)2 = 1.05 kW

34.64 Think of light going up and being absorbed by the bead which presents face area π r 2b .

If we take the bead to be perfectly absorbing, the light pressure is P = Sav

c= I

c= FlA

(a) Fl = Fg so I = Flc

Fgc

A= m gc

πrb2

From the definition of density,

ρ = mV

= m43

πrb3

1rb

= 43

π ρ m( )1/3

Substituting for rb , I = m gc

π4πρ3m

2/3

= gc4ρ3

2/3 mπ

1/3

4ρgc3

3m4π ρ

1/3

(b) P = IA = π r 24ρgc

4π ρ

1/3

318 Chapter 34 Solutions

34.65 The mirror intercepts power

P = I1A1 = (1.00 × 10 3 W/m2)π (0.500 m)2 = 785 W

In the image,

(a) I2 = P

A2= 785 W

π 0.0200 m( )2 = 625 kW/m2

(b) I2 = E 2max2µ0c so Emax = (2µ0c I2)1/2 = [2(4π × 10–7)(3.00 × 108)(6.25 × 105)]1/2= 21.7 kN/C

Bmax = Emax

c = 72.4 µT

0.400(785 W)t = (1.00 kg)

4186

Jkg · C° (100°C – 20.0°C)

t = 3.35 × 105 J

314 W = 1.07 × 103 s = 17.8 min

34.66 (a) λ = c

f= 3.00 × 108 m s

20.0 × 109 s−1 = 1.50 cm

(b) U = P ∆t( ) = 25.0 × 103

1.00 × 10−9 s( ) = 25.0 × 10−6 J = 25.0 µJ

(c)

uav = UV

= U

πr2( )l = U

πr2( )c ∆t( )= 25.0 × 10−6 J

π 0.0600 m( )2 3.00 × 108 m s( ) 1.00 × 10−9 s( )

uav = 7.37 × 10−3 J m3 = 7.37 mJ m3

(d) Emax = 2uav

e0=

2 7.37 × 10−3 J m3( )8.85 × 10−12 C2 N ⋅ m2 = 4.08 × 104 V m = 40.8 kV/m

Bmax = Emax

c= 4.08 × 104 V m

3.00 × 108 m s= 1.36 × 10−4 T = 136 µT

(e) F = PA = S

A = cuav

A = uavA

= 7.37 × 10−3

Jm3

π 0.0600 m( )2 = 8.33 × 10−5 N = 83.3 µN

Chapter 34 Solutions 319

34.67 (a) On the right side of the equation,

C2 m s2( )2

C2 N ⋅ m2( ) m s( )3 = N ⋅ m2 ⋅ C2 ⋅ m2 ⋅ s3

C2 ⋅ s4 ⋅ m3 = N ⋅ ms

= Js

= W

(b) F = ma = qE or a = qE

1.60 × 10−19 C( ) 100 N C( )9.11× 10−31 kg

= 1.76 × 1013 m s2

The radiated power is then:

P = q2a2

6πe0c3 =1.60 × 10−19( )2

1.76 × 1013( )2

6π 8.85 × 10−12( ) 3.00 × 108( )3 = 1 75 10 27. × − W

= qvB so

v = qBr

The proton accelerates at

a = v2

r= q2B2r

m2 =1.60 × 10−19( )2

0.350( )2 0.500( )1.67 × 10−27( )2 = 5.62 × 1014 m s2

The proton then radiates P

= q2a2

6πe0c3 =1.60 × 10−19( )2

5.62 × 1014( )2

6π 8.85 × 10−12( ) 3.00 × 108( )3 = 1.80 × 10−24 W

34.68 P = S

c= Power

Ac= P

2πrlc = 60.0 W

2π(0.0500 m)(1.00 m)(3.00 × 108 m \ s)= 6.37 × 10–7 Pa

34.69 F = PA = SA

P / A( )Ac

= P

τ = F

= Pl

2c, and τ = κ θ

Therefore,

θ = Pl

2cκ=

3.00 × 10−3( ) 0.0600( )2 3.00 × 108( ) 1.00 × 10−11( ) = 3.00 × 10– 2 deg

*34.70 We take R to be the planet’s distance from its star. The planet, of radius r , presents a

projected area πr2 perpendicular to the starlight. It radiates over area 4πr2.

At steady-state, P in = Pout : e Iin πr2( ) = eσ 4πr2( )T4

6.00 × 1023 W4πR2

πr2( ) = eσ 4πr2( )T4 so that 6.00 × 1023 W = 16π σR2 T4

R = 6.00 × 1023 W16π σT4 = 6.00 × 1023 W

16π 5.67 × 10−8 W m2 ⋅ K4( ) 310 K( )4 = 4.77 × 109 m = 4.77 Gm

320 Chapter 34 Solutions

34.71 The light intensity is I = Sav =

2µ0c

The light pressure is P = S

c= E2

2µ0c2 = 12 e0 E2

For the asteroid, PA = ma and a =

e0E2A2m

34.72 f = 90.0 MHz, Emax = 2.00 × 10−3 V / m = 200 mV/m

(a) λ = c

f= 3.33 m

T = 1

f= 1.11× 10−8 s = 11.1 ns

Bmax = Emax

c= 6.67 × 10−12 T = 6.67 pT

(b) E = (2.00 mV / m) cos 2π x

3.33 m− t

11.1 ns

j B = 6.67 pT( )k cos2π x

3.33 m− t

11.1 ns

2µ0c= (2.00 × 10−3 )2

2(4π× 10−7 )(3.00 × 108)= 5.31× 10−9 W / m2

(d) I =cuav so uav = 1.77 × 10−17 J / m3

(e) P = 2I

c= (2)(5.31× 10−9)

3.00 × 108 = 3.54 × 10−17 Pa

Chapter 35 Solutons

35.1 The Moon's radius is 1.74 × 10 6 m and the Earth's radius is 6.37 × 10 6 m. The total distancetraveled by the light is:

d = 2(3.84 × 10 8 m – 1.74 × 10 6 m – 6.37 × 10 6 m) = 7.52 × 10 8 m

This takes 2.51 s, so v = 7.52 × 10 8 m

2.51 s = 2.995 × 10 8 m/s = 299.5 Mm/s

35.2 ∆x = ct ; c = ∆xt =

2(1.50 × 10 8 km)(1000 m/km)(22.0 min)(60.0 s/min) = 2.27 × 10 8 m/s = 227 Mm/s

35.3 The experiment is most convincing if the wheel turns fast enough to pass outgoing lightthrough one notch and returning light through the next: t = 2l c

θ = ω t = ω 2l

ω = cθ

2l= (2.998 × 108)[2π/(720)]

2(11.45 × 10 3)= 114 rad/s

The returning light would be blocked by a tooth at one-half the angular speed, giving anotherdata point.

35.4 (a) For the light beam to make it through both slots, the time for the light to travel the distance dmust equal the time for the disk to rotate through the angle θ, if c is the speed of light,

= θω

, so c = dω

(b) We are given that

d = 2.50 m, θ π= °

= × −1 00

60 0 1802 91 10 4.

rad rad,

ω = 5555

revs

2π rad1.00 rev

= 3.49 × 104 rad s

c = dω

θ=

2.50 m( ) 3.49 × 10 4 rad s( )2.91× 10−4 rad

= 3.00 × 108 m s = 300 Mm/s

35.5 Using Snell's law, sin sinθ θ2

21= n

θ 2 = 25.5° λ

λ2

1= =

n 442 nm

Chapter 35 Solutions 323

35.6 (a) f

c= = ××

=−λ3 00 108

7. m/s

6.328 10 m 4.74 × 10 14 Hz

(b) λ

λglass

air nm1.50

= = =n

632 8. 422 nm

cnglassair m/s

1.50 m/s == = × = ×3 00 10

2 00 108

8.. 200 Mm/s

35.7 n n1 1 2 2sin sinθ θ=

sin θ1 = 1.333 sin 45.0°

sin θ1 = (1.33)(0.707) = 0.943

θ1 = 70.5° → 19.5° above the horizon Figure for Goal Solution

Goal Solution An underwater scuba diver sees the Sun at an apparent angle of 45.0° from the vertical. What is theactual direction of the Sun?

G : The sunlight refracts as it enters the water from the air. Because the water has a higher index ofrefraction, the light slows down and bends toward the vertical line that is normal to the interface.Therefore, the elevation angle of the Sun above the water will be less than 45° as shown in thediagram to the right, even though it appears to the diver that the sun is 45° above the horizon.

O : We can use Snell’s law of refraction to find the precise angle of incidence.

A : Snell’s law is: n n1 1 2 2sin sinθ θ=

which gives sin θ1 = 1.333 sin 45.0°

sinθ1 = (1.333)(0.707) = 0.943

The sunlight is at θ1 = 70.5° to the vertical, so the Sun is 19.5° above the horizon.

L : The calculated result agrees with our prediction. When applying Snell’s law, it is easy to mix up theindex values and to confuse angles-with-the-normal and angles-with-the-surface. Making a sketchand a prediction as we did here helps avoid careless mistakes.

324 Chapter 35 Solutions

*35.8 (a) n n1 1 2 2sin sinθ θ=

1.00 sin 30.0° = n sin 19.24°

n = 1.52

c= = ××

=−λ3 00 108

7. m/s

6.328 10 m 4.74 × 10 14 Hz in air and in syrup.

(d) v

= = × = ×3 00 101 98 10

88.

. m/s

1.52 m/s = 198 Mm/s

(b) λ = = ×

×=v

f1 98 104 74 10

14.. /

m/ss

417 nm

35.9 (a) Flint Glass: v

= = × = × =3 00 101 81 10

88.

. m s

1.66 m s 181 Mm/s

(b) Water: v

= = × = × =3 00 102 25 10

88.

. m s

1.333 m s 225 Mm/s

= = × = × =3 00 101 36 10

88.

. m s

2.20 m s 136 Mm/s

35.10 n n1 1 2 2sin sinθ θ= ; 1.333 sin 37.0° = n2 sin 25.0°

n2 = 1.90 = cv ; v =

c1.90 = 1.58 × 10 8 m/s = 158 Mm/s

35.11 n n1 1 2 2sin sinθ θ= ; θ 2 = sin–1

n1 sin θ 1

θ2

1 1 00 301 50

= °

=−sin( . )(sin )

19.5°

θ 2 and θ 3 are alternate interior angles formed by the ray cuttingparallel normals. So, θ 3 = θ 2 = 19.5° .

1.50 sin θ 3 = (1.00) sin θ4

θ4 = 30.0°

Chapter 35 Solutions 325

35.12 (a) Water λ λ= = =0 436

n nm

1.333 327 nm

(b) Glass λ λ= = =0 436

n nm

1.52 287 nm

*35.13 sin sinθ θ1 2= nw

sin

.sin

.sin( . . ) .θ θ2 1

11 333

90 0 28 0 0 662= = ° − ° =

θ21 0 662 41 5= = °−sin . .

d= =°

=tan

.θ2

3 00 mtan 41.5

3.39 m

35.14 (a) From geometry, 1.25 m = d sin 40.0°

so d = 1.94 m

(b) 50.0° above horizontal , or parallel to theincident ray

*35.15 The incident light reaches the left-hand mirror atdistance

(1.00 m) tan 5.00° = 0.0875 m

above its bottom edge. The reflected light first reachesthe right-hand mirror at height

2(0.0875 m) = 0.175 m

It bounces between the mirrors with this distancebetween points of contact with either.

Since 1.00 m

0.175 m = 5.72, the light reflects

five times from the right-hand mirror and six times from the left.

326 Chapter 35 Solutions

*35.16 At entry, n n1 1 2 2sin sinθ θ= or 1.00 sin 30.0° = 1.50 sin θ 2

θ 2 = 19.5°

The distance h the light travels in the medium is given by

cos θ 2 = (2.00 cm)

h or h =

°=( .

.2 00

2 12 cm)

cos 19.5 cm

The angle of deviation upon entry is α θ θ= − = ° − ° = °1 2 30 0 19 5 10 5. . .

The offset distance comes from sin α = d

h: d = (2.21 cm) sin 10.5° = 0.388 cm

*35.17 The distance, h, traveled by the light is h =

°=2 00

2 12.

. cm

cos 19.5 cm

The speed of light in the material is v

= = × = ×3 00 102 00 10

88.

. m/s

1.50 m/s

Therefore, t

= = ××

= ×−

−2 12 102 00 10

1 06 102

810.

m m/s

s = 106 ps

*35.18 Applying Snell's law at the air-oil interface,

n nair oilsin sin .θ = °20 0 yields θ = 30.4°

Applying Snell's law at the oil-water interface

n nw sin sin .′ = °θ oil 20 0 yields ′ = °θ 22 3.

*35.19 time difference = (time for light to travel 6.20 m in ice) – (time to travel 6.20 m in air)

∆t = 6.20 m

vice− 6.20 m

c but

v = c

∆t = (6.20 m)

1.309c

− 1c

= (6.20 m)

c(0.309) = 6.39 × 10−9 s = 6.39 ns

Chapter 35 Solutions 327

*35.20 Consider glass with an index of refraction of 1.5, which is 3 mm thick The speed of light i nthe glass is

3 101 5

2 108

8× = × m/s m/s

The extra travel time is

3 × 10−3 m2 × 108 m / s

− 3 × 10−3 m3 × 108 m / s

~ 10–11 s

For light of wavelength 600 nm in vacuum and wavelength

600400

nm1.5

nm= in glass,

the extra optical path, in wavelengths, is

3 104 10

3 106 10

7××

− ××

−

− m m

m m

~ 103 wavelengths

*35.21 Refraction proceeds according to 1 00 1 661 2. sin . sin( ) = ( )θ θ (1)

(a) For the normal component of velocity to be constant, v1 cos θ1 = v2 cos θ2

or c c( ) = ( )cos . cosθ θ1 21 66 (2)

We multiply Equations (1) and (2), obtaining: sin cos sin cosθ θ θ θ1 1 2 2=

or sin sin2 21 2θ θ=

The solution θ θ1 2 0= = does not satisfy Equation (2) and must be rejected. The physicalsolution is 2 180 21 2θ θ= ° − or θ θ2 190 0= ° −. . Then Equation (1) becomes:

sin . cosθ θ1 11 66= , or tan .θ1 1 66=

which yields θ1 = 58.9°

(b) Light entering the glass slows down and makes a smaller angle with the normal. Both effectsreduce the velocity component parallel to the surface of the glass, so that component cannotremain constant, or will remain constant only in the trivial case θ1 = θ2 = 0

35.22 See the sketch showing the path of the lightray. α and γ are angles of incidence atmirrors 1 and 2.

For triangle abca, 2α + 2γ + β = 180°

or β α γ= ° − +( )180 2 (1)

Now for triangle bcdb,

90.0° − α( ) + 90.0° − γ( ) + θ = 180°

or θ = α + γ (2)

328 Chapter 35 Solutions

Substituting Equation (2) into Equation (1) gives β = 180° − 2θ

Note: From Equation (2), γ = θ − α . Thus, the ray will follow a path like that shown only ifα < θ . For α > θ , γ is negative and multiple reflections from each mirror will occur beforethe incident and reflected rays intersect.

35.23 Let n(x) be the index of refraction at distance x below the top of the atmosphere and

n x = h( ) = n be its value at the planet surface. Then,

n x( ) = 1.000 + n − 1.000

(a) The total time required to traverse the atmosphere is

t = dx

h∫ = n x( )

cdx

h∫ = 1

c1.000 + n − 1.000

h∫ dx

= h

c+ n − 1.000( )

chh2

= h

cn + 1.000

t = ×

×+

=20 0 10 1 005 1 000

3. . . m3.00 10 m s8 66.8 µs

(b) The travel time in the absence of an atmosphere would be h / c . Thus, the time in thepresence of an atmosphere is

n + 1.0002

= 1.0025 times larger or 0.250% longer .

35.24 Let n(x) be the index of refraction at distance x below the top of the atmosphere and

n x = h( ) = n be its value at the planet surface. Then,

n x( ) = 1.000 + n − 1.000

(a) The total time required to traverse the atmosphere is

t = dx

h∫ = n x( )

cdx

h∫ = 1

c1.000 + n − 1.000

h∫ dx

= h

c+ n − 1.000( )

chh2

n + 1.0002

(b) The travel time in the absence of an atmosphere would be h / c . Thus, the time in thepresence of an atmosphere is

n +

1 0002.

times larger

Chapter 35 Solutions 329

35.25 From Fig. 35.20 nv = 1.470 at 400 nm and nr = 1.458 at 700 nm

Then ( . )sin . sin1 00 1 470θ θ= v and ( . )sin . sin1 00 1 458θ θ= r

δr − δv = θr − θv = sin−1 sin θ

1.458

− sin−1 sin θ

1.470

∆δ = sin−1 sin 30.0°

1.458

− sin−1 sin 30.0°

1.470

= 0.171°

35.26 n1 sin θ1 = n2 sin θ2 so θ2 = sin−1 n1 sin θ1

θ2 = sin−1 (1.00)(sin 30.0°)

1.50

= 19.5°

θ3 = (90.0° − 19.5°) + 60.0°[ ] − 180°( ) + 90.0° = 40.5°

n3 sin θ3 = n4 sin θ4 so θ4 = sin−1 n3 sin θ3

= sin−1 (1.50)(sin 40.5°)1.00

= 77.1°

35.27 Taking Φ to be the apex angle and δmin to be the angle of minimum deviation, fromEquation 35.9, the index of refraction of the prism material is

n =

sinΦ +δmin

sin Φ 2( )

Solving for δmin, δmin = 2 sin−1 n sin

Φ2

− Φ = 2 sin−1 2.20( ) sin 25.0°( )[ ] − 50.0° = 86.8°

35.28 n(700 nm) = 1.458

(a) (1.00) sin 75.0° = 1.458 sin θ 2; θ 2 = 41.5°

(b) Let θ3 + β = 90.0° , θ2 + α = 90.0° ; then α + β + 60.0° = 180°

So 60.0° − θ2 − θ3 = 0 ⇒ 60.0° − 41.5° = θ3 = 18.5°

(d) γ = (θ1 − θ2 ) + [β − (90.0° − θ4 )]

γ = 75.0° − 41.5° + (90.0° − 18.5°) − (90.0° − 27.6°) = 42.6°

330 Chapter 35 Solutions

35.29 For the incoming ray, sin

sinθ θ2

1=n

Using the figure to the right, (θ2 )violet = sin−1 sin 50.0°

1.66

= 27.48°

(θ2 )red = sin−1 sin 50.0°

1.62

= 28.22°

For the outgoing ray, ′θ3 = 60.0° – θ2 and sinθ4 = nsinθ3

(θ4 )violet = sin−1[1.66 sin 32.52°] = 63.17°

(θ4 )red = sin−1[1.62 sin 31.78°] = 58.56°

The dispersion is the difference ∆θ 4 = (θ4 )violet − (θ4 )red = 63.17° – 58.56° = 4.61°

35.30 n =

sinΦ +δmin

sin Φ 2( )

For small Φ, δmin ≈ Φ so

Φ +δmin

2 is also a small angle. Then, using the small angle

approximation ( sinθ ≈ θ when θ << 1 rad), we have:

n ≈

Φ +δmin( ) 2Φ 2

= Φ +δmin

Φ or δmin ≈ n − 1( )Φ where Φ is in radians.

35.31 At the first refraction, 1 00 1 2. sin sin( ) =θ θn

The critical angle at the second surface is given by

n sin θ3 = 1.00 , or θ3 = sin−1 1.00

1.50

= 41.8°.

But, θ2 = 60.0°−θ3 . Thus, to avoid total internal reflection at thesecond surface (i.e., have θ3 < 41.8°), it is necessary that θ2 > 18.2°.Since sin θ1 = n sin θ2 , this requirement becomes

sin θ1 > 1.50( )sin 18.2°( ) = 0.468, or θ 1 > 27.9°

Chapter 35 Solutions 331

35.32 At the first refraction, 1.00( )sin θ1 = n sin θ2 . The critical angle atthe second surface is given by

n sin θ3 = 1.00 , or θ3 = sin−1 1.00 n( )But 90.0° − θ2( ) + 90.0° − θ3( ) + Φ =180°, which gives θ2 = Φ −θ3 .

Thus, to have θ3 < sin−1 1.00 n( ) and avoid total internal reflection at the second surface, it is

necessary that θ2 > Φ −sin−1 1.00 n( ) . Since sin sinθ θ1 2= n , this requirement becomes

sin θ1 > n sin Φ − sin−1 1.00

or θ1 > sin−1 n sin Φ − sin−1 1.00

Through the application of trigonometric identities, θ1 > sin−1 n2 − 1 sin Φ − cos Φ

35.33 n =

sin δ + φ( )sin φ/ 2( ) so

1.544sin 1

2φ( ) = sin 5°+ 1

2φ( ) = cos 1

2φ( )sin 5°+sin 1

2φ( )cos5°

tan 1

2φ( ) = sin 5°

1.544 − cos5°and φ = 18.1°

*35.34 Note for use in every part: Φ + 90.0° − θ2( ) + 90.0° − θ3( ) = 180°

so θ3 = Φ −θ2

At the first surface is α = θ1 − θ2

At exit, the deviation is β = θ4 − θ3

The total deviation is therefore δ = α + β = θ1 + θ4 − θ2 − θ3 = θ1 + θ4 − Φ

(a) At entry: n1 sin θ1 = n2 sin θ2 or θ2 = sin−1 sin 48.6°

1.50

= 30.0°

Thus, θ3 60 0 30 0 30 0= ° − ° = °. . .

At exit: 1.50 sin 30.0°= 1.00 sin θ4 or θ4 = sin−1 1.50 sin 30.0°( )[ ] = 48.6°

so the path through the prism is symmetric when θ1 48 6= °. .

(b) δ = ° + ° − ° =48 6 48 6 60 0. . . 37.2°

sin ..

.θ θ2 245 6

1 5028 4= ° ⇒ = ° θ3 60 0 28 4 31 6= ° − ° = °. . .

At exit: sin . sin . .θ θ4 41 50 31 6 51 7= °( ) ⇒ = ° δ = ° + ° − ° =45 6 51 7 60 0. . . 37.3°

(d) At entry: sin

sin ..

.θ θ2 251 6

1 5031 5= ° ⇒ = ° θ3 60 0 31 5 28 5= ° − ° = °. . .

At exit: sin . sin . .θ θ4 41 50 28 5 45 7= °( ) ⇒ = ° δ = ° + ° − ° =51 6 45 7 60 0. . . 37.3°

332 Chapter 35 Solutions

35.35 n sin θ = 1. From Table 35.1,

(a) θ =

=−sin

.1 1

2 419 24.4°

(b) θ =

=−sin

.1 1

1 66 37.0°

=−sin

.1 1

1 309 49.8°

35.36 sin ; sinθ θc c

= =

−2

1 2

(a) Diamond: θc =

=−sin

.1 1 333

2 419 33.4°

(b) Flint glass: θc =

=−sin

1 1 3331 66

53.4°

35.37 sin θc

= 2

1 (Equation 35.10)

n2 = n1 sin 88.8° = (1.0003)(0.9998) = 1.000 08

*35.38 sin

..θc

= = =air

pipe

1 001 36

0 735 θ c = 47.3°

Geometry shows that the angle of refraction at the end is

θ r = 90.0° – θ c = 90.0° – 47.3° = 42.7°

Then, Snell's law at the end, 1.00 sin θ = 1.36 sin 42.7°

gives θ = 67.2°

35.39 For total internal reflection, n n1 1 2 90 0sin sin .θ = °

(1.50) sin θ1 = (1.33)(1.00) or θ1 = 62.4°

Chapter 35 Solutions 333

35.40 To avoid internal reflection and come out through thevertical face, light inside the cube must have

θ31 1< −sin ( / )n

So θ2190 0 1> ° − −. sin ( / )n

But θ θ1 290 0 1< ° <. sin and n

In the critical case, sin ( / ) . sin ( / )− −= ° −1 11 90 0 1n n

1/n = sin 45.0° n = 1.41

35.41 From Snell's law, n n1 1 2 2sin sinθ θ=

At the extreme angle of viewing, θ 2 = 90.0°

(1.59)(sin θ1) = (1.00) · sin 90.0°

So θ1 = 39.0°

Therefore, the depth of the air bubble is

tan θ1< d <

tan θ1

or 1.08 cm < d < 1.17 cm

*35.42 (a)

sinsin

θθ

1= v

v and θ2 90 0= °. at the critical angle

sin .sin

90 0 1850343

° =θc

m s m s

so θc = =−sin .1 0 185 10.7°

(b) Sound can be totally reflected if it is traveling in the medium where it travels slower: air

334 Chapter 35 Solutions

*35.43 For plastic with index of refraction n ≥ 1.42 surrounded by air, the critical angle for totalinternal reflection is given by

θc = sin−1 1

≤ sin−1 1

1.42

= 44.8°

In the gasoline gauge, skylight from above travels down the plastic. The rays close to thevertical are totally reflected from both the sides of the slab and from facets at the lower end ofthe plastic, where it is not immersed in gasoline. This light returns up inside the plastic andmakes it look bright. Where the plastic is immersed in gasoline, with index of refractionabout 1.50, total internal reflection should not happen. The light passes out of the lower endof the plastic with little reflected, making this part of the gauge look dark. To frustrate totalinternal reflection in the gasoline, the index of refraction of the plastic should be n < 2.12 ,since

θc = sin−1 1.50

2.12( ) = 45.0° .

*35.44 Assume the lifeguard’s path makes angle θ1 with the north-south normal to the shoreline, and angle θ2 with this normalin the water. By Fermat’s principle, his path should followthe law of refraction:

sin θ1

sin θ2= v1

v2= 7.00 m s

1.40 m s= 5.00 or

θ2 = sin−1 sin θ1

The lifeguard on land travels eastward a distance x = ( )16 0 1. tan m θ . Then in the water, hetravels 26 0 20 0 2. . tan m m− = ( )x θ further east. Thus, 26 0 16 0 20 01 2. . tan . tan m m m= ( ) + ( )θ θ

or 26.0 m = 16.0 m( )tan θ1 + 20.0 m( )tan sin−1 sin θ1

We home in on the solution as follows:

θ1 (deg) 50.0 60.0 54.0 54.8 54.81right-hand side 22.2 m 31.2 m 25.3 m 25.99 m 26.003 m

The lifeguard should start running at 54.8° east of north .

*35.45 Let the air and glass be medium 1 and 2, respectively. By Snell's law, n2 sin θ2 = n1 sin θ1

or 1.56 sin θ2 = sin θ1

But the conditions of the problem are such that θ1 = 2θ2. 1.56 sin θ2 = sin 2θ2

We now use the double-angle trig identity suggested. 1.56 sin θ2 = 2 sin θ2 cos θ2

or cos θ2 = 1.56

2= 0.780

Thus, θ2 = 38.7° and θ1 = 2θ2 = 77.5°

Chapter 35 Solutions 335

*35.46 (a) ′ = =θ θ1 1 30.0°

n n1 1 2 2sin sinθ θ=

(1.00) sin 30.0° = 1.55 sinθ2

θ2 = 18.8°

(b) ′θ1 = θ1 = 30.0°

θ2 = sin−1 n1 sin θ1

= °

=−sin

. sin .1 1 55 30 01

50.8°

incidence reflection refraction incidence reflection refraction0 0 0 0 0 0

10.0 10.0 6.43 10.0 10.0 15.620.0 20.0 12.7 20.0 20.0 32.030.0 30.0 18.8 30.0 30.0 50.840.0 40.0 24.5 40.0 40.0 85.150.0 50.0 29.6 50.0 50.0 none*60.0 60.0 34.0 60.0 60.0 none*70.0 70.0 37.3 70.0 70.0 none*80.0 80.0 39.4 80.0 80.0 none*90.0 90.0 40.2 90.0 90.0 none*

*total internal reflection

35.47 For water, sin θc = 1

4 / 3= 3

Thus θc = sin−1 (0.750) = 48.6°

and d = 2 (1.00 m)tan θc[ ]

d = (2.00 m)tan 48.6° = 2.27 m Figure for Goal Solution

336 Chapter 35 Solutions

Goal Solution A small underwater pool light is 1.00 m below the surface. The light emerging from the water forms acircle on the water's surface. What is the diameter of this circle?

G : Only the light that is directed upwards and hits the water’s surface at less than the critical angle willbe transmitted to the air so that someone outside can see it. The light that hits the surface fartherfrom the center at an angle greater than θc will be totally reflected within the water, unable to beseen from the outside. From the diagram above, the diameter of this circle of light appears to beabout 2 m.

O : We can apply Snell’s law to find the critical angle, and the diameter can then be found from thegeometry.

A : The critical angle is found when the refracted ray just grazes the surface (θ2 = 90°). The index ofrefraction of water is n2 = 1.33, and n1 = 1.00 for air, so

n nc1 2sin sinθ = ° 90 gives θc = sin−1 1

1.333

= sin−1 0.750( ) = 48.6°

The radius then satisfies tanθc = r

(1.00 m)

So the diameter is d = 2r = 2 1.00 m( )tan 48.6°= 2.27 m

L : Only the light rays within a 97.2° cone above the lamp escape the water and can be seen by an outsideobserver (Note: this angle does not depend on the depth of the light source). The path of a light rayis always reversible, so if a person were located beneath the water, they could see the wholehemisphere above the water surface within this cone; this is a good experiment to try the next timeyou go swimming!

*35.48 Call θ1 the angle of incidence and of reflectionon the left face and θ2 those angles on the rightface. Let α represent the complement of θ1 and βbe the complement of θ2. Now α = γ and β = δbecause they are pairs of alternate interiorangles. We have

A = γ + δ = α + β

and B = α + A + β = α + β + A = 2A

Chapter 35 Solutions 337

*35.49 (a) We see the Sun swinging around a circle in the extended plane of our parallel of latitude. Itsangular speed is

ω = ∆θ

∆t= 2π rad

86 400 s= 7.27 × 10−5 rad s

The direction of sunlight crossing the cell from the window changes at this rate, moving onthe opposite wall at speed

v = rω = 2.37 m( ) 7.27 × 10−5 rad s( ) = 1.72 × 10−4 m s = 0.172 mm s

(b) The mirror folds into the cell the motion that would occur in a room twice as wide:

v = rω = 2 0.174 mm s( ) = 0.345 mm s

As the Sun moves southward and upward at 50.0°, we may regard the corner of the windowas fixed, and both patches of light move northward and downward at 50.0° .

*35.50 By Snell's law, n1 sin θ1 = n2 sin θ2

With v = cn ,

cv1

sin θ1 = cv2

sin θ2 or

sin sinθ θ1

2v v=

This is also true for sound. Here,

sin . sin12 0340 1

2° = m/s 510 m/s

θ2 = arcsin (4.44 sin 12.0°) = 67.4°

*35.51 (a)

n = cv

= 2.998 × 108 m s

61.15 kmhr

1.00 h3600 s

1.00 × 103 m1.00 km

= 1.76 × 107

(b) n1 sin θ1 = n2 sin θ2 so 1 76 10 1 00 90 07

1. sin . sin .×( ) = ( ) °θ

θ1 = 3 25 10 6. × − degree

This problem is misleading. The speed of energy transport is slow, but the speed of thewavefront advance is normally fast. The condensate's index of refraction is not far fromunity.

338 Chapter 35 Solutions

*35.52 Violet light:

1 00 25 0 1 689 14 4902 2. sin . . sin .( ) ° = ⇒ = °θ θ

yv = ( ) = ( ) °5 00 5 00 14 4902. tan . tan . cm cmθ = 1.2622 cm

Red Light:

1 00 25 0 1 642 14 9152 2. sin . . sin .( ) ° = ⇒ = °θ θ

yR = ( ) ° =5 00 14 915 1 3318. tan . . cm cm

The emergent beams are both at 25.0° from the normal. Thus,

w y= °∆ cos .25 0 where ∆y = 1.3318 cm − 1.2622 cm = 0.0396 cm

w = ( ) ° =0 396 25 0. cos . mm 0.359 mm

35.53 Horizontal light rays from the settingSun pass above the hiker. The light raysare twice refracted and once reflected, asin Figure (b) below, by just the certainspecial raindrops at 40.0° to 42.0° fromthe hiker's shadow, and reach the hikeras the rainbow.

The hiker sees a greater percentage of theviolet inner edge, so we consider the redouter edge. The radius R of the circle ofdroplets is

Figure (a)

R = (8.00 km)(sin 42.0°) = 5.35 km

Then the angle φ, between the vertical and the radius wherethe bow touches the ground, is given by

cos

.φ = = =2 005 3

0 374 km 2.00 km

5 kmR or φ = 68.1°

The angle filled by the visible bow is 360° – (2 × 68.1°) = 224°,so the visible bow is

224°360° = 62.2% of a circle

Figure (b)

Chapter 35 Solutions 339

35.54 From Snell’s law, 1 00

431 2. sin sin( ) =θ θ

x R r= =sin sinθ θ2 1

= sin θ2

sin θ1= 3

θ 2

θ 1

rzd

xeye

Fish at depth dImage at depth z

Fish

apparent depthactual depth

= zd

= r cos θ1

R cos θ2= 3

4cos θ1

1 − sin2 θ2

But sin2 θ2 = 3

4sin θ1

= 916

1 − cos2 θ1( )

= 34

cos θ1

1 − 916

+ 916

cos2 θ1

= 34

cos θ1

7 + 9 cos2 θ1

or z = 3d cos θ1

7 + 9 cos2 θ1

35.55 As the beam enters the slab, 1 00 50 0 1 48 2. sin . . sin( ) ° = ( ) θ giving θ2 31 2= °. . The beam thenstrikes the top of the slab at x1 1 55= °( ). mm tan 31.2 from the left end. Thereafter, the beamstrikes a face each time it has traveled a distance of 2 1x along the length of the slab. Since theslab is 420 mm long, the beam has an additional 420 1 mm − x to travel after the first reflection.The number of additional reflections is

420 mm − x1

2x1= 420 mm − 1.55 mm tan 31.2°( )

3.10 mm tan 31.2°( ) = 81.5

or 81 reflections since the answer must be aninteger. The total number of reflections made i nthe slab is then 82 .

*35.56 (a)

′S1

S1= n2 − n1

n2 + n1

= 1.52 − 1.001.52 + 1.00

= 0.0426

(b) If medium 1 is glass and medium 2 is air,

′S1

S1= n2 − n1

n2 + n1

= 1.00 − 1.521.00 + 1.52

= 0.0426;

There is no difference

(c)

′S1

S1= 1.76 × 107 − 1.00

1.76 × 107 + 1.00

= 1.76 × 107 + 1.00 − 2.001.76 × 107 + 1.00

′S1

S1 = −

× +

≈ −× +

1 00

2 001 76 10 1 00

1 00 22 00

1 76 10 1 007

7..

. ..

.. . = 1.00 − 2.27 × 10−7 or 100%

This suggests he appearance would be very shiny, reflecting practically all incident light .See, however, the note concluding the solution to problem 35.51.

340 Chapter 35 Solutions

*35.57 (a) With n1 = 1 and n2 = n , the reflected fractional intensity is

′S1

S1= n − 1

n + 1

The remaining intensity must be transmitted:

S1= 1 − n − 1

n + 1

= n + 1( )2 − n − 1( )2

n + 1( )2 = n2 + 2n + 1 − n2 + 2n − 1

n + 1( )2 =

n + 1( )2

(b) At entry,

S1= 1 − n − 1

n + 1

= 4 2.419( )2.419 + 1( )2 = 0.828

At exit,

S2= 0.828

Overall,

S1= S3

= 0.828( )2 = 0.685 or 68.5%

*35.58 Define T = 4n

n + 1( )2 as the transmission coefficient for one

encounter with aninterface. For diamond and air, it is 0.828, as in problem57.

As shown in the figure, the total amount transmitted is

T2 + T2 1 − T( )2 + T2 1 − T( )4 + T2 1 − T( )6

+ . . . + T2 1 − T( )2n + . . .

We have 1 − T = 1 − 0.828 = 0.172 so the totaltransmission is

0.828( )2 1 + 0.172( )2 + 0.172( )4 + 0.172( )6 + . . .[ ]

To sum this series, define F = 1 + 0.172( )2 + 0.172( )4 + 0.172( )6 + . . . .

Note that 0.172( )2 F = 0.172( )2 + 0.172( )4 + 0.172( )6 + . . ., and

1 + 0.172( )2 F = 1 + 0.172( )2 + 0.172( )4 + 0.172( )6 + . . . = F .

Then, 1 = F − 0.172( )2 F or F = 1

1 − 0.172( )2

The overall transmission is then

0.828 2( )− ( )

=1 0 172

0 7062.. or 70.6%

Chapter 35 Solutions 341

35.59 n sin . sin .42 0 90 0° = ° so n =

°=1

42 01 49

sin ..

sin sin .θ1 18 0= °n and sin

sin .sin .

θ118 042 0

= °°

θ1 = 27.5°

Figure for Goal Solution

Goal Solution The light beam shown in Figure P35.59 strikes surface 2 at the critical angle. Determine the angle ofincidence θ1.

G : From the diagram it appears that the angle of incidence is about 40°.

O : We can find θ1 by applying Snell’s law at the first interface where the light is refracted. At surface 2,knowing that the 42.0° angle of reflection is the critical angle, we can work backwards to find θ1.

A : Define n1 to be the index of refraction of the surrounding medium and n2 to be that for the prismmaterial. We can use the critical angle of 42.0° to find the ratio n2 n1 :

n2 sin 42.0°= n1 sin 90.0°

So,

n1= 1

sin 42.0°= 1.49

Call the angle of refraction θ2 at the surface 1. The ray inside the prism forms a triangle withsurfaces 1 and 2, so the sum of the interior angles of this triangle must be 180°. Thus,

90.0°−θ2( ) + 60.0°+ 90.0°−42.0°( ) = 180°

Therefore, θ2 = 18.0°

Applying Snell’s law at surface 1, n1 sinθ1 = n2 sin 18.0°

sinθ1 = n2 n1( )sin θ2 = 1.49( )sin 18.0°

θ1 = 27.5°

L : The result is a bit less than the 40.0° we expected, but this is probably because the figure is not drawnto scale. This problem was a bit tricky because it required four key concepts (refraction, reflection,critical angle, and geometry) in order to find the solution. One practical extension of this problem isto consider what would happen to the exiting light if the angle of incidence were varied slightly.Would all the light still be reflected off surface 2, or would some light be refracted and pass throughthis second surface?

342 Chapter 35 Solutions

35.60 Light passing the top of the pole makes an angle ofincidence φ1 = 90.0° − θ . It falls on the water surface atdistance

s1 = (L − d)

tan θ from the pole,

and has an angle of refraction φ2 from (1.00)sin φ1 = n sin φ2 .Then s d2 2= tan φ and the whole shadow length is

s1 + s2 = L − d

tan θ+ d tan sin−1 sin φ1

s1 + s2 = L − d

tan θ+ d tan sin−1 cos θ

= 2.00 mtan 40.0°

+ 2.00 m( )tan sin−1 cos 40.0°1.33

= 3.79 m

35.61 (a) For polystyrene surrounded by air, internal reflection requires

θ3 = sin−1 1.00

1.49

= 42.2°

Then from the geometry, θ 2 = 90.0° – θ 3 = 47.8°

From Snell's law, sin θ1 = 1.49( )sin 47.8°= 1.10

This has no solution. Therefore, total internal reflection always happens .

(b) For polystyrene surrounded by water, θ3 = sin−1 1.33

1.49

= 63.2°

and θ2 = 26.8°

From Snell's law, θ1 = 30.3°

*35.62 δ θ θ= − = °1 2 10 0. and n1 sin θ1 = n2 sin θ2 with n1 = 1, n2 = 4

Thus, θ1 = sin−1(n2 sin θ2 ) = sin−1 n2 sin(θ1 − 10.0°)[ ]

Chapter 35 Solutions 343

(You can use a calculator to home in on an approximate solution to this equation, testingdifferent values of θ1 until you find that θ1 = 36.5° . Alternatively, you can solve for θ1

exactly, as shown below.)

We are given that sin θ1 = 4

3sin(θ1 − 10.0°)

This is the sine of a difference, so

sin θ1 = sin θ1 cos 10.0° − cos θ1 sin 10.0°

Rearranging, sin . cos cos . sin10 0 10 0

341 1° = ° −

θ θ

sin .cos . .

tan10 0

10 0 0 750 1°

° −= θ and θ1

1 0 740= =−tan . 36.5°

35.63 tan

.θ14 00= cm

h and

tan

.θ22 00= cm

tan tan tan21 2

2 22θ θ θ= ( )2.00 = 4.00

sin( sin )

.sin

( sin )

221

4 001

θθ

θθ−

=−

(1)

Snell's law in this case is: n n1 1 2 2sin sinθ θ=

sin . sinθ θ1 21 333=

Squaring both sides, sin2 θ1 = 1.777 sin2 θ2 (2)

Substituting (2) into (1),

1.777 sin2 θ2

1 − 1.777 sin2 θ2= 4.00

sin2 θ2

1 − sin2 θ2

Defining x = sin2 θ ,

0.4441 − 1.777x( ) = 1

1 − x( )

Solving for x, 0.444 − 0.444x = 1 − 1.777x and x = 0.417

From x we can solve for θ2 : θ2 = sin−1 0.417 = 40.2°

Thus, the height is h = =

°=( )

tan( )tan( . )

2.00 cm 2.00 cmθ2 40 2

2.37 cm

344 Chapter 35 Solutions

35.64 Observe in the sketch that the angle of incidence at point P is γ ,and using triangle OPQ: sin γ = L / R .

Also, cos γ = 1 − sin2 γ = R2 − L2

Applying Snell’s law at point P, 1.00( )sin γ = n sin φ

Thus, sin φ = sin γ

n= L

and cos φ = 1 − sin2 φ = n2R2 − L2

From triangle OPS, φ + α + 90.0°( ) + 90.0° − γ( ) = 180° or the angle of incidence at point S isα = γ − φ. Then, applying Snell’s law at point S gives 1.00( )sin θ = n sin α = n sin γ − φ( ) , or

sin θ = n sin γ cos φ − cos γ sin φ[ ] = n

n2R2 − L2

nR− R2 − L2

sinθ = L

R2 n2R2 − L2 − R2 − L2

and θ =

sin−1 L

R2 n2R2 − L2 − R2 − L2

35.65 To derive the law of reflection, locate point O so that the time oftravel from point A to point B will be minimum.

The total light path is L a b= +sec secθ θ1 2

The time of travel is t = 1

(a sec θ1 + b sec θ2 )

If point O is displaced by dx, then

dt = 1

(a sec θ1 tan θ1 dθ1 + b sec θ2 tan θ2 dθ2 ) = 0 (1)

(since for minimum time dt = 0).

Also, c + d = a tan θ1 + b tan θ2 = constant

so, a sec2 θ1 dθ1 + b sec2 θ2 dθ2 = 0 (2)

Divide equations (1) and (2) to find θ 1 = θ 2

Chapter 35 Solutions 345

35.66 As shown in the sketch, the angle of incidence at point A is:

θ = sin−1 d 2( )

= sin−1 1.00 m

2.00 m

= 30.0°

If the emerging ray is to be parallel to the incident ray,the path must be symmetric about the center line CB ofthe cylinder . In the isosceles triangle ABC , γ = α and

β θ= ° −180 . Therefore, α + β + γ = 180° becomes

2 180 180α θ+ ° − = ° or α = θ

2= 15.0°

Then, applying Snell’s law at point A , n sin . sinα θ= ( )1 00

or n = = °

°=sin

sinsin .sin .

θα

30 015 0

1.93

35.67 (a) At the boundary of the air and glass, the critical angle is givenby

sin θc = 1

Consider the critical ray PB ′B : tan θc = d 4

t or

sin θc

cos θc= d

Squaring the last equation gives:

sin2 θc

cos2 θc= sin2 θc

1 − sin2 θc= d

Since sin θc = 1

n, this becomes

1n2 − 1

= d4t

or n = 1 + 4t d( )2

(b) Solving for d , d = 4t

n2 − 1

Thus, if n = 1.52 and t = 0.600 cm ,

d = 4 0.600 cm( )1.52( )2 − 1

= 2.10 cm

(c) Since violet light has a larger index of refraction, it will lead to a smaller critical angleand the inner edge of the white halo will be tinged with violet light.

346 Chapter 35 Solutions

35.68 From the sketch, observe that the angle ofincidence at point A is the same as theprism angle θ at point O . Given that θ = 60.0° , application of Snell’s law at point A gives

1 50 1 00 60 0. sin . sin .β = ° or β = 35.3°

From triangle AOB , we calculate the angleof incidence (and reflection) at point B .

θ β γ+ ° −( ) + ° −( ) = °90 0 90 0 180. . so γ θ β= − = ° − ° = °60 0 35 3 24 7. . .

Now, using triangle BCQ : 90 0 90 0 90 0 180. . .° −( ) + ° −( ) + ° −( ) = °γ δ θ

Thus the angle of incidence at point C is δ θ γ= ° −( ) − = ° − ° = °90 0 30 0 24 7 5 30. . . .

Finally, Snell’s law applied at point C gives 1 00 1 50 5 30. sin . sin .φ = °

or φ = °( ) =−sin . sin .1 1 50 5 30 7.96°

35.69 (a) Given that θ θ1 245 0 76 0= ° = °. . and , Snell’s law atthe first surface gives

n sin . sin .α = ( ) °1 00 45 0 (1)

Observe that the angle of incidence at the secondsurface is β α= ° −90 0. . Thus, Snell’s law at thesecond surface yields

n nsin sin . . sin .β α= ° −( ) = ( ) °90 0 1 00 76 0 , or

n cos sin .α = °76 0 (2)

Dividing Equation (1) by Equation (2), tan

sin .sin .

.α = °°

=45 076 0

0 729 or α = 36.1°

Then, from Equation (1), n = ° = °

°=sin .

sinsin .sin .

45 0 45 036 1α

1.20

(b) From the sketch, observe that the distance the light travels in the plastic is d = L sin α . Also,the speed of light in the plastic is v = c n , so the time required to travel through the plastic is

t = dv

= nLc sin α

= 1.20( ) 0.500 m( )3.00 × 108 m s( )sin 36.1°

= 3.40 × 10−9 s = 3.40 ns

Chapter 35 Solutions 347

35.70 sin θ1 sin θ2 sin θ1/ sin θ20.174 0.131 1.33040.342 0.261 1.31290.500 0.379 1.31770.643 0.480 1.33850.766 0.576 1.32890.866 0.647 1.33900.940 0.711 1.32200.985 0.740 1.3315

The straightness of the graph linedemonstrates Snell's proportionality.The slope of the line is n = ±1 3276 0 01. .

and n = 1.328 ± 0.8%

Chapter 36 Solutions

*36.1 I stand 40 cm from my bathroom mirror. I scatter light which travels to the mirror and backto me in time

0.8 m3 × 108 m/s

~ 10–9 s

showing me a view of myself as I was at that look-back time. I'm no Dorian Gray!

*36.2 The virtual image is as far behind the mirror as the choiris in front of the mirror. Thus, the image is 5.30 m behindthe mirror.

The image of the choir is 0.800 m + 5.30 m = 6.10 m fromthe organist. Using similar triangles:

′h0.600 m

= 6.10 m0.800 m

′h = 0.600 m( ) 6.10 m0.800 m

= 4.58 m

h’ 0.600 m

5.30 m 0.800 m

Organist

mirror

image of choir

View Looking DownSouth

36.3 The flatness of the mirror is described by R = ∞,

f = ∞, and 1/ f = 0. By our general mirror equation,

+ 1q

= 1f

, or q = −p

Thus, the image is as far behind the mirror as theperson is in front. The magnification is then Figure for Goal Solution

M = – qp = 1 =

h'h so h' = h = 70.0"

The required height of the mirror is defined by the triangle from the person's eyes to the topand bottom of his image, as shown. From the geometry of the triangle, we see that the mirrorheight must be:

h '

p – q = h'

2p = h'2 Thus, the mirror must be at least 35.0" high .

Chapter 36 Solutions 351

Goal Solution Determine the minimum height of a vertical flat mirror in which a person 5'10" in height can see his orher full image. (A ray diagram would be helpful.)

G : A diagram with the optical rays that create the image of the person is shown above. From thisdiagram, it appears that the mirror only needs to be about half the height of the person.

O : The required height of the mirror can be found from the mirror equation, where this flat mirror isdescribed by

R = ∞, f = ∞, and 1/ f = 0.

A : The general mirror equation is

+ 1q

= 1f

, so with f = ∞, q = −p

Thus, the image is as far behind the mirror as the person is in front. The magnification is then

M = −q

p= 1 = ′h

so ′h = h = 70.0 in.

The required height of the mirror is defined by the triangle from the person's eyes to the top andbottom of the image, as shown. From the geometry of the similar triangles, we see that the length ofthe mirror must be:

L = ′h

pp − q

= ′hp

= ′h2

= 70.0 in2

= 35.0 in . Thus, the mirror must be at least 35.0 in high.

L : Our result agrees with our prediction from the ray diagram. Evidently, a full-length mirror onlyneeds to be a half-height mirror! On a practical note, the vertical positioning of such a mirror is alsoimportant for the person to be able to view his or her full image. To allow for some variation i npositioning and viewing by persons of different heights, most full-length mirrors are about 5’ i nlength.

36.4 A graphical construction produces 5 images, with images I1and I2 directly into the mirrors from the object O,

and (O, I3, I4) and (I1, I2, I5) forming the vertices of equilateraltriangles.

352 Chapter 36 Solutions

*36.5 (1) The first image in the left mirror is 5.00 ft behind the mirror, or 10.0 ft from the position ofthe person.

(2) The first image in the right mirror is located 10.0 ft behind the right mirror, but this location is25.0 ft from the left mirror. Thus, the second image in the left mirror is 25.0 ft behind themirror, or 30.0 ft from the person.

(3) The first image in the left mirror forms an image in the right mirror. This first image is 20.0 ftfrom the right mirror, and, thus, an image 20.0 ft behind the right mirror is formed. Thisimage in the right mirror also forms an image in the left mirror. The distance from thisimage in the right mirror to the left mirror is 35.0 ft. The third image in the left mirror is,thus, 35.0 ft behind the mirror, or 40.0 ft from the person.

*36.6 For a concave mirror, both R and f are positive. We also know that f = R2 = 10.0 cm

(a)1q =

1f –

1p =

110.0 cm –

140.0 cm =

340.0 cm , and q = 13.3 cm

M = qp = –

13.3 cm40.0 cm = – 0.333

The image is 13.3 cm in front of the mirror, is real, and inverted .

(b)1q =

1f –

1p =

110.0 cm –

120.0 cm =

120.0 cm , and q = 20.0 cm

M = qp = –

20.0 cm20.0 cm = –1.00

The image is 20.0 cm in front of the mirror, is real, and inverted .

1f –

1p =

110.0 cm –

110.0 cm = 0 Thus, q = infinity.

No image is formed. The rays are reflected parallel to each other.

*36.71q =

1f –

1p = –

10.275 m –

110.0 m gives q = – 0.267 m

Thus, the image is virtual .

M = – qp = –

– 0.26710.0 m = 0.0267

Thus, the image is upright (+M) and diminished ( M < 1( )

Chapter 36 Solutions 353

*36.8 With radius 2.50 m, the cylindrical wall is a highly efficient mirror for sound, with focallength

f = R2 = 1.25 m

In a vertical plane the sound disperses as usual, but that radiated in a horizontal plane isconcentrated in a sound image at distance q from the back of the niche, where

1p +

1q =

1f so

12.00 m +

1q =

11.25 m q = 3.33 m

36.9 (a)1p +

1q =

2R gives

1(30.0 cm) +

1q =

2(– 40.0 cm)

1q = –

2(40.0 cm) –

1(30.0 cm) = – 0.0833 cm–1 so q = –12.0 cm

M = – qp = –

(–12.0 cm)(30.0 cm) = 0.400

(b)1p +

1q =

2R gives

1(60.0 cm) +

1q = –

2(40.0 cm)

1q = –

2(40.0 cm) –

1(60.0 cm) = – 0.0666 cm–1 so q = –15.0 cm

M = – qp = –

(–15.0 cm)(60.0 cm) = 0.250

36.10 (a) M = − q

p. For a real image, q > 0 so in this case M = − 4.00

q pM cm= − =120 and from

+ 1q

= 2R

R = 2pq

(p + q)= 2(30.0 cm)(120 cm)

(150 cm)= 48.0 cm

354 Chapter 36 Solutions

(b)

36.11 (a)

+ 1q

= 2R

becomes

= 2(60.0 cm)

− 190.0 cm( )

q = 45.0 cm and M = −q

p= − (45.0 cm)

(90.0 cm)= – 0.500

(b)

+ 1q

= 2R

becomes

= 2(60.0 cm)

− 1(20.0 cm)

q = – 60.0 cm and M = − q

p= − (−60.0 cm)

(20.0 cm)= 3.00

(c) The image in (a) is real, inverted and diminished. That of(b) is virtual, upright, and enlarged. The ray diagrams aresimilar to Figures 36.15(a) and 36.15(b), respectively.

(a)

(b)

Figures for Goal Solution

Chapter 36 Solutions 355

Goal Solution A concave mirror has a radius of curvature of 60.0 cm. Calculate the image position and magnification ofan object placed in front of the mirror (a) at a distance of 90.0 cm and (b) at a distance of 20.0 cm. (c) Ineach case, draw ray diagrams to obtain the image characteristics.

G : It is always a good idea to first draw a ray diagram for any optics problem. This gives a qualitativesense of how the image appears relative to the object. From the ray diagrams above, we see thatwhen the object is 90 cm from the mirror, the image will be real, inverted, diminished, and locatedabout 45 cm in front of the mirror, midway between the center of curvature and the focal point.When the object is 20 cm from the mirror, the image is be virtual, upright, magnified, and locatedabout 50 cm behind the mirror.

O : The mirror equation can be used to find precise quantitative values.

A : (a) The mirror equation is applied using the sign conventions listed in the text.

+ 1q

= 2R

190.0 cm

+ 1q

= 260.0 cm

so q = 45.0 cm (real, in front of the mirror)

M = −q

p= − 45.0 cm

90.0 cm= −0.500 (inverted)

(b)

+ 1q

= 2R

120.0 cm

+ 1q

= 260.0 cm

so q = − 60.0 cm (virtual, behind the mirror)

M = − q

p= − −60.0 cm

20.0 cm= 3.00 (upright)

L : The calculated image characteristics agree well with our predictions. It is easy to miss a minus signor to make a computational mistake when using the mirror-lens equation, so the qualitative valuesobtained from the ray diagrams are useful for a check on the reasonableness of the calculated values.

36.12 For a concave mirror, R and f are positive. Also, for an erect image, M is positive. Therefore,

M = − q

p= 4 and q = – 4p.

1f =

1p +

1q becomes

140.0 cm =

1p –

14p =

34p ; from which, p = 30.0 cm

36.13 (a) q = (p + 5.00 m) and, since the image must be real, M = – qp = – 5

or q = 5p. Therefore, p + 5.00 = 5p or p = 1.25 m and q = 6.25 m.

From

+ 1q

= 2R

, R = 2pq

(q + p) = 2(1.25)(6.25)(6.25 + 1.25) = 2.08 m (concave)

(b) From part (a), p = 1.25 m; the mirror should be 1.25 m in frontof the object.

356 Chapter 36 Solutions

36.14 (a) The image is the trapezoid ′a ′b ′d ′e as shown in the ray diagram.

a b

de ′a

′b

′d ′e

′hL ′hR

qR - qL

pR = 40.0 cm

pL = 60.0 cm

(b) To find the area of the trapezoid, the image distances, qR and qL, along with the heights ′hRand ′hL , must be determined. The mirror equation,

+ 1q

= 2R

becomes

140.0 cm

+ 1qR

= 220.0 cm

or qR = 13.3 cm

hR = hMR = h

−qR

= 10.0 cm( ) −13.3 cm40.0 cm

= −3.33 cm

Also

160.0 cm

+ 1qL

= 220.0 cm

or qL = 12 0. cm

hL = hML = 10.0 cm( ) −12.0 cm

60.0 cm

= −2.00 cm

The area of the trapezoid is the sum of the area of a square plus the area of a triangle:

At = A1 + A2 = qR − qL( )hL + 1

2qR − qL( ) hR − hL( ) = 3.56 cm2

36.15 Assume that the object distance is the same in both cases (i.e., her face is the same distancefrom the hubcap regardless of which way it is turned). Also realize that the near image(q = – 10.0 cm) occurs when using the convex side of the hubcap. Applying the mirrorequation to both cases gives:

(concave side: R = R , q = −30.0 cm )

− 130.0

= 2R

, or

= 30.0 cm − p(30.0 cm)p

[1]

(convex side: R = − R , q = −10.0 cm)

− 110.0

= − 2R

, or

= p − 10.0 cm10.0 cm( )p [2]

(a) Equating Equations (1) and (2) gives:

30.0 cm −p3.00

= p − 10.0 cm or p = 15.0 cm Thus,

her face is 15.0 cm from the hubcap.

(b) Using the above result ( p = 15.0 cm ) in Equation [1] gives:

= 30.0 cm − 15.0 cm(30.0 cm)(15.0 cm)

= 130.0 cm

, and R = 60.0 cm

Chapter 36 Solutions 357

The radius of the hubcap is 60.0 cm .

36.16

+ 1q

= 1f

f = R2 = –1.50 cm

q = – 15.011.0 cm (behind mirror) M =

–qp =

111.0

36.17 (a) The image starts from a point whose height above the mirror vertex is given by

1 1 1 2p q f R

+ = =

13.00 m

+ 1q

= 10.500 m

Therefore, q = 0.600 m

As the ball falls, p decreases and q increases. Ball and image pass when q p1 1= . When this istrue,

1 1 10 500

1 1 1p p p+ = =

. m or p1 1 00= . m.

As the ball passes the focal point, the image switches from infinitely far above the mirror toinfinitely far below the mirror. As the ball approaches the mirror from above, the virtualimage approaches the mirror from below, reaching it together when p2 = q2 = 0.

(b) The falling ball passes its real image when it has fallen

3.00 m − 1.00 m = 2.00 m = 12 gt2, or when

t = ( ) =2 2 00

9 80.

. m

m s2 0.639 s .

The ball reaches its virtual image when it has traversed

3.00 m − 0 = 3.00 m = 12 gt2, or at

t = ( ) =2 3 00

9 80.

. m

m s2 0.782 s .

36.18 When R → ∞, Equation 36.8 for a spherical surface becomes q = −p n2 n1( ). We use this tolocate the final images of the two surfaces of the glass plate. First, find the image the glassforms of the bottom of the plate:

qB1 = − 1.33

1.66

(8.00 cm) = −6.41 cm

358 Chapter 36 Solutions

This virtual image is 6.41 cm below the top surface of the glass or 18.41 cm below the watersurface. Next, use this image as an object and locate the image the water forms of the bottomof the plate.

qB2 = − 1.00

1.33

(18.41 cm) = −13.84 cm or 13.84 cm below the water surface.

Now find image the water forms of the top surface of the glass.

q3 = − 1

1.33

(12.0 cm) = −9.02 cm, or 9.02 cm below the water surface.

Therefore, the apparent thickness of the glass is ∆t = 13.84 cm – 9.02 cm = 4.82 cm

36.19n1p +

n2q =

n2 – n1R = 0 and R → ∞

q = – n2n1

p = – 1

1.309 (50.0 cm) = – 38.2 cm

Thus, the virtual image of the dust speck is 38.2 cm below the top surface of the ice.

*36.20n1p +

n2q =

n2 – n1R so

1.00∞ +

1.4021.0 mm =

1.40 – 1.006.00 mm and 0.0667 =

0.0667

They agree. The image is inverted, real and diminished.

Chapter 36 Solutions 359

36.21 From Equation 36.8, n1p +

n2q =

(n2 – n1)R

Solve for q to find q = n2 R p

p(n2 – n1) – n1 R

In this case, n1 = 1.50, n2 = 1.00, R = –15.0 cm, and p = 10.0 cm,

So q = (1.00)(–15.0 cm)(10.0 cm)

(10.0 cm)(1.00 – 1.50) – (1.50)(–15.0 cm) = – 8.57 cm

Therefore, the apparent depth is 8.57 cm .

36.22 p = ∞ and q = +2R

1.00p +

n2q =

n2 – 1.00R

0 + n22R =

n2 – 1.00R so n2 = 2.00

36.23

p+ n2

q= (n2 − n1)

Rbecause

1.00p

+ 1.50q

= 1.50 − 1.006.00 cm

= 1.0012.0 cm

(a)

120.0 cm

+ 1.50q

= 1.0012.0 cm

q = 1.501.00

12.0 cm− 1.00

20.0 cm

= 45.0 cm

(b)

1.0010.0 cm

+ 1.50q

= 1.0012.0 cm

q = 1.501.00

12.0 cm− 1.00

10.0 cm

= – 90.0 cm

(c)

1.003.00 cm

+ 1.50q

= 1.0012.0 cm

q = 1.501.00

12.0 cm− 1.00

3.00 cm

= – 6.00 cm

36.24 For a plane surface,

p+ n2

q= n2 − n1

R becomes

q = − n2p

n1.

Thus, the magnitudes of the rate of change in the image and object positions are related by

dqdt

= n2

dpdt

If the fish swims toward the wall with a speed of 2.00 cm s, the speed of the image is given by

vimage = dq

dt= 1.00

1.332.00 cm s( ) = 1.50 cm/s

360 Chapter 36 Solutions

36.25

p+ n2

q= n2 − n1

R n1 = 1.33 n2 = 1.00 p = +10.0 cm R = −15.0 cm

q = −9.01 cm , or the fish appears to be 9.01 cm inside the bowl

*36.26 Let R1 = outer radius and R2 = inner radius

1f = (n – 1)

R1 –

1R2

= (1.50 – 1)

2.00 m – 1

2.50 cm = 0.0500

cm so f = 20.0 cm

36.27 (a)1f = (n – 1)

R1 –

1R2

= (0.440)

(12.0 cm) – 1

(–18.0 cm) : f = 16.4 cm

(b)1f = (0.440)

(18.0 cm) – 1

(–12.0 cm) : f = 16.4 cm

Figure for Goal Solution

Goal Solution The left face of a biconvex lens has a radius of curvature of magnitude 12.0 cm, and the right face has aradius of curvature of magnitude 18.0 cm. The index of refraction of the glass is 1.44. (a) Calculate thefocal length of the lens. (b) Calculate the focal length if the radii of curvature of the two faces areinterchanged.

G : Since this is a biconvex lens, the center is thicker than the edges, and the lens will tend to convergeincident light rays. Therefore it has a positive focal length. Exchanging the radii of curvatureamounts to turning the lens around so the light enters the opposite side first. However, this doesnot change the fact that the center of the lens is still thicker than the edges, so we should not expectthe focal length of the lens to be different (assuming the thin-lens approximation is valid).

O : The lens makers’ equation can be used to find the focal length of this lens.

A : The centers of curvature of the lens surfaces are on opposite sides, so the second surface has anegative radius:

(a)

= n − 1( ) 1R1

− 1R2

= 1.44 − 1.00( ) 112.0 cm

− 1−18.0 cm

so f = 16.4 cm

(b)

= 0.440( ) 118.0 cm

− 1−12.0 cm

so f = 16.4 cm

L : As expected, reversing the orientation of the lens does not change what it does to the light, as long asthe lens is relatively thin (variations may be noticed with a thick lens). The fact that light rays can betraced forward or backward through an optical system is sometimes referred to as the principle ofreversibility. We can see that the focal length of this biconvex lens is about the same magnitude asthe average radius of curvature. A few approximations, useful as checks, are that a symmetricbiconvex lens with radii of magnitude R will have focal length f ≈ R; a plano-convex lens withradius R will have f ≈ R / 2; and a symmetric biconcave lens has f ≈ −R . These approximationsapply when the lens has n ≈ 1.5, which is typical of many types of clear glass and plastic.

*36.28 For a converging lens, f is positive. We use 1p +

1q =

1f .

Chapter 36 Solutions 361

(a)1q =

1f –

1p =

120.0 cm –

140.0 cm =

140.0 cm q = 40.0 cm

M = – qp = –

40.040.0 = –1.00

The image is real, inverted , and located 40.0 cm past the lens.

(b)1q =

1f –

1p =

120.0 cm –

120.0 cm = 0 q = infinity

No image is formed. The rays emerging from the lens are parallel to each other.

1f –

1p =

120.0 cm –

110.0 cm = –

120.0 cm q = – 20.0 cm

M = – qp = –

–20.010.0 = 2.00

The image is upright, virtual , and 20.0 cm in front of the lens.

*36.29 (a)1q =

1f –

1p =

125.0 cm –

126.0 cm q = 650 cm

The image is real, inverted, and enlarged .

(b)1q =

1f –

1p =

125.0 cm –

124.0 cm q = – 600 cm

The image is virtual, upright, and enlarged .

36.30 (a)1p +

1q =

1f ⇒

1(32.0 cm) +

1(8.00 cm) =

1f so f = 6.40 cm

(b) M = – qp =

– (8.00 cm)(32.00 cm) = – 0.250

362 Chapter 36 Solutions

36.31 We are looking at an enlarged, upright, virtual image:

M = h'h = 2 = –

qp so p = –

q2 = –

– 2.84 cm2 = +1.42 cm

1p +

1q =

1f gives

11.42 cm +

1– 2.84 cm =

f = 2.84 cm

*36.32 To use the lens as a magnifying glass, we form an upright, virtual image:

M = +2.00 = – qp or

1p +

1q =

We eliminate q = – 2.00p::1p +

1– 2.00p =

115.0 cm or

– 2.00 + 1.00– 2.00p =

115.0 cm

Solving, p = 7.50 cm

36.33 (a) Note that q = 12.9 cm – p

so1p +

112.9 – p =

12.44

which yields a quadratic in p: – p2 + 12.9p = 31.5

which has solutions p = 9.63 cm or p = 3.27 cm

Both solutions are valid.

(b) For a virtual image, – q = p + 12.9 cm

1p –

112.9 + p =

12.44

or p2 + 12.9p = 31.8

from which p = 2.10 cm or p = – 15.0cm.

We must have a real object so the – 15.0 cm solution must berejected.

Chapter 36 Solutions 363

36.34 (a)1p +

1q =

1f :

1p +

1– 30.0 cm =

112.5 cm

p = 8.82 cm M = – qp = –

(– 30.0)8.82 = 3.40, upright

(b) See the figure to the right.

*36.351p +

1q =

1f : p–1 + q–1 = constant

We may differentiate through with respect to p: –1p–2 – 1q–2 dqd p = 0

dqd p = –

p2 = – M2

36.36 The image is inverted: M = h'h =

–1.80 m0.0240 m = – 75.0 =

– qp q = 75.0p

(b) q + p = 3.00 m = 75.0p + p p = 39.5 mm

(a) q = 2.96 m1f =

1p +

1q =

10.0395 m +

12.96 m

f = 39.0 mm

36.37 (a)

+ 1q

= 1f

1(20.0 cm)

1 1( 32.0 cm)

+ =−q

so q = − 1

20.0+ 1

32.0

−1

= –12.3 cm

The image is 12.3 cm to the left of the lens.

(b) M = − q

p= − (−12.3 cm)

(20.0 cm)= 0.615

364 Chapter 36 Solutions

36.38 (a)

= (n − 1)1

R1− 1

= (1.50 − 1)1

15.0 cm− 1

−12.0 cm( )

, or f = 13.3 cm

(b) Ray Diagram:

Fh = 10.0 cm

pR = 20.0 cm

pL = 30.0 cm

c d

a’

b’

c’ d’

h’R

h’L

qR qL

10.0 cm

1pR

+ 1qR

= 1f

becomes:

120.0 cm

+ 1qR

= 113.3 cm

or qR = 40 0. cm

′hR = hMR = h

− qR

= (10.0 cm)(−2.00) = −20.0 cm

130.0 cm

+ 1qL

= 113.3 cm

: qL = 24.0 cm

′hL = hML = (10.0 cm)(−0.800) = −8.00 cm

Thus, the area of the image is: Area = qR − qL ′hL + 12 qR − qL ′hR − ′hL = 224 cm2

36.39 (a) The distance from the object to the lens is p , so the image distance is q = 5.00 m − p .

Thus,

+ 1q

= 1f

becomes:

+ 15.00 m − p

= 10.800 m

This reduces to a quadratic equation: p2 − 5.00 m( )p + 4.00 m( ) = 0

which yields p = 4.00 m, or p = 1.00 m .

Thus, there are two possible object distances, both corresponding to real objects.

(b) For p = 4.00 m: q = 5.00 m − 4.00 m = 1.00 m: M = − 1.00 m

4.00 m= – 0.250 .

For p = 1.00 m: q = 5.00 m − 1.00 m = 4.00 m: M = − 4.00 m

1.00 m= – 4.00 .

Chapter 36 Solutions 365

Both images are real and inverted , but the magnifications are different, with one beinglarger than the object and the other smaller.

36.40 (a) The image distance is: q = d − p . Thus,

+ 1q

= 1f

becomes

+ 1d − p

= 1f

This reduces to a quadratic equation: p2 + −d( )p + f d( ) = 0

which yields: p =

d ± d2 − 4 f d2

= d2

± d2

4− f d

Since f < d 4 , both solutions are meaningful and the two solutions are not equal to eachother. Thus, there are two distinct lens positions that form an image on the screen.

(b) The smaller solution for p gives a larger value for q, with a real, enlarged, inverted image .

The larger solution for p describes a real, diminished, inverted image .

*36.41 To properly focus the image of a distant object, the lens must be at a distance equal to the focallength from the film ( q1 = 65.0 mm). For the closer object:

1p2

+ 1q2

= 1f

becomes

12000 mm

+ 1q2

= 165.0 mm

and q2 = 65.0 mm( ) 2000

2000 − 65.0

The lens must be moved away from the film by a distance

D = q2 − q1 = 65.0 mm( ) 2000

2000 − 65.0

− 65.0 mm = 2.18 mm

*36.42 (a) The focal length of the lens is given by

= n − 1( ) 1R1

− 1R2

= 1.53 − 1.00( ) 1−32.5 cm

− 142.5 cm

f = −34.7 cm

Note that R1 is negative because the center of curvature of the firstsurface is on the virtual image side.

R1 R2

366 Chapter 36 Solutions

When p = ∞, the thin lens equation gives q = f . Thus, the violet

image of a very distant object is formed at q = −34 7. cm . The

image is virtual, upright, and diminished .

(b) The same ray diagram and image characteristics apply for red light.Again, q = f , and now

= 1.51 − 1.00( ) 1−32.5 cm

− 142.5 cm

giving f = −36.1 cm .

⇑F F

36.43 Ray h1 is undeviated at the plane surface and strikes the second surface at angle of incidencegiven by

θ1 = sin−1 h1

= sin−1 0.500 cm

20.0 cm

= 1.43°

Then, (1.00)sinθ2 = (1.60)sinθ1 = (1.60)

0.50020.0 cm

so θ 2 = 2.29°

The angle this emerging ray makes with the horizontal is

θ 2 – θ1 = 0.860°

It crosses the axis at a point farther out by f1 where

f1 = h 1

tan(θ 2 – θ1) =

0.500 cmtan(0.860°) = 33.3 cm

The point of exit for this ray is distant axially from the lens vertex by

20.0 cm – (20.0 cm)2 – (0.500 cm)2 = 0.00625 cm

so ray h1 crosses the axis at this distance from the vertex:

x1 = 33.3 cm – 0.00625 cm = 33.3 cm

Now we repeat this calculation for ray h2: θ1 = sin−1 12.0 cm

20.0 cm

= 36.9°

(1.00)sinθ2 = (1.60)sinθ1 = (1.60)

12.0020.0

θ 2 = 73.7°

f2 = h 2

tan(θ 2 – θ1) =

12.0 cmtan(36.8°) = 16.0 cm

x2 = 16.0 cm( ) 20.0 cm − (20.0 cm)2 − (12.0 cm)2

= 12.0 cm

Now ∆x = 33.3 cm – 12.0 cm = 21.3 cm

Chapter 36 Solutions 367

36.44 For starlight going through Nick's glasses,

+ 1q

= 1f

1∞

+ 1−0.800 m( ) = 1

f= −1.25 diopters

For a nearby object,

+ 1−0.180 m( ) = −1.25 m-1, so p = 23.2 cm

36.45 P = 1

f= 1

p+ 1

q= 1

∞− 1

0.250 m= −4.00 diopters = −4.00 diopters, a diverging lens

36.46 Consider an object at infinity, imaged at the person's far point:

+ 1q

= 1f

1∞

+ 1q

= −4.00 m-1 q = −25.0 cm

The person's far point is 25.0 cm + 2.00 cm = 27.0 cm from his eyes. For the contact lenses wewant

1∞

+ 1−0.270 m( ) = 1

f= −3.70 diopters

36. 47 First, we use the thin lens equation to find the object distance:

+ 1−25.0 cm( ) = 1

10.0 cm

Then, p = 7.14 cm and Then, M = − q

p= −

−25.0 cm( )7.14 cm

= 3.50

36.48 (a) From the thin lens equation:

+ 1−25.0 cm( ) = 1

5.00 cmor p = 4.17 cm

(b) M = − q

p= 1 + 25.0 cm

f= 1 + 25.0 cm

5.00 cm= 6.00

36.49 Using Equation 36.20, M = − L

25.0 cmf e

= − 23.0 cm0.400 cm

25.0 cm2.50 cm

= – 575

368 Chapter 36 Solutions

36.50 M = M1me = M1

25.0 cmf e

⇒ f e = M1

25.0 cm( ) = −12.0

−140

25.0 cm( ) = 2.14 cm

36.51 f o = 20.0 m f e = 0.0250 m

(a) The angular magnification produced by this telescope is: m = − f o

f e= – 800

(b) Since m < 0, the image is inverted .

Chapter 36 Solutions 369

*36.52 (a) The lensmaker's equation

+ 1q

= 1f

gives

q = 11 f − 1 p

= 1p − ff p

= f pp − f

Then, M = ′h

h= − q

p= − f

p − f

gives

′h = hff − p

(b) For p >> f , f − p ≈ −p . Then, ′h = − hf

p .

′h = − hf

p= − 108.6 m( ) 4.00 m( )

407 × 103 m= −1.07 mm

*36.53 (b) Call the focal length of the objective f o and that of the eyepiece − f e . The distance betweenthe lenses is f o − f e . The objective forms a real diminished inverted image of a very distantobject at q1 = f o . This image is a virtual object for the eyepiece at p2 = − f e .

For it

+ 1q

= 1f

becomes

1− f e

+ 1q

= 1− f e

1q2

= 0

and q2 = ∞

(a) The user views the image as virtual . Letting ′h

represent the height of the first image, θo = ′h fo and

θ = ′h fe . The angular magnification is

m = θ

θo=

′h fe

′h fo= f o

f e

f o

f e= 3.00.

Thus,

f e = f o

3.00 and

23 f o = 10.0 cm.

θ 0

h’F0 θ 0

OFe

f o = 15.0 cm

f e = 5.00 cm and f e = −5 00. cm

370 Chapter 36 Solutions

*36.54 Let I0 represent the intensity of the light from the nebula and θo its angular diameter. Withthe first telescope, the image diameter ′h on the film is given by θo = − ′h fo as

′h = −θo 2000 mm( ).

The light power captured by the telescope aperture is P1 = I0A1 = I0 π 200 mm( )2 4[ ] , and the

light energy focused on the film during the exposure is E1 = P1t1 = I0 π 200 mm( )2 4[ ] 1.50 min( ).

Likewise, the light power captured by the aperture of the second telescope is

P2 = I0A2 = I0 π 60.0 mm( )2 4[ ] and the light energy is

E I t2 0

2260 0 4= ( )[ ]π . mm . Therefore, to

have the same light energy per unit area, it is necessary that

I0 π 60.0 mm( )2 4[ ] t2

π θo 900 mm( )2 4[ ] =I0 π 200 mm( )2 4[ ] 1.50 min( )

π θo 2000 mm( )2 4[ ]The required exposure time with the second telescope is

t2 = 200 mm( )2 900 mm( )2

60.0 mm( )2 2000 mm( )2 1.50 min( ) = 3.38 min

36.55 Only a diverging lens gives an upright diminished image. The image is virtual and

d = p − q = p + q: M = – qp so q = – Mp and d = p – Mp

p = d

1 – M :1p +

1q =

1f =

1p +

1– M p =

– M + 1– M p =

(1 – M)2

– M d

f = – M d

(1 – M)2 = – (0.500)(20.0 cm)

(1 – 0.500)2 = – 40.0 cm

36.56 If M < 1, the lens is diverging and the image is virtual. d = p – q = p + q

M = – qp so q = –Mp and d = p – Mp

p = d

1 – M :1p +

1q =

1f =

1p +

1– M p =

– M + 1– M p =

(1 – M)2

– M d f = – M d

(1 – M)2

If M > 1, the lens is converging and the image is still virtual.

Now d = – q – p. We obtain in this case f = M d

(M – 1)2

Chapter 36 Solutions 371

*36.57 Start with the first pass through the lens.

1q1

= 1f1

– 1

p1 =

180.0 cm –

1100 cm q1 = 400 cm to right of lens

For the mirror, p2 = – 300 cm

1q2

= 1f2

– 1

p2 =

1– 50.0 cm –

1– 300 cm q2 = – 60.0 cm

For the second pass through the lens, p3 = 160 cm

1q3

= 1f1

– 1

p3 =

180.0 cm –

1160 cm q3 = 160 cm to left of lens

M1 = – q1p1

= – 400 cm100 cm = – 4.00 M2 = –

q2p2

= – – 60.0 cm– 300 cm = –

M3 = – q3p3

= – 160 cm160 cm = –1 M = M1M2M3 = – 0.800

Since M < 0 the final image is inverted .

*36.58 (a)

= n − 1( ) 1R1

− 1R2

1−65.0 cm

= 1.66 − 1( ) 150.0 cm

− 1R2

1R2

= 150.0 cm

+ 142.9 cm

so R2 = 23.1 cm

2.00 cm

(b) The distance along the axis from B to A is

R1 − R12 − 2.00 cm( )2 = 50.0 cm − 50.0 cm( )2 − 2.00 cm( )2 = 0.0400 cm

Similarly, the axial distance from C to D is

23 1 23 1 2 00 0 08682 2. . . . cm cm cm cm− ( ) − ( ) =

Then, AD = 0.100 cm − 0.0400 cm + 0.0868 cm = 0.147 cm .

372 Chapter 36 Solutions

*36.591q1

= 1f1

– 1

p1 =

110.0 cm –

112.5 cm so q1 = 50.0 cm (to left of mirror)

This serves as an object for the lens (a virtual object), so

1q2

= 1f2

– 1

p2 =

1– 16.7 cm –

1– 25.0 cm q2 = – 50.3 cm (to right of lens)

Thus, the final image is located 25.3 cm to right of mirror .

M1 = – q1p1

= – 50.0 cm12.5 cm = – 4.00

M2 = – q2p2

= – – 50.3 cm– 25.0 cm = – 2.01

M = M1M2 = 8.05

Thus, the final image is virtual, upright , 8.05 times the size of object, and 25.3 cm to right ofthe mirror.

36.60 We first find the focal length of the mirror.

1f =

1p +

1q =

110.0 cm +

18.00 cm =

940.0 cm and f = 4.44 cm

Hence, if p = 20.0 cm,1q =

1f –

1p =

14.44 cm –

120.0 cm =

15.5688.8 cm

Thus, q = 5.71 cm , real

36.61 A hemisphere is too thick to be described as a thin lens.The light is undeviated on entry into the flat face. W enext consider the light's exit from the second surface, forwhich R = – 6.00 cm

The incident rays are parallel, so p = ∞.

Then, n1p +

n2q =

n2 – n1R becomes 0 +

1q =

(1.00 – 1.56)– 6.00 cm and q = 10.7 cm

Chapter 36 Solutions 373

*36.62 (a)

I = P

4πr2 = 4.50 W

4π 1.60 × 10-2 m( )2 = 1 40. kW m2

(b) I = P

4πr2 = 4.50 W

4π 7.20 m( )2 = 6 91. mW m2

(c)

+ 1q

= 1f

⇒ 1

7.20 m+ 1

q= 1

0.350 mso q = 0.368 m and

M = ′h

3.20 cm= − q

p= − 0.368 m

7.20 m ′h = 0.164 cm

(d) The lens intercepts power given by P = IA = 6.91× 10−3 W m2( ) 1

4π 0.150 m( )2[ ]

and puts it all onto the image where I = P

6.91× 10−3 W m2( ) π 15.0 cm( )2 4[ ]π 0.164 cm( )2 4

= 58.1 W m2

36.63 From the thin lens equation, q1 = f1p1

p1 – f1 =

(– 6.00 cm)(12.0 cm)12.0 cm – (– 6.00 cm) = – 4.00 cm

When we require that q2 → ∞, the thin lensequation becomes p2 = f2;

In this case, p2 = d – (– 4.00 cm)

Therefore, d + 4.00 cm = f2 = 12.0 cm

and d = 8.00 cm

*36.64 (a) For the light the mirror intercepts, P = I0A = I0πRa2

350 W = 1000 W m2( )πRa

2 and Ra = 0.334 m or larger

(b) In

+ 1q

= 1f

= 2R

we have p → ∞ so q = R

M = ′h

h= − q

p , so

′h = −q h p( ) = − R

0.533°

π rad180°

= − R2

9.30 m rad( )

where h p is the angle the Sun subtends. The intensity at the image is then

I = P

π ′h 2 4= 4I0πRa

π ′h 2 = 4I0Ra2

R 2( )2 9.30 × 10−3 rad( )2

120 × 103 W m2 =16 1000 W m2( )Ra

R2 9.30 × 10−3 rad( )2 so

R= 0.0255 or larger

374 Chapter 36 Solutions

36.65 For the mirror, f = R/2 = +1.50 m. In addition, because the distance to the Sun is so muchlarger than any other figures, we can take p = ∞. The mirror equation,

1p +

1q =

1f , then gives q = f = 1.50 m .

Now, in M = – qp =

h'h , the magnification is nearly zero, but we can be more precise:

hp is

the angular diameter of the object. Thus, the image diameter is

h ' = – h qp = (– 0.533°)

180 rad/deg (1.50 m) = – 0.140 m = –1.40 cm

36.66 (a) The lens makers' equation,

= (n − 1)1

R1− 1

, becomes:

15.00 cm

= (n − 1)1

9.00 cm− 1

−11.0 cm( )

giving n = 1.99 .

(b) As the light passes through the lens for the first time, the thin lens equation

1p1

+ 1q1

= 1f

becomes:

18.00 cm

+ 1q1

= 15.00 cm

or q1 = 13.3 cm , and M1 = − q1

p1= −13.3 cm

8.00 cm= −1.67

This image becomes the object for the concave mirror with:

pm = 20.0 cm − q1 = 20.0 cm − 13.3 cm = 6.67 cm , and f = R

2= +4.00 cm.

The mirror equation becomes:

16.67 cm

+ 1qm

= 14.00 cm

giving qm = 10.0 cm and M2 = − qm

pm= − 10.0 cm

6.67 cm= −1.50

The image formed by the mirror serves as a real object for the lens on the second pass of thelight through the lens with:

p3 = 20.0 cm − qm = +10.0 cm

The thin lens equation yields:

110.0 cm

+ 1q3

= 15.00 cm

or q3 = 10.0 cm, and M3 = − q3

p3= − 10.0 cm

10.0 cm= −1.00.

The final image is a real image located 10.0 cm to the left of the lens .

The overall magnification is Mtotal = M1M2M3 = −2 50. .

Chapter 36 Solutions 375

36.67 In the original situation, p1 + q1 = 1.50 m

In the final situation, p2 = p1 + 0.900 m

and q2 = q1 – 0.900 m.

Our lens equation is1p1

+ 1q1

= 1f =

1p2

+ 1q2

Substituting, we have1p1

+ 1

1.50 m – p1 =

1p1 + 0.900 +

10.600 – p1

Adding the fractions, 1.50 m – p1 + p1p1(1.50 m – p1) =

0.600 – p1 + p1 + 0.900(p1 + 0.900)(0.600 – p1)

Simplified, this becomes p1(1.50 m – p1) = (p1 + 0.900)(0.600 – p1)

(a) Thus, p1 = 0.5401.80 m = 0.300 m

p2 = p1 + 0.900 = 1.20 m

(b)1f =

10.300 m +

11.50 m – 0.300 m and f = 0.240 m

= – 0.250

36.68 As the light passes through, the lens attempts to form an image at distance q1 where

1q1

= 1f

− 1p1

or q1 = f p1

p1 − f

This image serves as a virtual object for the mirror with p2 = −q1. The plane mirror thenforms an image located at q2 = −p2 = +q1 above the mirror and lens.

This second image serves as a virtual object ( p3 = −q2 = −q1) for the lens as the light makes areturn passage through the lens. The final image formed by the lens is located at distance q3above the lens where

1q3

= 1f

− 1p3

= 1f

+ 1q1

= 1f

+ p1 − ff p1

= 2p1 − ff p1

or q3 = f p1

2p1 − f

If the final image coincides with the object, it is necessary to require q p3 1= , or

f p1

2p1 − f= p1.

This yields the solution p1 = f or the object must be located at the focal point of the lens .

376 Chapter 36 Solutions

36.69 For the objective:

1 1 1p q f

+ = becomes

13.40 mm

+ 1q

= 13.00 mm

so q = 25.5 mm

The objective produces magnification M1 = −q / p = − 25.5 mm

3.40 mm= −7.50

For the eyepiece as a simple magnifier, me = 25.0 cm

f= 25.0 cm

2.50 cm= 10.0

and overall M = M1me = −75.0

36.70 (a) Start with the second lens: This lens must form a virtual image located 19.0 cm to the left of it(i.e., q2 19 0= − . cm). The required object distance for this lens is then

p2 = q2 f2

q2 − f2=

−19.0 cm( ) 20.0 cm( )−19.0 cm − 20.0 cm

= 380 cm39.0

The image formed by the first lens serves as the object for the second lens. Therefore, theimage distance for the first lens is

q p1 250 0 50 0

380 1570= − = − =. . cm cm cm

39.0 cm

39.0

The distance the original object must be located to the left of the first lens is then given by

1p1

= 1f1

− 1q1

= 110.0 cm

− 39.01570 cm

= 157 − 39.01570 cm

= 1181570 cm

or p1

1570118

= = cm 13.3 cm

(b) M = M1M2 = − q1

− q2

= 1570 cm39.0

1181570 cm

−19.0 cm( ) 39.0( )380 cm

= −5 90.

36.71 (a) P

f p q= = + = +

∞=1 1 1 1

(0.0224 m)1

44.6 diopters

(b) P = 1

f= 1

p+ 1

q= 1

(0.330 m)+ 1

∞= 3.03 diopters

Chapter 36 Solutions 377

36.72 The object is located at the focal point of the upper mirror.Thus, the upper mirror creates an image at infinity (i.e.,parallel rays leave this mirror).

The lower mirror focuses these parallel rays at its focal point,located at the hole in the upper mirror. Thus, the

image is real, inverted, and actual size .

For the upper mirror:

+ 1q

= 1f

⇒ 1

7.50 cm+ 1

q1= 1

7.50 cm: q1 = ∞

For the lower mirror:

1∞

+ 1q2

= 17.50 cm

: q2 = 7.50 cm

Light directed into the hole in the upper mirror reflects asshown, to behave as if it were reflecting from the hole.

36.73 (a) Lens one:

140.0 cm

+ 1q1

= 130.0 cm

: q1 = 120 cm

qp1

12040 0

= − = − cm cm. = −3.00

This real image is a virtual object for the second lens, at

p2 = 110 cm − 120 cm = −10.0 cm

110 0

1 120 02−

+ =−. . cm cmq

: q2 = 20.0 cm

M2 = − q2

p2= − 20.0 cm

−10.0 cm( ) = +2.00

Moverall = M1M2 = −6.00

(b) Moverall < 0, so final image is inverted .

110 0

1 120 02−

+ =. . cm cmq

q2 = 6.67 cm

M2 = − 6.67 cm

(−10.0 cm)= +0.667

Moverall = M1M2 = −2 00.

Again, Moverall < 0 and the final image is inverted .

378 Chapter 36 Solutions

Chapter 37 Solutions

37.1 ∆ybright = λLd =

(632.8 × 10- 9)(5.00)

2.00 × 10- 4 m = 1.58 cm

37.2 y

mbright = λFor m = 1,

λ = =

×( ) ×( )=

− −ydL

3 40 10 5 00 10

3 30

3 4. .

m m

m 515 nm

37.3 Note, with the conditions given, the small angleapproximation does not work well. That is, sin θ, tan θ, and θare significantly different. The approach to be used isoutlined below.

(a) At the m = 2 maximum, tan θ = 400 m1000 m = 0.400

θ = 21.8° so λ = d sin θ

m = (300 m) sin 21.8°

2 = 55.7 m

(b) The next minimum encountered is the m = 2 minimum; and at that point,

d sin θ =

m +

12 λ which becomes d sin θ =

52 λ

or sin θ = 5λ2d =

5(55.7 m)2(300 m) = 0.464 and θ = 27.7°

so y = (1000 m) tan 27.7° = 524 m

Therefore, the car must travel an additional 124 m .

37.4 λ = vf =

354 m/s2000/s = 0.177 m

(a) d sin θ = mλ so (0.300 m) sin θ = 1(0.177 m) and θ = 36.2°

(b) d sin θ = mλ so d sin 36.2° = 1(0.0300 m) and d = 5.08 cm

f = c

λ =

3.00 × 108 m/s5.90 × 10–7 m

= 508 THz

2 Chapter 37 Solutions

37.5 For the tenth minimum, m = 9. Using Equation 37.3, sin θ = λd

9 +

Also, tan θ = yL . For small θ, sin θ ≈ tan θ. Thus,

d = 9.5λsin θ

= 9.5λ L

y = 9.5(589 × 10–9 m)(2.00 m)

7.26 × 10–3 m = 1.54 × 10–3 m = 1.54 mm

Goal Solution

Young's double-slit experiment is performed with 589-nm light and a slit-to-screen distance of 2.00 m.The tenth interference minimum is observed 7.26 mm from the central maximum. Determine thespacing of the slits.

G : For the situation described, the observed interference pattern is very narrow, (the minima are lessthan 1 mm apart when the screen is 2 m away). In fact, the minima and maxima are so closetogether that it would probably be difficult to resolve adjacent maxima, so the pattern might looklike a solid blur to the naked eye. Since the angular spacing of the pattern is inversely proportional tothe slit width, we should expect that for this narrow pattern, the space between the slits will be largerthan the typical fraction of a millimeter, and certainly much greater than the wavelength of the light(d >> λ = 589 nm).

O : Since we are given the location of the tenth minimum for this interference pattern, we should usethe equation for destructive interference from a double slit. The figure for Problem 7 shows thecritical variables for this problem.

A : In the equation dsinθ = m + 12( )λ ,

The first minimum is described by m = 0 and the tenth by m = 9 : sinθ = λ

d9 + 1

2( )Also, tanθ = y L , but for small θ , sinθ ≈ tanθ . Thus,

d =

9.5λsinθ

=9.5λL

d =

9.5(5890· 10- 10 m)(2.00 m)7.26· 10-3 m

= 1.54· 10- 3 m = 1.54 mm = 1.54 mm

L : The spacing between the slits is relatively large, as we expected (about 3 000 times greater than thewavelength of the light). In order to more clearly distinguish between maxima and minima, thepattern could be expanded by increasing the distance to the screen. However, as L is increased, theoverall pattern would be less bright as the light expands over a larger area, so that beyond somedistance, the light would be too dim to see.

Chapter 37 Solutions 3

*37.6 λ = 340 m/s2000 Hz = 0.170 m

Maxima are at d sin θ = mλ:

m = 0 gives θ = 0°

m = 1 gives sin θ = λd =

0.170 m0.350 m θ = 29.1°

m = 2 gives sin θ = 2λd = 0.971 θ = 76.3°

m = 3 gives sin θ = 1.46 No solution.

Minima at d sin θ = m + 12 λ:

m = 0 gives sin θ = λ2d = 0.243 θ = 14.1°

m = 1 gives sin θ = 3λ2d = 0.729 θ = 46.8°

m = 2 gives No solution.

So we have maxima at 0°, 29.1°, and 76.3° and minima at 14.1° and 46.8°.

37.7 (a) For the bright fringe,

ybright = mλ L

d where m = 1

y = (546.1× 10−9 m)(1.20 m)

0.250 × 10−3 m = 2.62 × 10−3 m = 2.62 mm

(b) For the dark bands, ydark = λ L

dm + 1

; m = 0, 1, 2, 3, . . .

y2 − y1 = λ L

d1 + 1

− 0 + 1

= λ Ld

1( ) = (546.1× 10−9 m)(1.20 m)

0.250 × 10−3 m

∆y = 2.62 mm

Figures for Goal Solution

4 Chapter 37 Solutions

Goal Solution A pair of narrow, parallel slits separated by 0.250 mm is illuminated by green light (λ = 546.1 nm). Theinterference pattern is observed on a screen 1.20 m away from the plane of the slits. Calculate the distance(a) from the central maximum to the first bright region on either side of the central maximum and (b)between the first and second dark bands.

G : The spacing between adjacent maxima and minima should be fairly uniform across the pattern aslong as the width of the pattern is much less than the distance to the screen (so that the small angleapproximation is valid). The separation between fringes should be at least a millimeter if the patterncan be easily observed with a naked eye.

O : The bright regions are areas of constructive interference and the dark bands are destructiveinterference, so the corresponding double-slit equations will be used to find the y distances.

It can be confusing to keep track of four different symbols for distances. Three are shown in thedrawing to the right. Note that:

y is the unknown distance from the bright central maximum(m = 0) to another maximum or minimum on either side of the center of theinterference pattern.

λ is the wavelength of the light, determined by the source.

A : (a) For very small θ sin θ ≈ tan θ and tan θ = y / L

and the equation for constructive interference sinθ = mλ d (Eq. 37.2)

becomes ybright ≈ λL d( )m (Eq. 37.5)

Substituting values, ybright = (546 × 10−9 m)(1.20 m)

0.250 × 10−3 m1( ) = 2.62 mm

(b) If you have trouble remembering whether Equation 37.5 or Eq. 37.6 applies to a given situation,you can instead remember that the first bright band is in the center, and dark bands are halfwaybetween bright bands. Thus, Eq. 37.5 describes them all, with m = 0, 1, 2 . . . for bright bands, andwith m = 0.5, 1.5, 2.5 . . . for dark bands. The dark band version of Eq. 37.5 is simply Eq. 37.6:

ydark = λL

dm + 1

2( )

∆ydark = 1 + 1

2( ) λLd

− 0 + 12( ) λL

d= λL

d= 2.62 mm

L : This spacing is large enough for easy resolution of adjacent fringes. The distance between minima isthe same as the distance between maxima. We expected this equality since the angles are small:

θ = 2.62 mm( ) 1.20 m( ) = 0.00218 rad = 0.125°

When the angular spacing exceeds about 3°, then sin θ differs from tan θ when written to threesignificant figures.

Chapter 37 Solutions 5

37.8 Taking m = 0 and y = 0.200 mm in Equation 37.6 gives

dy≈ 2λ

= × ×

×=

− −

−2 0 400 10 10

442 100 362

3 3

9( .

.m)(0.200 m)

L ≈ 36.2 cm

Geometric optics incorrectly predicts bright regions oppositethe slits and darkness in between. But, as this example shows,interference can produce just the opposite.

37.9 Location of A = central maximum,

Location of B = first minimum.

So, ∆y y y

= − = +

− = =[ ] .min max

λ λ0

012

20 0 m

Thus, d

L= ( ) = ( )( ) =λ2 20 0

3 00 15040 0.

.. m

m m m

11.3 m

37.10 At 30.0°, d msin θ λ=

(3.20 × 10−4 m) sin 30.0°= m(500 × 10−9 m) so m = 320

There are 320 maxima to the right, 320 to the left, and one for m = 0 straight ahead.

There are 641 maxima .

37.11 φ = 2π

λdsin θ ≈ 2π

λd

(a) φ π=

×× ( ) =−

− °25 00 10

1 20 10 0 50074

( .( . sin .

m) m) 13.2 rad

(b) φ = 2π

(5.00 × 10−7 m)(1.20 × 10−4 m)

5.00 × 10−3 m1.20 m

= 6.28 rad

6 Chapter 37 Solutions

λ,

θ = sin−1 λ φ

2πd

= sin−1 (5.00 × 10−7 m)(0.333 rad)2π(1.20 × 10−4 m)

θ = 1 27 10 2. × − deg

(d) If dsin θ λ=

θ = sin−1 λ

= sin−1 5 × 10−7 m

4(1.20 × 10−4 m)

θ = 5 97 10 2. × − deg

37.12 The path difference between rays 1 and 2 is: δ = dsin θ1 − dsin θ2

For constructive interference, this path difference must be equal to an integral number ofwavelengths: dsin θ1 − dsin θ2 = mλ , or

d(sin θ1 − sin θ2 ) = mλ

37.13 (a) The path difference δ = dsin θ and when L >> y

δ ≈ = × × = × =

− −−yd

L( . )( .

1 80 10 1 50 101 40

1 93 102 4

6 m m) m

m 1.93 µm

(b)

δλ

= ××

=−

−1 93 106 43 10

3 006

7..

. m m

, or δ λ= 3 00.

37.14 (a)I

Imax = cos2

2 (Equation 37.11)

Therefore, φ = 2 cos–1

Imax

1/2 = 2 cos–1 (0.640)1/2 = 1.29 rad

(b) δ = λφ2π

= (486 nm)(1.29 rad)

2π = 99.8 nm

Chapter 37 Solutions 7

37.15 Iav = Imax cos2

π d sin θ

For small θ, sin θ = yL and Iav = 0.750 Imax

y = λ Lπ d

cos–1

Iav

Imax

1/2

y = (6.00 × 10–7)(1.20 m)

π (2.50 × 10–3 m) cos–1

0.750 Imax

Imax

1/2 = 48.0 µm

37.16 I I

ydL

max cos2 π

IImax

cos. .

. .=

×( ) ×( )×( )( )

=− −

−2

3 4

6 00 10 1 80 10

656 3 10 0 800

π m m

m m 0.987

37.17 (a) From Equation 37.8,

φ = 2πd

λsin θ = 2πd

λ⋅ y

y2 + D2

φ ≈ 2πydλ D

=2π 0.850 × 10−3 m( ) 2.50 × 10−3 m( )

600 × 10−9 m( ) 2.80 m( )= 7.95 rad

(b)

IImax

=cos2 πd

λsin θ

cos2 πdλ

sin θmax

=cos2 φ

2cos2 mπ

IImax

= cos2 φ2

= cos2 7.95 rad2

= 0.453

8 Chapter 37 Solutions

Goal Solution Two narrow parallel slits separated by 0.850 mm are illuminated by 600-nm light, and the viewing screenis 2.80 m away from the slits. (a) What is the phase difference between the two interfering waves on ascreen at a point 2.50 mm from the central bright fringe? (b) What is the ratio of the intensity at this pointto the intensity at the center of a bright fringe?

G : It is difficult to accurately predict the relative intensity at the point of interest without actually doingthe calculation. The waves from each slit could meet in phase ( φ = 0) to produce a bright spot ofconstructive interference, out of phase ( φ = 180°) to produce a dark region of destructive interference,or most likely the phase difference will be somewhere between these extremes, 0 < φ < 180° , so thatthe relative intensity will be 0 < I Imax < 1.

O : The phase angle depends on the path difference of the waves according to Equation 37.8. This phasedifference is used to find the average intensity at the point of interest. Then the relative intensity issimply this intensity divided by the maximum intensity.

A : (a) Using the variables shown in the diagram for problem 7 we have,

φ = 2πdλ

sinθ = 2πdλ

y2 + L2

≅ 2πyd

λL

=2π 0.850· 10- 3 m( ) 0.00250 m( )

600· 10- 9 m( ) 2.80 m( )= 7.95 rad = 2π+ 1.66 rad = 95.5

(b)

IImax

=cos2 πd

λsinθ

cos2 πdλ

sinθmax

=cos2 φ

cos2 mπ( )= cos2 φ

= cos2 95.52Ł ł

= 0.452

L : It appears that at this point, the waves show partial interference so that the combination is about halfthe brightness found at the central maximum. We should remember that the equations used in thissolution do not account for the diffraction caused by the finite width of each slit. This diffractioneffect creates an “envelope” that diminishes in intensity away from the central maximum as shownby the dotted line in Figures 37.13 and P37.60. Therefore, the relative intensity aty = 2.50 mm will actually be slightly less than 0.452.

37.18 (a) The resultant amplitude is

Er = E0 sin ωt + E0 sin ωt + φ( ) + E0 sin ωt + 2φ( ), where φ = 2π

λdsin θ .

Er = E0 sin ωt + sin ωt cos φ + cos ωt sin φ + sin ωt cos 2φ + cos ωt sin 2φ( )

Er = E0(sin ωt) 1 + cos φ + 2 cos2 φ − 1( ) + E0(cos ωt)(sin φ + 2 sin φ cos φ)

Er = E0(1 + 2 cos φ)(sin ωt cos φ + cos ωt sin φ) = + +E t0 (1 2 cos ) sin ( )φ ω φ

Chapter 37 Solutions 9

Then the intensity is I ∝ Er

2 = E02 (1 + 2 cos φ)2 1

where the time average of sin2 (ωt + φ) is 1 2.

From one slit alone we would get intensity Imax ∝ E0

2 12

I = Imax 1 + 2 cos

2πdsin θλ

(b) Look at the N = 3 graph in Figure 37.13. Minimum intensity is zero, attained where

cos 1/2φ = − . One relative maximum occurs at cos 1.00φ = − , where I I= max .

The larger local maximum happens where cos φ = +1.00, giving I = 9.00 I0.

The ratio of intensities at primary versus secondary maxima is 9.00 .

*37.19 (a) We can use sin A + sin B = 2 sin A 2 + B 2( ) cos A 2 − B 2( ) to find the sum of the two sinefunctions to be

E1 + E2 = 24.0 kN C( ) sin 15x − 4.5t + 35.0°( ) cos 35.0°

E1 + E2 = 19.7 kN C( ) sin 15x − 4.5t + 35.0°( )

Thus, the total wave has amplitude 19.7 kN/C and has a constant phase difference of 35.0° from the first wave.

(b) In units of kN/C, the resultant phasor is

ER = E1 + E2 = ( ) + ° + °( )12 0 12 0 70 0 12 0 70 0. . cos . . sin .i i j = +16 1 11 3. .i j

ER = 16.1( )2 + 11.3( )2 at tan−1 11.3 16.1( ) = 19.7 kN/C at 35.0° E1

70.0°

αx

y kx - ωt

+15.5cos80.0°i − 15.5sin80.0° j

+17.0 cos 160°i + 17.0 sin 160° j

E i jR = − + =9 18 1 83. . 9.36 kN/C at 169°

E2ER

ykx - ωt

The wave function of the total wave is EP = 9.36 kN C( ) sin 15x − 4.5t + 169°( )

10 Chapter 37 Solutions

37.20 (a) E i i jR E= + ° + °( )[0 20 0 20 0cos . sin . + i cos 40.0°+j sin 40.0°( )]

E i jR E= +[ ]0 2 71 0 985. . = ° =2 88 0. E at 20.0 2.88E0 at 0.349 rad

EP = 2.88E0 sin ωt + 0.349( ) E1

(b) E i i jR E= + ° + °( )[0 60 0 60 0cos . sin . + i cos120°+ j sin 120°( )]

ER = E0 1.00i + 1.73 j[ ] = 2.00E0 at 60.0°= 2.00E0 at π 3 rad

EP = 2.00E0 sin ωt + π 3( ) E1

E i jR E= +[ ] =0 0 0 0

EP = 0 E1

(d) E i i jR E= + +( )[0 3 2 3 2cos sinπ π + i cos 3π + j sin 3π( )]

E i jR E E= −[ ] = ° =0 00 1 00. at 270 E0 3 at 2 radπ

EP = E0 sin ωt + 3π 2( )

E1y

37.21 E i jR = + = ( ) + ( ) ( )−6 00 8 00 6 00 8 00 8 00 6 002 2 1. . . . tan . . at

ER = 10.0 at 53.1°= 10.0 at 0.927 rad

EP = 10.0 sin 100πt + 0.927( ) 6.00

8.00

π 2α

37.22 If E E t1 10= sin ω and E E t2 20= +( )sin ω φ , then by phasor addition,the amplitude of E is

E E E E0 10 202

202= +( ) + ( )cos sinφ φ = E10

2 + 2E10E20 cos φ + E202

and the phase angle is found from sinθ = E20 sin φ

E10

E20E0

θ φ

Chapter 37 Solutions 11

37.23 E i i jR = + ° + °( )12 0 18 0 60 0 18 0 60 0. . cos . . sin .

E i jR = + = °21 0 15 6 26 2. . . at 36.6

EP = 26 2 36 6. sin .ωt + °( ) 12.0

18.0ER

θ 60.0°

37.24 Constructive interference occurs where m = 0, 1, 2, 3, . . ., for

2πx1

λ− 2π f t + π

− 2πx2

λ− 2π f t + π

= 2πm

2π(x1 − x2 )λ

+ π6

− π8

= 2πm

(x1 − x2 )λ

+ 112

− 116

= m x1 − x2 = m − 1

λ m = 0, 1, 2, 3, . . .

37.25 See the figure to the right:

φ = π/2

37.26 ER2 = E1

2 + E22 − 2E1E2 cos β , where β = 180 − φ.

Since I ∝ E2 ,

IR = I1 + I2 + 2 I1I2 cos φ

37.27 Take φ = 360° N where N defines the

number of coherent sources. Then,

ER = E0 sin ωt + mφ( )

m=1

∑ = 0

In essence, the set of N electric fieldcomponents complete a full circle and returnto zero. 12.0 x

The N = 6 case

φ =

360°

6= 60.0°

60.0°

12 Chapter 37 Solutions

*37.28 Light reflecting from the first surface suffers phase reversal. Light reflecting from the secondsurface does not, but passes twice through the thickness t of the film. So, for constructiveinterference, we require

λ n2 + 2t = λ n

where λ n = λn is the wavelength in the material. Then 2t =

λ n2 =

λ2n

λ = 4nt = 4 × 1.33 × 115 nm = 612 nm

37.29 (a) The light reflected from the top of the oilfilm undergoes phase reversal. Since 1.45 >1.33, the light reflected from the bottomundergoes no reversal. For constructiveinterference of reflected light, we then have

2nt =

m +

12 λ or λ m =

2nt

m +

= 2(1.45)(280 nm)

m +

Substituting for m, we have m = 0: λ 0 = 1620 nm (infrared)

m = 1: λ 1 = 541 nm (green)

m = 2: λ 2 = 325 nm (ultraviolet)

Both infrared and ultraviolet light are invisible to the human eye, so the dominant color i nthe reflected light is green .

(b) The dominant wavelengths in the transmitted light are those that produce destructiveinterference in the reflected light. The condition for destructive interference upon reflectionis

2nt m= λ or λ m

ntm m

= =2 812 nm

Substituting for m gives: m = 1, λ 1 812= nm (near infrared)

m = 2, λ 2 406= nm (violet)

m = 3, λ 3 271= nm (ultraviolet)

Of these. the only wavelength visible to the human eye (and hence the dominate wavelengthobserved in the transmitted light) is 406 nm. Thus, the dominant color in the transmittedlight is violet .

Chapter 37 Solutions 13

37.30 Since 1 < 1.25 < 1.33, light reflected both from the top and from the bottom surface of the oilsuffers phase reversal.

For constructive interference we require 2t = mλ cons

and for destructive interference, 2t = m + 1

2 λ des

Then λ cons

λ dest = 1 +

12m =

640 nm512 nm = 1.25 and m =

Therefore, t = 2(640 nm)

2(1.25) = 512 nm

37.31 Treating the anti-reflectance coating like a camera-lens coating, 2t =

m +

λn

Let m = 0: t = λ4n =

3.00 cm4(1.50) = 0.500 cm

This anti-reflectance coating could be easily countered by changing the wavelength of theradar—to 1.50 cm—now creating maximum reflection!

37.32 2nt = m + 1

λ so

t = m + 1

λ2n

Minimum t = 1

(500 nm)2(1.30)

= 96.2 nm

37.33 Since the light undergoes a 180° phase change at each surface of the film, the condition forconstructive interference is 2nt = mλ , or λ = 2nt m. The film thickness is

t = × = × =− −1 00 10 1 00 10 1005 7. . cm m nm. Therefore, the wavelengths intensified in thereflected light are

λ = 2 1.38( ) 100 nm( )

m= 276 nm

m where m = 1, 2, 3, . . .

or λ 1 276= nm , λ 2 138= nm , . . . . All reflection maxima are in the ultraviolet and beyond.

No visible wavelengths are intensified.

14 Chapter 37 Solutions

*37.34 (a) For maximum transmission, we want destructive interference in the light reflected from thefront and back surfaces of the film.

If the surrounding glass has refractive index greater than 1.378, light reflected from the frontsurface suffers no phase reversal and light reflected from the back does undergo phasereversal. This effect by itself would produce destructive interference, so we want the distancedown and back to be one whole wavelength in the film: 2t = λ n .

n= = ( ) =λ

2656 3. nm2 1.378

238 nm

(b) The filter will expand. As t increases in 2nt = λ , so does λ increase .

or λ = ( ) =1 378 238. nm 328 nm (near ultraviolet) .

37.35 If the path length ∆ = λ , the transmitted light will be bright. Since ∆ = =2d λ ,

dmin

nm2

= = =λ2

580 290 nm

37.36 The condition for bright fringes is

2t + λ

2n= m

λn

m = 1, 2, 3, . . .

From the sketch, observe that

t = R 1 − cos θ( ) ≈ R 1 − 1 + θ2

= R

2rR

= r2

Rθ

The condition for a bright fringe becomes

R= m − 1

λn

Thus, for fixed m and λ , nr2 = constant.

Therefore, nliquidr f2 = nairri

2 and nliquid

cm= ( ) ( )

( )=1 00

1 50

1 31

2..

. 1.31

Chapter 37 Solutions 15

37.37 For destructive interference in the air, 2t = mλ .

For 30 dark fringes, including the one where theplates meet,

t = 29(600 nm)

2 = 8.70 × 10– 6 m

Therefore, the radius of the wire is

r = d2 =

8.70 µm2 = 4.35 µm

Goal Solution An air wedge is formed between two glass plates separated at one edge by a very fine wire as shown i nFigure P37.37. When the wedge is illuminated from above by 600-nm light, 30 dark fringes are observed.Calculate the radius of the wire.

G : The radius of the wire is probably less than 0.1 mm since it is described as a “very fine wire.”

O : Light reflecting from the bottom surface of the top plate undergoes no phase shift, while lightreflecting from the top surface of the bottom plate is shifted by π, and also has to travel an extradistance 2t , where t is the thickness of the air wedge.

For destructive interference, 2t = mλ m = 0, 1, 2, 3, . . .( )

The first dark fringe appears where m = 0 at the line of contact between the plates. The 30th darkfringe gives for the diameter of the wire 2t = 29λ , and t = 14.5λ .

A : r = t

2= 7.25λ = 7.25 600 × 10−9 m( ) = 4.35 µm

L : This wire is not only less than 0.1 mm; it is even thinner than a typical human hair ( ~ 50 µm).

37.38 For destructive interference, 2t = λm

At the position of the maximum thickness of the air film,

m = 2tn

λ= 2(4.00 × 10−5 m)(1.00)

5.461× 10−7 m= 146.5

The greatest integer value is m = 146.

Therefore, including the dark band at zero thickness, there are 147 dark fringes .

16 Chapter 37 Solutions

*37.39 For total darkness, we want destructive interference for reflected light for both 400 nm and 600nm. With phase reversal at just one reflecting surface, the condition for destructiveinterference is

2nairt = mλ m = 0, 1, 2, . . .

The least common multiple of these two wavelengths is 1200 nm, so we get no reflected lightat 2 1 00 3 400 2 600 1200.( ) = ( ) = ( ) =t nm nm nm, so t = 600 nm at this second dark fringe.

By similar triangles,

600 0 0500 nm mm10.0 cmx

= .,

or the distance from the contact point is x = ×( ) ×

=−

−600 10 9 m0.100 m

5.00 10 m5 1.20 mm

37.40 2

2 2 1 80 10550 5 10

9t m mt

= = =( . m)

m=λ

λ⇒ ×

−

−. 654 dark fringes

37.41 When the mirror on one arm is displaced by ∆ l , the path difference increases by 2∆ l . A shiftresulting in the formation of successive dark (or bright) fringes requires a path length changeof one-half wavelength. Therefore, 2∆ l = mλ/2, where in this case, m = 250.

∆l = m

λ4

=250( ) 6.328 × 10−7 m( )

4= 39.6 µm

37.42 Distance = 2(3.82 × 10– 4 m) = 1700λ λ = 4.49 × 10– 7 m = 449 nm

The light is blue

37.43 Counting light going both directions, the number of wavelengths originally in the cylinder is

m1 = 2L

λ. It changes to

m2 = 2L

λ n= 2nL

λ as the cylinder is filled with gas. If N is the number of

bright fringes passing, N = m2 − m1 = 2L

λn − 1( ) , or the index of refraction of the gas is

n = 1 + Nλ

2L= 1 +

35 633 × 10-9 m( )2 0.0300 m( ) = 1.000 369

Chapter 37 Solutions 17

37.44 Counting light going both directions, the number of wavelengths originally in the cylinder is

m1 = 2L

λ. It changes to

m2 = 2L

λ n= 2nL

λ as the cylinder is filled with gas. If N is the number of

bright fringes passing, N = m2 − m1 = 2L

λn − 1( ) , or the index of refraction of the gas is

n = 1 + Nλ

37.45 The wavelength is λ = c

f= 3.00 × 108 m s

60.0 × 106 s−1 = 5.00 m .

Along the line AB the two traveling waves going in opposite directions add to give a standingwave. The two transmitters are exactly 2.00 wavelengths apart and the signal from B, when itarrives at A, will always be in phase with transmitter B. Since B is 180° out of phase with A ,the two signals always interfere destructively at the position of A.

The first antinode (point of constructive interference) is located at distance

λ4

5 00= =. m4

1.25 m from the node at A.

*37.46 My middle finger has width d = 2 cm.

(a) Two adjacent directions of constructive interference for 600-nm light are described by

d sin θ = mλ θ0 = 0 (2 × 10– 2 m) sin θ1 = 1 ( 6 × 10– 7 m)

Thus, θ1 = 2 × 10– 3 degree

and θ1 – θ0 ~ 10– 3 degree

(b) Choose θ1 = 20° 2 × 10– 2 m sin 20° = 1λ λ = 7 mm

Millimeter waves are microwaves

f = c

λ =

3 × 108 m/s7 × 10– 3 m

~ 1011 Hz

37.47 If the center point on the screen is to be a dark spot rather than bright, passage through theplastic must delay the light by one-half wavelength. Calling the thickness of the plastic t.

18 Chapter 37 Solutions

tλ

+ 12

= tλ n( ) = nt

λor t =

λ2( 1)n −

where n is the index of refraction for the

plastic.

Chapter 37 Solutions 19

*37.48 No phase shift upon reflection from the upper surface (glass to air) of the film, but there willbe a shift of λ 2 due to the reflection at the lower surface of the film (air to metal). The totalphase difference in the two reflected beams is then δ λ= +2 2nt . For constructiveinterference, δ λ= m , or 2 1 00 2( . )t m+ =λ λ . Thus, the film thickness for the m th order brightfringe is:

tm = m − 1

λ2

= mλ2

− λ

and the thickness for the m – 1 bright fringe is: tm−1 = (m − 1)

λ2

− λ

Therefore, the change in thickness required to go from one bright fringe to the next is

∆t = tm − tm−1 = λ 2. To go through 200 bright fringes, the change in thickness of the air filmmust be: 200 λ 2( ) = 100λ . Thus, the increase in the length of the rod is

∆L = 100λ = 100 5.00 × 10−7 m( ) = 5.00 × 10−5 m,

From ∆ ∆L L Ti= ( )α , we have: α = ∆L

Li ∆T( ) = 5.00 × 10−5 m0.100 m( ) 25.0°C( ) = 20 0 10 6 1. × °− −C

*37.49 Since 1 < 1.25 < 1.34, light reflected from top and bottom surfaces of the oil undergoes phasereversal. The path difference is then 2t, which must be equal to

m λ n = m λn

for maximum reflection, with m = 1 for the given first-order condition and n = 1.25. So

t = mλ2n =

1(500 nm)2(1.25) = 200 nm

The volume we assume to be constant: 1.00 m3 = (200 nm)A

A = 1.00 m3

200(10– 9 m) = 5.00 × 106 m2 = 5.00 km2

37.50 For destructive interference, the path length must differ by mλ . We may treat this problem asa double slit experiment if we remember the light undergoes a π/2-phase shift at the mirror.The second slit is the mirror image of the source, 1.00 cm below the mirror plane. UsingEquation 37.5,

ydark = mλ L

d = 1(5.00 × 10– 7 m)(100 m)

(2.00 × 10– 2 m) = 2.50 mm

20 Chapter 37 Solutions

37.51 One radio wave reaches the receiver R directly from thedistant source at an angle θ above the horizontal. Theother wave undergoes phase reversal as it reflects fromthe water at P.

Constructive interference first occurs for a path differenceof

d = λ2 (1)

The angles θ in the figure are equal because they each form part of a right triangle with ashared angle at R'.

It is equally far from P to R as from P to R', the mirror image of the telescope.

So the path difference is d = 2(20.0 m) sin θ = (40.0 m) sin θ

The wavelength is λ = cf =

3.00 × 108 m/s60.0 × 106 Hz

= 5.00 m

Substituting for d and λ in Equation (1), (40.0 m) sin θ = 5.00 m

Solving for the angle θ, sin θ = 5.00 m80.0 m and θ = 3.58°

37.52 2 (15.0 km)2 + h2 = 30.175 km

(15.0 km)2 + h2 = 227.63 h = 1.62 km

37.53 From Equation 37.13, I

Imax = cos2

π yd

λL

Let λ 2 equal the wavelength for which I

Imax →

I2Imax

= 0.640

Then λ 2= π yd/L

cos– 1 (I2/Imax)1/2

But π yd

L = λ 1 cos– 1

Imax

1/2 = (600 nm) cos– 1 (0.900) = 271 nm

Substituting this value into the expression for λ 2, λ 2 = 271 nm

cos– 1 (0.6401/2) = 421 nm

Note that in this problem, cos– 1

Imax

1/2 must be expressed in radians.

Chapter 37 Solutions 21

37.54 For Young's experiment, use δ θ λ= =d msin . Then, at the point where the two bright linescoincide,

d m msin θ λ λ= =1 1 2 2 so

λλ

540450

= = =mm

sin θ = 6λ 2

d= 6(450 nm)

0.150 mm= 0.0180

Since sin θ θ≈ and L = 1 40. m, x L= = ( )( ) =θ 0 0180 1 40. . m 2.52 cm

37.55 For dark fringes, 2nt = mλ

and at the edge of the wedge, t = 84(500 nm)

2 .

When submerged in water, 2nt = mλ

m = 2(1.33)(42)(500 nm)

500 nm so m + 1 = 113 dark fringes

*37.56 At entrance, 1.00 sin 30.0° = 1.38 sin θ2 θ2 = 21.2°

Call t the unknown thickness. Then

cos 21.2° = ta a =

tcos 21.2°

tan 21.2° = ct c = t tan 21.2°

sin θ1 = b2c b = 2t tan 21.2° sin 30.0°

The net shift for the second ray, including the phase reversal on reflection of the first, is

2an – b – λ2

where the factor n accounts for the shorter wavelength in the film. For constructiveinterference, we require

2an – b – λ2 = mλ

The minimum thickness will be given by 2an – b – λ2 = 0.

λ2 = 2an – b = 2

ntcos 21.2° – 2t(tan 21.2°) sin 30.0°

590 nm2 =

2 × 1.38

cos 21.2° – 2 tan 21.2° sin 30.0° t = 2.57t t = 115 nm

22 Chapter 37 Solutions

37.57 The shift between the two reflected waves is

δ = 2na − b − λ 2 where a and b are as shown in the raydiagram, n is the index of refraction, and the factor of λ 2 isdue to phase reversal at the top surface. For constructiveinterference, δ λ= m where m has integer values. Thiscondition becomes

φ 1φ 1

aθ 2

t n

2na − b = m + 1

λ (1)

From the figure's geometry, a

t=cos θ2

, c = asin θ2 = tsin θ2

cos θ2,

b c

t= =22

21sin

sincos

sinφ θθ

Also, from Snell’s law, sin sinφ θ1 2= n . Thus, b = 2nt sin2 θ2

cos θ2

With these results, the condition for constructive interference given in Equation (1) becomes:

tcos θ2

− 2nt sin2 θ2

cos θ2= 2nt

cos θ21 − sin2 θ2( ) = m + 1

λ or

2nt cos θ2 = m + 1

37.58 (a) Minimum: 2nt = mλ2 m = 0, 1, 2, . . .

Maximum: 2nt =

m' +

12 λ 1 m' = 0, 1, 2, . . .

for λ 1 > λ 2,

m' +

12 < m so m' = m – 1

Then 2nt = mλ 2 =

m –

12 λ 1

2m λ 2 = 2mλ 1 – λ 1 so m = λ 1

2(λ 1 – λ 2)

(b) m = 500

2(500 – 370) = 1.92 → 2 (wavelengths measured to ± 5 nm)

[Minimum]: 2nt = mλ 2 2(1.40)t = 2(370 nm) t = 264 nm

[Maximum]: 2nt =

m – 1 +

12 λ = 1.5λ 2(1.40)t = (1.5)500 nm t = 268 nm

Film thickness = 266 nm

Chapter 37 Solutions 23

37.59 From the sketch, observe that

x = h2 + d 2( )2 = 4h2 + d2

Including the phase reversal due to reflection fromthe ground, the total shift between the two wavesis δ = 2x − d − λ 2.

d/2

x x h

(a) For constructive interference, the total shift must be an integral number of wavelengths, or δ = mλ where m = 0, 1, 2, 3, . . . .

Thus, 2x − d = m + 1

λ or

λ = 4x − 2d

2m + 1

For the longest wavelength, m = 0, giving λ = − =4 2x d 2 4h2 + d2 − 2d

(b) For destructive interference, δ = m − 1

λ where m = 1, 2, 3, . . . .

Thus, 2x − d = mλ or λ = 2x − d

For the longest wavelength, m = 1 giving λ = − =2x d 4h2 + d2 − d

37.60 Call t the thickness of the sheet. The central maximumcorresponds to zero phase difference. Thus, the addeddistance ∆r traveled by the light from the lower slit mustintroduce a phase difference equal to that introduced bythe plastic film. As light advances through distance t i nair, the number of cycles it goes through is t / λ a .

The number of cycles in the sheet is t

(λa /n) =

λa

Thus, the sheet introduces phase difference φ = 2π

λa –

λa

The corresponding difference in path length is ∆r = φ

λa

2π =

2πλa

(nt – t) λa

2π = (n – 1)t

Note that the wavelength of the light does not appear in this equation. In the figure, the tworays from the slits are essentially parallel, so the angle θ may be expressed as

tanθ = ∆r d = ′y L .

Substituting for ∆r and solving for y' gives

y' = ∆r

d = t(n – 1)L

d = (5.00 × 10– 5 m)(1.50 – 1)(1.00 m)

(3.00 × 10– 4 m) = 0.0833 m = 8.33 cm

24 Chapter 37 Solutions

37.61 Call t the thickness of the film. The central maximumcorresponds to zero phase difference. Thus, the addeddistance ∆r traveled by the light from the lower slit mustintroduce a phase difference equal to that introduced bythe plastic film. The phase difference φ is

φ = 2π

λa (n – 1)

The corresponding difference in path length ∆r is ∆r = φ

λa

2π = 2π

λa (n – 1)

λa

2π = t(n – 1)

Note that the wavelength of the light does not appear in this equation. In the figure, the tworays from the slits are essentially parallel.

Thus the angle θ may be expressed as tan θ = ∆rd =

y'L

Eliminating ∆r by substitution,y'L =

t(n – 1)d gives y' =

t(n – 1)Ld

Goal Solution Consider the double-slit arrangement shown in Figure P37.60, where the slit separation is d and the slit toscreen distance is L. A sheet of transparent plastic having an index of refraction n and thickness t isplaced over the upper slit. As a result, the central maximum of the interference pattern moves upward adistance y'. Find y'.

G : Since the film shifts the pattern upward, we should expect ′y to be proportional to n, t , and L .

O : The film increases the optical path length of the light passing through the upper slit, so the physicaldistance of this path must be shorter for the waves to meet in phase ( φ = 0) to produce the centralmaximum. Thus, the added distance ∆r traveled by the light from the lower slit must introduce aphase difference equal to that introduced by the plastic film.

A : First calculate the additional phase difference due to the plastic. Recall that the relation betweenphase difference and path difference is φ = 2π δ λ. The presence of plastic affects this by changing thewavelength of the light, so that the phase change of the light in air and plastic, as it travels over thethickness t is

φair = 2πt

λ airand

φplastic =

2πtλ air / n

Thus, plastic causes an additional phase change of ∆φ = 2πt

λ airn − 1( )

Next, in order to interfere constructively, we must calculate the additional distance that the lightfrom the bottom slit must travel.

∆r = ∆φ λair

2π= t n − 1( )

In the small angle approximation we can write ∆r = ′y d L , so

¢y =t n - 1( ) L

L : As expected, ′y is proportional to t and L . It increases with increasing n, being proportional to

n − 1( ). It is also inversely proportional to the slit separation d , which makes sense since slits that arecloser together make a wider interference pattern.

Chapter 37 Solutions 25

37.62 λ = = ×

×=c

f3 00 101 50 10

2008

6..

m s Hz

For destructive interference, the path difference is one-half wavelength.

Thus,

λ2

= 100 m = x + x2 + 2.00 × 104 m( )2− 2.00 × 104 m ,

or 2.01× 104 m − x = x2 + 2.00 × 104 m( )2

Squaring and solving, x = 99.8 m

37.63 (a) Constructive interference in the reflected light requires 2t = m + 1

2( )λ . The first bright ring

has m = 0 and the 55th has m = 54, so at the edge of the lens

t =

54.5(650 10 m)2

= 17.7 m9× −

Now from the geometry in Figure 37.18, the distance from the center of curvature down to theflat side of the lens is

R 2 − r 2 = R − t or R2 − r2 = R2 − 2Rt + t2

R =

r 2 + t 2

2t=

(5.00 × 10−2 m)2 + (1.77 × 10−5 m)2

2(1.77 × 10−5 m)= 70.6 m

(b)

1= (n 1)

1 1= 0.520

1 170.6 m2 2f R R

− −

∞

−−

so f = 136 m

37.64 Bright fringes occur when 2t = λ

nm + 1

and dark fringes occur when 2t = λ

The thickness of the film at x is t = h

x .

Therefore, xbright = λ l

2hnm + 1

and

xdark = λ l m

2hn

26 Chapter 37 Solutions

37.65 E E E ER = + +1 2 3 = + +

cos . cos . cosπ π π6

3 0072

6 0043

i + + +

sin . sin . sinπ π π6

3 0072

6 0043

E i jR = − −2 13 7 70. .

ER = −( ) + −( ) −

−

=2 13 7 707 702 13

7 992 2. ...

. at tan at 4.44 rad-1

Thus, EP = 7 99 4 44. sin .ωt +( ) rad

yE1

37.66 For bright rings the gap t between surfaces is given by

2t = m + 1

2( )λ The first bright ring has m = 0 and the hundredth

has m = 99.

So, t = ( ) ×( ) =−1

2 99 5 500 10 24 99. m . mµ .

Call rb the ring radius. From the geometry of the figure at theright,

t = r − r2 − rb

2 − R − R2 − rb2

Since r rb << , we can expand in series: t = r − r 1 − 1

2rb

r 2

− R + R 1 − 1

2rb

= 1

2rb

r− 1

2rb

rb = 2t1/ r − 1/ R

1/2

=2 24.9 × 10−6 m( )

1/ 4.00 m − 1/12.0 m

1/2

= 1.73 cm

*37.67 The shift between the waves reflecting from the top and bottomsurfaces of the film at the point where the film has thickness t is

δ = 2tnfilm + λ 2( ), with the factor of λ 2 being due to a phasereversal at one of the surfaces.

For the dark rings (destructive interference), the total shift

should be δ = m + 1

2( )λ with m = 0, 1, 2, 3, . . . . This requires

that t = mλ 2nfilm .

Rθ

tnfilm

To find t in terms of r and R, R2 = r2 + (R − t)2 so r

2 = 2Rt + t2

Since t is much smaller than R, t2 << 2Rt and

r2 ≈ 2Rt = 2R

mλ2nfilm

Thus, where m is an integer, r ≈ mλR

nfilm

Chapter 37 Solutions 27

37.68 (a) Bright bands are observed when 2nt = m + 1

Hence, the first bright band (m = 0) corresponds to nt = λ 4.

Since

2= , we have

x x

x2 12

13 00=

= ( ) =λ

λ. cm

680 nm420 nm

4.86 cm

(b) t

n11

4420= = =λ nm4(1.33)

78.9 nm t

n22

4680= = =λ nm4(1.33)

128 nm

.tx1

78 93 00

nm. cm

2 63 10 6. × − rad

37.69 2h sin θ = m + 1

λ bright

∆y2L

= 1

2λ so

h = Lλ2∆y

=2.00 m( ) 606 × 10−9 m( )

2 1.2 × 10−3 m( ) = 0.505 mm

37.70 Superposing the two vectors, ER = E1 + E2

ER = E1 + E2 = E0 + E0

3cos φ

+ E0

3sin φ

= E0

2 + 23

E02 cos φ + E0

9cos2 φ + E0

9sin2 φ

ER = 10

9E0

2 + 23

E02 cos φ

Since intensity is proportional to the square of the amplitude,

I = 10

9Imax + 2

3Imax cos φ

Using the trigonometric identity cos cosφ φ= −2

212 , this becomes

I = 10

9Imax + 2

3Imax 2 cos2 φ

2− 1

= 4

9Imax + 4

3Imax cos2 φ

or I = 4

9Imax 1 + 3 cos2 φ

CHAPTER 38

38.1 sin θ = λa =

6.328 × 10– 7

3.00 × 10– 4 = 2.11 × 10– 3

y1.00 m = tan θ ≈ sin θ = θ (for small θ)

2y = 4.22 mm

38.2 The positions of the first-order minima are y L ≈ sin θ = ±λ a . Thus, the spacing betweenthese two minima is ∆y = 2 λ a( )L and the wavelength is

λ = ∆y

= 4.10 × 10−3 m

0.550 × 10−3 m

2.06 m

= 547 nm

38.3yL = sin θ =

mλa ∆y = 3.00 × 10– 3 m ∆m = 3 – 1 = 2 and a =

∆mλ L∆y

a = (2)(690 × 10– 9 m)(0.500 m)

3.00 × 10– 3 m = 2.30 × 10– 4 m

*38.4 For destructive interference,

sin θ = m λa =

λa =

5.00 cm36.0 cm = 0.139 and θ = 7.98°

dL = tan θ gives d = L tan θ = (6.50 m) tan 7.98° = 0.912 m

d = 91.2 cm

Chapter 38 Solutions 407

*38.5 If the speed of sound is 340 m/s,

λ = vf =

340 m/s650 /s = 0.523 m

Diffraction minima occur at angles described by a sin θ = mλ

1.10 m sin θ 1 = 1(0.523 m) θ 1 = 28.4°

1.10 m sin θ 2 = 2(0.523 m) θ 2 = 72.0°

1.10 m sin θ 3 = 3(0.523 m) θ 3 nonexistent

Maxima appear straight ahead at 0° and left and right at an angle given approximately by

(1.10 m) sin θ x = 1.5(0.523 m) θ x ≈ 46°

There is no solution to a sin θ = 2.5λ, so our answer is already complete, with three soundmaxima.

38.6 (a) sin θ λ= =y

Lma

Therefore, for first minimum, m = 1 and

La ym

= =×( ) ×( )( ) ×( ) =

− −

−λ

7 50 10 8 50 10

1 587 5 10

4 4

. m . m

. m 1.09 m

(b) w = 2y1 yields y1 = 0.850 mm

w = 2 0.850 × 10−3 m( ) = 1.70 mm

38.7 sin θ ≈ = × −y

L4.1 10 m

1.20 m

β θλ2

4 00 10

546 1 104 10 10

1 207 86

3= =

×( )×

−

−π πasin . m

. m. m

. m. rad

IImax

sin //

sin ..

= ( )

=β

β2

27 86

7 86

2 2

1 62 10 2. × −

408 Chapter 38 Solutions

38.8 Bright fringes will be located approximately midway between adjacent dark fringes.Therefore, for the second bright fringe, let m = 2.5 and use

sin θ = mλ a ≈ y L .

The wavelength will be λ ≈ ay

mL= (0.800 × 10−3 m)(1.40 × 10−3 m)

2.5(0.800 m)= 5.60 × 10−7 m = 560 nm

38.9 Equation 38.1 states that sin θ = mλ a , where m = ± 1, ± 2, ± 3, . . . .The requirement for m = 1 is from an analysis of the extra path distancetraveled by ray 1 compared to ray 3 in Figure 38.5. This extra distancemust be equal to λ / 2 for destructive interference. When the sourcerays approach the slit at an angle β , there is a distance added to the pathdifference (of ray 1 compared to ray 3) of a / 2( )sinβ Then, fordestructive interference,

sin β +

sin θ =

λ2

so sin θ =

λa

− sin β.

Dividing the slit into 4 parts leads to the 2nd order minimum: sin θ =

2λa

− sin β

Dividing the slit into 6 parts gives the third order minimum: sin θ =

3λa

− sin β

Generalizing, we obtain the condition for the mth order minimum: sin θ =

maλ

− sin β

*38.10 (a) Double-slit interference maxima are at angles given by d msin θ λ= .

For m = 0 , θ0 = 0°

For m = ( ) = ( )1 2 1 0 5015, sin . .80 m mµ θ µ : θ11 0 179= ( ) =−sin . 10.3°

Similarly, for m = 2, 3, 4, 5 and 6, θ2 = 21.0° , θ3 = 32.5° , θ4 = 45.8° ,

θ5 = 63.6° , and θ6 = sin−1 1.07( ) = nonexistent .

Thus, there are 5 + 5 + 1 = 11 directions for interference maxima .

(b) We check for missing orders by looking for single-slit diffraction minima, at a msin θ λ= .

For m = 1, 0 700 1 0 5015. sin . m mµ θ µ( ) = ( ) and θ1 45 8= °. .

Thus, there is no bright fringe at this angle. There are only nine bright fringes , at

θ = ° ± ° ± ° ± ° ± °0 32 5, , . ,10.3 , 21.0 and 63.6 .

Chapter 38 Solutions 409

sin πasin θ λ( )πasin θ λ

At θ = 0°,

sin θθ

→ 1 and

IImax

→ 1.00

At θ = 10.3°,

πasin θλ

=π 0.700 µm( ) sin 10.3°

0.5015 µm= 0.785 rad = 45.0°

IImax

= sin 45.0°0.785

= 0.811

Similarly, at θ = 21.0° ,

πasin θλ

= 1.57 rad = 90.0° and

IImax

= 0.405

At θ = 32.5°,

πasin θλ

= 2.36 rad = 135° and

IImax

= 0.0901

At θ = 63.6° ,

πasin θλ

= 3.93 rad = 225° and

IImax

= 0.0324

38.11 sin θ = λa =

5.00 × 10– 7 m5.00 × 10– 4 = 1.00 × 10– 3 rad

38.12 θ min = yL = 1.22

λD

y = (1.22)(5.00 × 10– 7)(0.0300)

7.00 × 10– 3 = 2.61 µm

y = radius of star-imageL = length of eye

λ = 500 nmD = pupil diameter

θ = half angle

38.13 Following Equation 38.9 for diffraction from a circular opening, the beam spreads into a coneof half-angle

θ min = 1.22 λD = 1.22

(632.8 × 10– 9 m)(0.00500 m) = 1.54 × 10– 4 rad

The radius of the beam ten kilometers away is, from the definition of radian measure,

r beam = θ min (1.00 × 104 m) = 1.544 m

and its diameter is d beam = 2r beam = 3.09 m

410 Chapter 38 Solutions

Goal Solution A helium-neon laser emits light that has a wavelength of 632.8 nm. The circular aperture through whichthe beam emerges has a diameter of 0.500 cm. Estimate the diameter of the beam 10.0 km from the laser.

G : A typical laser pointer makes a spot about 5 cm in diameter at 100 m, so the spot size at 10 km wouldbe about 100 times bigger, or about 5 m across. Assuming that this HeNe laser is similar, we couldexpect a comparable beam diameter.

O : We assume that the light is parallel and not diverging as it passes through and fills the circularaperture. However, as the light passes through the circular aperture, it will spread from diffractionaccording to Equation 38.9.

A : The beam spreads into a cone of half-angle θmin = 1.22

λD

= 1.22632.8 × 10−9 m( )

0.00500 m( ) = 1.54 × 10−4 rad

The radius of the beam ten kilometers away is, from the definition of radian measure,

rbeam = θmin 1.00 × 104 m( ) = 1.54 m

and its diameter is dbeam = 2rbeam = 3.09 m

L : The beam is several meters across as expected, and is about 600 times larger than the laser aperture.Since most HeNe lasers are low power units in the mW range, the beam at this range would be sospread out that it would be too dim to see on a screen.

38.14 θ min = 1.22 λD =

dL 1.22

5.80 × 10– 7 m

4.00 × 10– 3 m =

d1.80 mi

1 mi

1609 m d = 0.512 m

The shortening of the wavelength inside the patriot's eye does not change the answer.

38.15 By Rayleigh's criterion, two dots separated center-to-center by 2.00 mm would overlap when

θmin = d

L= 1.22

λD

Thus,

L = dD1.22λ

=2.00 × 10−3 m( ) 4.00 × 10−3 m( )

1.22 500 × 10−9 nm( ) = 13.1 m

38.16 D = 1.22 λ

θ min =

1.22(5.00 × 10– 7)1.00 × 10– 5 m = 6.10 cm

Chapter 38 Solutions 411

38.17 θmin = 1.22

wavelengthpupil diameter

(distance between sources)

Thus,

1.22λd

= wvt

, or w =

1.22λ vt( )d

Taillights are red. Take λ ≈ 650 nm: w ≈

1.22 650 × 10−9 m( ) 20.0 m s( ) 600 s( )5.00 × 10−3 m

= 1.90 m

38.18 θmin = 1.22

wavelengthpupil diameter

= (distance between sources)L

1.22λd

= wvt

w =

1.22λ vt( )d

where λ ≈ 650 nm is the average wavelength radiated by the red taillights.

38.191.22λ

D = dL λ =

cf = 0.0200 m D = 2.10 m L = 9000 m

d = 1.22 (0.0200 m)(9000 m)

2.10 m = 105 m

38.20 Apply Rayleigh's criterion, θmin = x

D= 1.22

λd

where θmin = half-angle of light cone, x = radius of spot, λ =wavelength of light, d = diameter of telescope, D = distance to Moon.

Then, the diameter of the spot on the Moon is

2x = 2 1.22

λDd

2 1.22( ) 694.3 × 10−9 m( ) 3.84 × 108 m( )2.70 m

= 241 m

38.21 For 0.100° angular resolution, 1.22

3.00 × 10−3 m( )D

= 0.100°( ) π180°

D = 2.10 m

38.22 L = 88.6 × 109 m, D = 0.300 m, λ = 590 × 10−9 m

(a) 1.22

λD

= θmin = 2 40 10 6. × − rad

(b) d = θminL = 213 km

412 Chapter 38 Solutions

38.23 d = 1.00 cm

2000= 1.00 × 10−2 m

2000= 5.00 µm

sin θ = mλ

1 640 × 10−9 m( )5.00 × 10−6 m

= 0.128 θ = 7.35°

38.24 The principal maxima are defined by

d sin θ = mλ m = 0, 1, 2, . . .

For m = 1, λ = d sin θ

where θ is the angle between the central (m = 0) and the first order(m = 1) maxima. The value of θ can be determined from theinformation given about the distance between maxima and thegrating-to-screen distance. From the figure,

tan θ = 0.488 m1.72 m = 0.284 so θ = 15.8° and sin θ = 0.273

The distance between grating "slits" equals the reciprocal of the number of grating lines percentimeter

d = 1

5310 cm– 1 = 1.88 × 10– 4 cm = 1.88 × 103 nm

The wavelength is λ = d sin θ = (1.88 × 103 nm)(0.273) = 514 nm

38.25 The grating spacing is d = (1.00 × 10– 2 m)

4500 = 2.22 × 10– 6 m

In the 1st-order spectrum, diffraction angles are given by

sin θ = λd : sin θ 1 =

656 × 10– 9 m2.22 × 10– 6 m

= 0.295

so that for red θ1 = 17.17°

and for violet sin θ 2 = 434 × 10– 9 m2.22 × 10–6 m

= 0.195

so that θ 2 = 11.26°

Figure for Goal Solution

Chapter 38 Solutions 413

The angular separation is in first-order, ∆θ = 17.17° – 11.26° = 5.91°

In the second-order spectrum, ∆θ = sin– 1

2λ 1

d – sin– 1

2λ 2

d = 13.2°

Again, in the third order, ∆θ = sin– 1

3λ 1

d – sin– 1

3λ 2

d = 26.5°

Since the red line does not appear in the fourth-order spectrum, the answer is complete.

Goal Solution The hydrogen spectrum has a red line at 656 nm and a violet line at 434 nm. What is the angularseparation between two spectral lines obtained with a diffraction grating that has 4500 lines/cm?

G : Most diffraction gratings yield several spectral orders within the 180° viewing range, which meansthat the angle between red and violet lines is probably 10° to 30°.

O : The angular separation is the difference between the angles corresponding to the red and violetwavelengths for each visible spectral order according to the diffraction grating equation, dsinθ = mλ .

A : The grating spacing is d = 1.00 × 10−2 m( ) 4500 lines = 2.22 × 10−6 m

In the first-order spectrum (m = 1), the angles of diffraction are given by sin θ = λ d :

sinθ1r = 656 × 10−9 m

2.22 × 10−6 m= 0.295 so θ1r = 17.17°

sinθ1v = 434 × 10−9 m

2.22 × 10−6 m= 0.195 so θ1v = 11.26°

The angular separation is ∆θ1 = θ1r - θ1v = 17.17 - 11.26 = 5.91

In the 2nd-order ( m = 2) ∆θ2 = sin−1 2λ r

− sin−1 2λ v

= 13.2°

In the third order ( m = 3), ∆θ3 = sin−1 3λ r

− sin−1 3λ v

= 26.5°

In the fourth order, the red line is not visible: θ4r = sin−1 4λ r / d( ) = sin−1 1.18( ) does not exist

L : The full spectrum is visible in the first 3 orders with this diffraction grating, and the fourth ispartially visible. We can also see that the pattern is dispersed more for higher spectral orders sothat the angular separation between the red and blue lines increases as m increases. It is also worthnoting that the spectral orders can overlap (as is the case for the second and third order spectraabove), which makes the pattern look confusing if you do not know what you are looking for.

414 Chapter 38 Solutions

38.26 sin θ = 0.350: d = λ

sin θ =

632.8 nm0.350 = 1.81 × 103 nm

Line spacing = 1.81 µm

*38.27 (a) d = 1

3660 lines/cm = 2.732 × 10– 4 cm = 2.732 × 10– 6 m = 2732 nm

λ = d sin θ

m : At θ = 10.09° λ = 478.7 nm

At θ = 13.71°, λ = 647.6 nm

At θ = 14.77°, λ = 696.6 nm

(b) d = λ

sin θ 1 and λ = d sin θ 2 so sin θ 2 =

2λd =

2λ

sin θ 1

= 2 sin θ 1

Therefore, if θ 1 = 10.09° then sin θ 2 = 2 sin (10.09°) gives θ 2 = 20.51°

Similarly, for θ 1 = 13.71°, θ 2 = 28.30° and for θ 1 = 14.77°, θ 2 = 30.66°

38.28 d = 1

800/mm = 1.25 × 10– 6 m

The blue light goes off at angles sin θm = mλd : θ 1 = sin– 1

1 × 5.00 × 10– 7 m

1.25 × 10– 6 m = 23.6°

θ 2 = sin– 1 (2 × 0.400) = 53.1°

θ 3 = sin– 1 (3 × 0.400) = nonexistent

The red end of the spectrum is at θ 1 = sin– 1

1 × 7.00 × 10– 7 m

1.25 × 10– 6 m = 34.1°

θ 2 = sin– 1 (2 × 0.560) = nonexistent

So only the first-order spectrum is complete, and it does not overlap the second-orderspectrum.

Chapter 38 Solutions 415

38.29 (a) From Equation 38.12, R = Nm where N = 3000 lines cm( ) 4.00 cm( ) = 1.20 × 104 lines.

In the 1st order, R = (1)(1.20 × 104 lines) = 1 20 104. ×

In the 2nd order, R = (2)(1.20 × 104 lines) = 2 40 104. ×

In the 3rd order, R = (3)(1.20 × 104 lines) = 3 60 104. ×

(b) From Equation 38.11, R = λ

∆λ:

In the 3rd order, ∆λ = λ

R= 400 nm

3.60 × 104 = 0.0111 nm = 11.1 pm

38.30 sin θ = mλ

Therefore, taking the ends of the visible spectrum to be λ v = 400 nm and λ r = 750 nm, theends the different order spectra are:

End of second order: sin θ2r =

2λ r

d= 1500 nm

Start of third order: sin θ3v =

2λ v

d= 1200 nm

Thus, it is seen that θ2r > θ3v and these orders must overlap regardless of the value of the

grating spacing d.

38.31 (a) Nm = λ

∆λ N(1) =

531.7 nm0.19 nm = 2800

(b)1.32 × 10– 2 m

2800 = 4.72 µm

38.32 dsin θ = mλ and, differentiating, d(cos θ)dθ = mdλ or d 1 − sin2 θ ∆θ ≈ m∆λ

d 1 − m2λ2 / d2 ∆θ ≈ m∆λ so ∆θ ≈ ∆λ

d2 / m2 − λ2

416 Chapter 38 Solutions

38.33 d = 1.00 × 10−3 m mm

250 lines mm= 4.00 × 10−6 m = 4000 nm

dsin θ = mλ ⇒ m = dsin θ

(a) The number of times a complete order is seen is the same as the number of orders in whichthe long wavelength limit is visible.

mmax = dsin θmax

λ= 4000 nm( )sin 90.0°

700 nm= 5.71 or 5 orders is the maximum .

(b) The highest order in which the violet end of the spectrum can be seen is:

mmax = dsin θmax

λ= 4000 nm( )sin 90.0°

400 nm= 10.0 or 10 orders in the short - wavelength region

38.34 d = = × =−1

4200 cm2.38 10 m nm6 2380

dsin θ = mλ or θ = sin−1 mλ

and

y = L tan θ = L tan sin−1 mλ

Thus, ∆y = L tan sin−1 mλ 2

− tan sin−1 mλ 1

For m = 1, ∆y = 2.00 m( ) tan sin−1 589.6

2380

− tan sin−1 5892380

= 0.554 mm

For m = 2, ∆y = 2.00 m( ) tan sin−1 2 589.6( )

2380

− tan sin−1 2 589( )

2380

= 1.54 mm

For m = 3, ∆y = 2.00 m( ) tan sin−1 3 589.6( )

2380

− tan sin−1 3 589( )

2380

= 5.04 mm

Thus, the observed order must be m = 2 .

38.35 2d sin θ = mλ: λ = 2d sin θ

m = 2(0.353 × 10– 9 m) sin (7.60°)

(1) = 9.34 × 10– 11 m = 0.0934 nm

38.36 2d sin θ = mλ ⇒ d = m λ

2 sin θ =

(1)(0.129 nm)2 sin (8.15°) = 0.455 nm

Chapter 38 Solutions 417

38.37 2d sin θ = mλ so sin θ = mλ2d =

1(0.140 × 10– 9 m)2(0.281 × 10– 9 m)

= 0.249 and θ = 14.4°

38.38 sin θm = mλ2d : sin 12.6° =

1λ2d = 0.218

sin θ 2 = 2λ2d = 2(0.218) so θ 2 = 25.9°

Three other orders appear: θ 3 = sin–1 (3 × 0.218) = 40.9°

θ 4 = sin–1 (4 × 0.218) = 60.8°

θ 5 = sin–1 (5 × 0.218) = nonexistent

38.39 2dsin θ = mλ θ = sin−1 mλ

= sin−1 2 × 0.1662 × 0.314

= 31.9°

*38.40 Figure 38.25 of the text shows the situation. 2dsin θ = mλ or λ = 2dsin θ

m = ⇒ = ( ) °

=12 2 80 80 0

m1λ

. sin . 5.51 m

m = ⇒ = ( ) °

=22 2 80 80 0

m2λ

. sin . 2.76 m

m = ⇒ = ( ) °

=32 2 80 80 0

m3λ

. sin . 1.84 m

*38.41 The average value of the cosine-squared function is

one-half, so the first polarizer transmits 12 the light.

The second transmits cos2 30.0° = 34 .

If = 12 ×

34 Ii =

38 Ii

418 Chapter 38 Solutions

38.42 (a) θ 1 = 20.0°, θ 2 = 40.0°, θ 3 = 60.0°

If = Ii cos2(θ 1 – 0°) cos2(θ 2 – θ 1) cos2(θ 3 – θ 2)

If = (10.0 units) cos2(20.0°) cos2(20.0°) cos2(20.0°) = 6.89 units

(b) θ 1 = 0°, θ 2 = 30.0°, θ 3 = 60.0°

If = (10.0 units) cos2(0°) cos2(30.0°) cos2(30.0°) = 5.63 units

38.43 I = Imax cos2 θ ⇒ θ = cos– 1

Imax

1/2

(a)I

Imax =

13.00 ⇒ θ = cos– 1

3.00 1/2

= 54.7°

(b)I

Imax =

15.00 ⇒ θ = cos– 1

5.00 1/2

= 63.4°

(c)I

Imax =

110.0 ⇒ θ = cos– 1

10.0 1/2

= 71.6°

38.44 By Brewster's law, n p= tan = tan(48.0 ) =θ ° 1.11

38.45 sin θc n

= 1 or

c= =

°=1 1

34 41 77

sin sin ..

Also, tan θp = n. Thus, θp = tan−1 n( ) = tan−1 1.77( ) = 60.5°

38.46 sin θc = 1

nand tan θp = n

Thus, sin θc = 1

tan θpor cot sinθ θp c=

Chapter 38 Solutions 419

38.47 Complete polarization occurs at Brewster's angle tan θp = 1.33 θp = 53.1°

Thus, the Moon is 36.9° above the horizon.

38.48 For incident unpolarized light of intensity Imax :

After transmitting 1st disk: I = 1

Imax

After transmitting 2nd disk: I = 1

Imax cos2 θ

After transmitting 3rd disk: I = 1

Imax cos2 θ

cos2 90°−θ( )

where the angle between the first and second disk is θ = ωt .

Using trigonometric identities cos2 θ = 1

2(1 + cos 2θ) and

cos2 90 − θ( ) = sin2 θ = 1

21 − cos 2θ( )

we have I = 1

2Imax

(1 + cos 2θ)2

(1 − cos 2θ)2

= 18

Imax(1 − cos2 2θ) = 18

Imax12

(1 − cos 4θ)

Since θ = ωt , the intensity of the emerging beam is given by I = 1

16Imax(1 − cos 4ωt)

38.49 Let the first sheet have its axis at angle θ to the original plane of polarization, and let eachfurther sheet have its axis turned by the same angle.

The first sheet passes intensity I max cos2 θ.

The second sheet passes I max cos4 θ,

nd the nth sheet lets through I max cos2n θ ≥ 0.90I max where θ = 45° n

Try different integers to find cos2 × 5

45°

5 = 0.885, cos2 × 6

45°

6 = 0.902,

(a) So n = 6

(b) θ = 7.50°

420 Chapter 38 Solutions

*38.50 Consider vocal sound moving at 340 m s and of frequency 3000 Hz. Its wavelength is

λ = v

f= 340 m s

3000 Hz= 0.113 m

If your mouth, for horizontal dispersion, behaves similarly to a slit 6.00 cm wide, then

a msin θ λ= predicts no diffraction minima. You are a nearly isotropic source of this sound. Itspreads out from you nearly equally in all directions. On the other hand, if you use amegaphone with width 60.0 cm at its wide end, then a msin θ λ= predicts the first diffractionminimum at

θ = sin−1 mλ

= sin−1 0.113 m

0.600 m

= 10.9°

This suggests that the sound is radiated mostly toward the front into a diverging beam ofangular diameter only about 20°. With less sound energy wasted in other directions, more isavailable for your intended auditors. We could check that a distant observer to the side orbehind you receives less sound when a megaphone is used.

38.51 The first minimum is at a sin θ = 1λ.

This has no solution if λa > 1

or if a < λ = 632.8 nm

38.52 x = 1.22

λd

D = 1.225.00 × 10−7 m5.00 × 10−3 m

250 × 103 m( ) = 30.5 m

D = 250 × 103 mλ = 5.00 × 10– 7 md = 5.00 × 10– 3 m

38.53 d = 1

400/mm = 2.50 × 10– 6 m

(a) d sin θ = mλ θa = sin– 1

2 × 541 × 10– 9 m

2.50 × 10– 6 m = 25.6°

(b) λ = 541 × 10– 9 m

1.33 = 4.07 × 10– 7 m

θb = sin– 1

2 × 4.07 × 10– 7 m

2.50 × 10– 6 m = 19.0°

Chapter 38 Solutions 421

*38.54 (a) λ = v

f= 3.00 × 108 m s

1.40 × 109 s−1 = 0.214 m

θmin = 1.22

λD

= 1.220.214 m

3.60 × 104 m

= 7 26. radµ

= 7.26 µ rad

180 × 60 × 60 sπ

= 1.50 arc seconds

(b) θmin = d

d = θminL = 7.26 × 10−6 rad( ) 26 000 ly( ) = 0.189 ly

min . ..

= = ××

−

−1 22 1 22500 1012 0 10

3D m m

50 8. radµ 10.5 seconds of arc( )

(d) d = θminL = 50.8 × 10−6 rad( ) 30.0 m( ) = 1.52 × 10−3 m = 1.52 mm

38.55 θ λ

min2.00 m10.0 m

= = ( )( ) =1 22 1 22. .

D 0.244 rad = 14.0°

38.56 With a grazing angle of 36.0°, the angle of incidence is 54.0°

tan θp = n = tan 54.0° = 1.38

In the liquid, λ n = λ / n = 750 nm /1.38 = 545 nm

38.57 (a) dsin θ = mλ , or d = mλ

sin θ=

3 500 × 10−9 m( )sin 32.0°

= 2.83 µm

Therefore, lines per unit length =

1 md

=× −

12 83 10 6.

or lines per unit length = × =3 53 105. m 3 53 103. × cm .

(b) sin θ = mλ

m 500 × 10−9 m( )2.83 × 10−6 m

= m 0.177( )

For sin .θ ≤ 1 00, we must have m 0.177( ) ≤ 1.00 or m ≤ 5.65

Therefore, the highest order observed is m = 5

Total number primary maxima observed is 2m + 1 = 11

422 Chapter 38 Solutions

Goal Solution Light of wavelength 500 nm is incident normally on a diffraction grating. If the third-order maximum ofthe diffraction pattern is observed at 32.0°, (a) what is the number of rulings per centimeter for thegrating? (b) Determine the total number of primary maxima that can be observed in this situation.

G : The diffraction pattern described in this problem seems to be similar to previous problems that havediffraction gratings with 2 000 to 5 000 lines/mm. With the third-order maximum at 32°, there areprobably 5 or 6 maxima on each side of the central bright fringe, for a total of 11 or 13 primarymaxima.

O : The diffraction grating equation can be used to find the grating spacing and the angles of the othermaxima that should be visible within the 180° viewing range.

A : (a) Use Equation 38.10, d msinθ λ=

d = mλ

sinθ= (3)(5.00 × 10−7 m)

sin(32.0°)= 2.83 × 10−6 m

Thus, the grating gauge is

= 3.534 × 105 lines / m = 3530 lines / cm ◊

(b) sinθ = m

λd

= m(5.00 × 10−7 m)

2.83 × 10-6 m= m(0.177)

For sin θ ≤ 1, we require that m 1.77( ) ≤ 1 or m ≤ 5.65. Since m must be an integer, its maximumvalue is really 5. Therefore, the total number of maxima is 2m + 1 = 11

L : The results agree with our predictions, and apparently there are 5 maxima on either side of thecentral maximum. If more maxima were desired, a grating with fewer lines/cm would be required;however, this would reduce the ability to resolve the difference between lines that appear closetogether.

38.58 For the air-to-water interface,

tan θp = nwater

nair= 1.33

1.00 θp = 53.1°

and 1.00 sin sin 2( ) = ( )θ θp 1 33.

θ2

1 53 11 33

36 9= °

= °−sin

sin ..

θ3θ2

For the water-to-glass interface, tan θp = tan θ3 =

nglass

nwater= 1.50

1.33 so θ3 48 4= °.

The angle between surfaces is θ θ θ= − =3 2 11.5°

Chapter 38 Solutions 423

38.59 The limiting resolution between lines

θ λmin . .

..= =

×( )×( ) = ×

−

−−1 22 1 22 1 34 10 4

550 10 m

5 00 10 m rad

Assuming a picture screen with vertical dimension l , the minimum viewing distance for novisible lines is found from θmin = l 485( ) L . The desired ratio is then

= 1485θmin

= 1

485 1.34 × 10−4 rad( ) = 15.4

38.60 (a) Applying Snell's law gives n2 sin φ = n1 sin θ . From the sketch, wealso see that:

θ + φ+ β = π, or φ = π - (θ + β)

Using the given identity: sin φ = sin π cos(θ + β) − cos π sin(θ + β),

which reduces to: sinφ = sin(θ + β).

Applying the identity again: sin φ = sin θ cos β + cos θ sin β

Snell's law then becomes: n2 sin θ cos β + cos θ sin β( ) = n1 sin θ

or (after dividing by cos θ ): n2(tan θ cos β + sin β) = n1 tan θ .

Solving for tan θ gives: tan θ = n2 sin β

n1 − n2 cos β

(b) If β = 90.0° , n1 = 1.00, and n2 = n , the above result becomes:

tan θ = n(1.00)

1.00 − 0 , or n = tan θ , which is Brewster's law.

38.61 (a) From Equation 38.1, θ = sin−1 mλ

In this case m = 1 and λ = = ×

×= × −c

f3.00 10 m/s7.50 10 Hz

4.00 10 m8

Thus, θ = ×

=−

−

−sin..

24 00 106 00 10

m m

41.8°

(b) From Equation 38.4,

IImax

=sin β 2( )

β 2

where β = 2πasin θ

424 Chapter 38 Solutions

When θ = 15.0°, β = 2π 0.0600 m( )sin 15.0°

0.0400 m= 2.44 rad

and

IImax

= sin 1.22 rad( )1.22 rad

= 0.593

a so θ = 41.8° :

This is the minimum angle subtended by the two sources at the slit. Letα be the half angle between the sources, each a distance l = 0.100 mfrom the center line and a distance L from the slit plane. Then,

L = l cot α = 0.100 m( ) cot 41.8 2( ) = 0.262 m

38.62

IImax

=12

(cos2 45.0°)(cos2 45.0°) = 18

38.63 (a) The E and O rays, in phase at the surface of the plate, will have a phase difference

θ = 2π λ( )δ

after traveling distance d through the plate. Here δ is the difference in the optical path lengthsof these rays. The optical path length between two points is the product of the actual pathlength d and the index of refraction. Therefore,

δ = dnO − dnE

The absolute value is used since nO nE may be more or less than unity. Therefore,

θ = 2π

dnO − dnE =

2πλ

d nO − nE

(b) d = λ θ

2π nO − nE=

550 × 10−9 m( ) π 2( )2π 1.544 − 1.553

= 1.53 × 10−5 m = 15.3 µm

*38.64 For a diffraction grating, the locations of the principal maxima for wavelength λ are given by

sin θ = mλ d ≈ y L . The grating spacing may be expressed as d = a N where a is the width ofthe grating and N is the number of slits. Thus, the screen locations of the maxima become

Chapter 38 Solutions 425

y = NLmλ / a . If two nearly equal wavelengths are present, the difference in the screenlocations of corresponding maxima is

∆y =

NLm ∆λ( )a

For a single slit of width a, the location of the first diffraction minimum is sinθ = λ a ≈ y L ,or y = L / a( )λ . If the two wavelengths are to be just resolved by Rayleigh’s criterion, y = ∆yfrom above. Therefore,

λ =

NLm ∆λ( )a

or the resolving power of the grating is R”

λ∆λ

= Nm .

38.65 The first minimum in the single-slitdiffraction pattern occurs at

sin θ =

λa

»ymin

Thus, the slit width is given by

a =

λLymin

For a minimum located at ymin . .= ±6 36 0 08 mm mm ,

the width is a =

632.8· 10- 9 m( ) 1.00 m( )

6.36· 10- 3 m= 99.5 µm ± 1%

38.66 (a) From Equation 38.4,

IImax

=sin β 2( )

β 2( )

If we define φ ≡ β 2 this becomes

IImax

sin=

φφ

Therefore, when

IImax

=12

we must have

sin φφ

= 12

, or sin φ φ=

426 Chapter 38 Solutions

(b) Let y1 = sin φ and y2 =

φ2

A plot of y y1 2 and in the range 1.00 ≤ φ ≤ π 2 is shown tothe right.

The solution to the transcendental equation is found tobe φ = 1.39 rad .

2πasin θλ

= 2φ

gives sin θ =

φπŁ ł

λa

= 0.443

λa

is small, then θ ≈ 0.443

λa

This gives the half-width, measured away from the maximum at θ = 0. The pattern issymmetric, so the full width is given by

∆θ = 0.443

λa

− −0.443λa

0.886λa

38.67 φ 2 sin φ1 1.19 bigger than φ2 1.29 smaller than φ1.5 1.41 smaller1.4 1.3941.39 1.391 bigger1.395 1.3921.392 1.3917 smaller1.3915 1.39154 bigger1.39152 1.39155 bigger1.3916 1.391568 smaller1.39158 1.3915631.39157 1.3915601.39156 1.3915581.391559 1.39155781.391558 1.39155751.391557 1.39155731.3915574 1.3915574

We get the answer to seven digits after 17 steps. Clever guessing, like using the value of 2 sin φas the next guess for φ, could reduce this to around 13 steps.

Chapter 38 Solutions 427

*38.68 In I = Imax

sin β 2( )β 2( )

ºŒ

ßœ

find

dIdβ

= Imax2sin β 2( )

β 2( )

β 2( )cos β 2( ) 1 2( ) − sin β 2( ) 1 2( )

β 2( )2

and require that it be zero. The possibility sin β 2( ) = 0 locates all of the minima and thecentral maximum, according to

β 2 = 0, π, 2π, . . . ; β =

2πasin θλ

= 0, 2π, 4π, . . . ; asin θ = 0, λ , 2λ , . . . .

The side maxima are found from

β2

cosβ2Ł ł

- sinβ2Ł ł

= 0, or tan

β2Ł ł

=β2

This has solutions

β2

= 4.4934 ,

β2

= 7.7253 , and others, giving

(a) πasin θ = 4.4934λ asin θ = 1.4303λ

(b) πasin θ = 7.7253λ asin θ = 2.4590λ

*38.69 (a) We require θmin = 1.22

λD

= radius of diffraction diskL

= D2L

Then D2 = 2.44λ L

(b) D = ×( )( ) =−2 44 500 10 0 1509. . m m 428 µm

Chapter 39 Solutions

39.1 In the rest frame,

pi = m1v1i + m2v2i = (2000 kg)(20.0 m/s) + (1500 kg)(0 m/s) = 4.00 × 104 kg · m/s

pf = (m1 + m2)vf = (2000 kg + 1500 kg)vf

Since pi = pf, vf = 4.00 × 104 kg · m/s2000 kg + 1500 kg = 11.429 m/s

In the moving frame, these velocities are all reduced by +10.0 m/s.

′v1i = v1i − ′v = 20.0 m/s – (+10.0 m/s) = 10.0 m/s

′v2i = v2i − ′v = 0 m/s – (+10.0 m/s) = –10.0 m/s

′vf = 11.429 m/s – (+10.0 m/s) = 1.429 m/s

Our initial momentum is then

′pi = m1 ′v1i + m2 ′v2i = (2000 kg)(10.0 m/s) + (1500 kg)(–10.0 m/s) = 5000 kg · m/s

and our final momentum is

′pf = (2000 kg + 1500 kg) ′vf = (3500 kg)(1.429 m/s) = 5000 kg · m/s

39.2 (a) v = vT + vB = 60.0 m/s

(b) v = vT – vB = 20.0 m/s

2 = 202 + 402 = 44.7 m/s

39.3 The first observer watches some object accelerate under applied forces. Call the instantaneousvelocity of the object v1. The second observer has constant velocity v21 relative to the first,and measures the object to have velocity v2 = v1 − v21.

The second observer measures an acceleration of a2 = dv2dt =

dv1dt

This is the same as that measured by the first observer. In this nonrelativistic case, theymeasure the same forces as well. Thus, the second observer also confirms that ΣF = ma.

2 Chapter 39 Solutions

39.4 The laboratory observer notes Newton's second law to hold: F1 = ma1

(where the subscript 1 implies the measurement was made in the laboratory frame ofreference). The observer in the accelerating frame measures the acceleration of the mass as a2 = a1 – ′a

(where the subscript 2 implies the measurement was made in the accelerating frame ofreference, and the primed acceleration term is the acceleration of the accelerated frame withrespect to the laboratory frame of reference). If Newton's second law held for the acceleratingframe, that observer would then find valid the relation

F2 = ma2 or F1 = ma2

(since F1 = F2 and the mass is unchanged in each). But, instead, the accelerating frameobserver will find that F2 = ma2 – m ′a which is not Newton's second law.

*39.5 L = Lp 1 − v2 c2 ⇒ v = c 1 − L Lp( )2

Taking L = Lp / 2 where Lp = 1.00 m gives v = c 1 −

Lp 2

= c 1 − 14

= 0.866 c

39.6

∆t =∆tp

1 − v c( )2[ ] 1 2 so

v = c 1 −∆tp

∆t

1 2

For

∆t = 2∆tp ⇒ v = c 1 −∆tp

2∆tp

1/2

= c 1 − 14

1/2

= 0.866 c

*39.7 (a)

γ = 1

1 − v c( )2= 1

1 − 0.500( )2= 2

The time interval between pulses as measured by the Earth observer is

∆t = γ ∆tp = 2

360.0 s75.0

= 0.924 s

Thus, the Earth observer records a pulse rate of

60 00 924.. s min

s= 64.9/min

(b) At a relative speed v = 0.990c , the relativistic factor γ increases to 7 09. and the pulse rate

recorded by the Earth observer decreases to 10 6. min . That is, the life span of the astronaut

(reckoned by the total number of his heartbeats) is much longer as measured by an Earth clockthan by a clock aboard the space vehicle.

Chapter 39 Solutions 3

39.8 The observed length of an object moving at speed v is L = Lp 1 − v2 / c2 with Lp as the properlength. For the two ships, we know L2 = L1, L2 p = 3L1p , and v1 = 0.350c

Thus, L 22 = L1

2 and 9L1p

2 1 − v22

= L1p2 1 − 0.350( )2[ ]

giving 9 − 9

v 22

c2 = 0.878, or v2 = 0 950. c

*39.9 ∆t = γ ∆tp =

∆tp

1 − v2 / c2so

∆tp = 1 − v2 / c2

∆t ≈ 1 − v2

2c2

∆t and

∆t − ∆tp = v2

2c2

∆t

If v = 1000 km h = 1.00 × 106 m

3600 s= 277.8 m s, then

= 9.26 × 10−7

and ∆t − ∆tp( ) = (4.28 × 10−13 )(3600 s) = 1.54 × 10−9 s = 1.54 ns

39.10 γ −1 = 1 − v2

c2 = 1 − 0.950( )2 = 0.312

(a) astronauts' time: ∆tp = γ −1 ∆t = 0.312( ) 4.42 yr( ) = 1.38 yr

(b) astronauts' distance: L = γ −1 ∆Lp = 0.312( ) 4.20 ly( ) = 1.31 ly

39.11 The spaceship appears length-contracted to the Earth observer as given by

L = Lp 1 − v2 c2 or L2 = Lp

2 1 − v2 c2( )Also, the contracted length is related to the time required to pass overhead by:

L = vt or L2 = v2t2 = v2

c2 ct( )2

Equating these two expressions gives Lp

2 − Lp2 v2

c2 = (ct)2 v2

c2 or Lp

2 + (ct)2[ ] v2

c2 = Lp2

Using the given values: Lp = 300 m and t = 7.50 × 10– 7 s

this becomes (1.41 × 105 m2) v2

c 2 = 9.00 × 104 m2

giving v = 0.800 c

4 Chapter 39 Solutions

Goal Solution A spaceship with a proper length of 300 m takes 0.750 µs seconds to pass an Earth observer. Determine itsspeed as measured by the Earth observer.

G : We should first determine if the spaceship is traveling at a relativistic speed: classically,v = (300m)/(0.750 µs) = 4.00 × 108 m/s, which is faster than the speed of light (impossible)! Quiteclearly, the relativistic correction must be used to find the correct speed of the spaceship, which wecan guess will be close to the speed of light.

O : We can use the contracted length equation to find the speed of the spaceship in terms of the properlength and the time. The time of 0.750 µs is the proper time measured by the Earth observer, becauseit is the time interval between two events that she sees as happening at the same point in space. Thetwo events are the passage of the front end of the spaceship over her stopwatch, and the passage of theback end of the ship.

A : L = Lp / γ , with L = v∆t : v∆t = Lp 1 − v2 / c2( )1/2

Squaring both sides, v2∆t2 = Lp

2 1 − v2 / c2( )

v2c2 = Lp

2c2 / ∆t2 − v2Lp2 / ∆t2

Solving for the velocity,

v =c Lp / ∆t

c2 + Lp2 / ∆t2

v =3.00 × 108( ) 300 m( ) 0.750 × 10−6 s( )

3.00 × 108( )2+ 300 m( )2 0.750 × 10−6 s( )2

= 2.40 × 108 m / s

L : The spaceship is traveling at 0.8c. We can also verify that the general equation for the speed reducesto the classical relation v = Lp / ∆t when the time is relatively large.

39.12 The spaceship appears to be of length L to Earth observers,

where L = Lp 1 − v2

1/2

and L = vt

vt = Lp 1 − v2

1/2

so v2t2 = Lp

2 1 − v2

Solving for v, v2 t2 +

Lp2

= Lp

= Lp c2t2 + Lp2

−1/2

Chapter 39 Solutions 5

*39.13 For

= 0.990, γ = 7.09

(a) The muon’s lifetime as measured in the Earth’s rest frame is ∆t = 4.60 km

0.990c

and the lifetime measured in the muon’s rest frame is

∆tp = ∆tγ

= 17.09

4.60 × 103 m

0.990 3.00 × 108 m s( )

= 2.18 µs

(b) L = Lp 1 − v c( )2 =

γ= 4.60 × 103 m

7.09= 649 m

39.14 We find Carpenter's speed:

GMmr2 = mv2

v = GM

(R + h)

1/2

= (6.67 × 10−11)(5.98 × 1024 )(6.37 × 106 + 0.160 × 106 )

1/2

= 7.82 km / s

Then the time period of one orbit, T = 2π(R + h)

v= 2π(6.53 × 106)

7.82 × 103 = 5.25 × 103 s

(a) The time difference for 22 orbits is ∆t − ∆tp = (γ − 1)∆tp = 1 − v2 c2( )−1/2

− 1

22T( )

∆t − ∆tp ≈ 1 + 1

2v2

c2 − 1

22T( ) = 1

27.82 × 103 m / s

3 × 108 m / s

22 5.25 × 103 s( )= 39.2 µs

(b) For one orbit, ∆t − ∆tp = 39.2 µs

22= 1.78 µs . The press report is accurate to one digit .

39.15 For pion to travel 10.0 m in ∆t in our frame, 10.0 m = v ∆t = v(γ ∆tp ) =

v(26.0 × 10−9 s)

1 − v 2 / c2

Solving for the velocity, (3.85 × 108 m / s)2(1 − v2 / c2 ) = v 2

1.48 × 1017 m2 / s2 = v2(1+ 1.64)

v = 2.37 × 108 m / s = 0.789 c

*39.16

γ = 1

1 − v2

= 1.01 so v = 0.140 c

6 Chapter 39 Solutions

*39.17 (a) Since your ship is identical to his, and you are at rest with respect to your own ship, its lengthis 20.0 m .

(b) His ship is in motion relative to you, so you see its length contracted to 19.0 m .

from which

LLp

= 19.0 m20.0 m

= 0.950 = 1 − v2

c2 and v = 0.312 c

*39.18 (a)

∆t = γ ∆tp =∆tp

1 − v c( )2= 15.0 yr

1 − 0.700( )2= 21.0 yr

(b) d = v ∆t( ) = 0.700c[ ] 21.0 yr( ) = 0.700( ) 1.00 ly yr( )[ ] 21.0 yr( ) = 14.7 ly

d = v ∆tp( ) = 0.700c[ ] 15.0 yr( ) = 0.700( ) 1.00 ly yr( )[ ] 15.0 yr( ) = 10.5 ly

(d) Mission control gets signals for 21.0 yr while the battery is operating, and then for 14.7 yearsafter the battery stops powering the transmitter, 14.7 ly away: 21.0 yr + 14.7 yr = 35.7 yr

*39.19 The orbital speed of the Earth is as described by ΣF = ma:

GmSmE

r2 = mEv2

v = GmS

6.67 × 10−11 N ⋅ m2 kg2( ) 1.99 × 1030 kg( )1.496 × 1011 m

= 2.98 × 104 m s

The maximum frequency received by the extraterrestrials is

fobs = fsource1 + v c1 − v c

= 57.0 × 106 Hz( ) 1 + 2.98 × 104 m s( ) 3.00 × 108 m s( )1 − 2.98 × 104 m s( ) 3.00 × 108 m s( ) = 57.005 66 × 106 Hz

The minimum frequency received is

fobs = fsource1 − v c1 + v c

= 57.0 × 106 Hz( ) 1 − 2.98 × 104 m s( ) 3.00 × 108 m s( )1 + 2.98 × 104 m s( ) 3.00 × 108 m s( ) = 56.994 34 × 106 Hz

The difference, which lets them figure out the speed of our planet, is

57 005 66 56 994 34 106. .−( ) × = Hz 1 13 104. × Hz

Chapter 39 Solutions 7

39.20 (a) Let fc be the frequency as seen by the car. Thus, fc = fsource

c + vc − v

and, if f is the frequency of the reflected wave, f = fc

c + vc − v

Combining gives f = fsource

(c + v)(c − v)

(b) Using the above result, f c − v( ) = fsource(c + v)

which gives ( f − fsource )c = ( f + fsource )v ≈ 2 fsourcev

The beat frequency is then fb = f − fsource = 2 fsourcev

2vλ

3.00 × 108 m s= (2)(30.0 m s)

(0.0300 m)= 2000 Hz = 2.00 kHz

λ = c

fsource= 3.00 × 108 m s

10.0 × 109 Hz= 3.00 cm

(d) v = fb λ

2so

∆v = ∆ fb λ

2= 5 Hz( )(0.0300 m)

2= 0.0750 m s ≈ 0.2 mi / h

39.21 (a) When the source moves away from an observer, the observed frequency is

fobs = fsource

c − vs

c + vs

1/2

where vs = vsource

When vs << c , the binomial expansion gives

c − vs

c + vs

1 2

= 1 − vs

1 2

1 + vs

−1 2

≈ 1 − vs

1 − vs

≈ 1 − vs

So, fobs ≈ fsource 1 − vs

The observed wavelength is found from c = λ obs fobs = λ fsource :

λ obs = λ fsource

fobs≈ λ fsource

fsource 1 − vs c( ) = λ1 − vs c

∆λ = λ obs − λ = λ 1

1 − vs c− 1

= λ 11 − vs c

− 1

= λ vs c1 − vs c

Since 1 − vs c ≈ 1,

∆λλ

≈ vsource

(b) vsource = c

∆λλ

= c20.0 nm397 nm

= 0.050 4 c

8 Chapter 39 Solutions

39.22

′ux = ux − v1 − uxv / c2 = 0.950c − 0.750c

1 − 0.950 × 0.750= 0.696 c

39.23

′ux = ux − v1 − uxv c2 = −0.750c − 0.750c

1 − (−0.750)(0.750)= – 0.960 c

*39.24 γ = 10.0 We are also given: L1 = 2.00 m, and θ1 = 30.0° (bothmeasured in a reference frame moving relative to the rod).

Thus, L1x = L1 cos θ1 = (2.00 m)(0.867) = 1.73 m

and L1y = L1 sin θ1 = (2.00 m)(0.500) = 1.00 m

L2x = a "proper length" is related to L1x

by L1x = L2x γ

Therefore, L2x = 10.0L1x = 17.3 m and L2y = L1y = 1.00 m

(Lengths perpendicular to the motion are unchanged).

(a) L2 = (L2x )2 + (L2y )2 gives L2 = 17.4 m

(b) θ2 = tan−1 L2y

L2x gives θ 2 = 3.30°

39.25 ux = Enterprise velocity

v = Klingon velocity

From Equation 39.16,

′ux = ux − v

1 − ux vc2

= 0.900c − 0.800c1 − 0.900( ) 0.800( ) = 0.357c

Chapter 39 Solutions 9

*39.26 (a) From Equation 39.13, ∆ ′x = γ ∆x − v ∆t( ) ,

0 = γ 2.00 m − v 8.00 × 10−9 s( )[ ]

v = 2.00 m

8.00 × 10−9 s= 2.50 × 108 m s

γ = 1

1 − 2.50 × 108 m s( )23.00 × 108 m s( )2

= 1.81

(b) From Equation 39.11,

′x = γ x − vt( ) = 1.81 3.00 m − 2.50 × 108 m s( ) 1.00 × 10−9 s( )[ ] = 4.97 m

(c)

′t = γ t − vc2 x

= 1.81 1.00 × 10−9 s −

2.50 × 108 m s( )3.00 × 108 m s( )2 3.00 m( )

′t = −1.33 × 10−8 s

39.27 p = γmu

(a) For an electron moving at 0.0100 c,

γ = 1

1 − u c( )2= 1

1 − (0.0100)2= 1.00005 ≈ 1.00

Thus, p = 1.00 9.11× 10−31 kg( ) 0.0100( ) 3.00 × 108 m / s( ) = 2.73 × 10−24 kg ⋅ m s

(b) Following the same steps as used in part (a), we find at 0.500 c

γ = 1.15 and p = 1.58 × 10−22 kg ⋅ m s

*39.28 Using the relativistic form,

p = mu

1 − u c( )2= γmu ,

we find the difference ∆p from the classical momentum, mu : ∆ p = γmu − mu = (γ − 1)mu

(a) The difference is 1.00% when (γ − 1)mu = 0.0100γmu:

γ = 10.990

= 1

1 − u c( )2 ⇒ 1 − u c( )2 = 0.990( )2 or u = 0.141 c

(b) The difference is 10.0% when (γ − 1)mu = 0.100γmu :

γ = 10.900

= 1

1 − u c( )2 ⇒ 1 − u c( )2 = 0.900( )2 or u = 0.436 c

10 Chapter 39 Solutions

*39.29

p − mumu

= γmu − mumu

= γ − 1

γ − 1 = 1

1 − u c( )2− 1 ≈ 1 + 1

2uc

− 1 = 12

p − mumu

= 12

90.0 m s3.00 × 108 m s

= 4.50 × 10−14

39.30

p = mu

1 − u c( )2 becomes

1 − u2

c2 = m2u2

which gives: 1 = u2 m2

p2 + 1c2

or c2 = u2 m2c2

p2 + 1

and

u = c

m2c2

p2 + 1

*39.31 Relativistic momentum must be conserved:

For total momentum to be zero after as it was before, we must have, with subscript 2 referringto the heavier fragment, and subscript 1 to the lighter, p2 = p1

or γ2m2u2 = γ1m1u1 = 2.50 × 10−28 kg

1 − (0.893)2× (0.893c)

(1.67 × 10−27 kg)u2

1 − u2 c( )2= (4.960 × 10−28 kg)c

and u2 = 0.285 c

Chapter 39 Solutions 11

Goal Solution An unstable particle at rest breaks into two fragments of unequal mass. The rest mass of the lighterfragment is 2.50 × 10−28 kg, and that of the heavier fragment is 1.67 × 10−27 kg. If the lighter fragment hasa speed of 0.893c after the breakup, what is the speed of the heavier fragment?

G : The heavier fragment should have a speed less than that of the lighter piece since the momentum ofthe system must be conserved. However, due to the relativistic factor, the ratio of the speeds will notequal the simple ratio of the particle masses, which would give a speed of 0.134c for the heavierparticle.

O : Relativistic momentum of the system must be conserved. For the total momentum to be zero afterthe fission, as it was before, p1 + p2 = 0, where we will refer to the lighter particle with the subscript'1', and to the heavier particle with the subscript '2.'

A : γ2m2v2 + γ1m1v1 = 0 so γ2m2v2 + 2.50 × 10−28 kg

1 - 0.8932

0.893c( ) = 0

Rearranging,

1.67 × 10−27 kg

1 − v22 c2

c= −4.96 × 10−28 kg

Squaring both sides, 2.79 × 10−54( ) v2

= 2.46 × 10−55( ) 1 − v22

and v2 = −0.285c

We choose the negative sign only to mean that the two particles must move in opposite directions.The speed, then, is v2 = 0.285c

L : The speed of the heavier particle is less than the lighter particle, as expected. We can also see that forthis situation, the relativistic speed of the heavier particle is about twice as great as was predicted by asimple non-relativistic calculation.

39.32 ∆E = (γ1 − γ2 )mc2 . For an electron, mc2 = 0.511 MeV.

(a) ∆E = 1

(1 − 0.810)− 1

(1 − 0.250)

mc2 = 0.582 MeV

(b) ∆E = 1

1 − (0.990)2 − 11 − 0.810

mc2 = 2.45 MeV

39.33 E = γmc2 = 2mc2, or γ = 2

Thus,

= 1 − 1 γ( )2 = 32

, or u = c 3

The momentum is then p = γmu = 2m

c 32

= mc2

3 = 938.3 MeV

3 =

1.63 × 103

MeVc

12 Chapter 39 Solutions

*39.34 The relativistic kinetic energy of an object of mass m and speed u is Kr = 1

1 − u2 / c2− 1

mc2

For u = 0.100 c, Kr = 1

1 − 0.0100− 1

mc2 = 0.005038mc2

The classical equation Kc = 12 mu2 gives Kc = 1

2 m(0.100c)2 = 0.005000mc2

different by 0.005038 – 0.005000

0.005038 = 0.751%

For still smaller speeds the agreement will be still better.

39.35 (a) ER = mc2 = (1.67 × 10−27 kg)(2.998 × 108 m / s)2 = 1.50 × 10−10 J = 938 MeV

(b) E = γmc2 = 1.50 × 10−10 J

[1 − (0.95c / c)2]1/2 = 4.81× 10−10 J = 3.00 × 103 MeV

*39.36 (a) KE = E – ER = 5ER

E = 6ER = 6(9.11× 10−31 kg)(3.00 × 108 m / s)2 = 4.92 × 10−13 J = 3.07 MeV

(b) E = γ mc2 = γ ER

Thus, γ = E

Er= 6 = 1

1 − u2 c2 which yields u = 0.986 c

39.37 The relativistic density is

c2 V= mc2

c2 V= m

V= m

Lp( ) Lp( ) Lp 1 − u c( )2

= 8.00 g

1.00 cm( )3 1 − 0.900( )2= 18.4 g/cm3

Chapter 39 Solutions 13

*39.38 We must conserve both mass-energy and relativistic momentum. With subscript 1 referringto the 0.868c particle and subscript 2 to the 0.987c particle,

γ1 = 1

1 − 0.868( )2= 2.01 and

γ2 = 1

1 − 0.987( )2= 6.22

Conservation of mass-energy gives E1 + E2 = Etotal which is γ1m1c2 + γ2m2c2 = mtotalc

or 2.01m1 + 6.22m2 = 3.34 × 10−27 kg

This reduces to: m1 + 3.09m2 = 1.66 × 10−27 kg [1]

Since the momentum after must equal zero, p1 = p2 gives γ1m1u1 = γ2m2u2

or (2.01)(0.868c)m1 = (6.22)(0.987 c)m2

which becomes m1 = 3.52m2 [2]

Solving [1] and [2] simultaneously, m1 = 8.84 × 10−28 kg and m2 = 2.51× 10−28 kg

39.39 E = γ mc2, p = γ mu; E2 = (γ mc2)2; p2 = (γ mu )2;

E2 − p2c2 = (γmc2 )2 − (γmu)2 c2 = γ 2 (mc2 )2 − (mc)2 u2

= (mc2 )2 1 − u2

1 − u2

−1

= (mc2 )2 Q.E.D.

39.40 (a) K = 50.0 GeV

mc2 = 1.67 × 10−27 kg( ) 2.998 × 108 m s( )2 1

1.60 × 10−10 J Ge V

= 0.938 GeV

E = K + mc2 = 50.0 GeV + 0.938 GeV = 50.938 GeV

E2 = p2c2 + mc2( )2

⇒ p =E2 − mc2( )2

c2 = 50.938 GeV( )2 − 0.938 GeV( )2

p = 50.9

GeVc

= 50.9 GeV3.00 × 108 m s

1.60 × 10−10 J

1 GeV

= 2.72 × 10−17 kg ⋅ m s

(b)

E = γmc2 = mc2

1 − u c( )2 ⇒ u = c 1 − mc2 E( )2

v = 3.00 × 108 m s( ) 1 − 0.938 GeV

50.938 GeV

= 2 9995 108. × m s

39.41 (a) q ∆V( ) = K = γ − 1( )mec2

14 Chapter 39 Solutions

Thus,

γ = 1

1 − u c( )2= 1 + q ∆V( )

mec2 from which u = 0.302 c

(b) K = γ − 1( )mec

2 = q ∆V( ) = 1.60 × 10−19 C( ) 2.50 × 104 J C( ) = 4.00 × 10−15 J

39.42 (a) E = γmc2 = 20.0 GeV with mc2 = 0.511 MeV for electrons. Thus, γ = 20.0 × 109 eV

0.511× 106 eV= 3.91× 104

(b)

γ = 1

1 − u c( )2= 3.91× 104 from which u =0.999 999 999 7 c

γ= 3.00 × 103 m

3.91× 104 = 7.67 × 10−2 m = 7.67 cm

39.43 Conserving total momentum, pBefore decay = pafter decay = 0 : pν = pµ = γmµu = γ 206me( )u

Conservation of mass-energy gives: Eµ + Eν = Eπ

γmµc2 + pνc = mπc2

γ 206me( ) + pν

c= 270me

Substituting from the momentum equation above, γ 206me( ) + γ 206me( ) u

c= 270me

or γ 1 + u

= 270

206= 1.31 ⇒

= 0.264

Then, Kµ = γ − 1( )mµc2 = γ − 1( )206 mec

2( )

= 1

1 − 0.264( )2− 1

206 0.511 MeV( ) = 3.88 MeV

Also, Eν = Eπ − Eµ = mπc2 − γmµc2 = 270 − 206γ( )mec2

Eν = 270 − 206

1 − 0.264( )2

0.511 MeV( ) = 28.8 MeV

*39.44 Let a 0.3-kg flag be run up a flagpole 7 m high.

We put into it energy mgh = 0.3 kg(9.8 m/s2) 7 m ≈ 20 J

Chapter 39 Solutions 15

So we put into it extra mass ∆m = Ec 2

= 20 J

(3 × 108 m/s)2 = 2 × 10–16 kg

for a fractional increase of2 × 1016 kg

0.3 kg ~10–15

*39.45 E = 2.86 × 105 J. Also, the mass-energy relation says that E = mc2.

Therefore, m = E

c2 = 2.86 × 105 J

(3.00 × 108 m / s)2 = 3.18 × 10–12 kg

No, a mass loss of this magnitude (out of a total of 9.00 g) could not be detected .

39.46 (a) K = (γ − 1)mc2 = 1

1 − u2 / c2− 1

mc2 = 0.25mc2 = 2.25 × 1022 J

(b) E = mfuel c2 so mfuel = 2.25 × 1022

9.00 × 1016 = 2.50 × 105 kg

39.47

∆m = Ec2 = P t

c2 =0.800 1.00 × 109 J s( ) 3.00 yr( ) 3.16 × 107 s yr( )

3.00 × 108 m s( )2 = 0.842 kg

39.48 Since the total momentum is zero before decay, it is necessary that after the decay

pnucleus = pphoton =

Eγ

c= 14.0 keV

Also, for the recoiling nucleus, E2 = p2c2 + mc2( )2

with mc2 = 8.60 × 10−9 J = 53.8 GeV

Thus, mc2 + K( )2

= 14.0 keV( )2 + mc2( )2 or

1 + K

mc2

= 14.0 keVmc2

+ 1

So 1 + K

mc2 = 1 + 14.0 keVmc2

≈ 1 + 12

14.0 keVmc2

Binomial Theorem( )

and

K ≈ 14.0 keV( )2

2mc2 =14.0 × 103 eV( )2

2 53.8 × 109 eV( ) = 1 82 10 3. × − eV

39.49 P = dE

dt=

d mc2( )dt

= c2 dmdt

= 3.77 × 1026 W

16 Chapter 39 Solutions

Thus,

dmdt

= 3.77 × 1026 J s

3.00 × 108 m s( )2 = 4 19 109. × kg s

39.50 2mec2 = 1.02 MeV : Eγ ≥ 1.02 MeV

39.51 The moving observer sees the charge as stationary, so she says it feels no magnetic force.

q(E + v × B) = q( ′E + 0) and ′E = E + v × B

*39.52 (a) When Ke = Kp , mec

2 γe − 1( ) = mpc2 γp − 1( )

In this case, mec2 = 0.511 MeV, mpc2 = 938 MeV and

γe = 1 − 0.750( )2[ ]−1/2

= 1.5119

Substituting, γp = 1 + mec

2(γe − 1)mpc2 = 1 + 0.511 MeV( ) 1.5119 − 1( )

938 MeV= 1.000279

but

γp = 1

1 − up c

1/2 . Therefore, up = c 1 − γp

−2 = 0.0236 c

(b) When pe = pp , γpmpup = γemeue or γpup = γemeue

mp.

Thus, γpup =

1.5119( ) 0.511 MeV c2( ) 0.750c( )938 MeV c2 = 6.1772 × 10−4 c

and

c= 6.1772 × 10−4 1 −

which yields up = 6.18 × 10−4 c = 185 km s

39.53 (a) 1013 MeV = (γ – 1)mpc 2 so γ ≈ 1010 vp ≈ c

′t = t

γ= 105 yr

1010 = 10−5 yr ~ 10 2 s

(b) ′d = c ′t ~1011 m

Chapter 39 Solutions 17

Goal Solution The cosmic rays of highest energy are protons, which have kinetic energy on the order of 1013 MeV.(a) How long would it take a proton of this energy to travel across the Milky Way galaxy, having adiameter on the order of ~105 light-years, as measured in the proton's frame? (b) From the point of viewof the proton, how many kilometers across is the galaxy?

G : We can guess that the energetic cosmic rays will be traveling close to the speed of light, so the time ittakes a proton to traverse the Milky Way will be much less in the proton’s frame than 105 years. Thegalaxy will also appear smaller to the high-speed protons than the galaxy’s proper diameter of 105

light-years.

O : The kinetic energy of the protons can be used to determine the relativistic γ-factor, which can then beapplied to the time dilation and length contraction equations to find the time and distance in theproton’s frame of reference.

A : The relativistic kinetic energy of a proton is K = γ − 1( )mc2 = 1013 MeV

Its rest energy is mc2 = 1.67 × 10−27 kg( ) 2.998 × 108

2 1 eV1.60 × 10−19 kg ⋅ m2/ s2

= 938 MeV

So 1013 MeV = γ − 1( ) 938 MeV( ) , and therefore γ = 1.07 × 1010

The proton's speed in the galaxy’s reference frame can be found from γ = 1 1 − v2 / c2 :

1 − v2 c2 = 8.80 × 10−21 and v = c 1 − 8.80 × 10−21 = 1 − 4.40 × 10−21( )c ≈ 3.00 × 108 m / s

The proton’s speed is nearly as large as the speed of light. In the galaxy frame, the traversal time is

∆t = x / v = 105 light - years / c = 105 years

(a) This is dilated from the proper time measured in the proton's frame. The proper time is foundfrom ∆t = γ∆tp :

∆tp = ∆t / γ = 105 yr 1.07 × 1010 = 9.38 × 10−6 years = 296 s ~ a few hundred seconds

(b) The proton sees the galaxy moving by at a speed nearly equal to c, passing in 296 s:

∆Lp = v∆tp = 3.00 × 108( ) 296 s( ) = 8.88 × 107 km ~ 108 km

∆Lp = 8.88 × 1010 m( ) 9.46 × 1015 m / ly( ) = 9.39 × 10−6 ly ~ 10-5 ly

L : The results agree with our predictions, although we may not have guessed that the protons would betraveling so close to the speed of light! The calculated results should be rounded to zero significantfigures since we were given order of magnitude data. We should also note that the relative speed ofmotion v and the value of γ are the same in both the proton and galaxy reference frames.

18 Chapter 39 Solutions

39.54 Take the primed frame as:

(a) The mother ship: ux = ′u ′x + v

1 + ′u ′x v c2 = v + v1 + v2 c2 = 2v

1 + v2 c2 = 2(0.500c)

1 + (0.500)2 = 0.800 c

(b) The shuttle:

ux =v + 2v

1 + v2 / c2

1 + vc2

2v1 + v2 / c2

= 3v + v3 / c2

1 + 3v2 / c2 = 3(0.500c) + (0.500c)3 c2

1 + 3(0.500)2 = 0.929 c

39.55

∆ mc2

mc2 = 4 938.78 MeV( ) − 3728.4 MeV4 938.78 MeV( ) × 100% = 0.712%

39.56 dearth = vtearth = vγ tastro so 2.00 × 106 yr ⋅ c = v

1 − v2 / c230.0 yr

1 − v2 / c2 = v / c( )(1.50 × 10−5) 1 − v2

c2 = v2

c2 (2.25 × 10−10 )

1 = v2

c2 (1 + 2.25 × 10−10 ) so

= 1 + 2.25 × 10−10( )−1/2= 1 − 1

2 (2.25 × 10−10 )

= 1 − 1.12 × 10−10

*39.57 (a) Take the spaceship as the primed frame, moving toward the right at v c= +0 600. . Then

′ = +u cx 0 800. , and

ux = ′ux + v

1 + ′ux v( ) c2 = 0.800c +0.600c1 + 0.800( ) 0.600( ) = 0.946 c

(b) L =

γ= 0.200 ly( ) 1 − 0.600( )2 = 0.160 ly

(c) The aliens observe the 0.160-ly distance closing because the probe nibbles into it from one endat 0.800c and the Earth reduces it at the other end at 0.600c . Thus,

time

ly=+

=0 1600 800 0 600

.. .c c

0.114 yr

(d)

K = 1

1 − u2 c2− 1

mc2

= 1

1 − 0.946( )2− 1

4.00 × 105 kg( ) 3.00 × 108 m s( )2= 7 50 1022. × J

Chapter 39 Solutions 19

39.58 In this case, the proper time is T0 (the time measured by the students on a clock at rest relativeto them). The dilated time measured by the professor is: ∆t = γ T0

where ∆t = T + t. Here T is the time she waits before sending a signal and t is the timerequired for the signal to reach the students.

Thus, we have: T + t = γ T0 (1)

To determine the travel time t, realize that the distance the students will have moved beyondthe professor before the signal reaches them is: d = v(T + t)

The time required for the signal to travel this distance is: t = d

c= v

T + t( )

Solving for t gives: t =

v c( )T1 − v c( )

Substituting this into equation (1) yields: T +

v c( )T1 − v c( ) = γT0

or T = 1 − v c( )−1 = γT0

Then

T = T01 − v c( )

1 − v2 c2( ) = T0

1 − v / c( )1 + v / c( )[ ] 1 − v / c( )[ ]

= T0

1 − v c( )1 + v c( )

39.59 Look at the situation from the instructor's viewpoint since they are at rest relative to theclock, and hence measure the proper time. The Earth moves with velocity v = – 0.280 crelative to the instructors while the students move with a velocity ′u = – 0.600 c relative toEarth. Using the velocity addition equation, the velocity of the students relative to theinstructors (and hence the clock) is:

u = v + ′u

1 + v ′u c2 = (−0.280c) − (0.600c)1 + (−0.280c)(−0.600c) c2 = −0.753c (students relative to clock)

(a) With a proper time interval of ∆ tp = 50.0 min, the time interval measured by the students is:

∆t = γ∆tp with

γ = 1

1 − 0.753c( )2 / c2= 1.52

Thus, the students measure the exam to last T = 1.52(50.0 min) = 76.0 minutes

(b) The duration of the exam as measured by observers on Earth is:

∆t = γ∆tp with

γ = 1

1 − 0.280c( )2 c2so T = 1.04(50.0 min) = 52.1 minutes

20 Chapter 39 Solutions

*39.60 The energy which arrives in one year is E = P t = 1.79 × 1017 J / s( ) 3.16 × 107 s( ) = 5.66 × 1024 J

Thus,

m = Ec2 = 5.66 × 1024 J

3.00 × 108 m / s( )2 = 6.28 × 107 kg

*39.61 The observer sees the proper length of the tunnel, 50.0 m, but sees the train contracted tolength

L = Lp 1 − v2 c2 = 100 m 1 − (0.950)2 = 31.2 m

shorter than the tunnel by 50.0 – 31.2 = 18.8 m so it is completely within the tunnel.

*39.62 If the energy required to remove a mass m from the surface is equal to its mass energy mc2,then

GMsmRg

= mc2

and Rg = GMs

c2 = (6.67 × 10−11 N ⋅ m2 / kg2)(1.99 × 1030 kg)(3.00 × 108 m / s)2 = 1.47 × 103 m = 1.47 km

39.63 (a) At any speed, the momentum of the particle is given by

p = γmu = mu

1 − u c( )2

Since F = qE = dp

dt qE = d

dtmu 1 − u2 c2( )−1/2

qE = m 1 − u2 c2( )−1/2 du

dt+ 1

2mu 1 − u2 c2( )−3/2

2u c2( ) dudt

qEm

= dudt

1 − u2 c2 + u2 c2

1 − u2 c2( )3/2

and

a = du

dt= qE

m1 − u2

3/2

(b) As u → c, a → 0

(c)

1 − u2 / c2

3/20

v∫ = qE

mdt

t=0

t∫ so

u = qEct

m2c2 + q2E2t2

x = udt

t∫ = qEc

tdt

m2c2 + q2E2t20

t∫ =

cqE

m2c2 + q2E2t2 − mc

Chapter 39 Solutions 21

*39.64 (a) fobserved = fsource

1 + v c1 − v c

implies

cλ + ∆λ

= cλ

1 + v c1 − v c

1 − v c1 + v c

= λ + ∆λλ

and 1 + ∆λ

λ= 1 − v c

1 + v c

(b) 1 + 550 nm − 650 nm

650 nm= 1 − v c

1 + v c= 0.846

1 − v

c= 0.846( )2 1 + v

= 0.716 + 0.716

v = 0.166c = 4 97 107. × m s

39.65 (a) An observer at rest relative to the mirror sees the light travel a distance

D = 2d − x = 2 1.80 × 1012 m( ) − 0.800c( )t

where x = 0.800c( )t is the distance the ship moves toward the mirror in time t . Since thisobserver agrees that the speed of light is c , the time for it to travel distance D is:

t = D

c= 2(1.80 × 1012 m)

3.00 × 108 m / s− 0.800t = 6 67 103. × s

(b) The observer in the rocket sees a length-contracted initial distance to the mirror of:

L = d 1 − v2

c2 = 1.80 × 1012 m( ) 1 − (0.800c)2

c2 = 1.08 × 1012 m,

and the mirror moving toward the ship at speed v = 0.800c . Thus, he measures the distancethe light travels as:

D = 2 1.08 × 1012 m − y( )where y = (0.800c) t / 2( ) is the distance the mirror moves toward the ship before the lightreflects off it. This observer also measures the speed of light to be c , so the time for it to traveldistance D is:

t = D

c= 2

c1.08 × 1012 m − 0.800c( ) t

, which gives t = 4 00 103. × s

22 Chapter 39 Solutions

39.66 (a) An observer at rest relative to the mirror sees the light travel a distance D = 2d − x , where

x = vt is the distance the ship moves toward the mirror in time t . Since this observer agreesthat the speed of light is c , the time for it to travel distance D is

t = D

c= 2d − vt

c =

2dc + v

(b) The observer in the rocket sees a length-contracted initial distance to the mirror of

L = d 1 − v2

and the mirror moving toward the ship at speed v . Thus, he measures the distance the lighttravels as

D = 2 L − y( )where y = vt 2 is the distance the mirror moves toward the ship before the light reflects off it.This observer also measures the speed of light to be c , so the time for it to travel distance D is:

t = D

c= 2

cd 1 − v2

c2 − vt2

c + v( )t = 2d

cc + v( ) c − v( ) or

t = 2d

cc − vc + v

39.67 (a) Since Mary is in the same reference frame, ′S , as Ted, she observes the ball to have the samespeed Ted observes, namely ′ux = 0.800c .

(b)

∆ ′t =Lp

′ux= 1.80 × 1012 m

0.800 3.00 × 108 m s( ) = 7.5 10 s3 0 ×

c2 = 1.80 × 1012 m( ) 1 − (0.600c)2

c2 = 1.44 10 m12 ×

Since v = 0.600c and ′ux = −0.800c , the velocity Jim measures for the ball is

ux = ′ux + v

1 + ′uxv c2 =−0.800c( ) + 0.600c( )1 + −0.800( ) 0.600( ) = −0.385c

(d) Jim observes the ball and Mary to be initially separated by 1.44 × 1012 m. Mary's motion at0.600c and the ball's motion at 0.385c nibble into thi distance from both ends. The gap closesat the rate 0.600c + 0.385c = 0.985c, so the ball and catcher meet after a time

∆t = 1.44 × 1012 m

0.985 3.00 × 1018 m / s( ) = 4.88 × 103 s

Chapter 39 Solutions 23

39.68 (a) L 02 = L 0x

2 + L 0y2 and L

2 = L x2 + L y

The motion is in the x direction: Ly = L0y = L0 sin θ0

Lx = L0x 1 − v c( )2 = L0 cos θ0( ) 1 − v c( )2

Thus, L2 = L0

2 cos2 θ0 1 − vc

+ L02 sin2 θ0 = L 0

2 1 − vc

cos2 θ0

or L = L0 1 − v c( )2 cos2 θ0[ ] 1 2

(b)

tan θ =Ly

Lx=

L0y

L0x 1 − v c( )2= γ tanθ0

39.69 (a) First, we find the velocity of the stick relative to ′S using L = Lp 1 − ′ux( )2 c2

Thus

′ux = ± c 1 − L Lp( )2

Selecting the negative sign because the stick moves in the negative x direction in ′S gives:

′ux = −c 1 − 0.500 m

1.00 m

= −0.866c so the speed is ′ux = 0.866 c

Now determine the velocity of the stick relative to S, using the measured velocity of the stickrelative to ′S and the velocity of ′S relative to S. From the velocity addition equation, wehave:

ux = ′ux + v

1 + v ′ux c2 =−0.866c( ) + 0.600c( )

1 + 0.600c( ) −0.866c( ) = −0.554c and the speed is ux = 0.554 c

(b) Therefore, the contracted length of the stick as measured in S is:

L = Lp 1 − ux c( )2 = 1.00 m( ) 1 − 0.554( )2 = 0.833 m

24 Chapter 39 Solutions

39.70 (b) Consider a hermit who lives on an asteroid halfway between the Sun and Tau Ceti, stationarywith respect to both. Just as our spaceship is passing him, he also sees the blast waves fromboth explosions. Judging both stars to be stationary, this observer concludesthat the two stars blew up simultaneously .

(a) We in the spaceship moving past the hermit do not calculate the explosions to besimultaneous. We see the distance we have traveled from the Sun as

L = Lp 1 − v c( )2 = 6.00 ly( ) 1 − 0.800( )2 = 3.60 ly

We see the Sun flying away from us at 0.800c while the light from the Sun approaches at

1.00c. Thus, the gap between the Sun and its blast wave has opened at 1.80c, and the time wecalculate to have elapsed since the Sun exploded is

3.60 ly 1.80c = 2.00 yr.

We see Tau Ceti as moving toward us at 0.800c , while its light approaches at 1.00c, only

0.200c faster. We see the gap between that star and its blast wave as 3.60 ly and growing at

0.200c . We calculate that it must have been opening for

3.60 ly 0.200c = 18.0 yr

and conclude that Tau Ceti exploded 16.0 years before the Sun .

*39.71 The unshifted frequency is fsource = c

λ= 3.00 × 108 m s

394 × 10−9 m= 7.61× 1014 Hz

We observe frequency f = 3.00 × 108 m s

475 × 10−9 m= 6.32 × 1014 Hz

Then f = fsource

1 + v c1 − v c

gives: 6.32 = 7.61

1 + v c1 − v c

= 0.829( )2

Solving for v yields: v = −0.185c = 0.185c away( )

Chapter 39 Solutions 25

39.72 Take m = 1.00 kg.

The classical kinetic energy is Kc = 1

2 mu2 = 12 mc2 u

= 4.50 × 1016 J( ) uc

and the actual kinetic energy is

Kr = 1

1 − u c( )2− 1

mc2 = 9.00 × 1016 J( ) 1

1 − u c( )2− 1

u c Kc J( ) Kr J( )

0.000 0.000 0.0000.100 0.045 × 1016 0.0453 × 1016

0.200 0.180 × 1016 0.186 × 1016

0.300 0.405 × 1016 0.435 × 1016

0.400 0.720 × 1016 0.820 × 1016

0.500 1.13 × 1016 1.39 × 1016

0.600 1.62 × 1016 2.25 × 1016

0.700 2.21 × 1016 3.60 × 1016

0.800 2.88 × 1016 6.00 × 1016

0.900 3.65 × 1016 11.6 × 1016

0.990 4.41 × 1016 54.8 × 1016

Kc = 0.990Kr when

12 u c( )2 = 0.990

1 − u c( )2− 1

, yielding u = 0.115 c

Similarly, Kc = 0.950Kr when u = 0.257 c

and Kc = 0.500Kr when u = 0.786 c

39.73 ∆m = E

c2 = mc ∆T( )c2 =

ρVc ∆T( )c2

=1030 kg / m3( ) 1.40 × 109( ) 103 m( )3

4186 J / kg ⋅°C( ) 10.0 °C( )3.00 × 108 m / s( )2

∆m = 6.71 10 kg8×

Chapter 40 Solutions

40.1 T = 2.898 × 10−3 m ⋅ K

560 × 10−9 m= 5 18 103. × K

*40.2 (a) λ max = 2.898 × 10−3 m ⋅ K

2.898 × 10−3 m ⋅ K104 K

~ 10 7− m ultraviolet .

(b) λ max ~

2.898 × 10−3 m ⋅ K107 K

~ 10 10− m . γ − ray

40.3 (a) Using λ maxT = 2.898 10 m K3 × ⋅−

we getλ max =

2.898 10 m2900 K

× −3 = 9.99 × 10– 7 m = 999 nm

(b) The peak wavelength is in the infrared region of the electromagnetic spectrum, which ismuch wider than the visible region of the spectrum.

40.4 Planck's radiation law gives intensity-per-wavelength. Taking E to be the photon energy andn to be the number of photons emitted each second, we multiply by area and wavelengthrange to have energy-per-time leaving the hole:

P =2πhc2 (λ 2 − λ 1)π(d / 2)2

λ 1 + λ 2

2hc(λ 1+λ 2 )kBT − 1

= En = nhf where E = hf = 2hc

λ 1 + λ 2

n = P

8π2c d2(λ 2 − λ 1)

(λ 1 + λ 2 )4 e2hc (λ 1+λ 2 )kBT − 1( )

=8π2 3.00 × 108 m s( ) 5.00 × 10−5 m( )2

1.00 × 10−9 m( )

1001× 10−9 m( )4e

2 6.626 × 10− 34 J⋅s( ) 3.00×108 m s( )1001×10− 9 m( ) 1.38×10− 23 J K( ) 7.50×103 K( ) − 1

= ×−( ) =5.90 10 s

16 /.3 84

1.30 × 1015 / s

2 Chapter 40 Solutions

*40.5 (a) P = eAσT4 = 1 20.0 × 10−4 m2( ) 5.67 × 10−8 W m2 ⋅ K4( ) 5000 K( )4 = 7 09 104. × W

(b) λ λ λmax max max.T = ( ) = × ⋅ ⇒ =−5000 2 898 10 3 K m K 580 nm

hckBT

=6.626 × 10−34 J ⋅ s( ) 3.00 × 108 m s( )

1.38 × 10−23 J K( ) 5000 K( )= 2.88 × 10−6 m

The power per wavelength interval is P λ( ) = A I λ( ) = 2πhc2A

λ5 exp hc λ kBT( ) − 1[ ] , and

2πhc2A = 2π 6.626 × 10−34( ) 3.00 × 108( )2

20.0 × 10−4( ) = 7.50 × 10−19 J ⋅ m4

P 580 nm( ) = 7.50 × 10−19 J ⋅ m4 s

580 × 10−9 m( )5exp 2.88 µm 0.580 µm( ) − 1[ ]

= 1.15 × 1013 J m ⋅ se4.973 − 1

= 7 99 1010. × W m

(d) - (i) The other values are computed similarly:

λ hc/kBT ehc/λkBT –1 2π hc2A/λ5 P (λ), W/m

(d) 1.00 nm 2882.6 7.96 × 101251 7.50 × 1026 9.42 × 10–1226

(e) 5.00 nm 576.5 2.40 × 10250 2.40 × 1023 1.00 × 10–227

(f) 400 nm 7.21 1347 7.32 × 1013 5.44 × 1010

(g) 700 nm 4.12 60.4 4.46 × 1012 7.38 × 1010

(h) 1.00 mm 0.00288 0.00289 7.50 × 10– 4 0.260

(i) 10.0 cm 2.88 × 10–5 2.88 × 10–5 7.50 × 10–14 2.60 × 10–9

(j) We approximate the area under the P λ( ) versus λ curve, between 400 nm and 700 nm, as twotrapezoids:

P ≈

5.44 + 7.99( ) × 1010 Wm

580 − 400( ) × 10−9 m[ ]2

+ 7.99 + 7.38( ) × 1010 W

700 − 580( ) × 10−9 m[ ]2

P = 2.13 × 104 W so the power radiated as visible light is approximately 20 kW .

Chapter 40 Solutions 3

40.6 (a) P = eAσT4, so

T = P

eAσ

1 4

= 3.77 × 1026 W

1 4π 6.96 × 108 m( )2

5.67 × 10−8 W

m2 ⋅ K4

1 4

= 5 75 103. × K

(b) λ max

. ..

.= × ⋅ = × ⋅×

= × =− −

−2 898 10 2 898 105 75 10

5 04 103 3

37 m K m K

K m

T 504 nm

40.7 (a) E = hf = 6.626 × 10−34 J ⋅ s( ) 620 × 1012 s−1( ) 1.00 eV

1.60 × 10−19 J

= 2.57 eV

(b) E hf= = × ⋅( ) ×( ) ×

=− −−6 626 10 3 10 10

1 001 60 10

34 919. .

J s s eV

J1 1 28 10 5. × − eV

1.60 × 10−19 J

= 1 91 10 7. × − eV

(d) λ = c

f= 3.00 × 108 m s

620 × 1012 Hz= 4.84 × 10−7 m = 484 nm, visible light (blue)

λ = c

f= 3.00 × 108 m s

3.10 × 109 Hz= 9.68 × 10−2 m = 9.68 cm, radio wave

λ = c

f= 3.00 × 108 m s

46.0 × 106 Hz= 6.52 m, radio wave

40.8 E = hf = hc

λ=

6.626 × 10−34 J ⋅ s( ) 3.00 × 108 m s( )589.3 × 10−9 m

= 3.37 × 10−19 J photon

n = P

E= 10.0 J s

3.37 × 10−19 J photon= 2 96 1019. × photons s

40.9 Each photon has an energy E = hf = (6.626 × 10– 34)(99.7 × 106) = 6.61 × 10– 26 J

This implies that there are150 × 103 J/s

6.61 × 10– 26 J/photons = 2.27 × 1030 photons/s

4 Chapter 40 Solutions

*40.10 Energy of a single 500-nm photon:

Eγ = hf = hc

λ =

(6.626 × 10 – 34 J · s)(3.00 × 108 m/s)500 × 10 – 9 m

= 3.98 × 10 – 19 J

The energy entering the eye each second

E = P t= (IA)t = (4.00 × 10 – 11 W/m2) π4 (8.50 × 10 – 3 m) 2(1.00 s) = 2.27 × 10 – 15 J

The number of photons required to yield this energy

n = EEγ

= 2.27 × 10 – 15 J

3.98 × 10 – 19 J/photon = 5.71 × 103 photons

40.11 We take θ = 0.0300 radians. Then the pendulum's total energy is

E = mgh = mg(L – L cos θ)

E = (1.00 kg)(9.80 m/s2)(1.00 – 0.9995) = 4.41 × 10–3 J

The frequency of oscillation is f = ω

2π= 1

2πg L = 0.498 Hz

The energy is quantized, E = nhf

Therefore, n = Eh f =

4.41 × 10–3 J(6.626 × 10–34 J · s)(0.498 s–1)

= 1.34 × 1031

40.12 The radiation wavelength of λ ′= 500 nm that is observed by observers on Earth is not the truewavelength, λ , emitted by the star because of the Doppler effect. The true wavelength isrelated to the observed wavelength using:

c′λ

= cλ

1 − v c( )1 + v c( )

λ = λ ′

1 − v c( )1 + v c( ) = 500 nm( ) 1 − 0.280( )

1 + 0.280( ) = 375 nm

The temperature of the star is given by λ maxT = 2.898 × 10−3 m ⋅ K:

T = 2.898 × 10−3 m ⋅ K

λ max= 2.898 × 10−3 m ⋅ K

375 × 10−9 = 7.73 10 K3×

Chapter 40 Solutions 5

40.13 This follows from the fact that at low T or long λ , the exponential factor in the denominatorof Planck's radiation law is large compared to 1, so the factor of 1 in the denominator can beneglected. In this approximation, one arrives at Wien's radiation law.

*40.14 Planck’s radiation law is

I λ ,T( ) = 2πhc2

λ5 ehc λ kBT − 1( )Using the series expansion

ex = 1 + x + x2

2!+ x3

3!+ . . .

Planck’s law reduces to I λ ,T( ) = 2πhc2

λ5 1 + hc λ kBT + . . .( ) − 1[ ] ≈ 2πhc2

λ5 hc λ kBT( ) = 2πckBTλ4

which is the Rayleigh-Jeans law, for very long wavelengths.

40.15 (a) λ c = hc

φ =

(6.626 × 10– 34 J · s)(3.00 × 108 m/s)(4.20 eV)(1.60 × 10–19 J/eV)

= 296 nm

fc = c

λ c =

3.00 × 108 m/s296 × 10– 9 m

= 1.01 × 1015 Hz

(b)h c

λ = φ + e(∆VS):

(6.626 × 10– 34)(3.00 × 108)180 × 10– 9 = (4.20 eV)(1.60 × 10– 19 J/eV) + (1.60 × 10–19)(∆VS)

Therefore, ∆VS = 2.71 V

40.16 K mvmax max= 12

2 = 12 (9.11 × 10– 31)(4.60 × 105) 2 = 9.64 × 10 – 20 J = 0.602 eV

(a) φ = E – Kmax = 1240 eV · nm

625 nm – 0.602 eV = 1.38 eV

(b) fc = φh =

1.38 eV6.626 × 10– 34 J · s

1.60 × 10– 19 J

1 eV = 3.34 × 1014 Hz

6 Chapter 40 Solutions

40.17 (a) λ c = hc

φ Li:

λ c =6.626 × 10−34 J ⋅ s( ) 3.00 × 108 m / s( )

2.30 eV( ) 1.60 × 10−19 J / eV( ) = 540 nm

Be:

λ c =6.626 × 10−34 J ⋅ s( ) 3.00 × 108 m / s( )

3.90 eV( ) 1.60 × 10−19 J / eV( ) = 318 nm

Hg:

λ c =6.626 × 10−34 J ⋅ s( ) 3.00 × 108 m / s( )

4.50 eV( ) 1.60 × 10−19 J / eV( ) = 276 nm

λ < λ c for photo current. Thus, only lithium will exhibit the photoelectric effect.

(b) For lithium,

hcλ

= φ + Kmax

6.626 × 10−34 J ⋅ s( ) 3.00 × 108 m / s( )400 × 10−9 m

= 2.30 eV( ) 1.60 × 10−19( ) + Kmax

Kmax = 1.29 × 10−19 J = 0.808 eV

40.18 From condition (i), hf = e(∆VS 1) + φ1 and hf = e(∆VS 2) + φ2

(∆VS 1) = (∆VS 2) + 1.48 V

Then φ2 – φ1 = 1.48 eV

From condition (ii), h fc 1 = φ1 = 0.600hfc 2 = 0.600φ2

φ2 – 0.600φ2 = 1.48 eV

φ2 = 3.70 eV φ1 = 2.22 eV

40.19 (a) e ∆VS( ) = hc

λ− φ → φ = 1240 nm ⋅ eV

546.1 nm− 0.376 eV = 1.90 eV

(b) e ∆VS( ) = hc

λ− φ = 1240 nm ⋅ eV

587.5 nm− 1.90 eV → ∆VS = 0.216 V

Chapter 40 Solutions 7

Goal Solution Two light sources are used in a photoelectric experiment to determine the work function for a particularmetal surface. When green light from a mercury lamp (λ = 546.1 nm) is used, a retarding potential of0.376 V reduces the photocurrent to zero. (a) Based on this measurement, what is the work function forthis metal? (b) What stopping potential would be observed when using the yellow light from a heliumdischarge tube (λ = 587.5 nm)?

G : According to Table 40.1, the work function for most metals is on the order of a few eV, so this metalis probably similar. We can expect the stopping potential for the yellow light to be slightly lowerthan 0.376 V since the yellow light has a longer wavelength (lower frequency) and therefore lessenergy than the green light.

O : In this photoelectric experiment, the green light has sufficient energy hf to overcome the workfunction of the metal φ so that the ejected electrons have a maximum kinetic energy of 0.376 eV.With this information, we can use the photoelectric effect equation to find the work function, whichcan then be used to find the stopping potential for the less energetic yellow light.

A : (a) Einstein’s photoelectric effect equation is Kmax = hf − φ, and the energy required to raise anelectron through a 1 V potential is 1 eV, so that Kmax = eVs = 0.376 eV.

A photon from the mercury lamp has energy: hf = hc

λ=

4.14 × 10−15 eV ⋅ s( ) 3.00 × 108 m s( )546.1× 10−9 m

E = hf = 2.27 eV

Therefore, the work function for this metal is: φ = hf − Kmax = 2.27 eV − 0.376 eV( ) = 1.90 eV

(b) For the yellow light, λ = 587.5 nm, and hf = hc

λ=

4.14 × 10−15 eV ⋅ s( ) 3.00 × 108 m / s( )587.5 × 10−9 m

E = 2.11 eV

Therefore, Kmax = hf − φ = 2.11 eV − 1.90 eV = 0.216 eV, so Vs = 0.216 V

L : The work function for this metal is lower than we expected, and does not correspond with any of thevalues in Table 40.1. Further examination in the CRC Handbook of Chemistry and Physics revealsthat all of the metal elements have work functions between 2 and 6 eV. However, a single metal’swork function may vary by about 1 eV depending on impurities in the metal, so it is just barelypossible that a metal might have a work function of 1.90 eV.

The stopping potential for the yellow light is indeed lower than for the green light as we expected.An interesting calculation is to find the wavelength for the lowest energy light that will ejectelectrons from this metal. That threshold wavelength for Kmax = 0 is 658 nm, which is red light in thevisible portion of the electromagnetic spectrum.)

8 Chapter 40 Solutions

40.20 From the photoelectric equation, we have: e ∆VS1( ) = Eγ1 − φ and e ∆VS2( ) = Eγ 2 − φ

Since ∆VS2 = 0.700 ∆VS1( ), then e ∆VS2( ) = 0.700(Eγ1 − φ) = Eγ 2 − φ

or (1 − 0.700)φ = Eγ 2 − 0.700Eγ1

and the work function is: φ =

Eγ 2 − 0.700Eγ1

0.300

The photon energies are: Eγ1 =

hcλ 1

= 1240 nm ⋅ eV410 eV

= 3.03 eV

and E

hcγ λ2

1240445

2 79= = ⋅ = nm eV eV

eV.

Thus, the work function is φ = 2.79 eV − 0.700 3.03 eV( )

0.300= 2.23 eV

and we recognize this as characteristic of potassium .

*40.21 The energy needed is E = 1.00 eV = 1.60 × 10– 19 J

The energy absorbed in time t is E = Pt = (IA)t

so t = E

IA = 1.60 × 10– 19 J

(500 J/s · m2)[π (2.82 × 10– 15 m)2] = 1.28 × 107 s = 148 days

The gross failure of the classical theory of the photoelectric effect contrasts with the success ofquantum mechanics.

*40.22 Ultraviolet photons will be absorbed to knock electrons out of the sphere with maximumkinetic energy Kmax = hf − φ, or

Kmax =

6.626 × 10−34 J ⋅ s( ) 3.00 × 108 m s( )200 × 10−9 m

1.00 eV1.60 × 10−19 J

−4.70 eV = 1.51 eV

The sphere is left with positive charge and so with positive potential relative to V = 0 at r = ∞.As its potential approaches 1.51 V, no further electrons will be able to escape, but will fall backonto the sphere. Its charge is then given by

V = keQ

r or

Q = rV

ke=

5.00 × 10−2 m( ) 1.51 N ⋅ m C( )8.99 × 109 N ⋅ m2 C2 = 8 41 10 12. × − C

Chapter 40 Solutions 9

40.23 (a) By having the photon source move toward the metal, the incident photons are Dopplershifted to higher frequencies, and hence, higher energy.

(b) If v = 0.280c ,

′f = f1 + v / c1 − v / c

= 7.00 × 1014( ) 1.280.720 = 9.33 × 1014 Hz

Therefore, φ = 6.626 × 10−34 J ⋅ s( ) 9.33 × 1014 Hz( ) = 6.18 × 10−19 J = 3.87 eV

and Kmax = hf − φ = 6.626 × 10−34 J ⋅ s( ) 3.05 × 1015 Hz( ) 1.00 eV

1.60 × 10−19 J

− 3.87 eV = 8.78 eV

*40.24 E = hc

λ =

(6.626 × 10– 34 J · s)(3.00 × 108 m/s)700 × 10– 9 m

= 2.84 × 10– 19 J = 1.78 eV

p = h

λ =

6.626 × 10– 34 J · s700 × 10– 9 m

= 9.47 × 10– 28 kg · m/s

40.25 (a) ∆λ = h

mec(1 − cos θ) =

6.626 × 10 – 34

(9.11 × 10 – 31)(3.00 × 108) (1 – cos 37.0°) = 4.88 × 10 – 13 m

(b) E0 = hc / λ 0 : 300 × 103 eV( ) 1.60 × 10−19 J / eV( ) = 6.626 × 10−34( ) 3.00 × 108 m / s( ) λ 0

λ 0 = 4.14 × 10 – 12 m and ′λ = λ 0 + ∆λ = 4.63 × 10−12 m

′E = hc

′λ=

6.626 × 10−34 J ⋅ s( ) 3.00 × 108 m / s( )4.63 × 10−12 m

= 4.30 × 1014 J = 268 keV

40.26 This is Compton scattering through 180°:

E0 = hc

λ 0 =

(6.626 × 10 – 34 J · s)(3.00 × 108 m/s)(0.110 × 10 – 9 m)(1.60 × 10–19 J/eV)

= 11.3 keV

∆λ = h

mec(1 − cos θ) = (2.43 × 10 – 12 m)(1 – cos 180°) = 4.86 × 10 – 12 m

′λ = λ 0 + ∆λ = 0.115 nm so

′E = hc′λ

= 10.8 keV

Momentum conservation:h

λ 0 i =

h′λ (–i) + pe (i) and

pe = h

1λ 0

− 1′λ

pe = 6.626 × 10−34 J ⋅ s( ) 3.00 × 108 m / s( )/ c

1.60 × 10−19 J / eV

10.110 × 10−9 m

+ 10.115 × 10−9 m

= 22.1 keV/c

10 Chapter 40 Solutions

Energy conservation: 11.3 keV = 10.8 keV + Ke so that Ke = 478 eV

Check: E 2 = p 2 c 2 + m ce2 4 or (mec

2 + Ke )2 = (pc)2 + (mec2 )2

(511 keV + 0.478 keV)2 = (22.1 keV)2 + (511 keV)2

2.62 × 1011 = 2.62 × 1011

40.27 Ke = E0 – ′E

With Ke = ′E , ′E = E0 – ′E : ′E = E02

′λ = hc ′E =

h c12 E0

= 2 hcE0

= 2λ 0 ′λ = λ 0 + λ C (1 – cos θ)

2λ 0 = λ 0 + λ C (1 – cos θ) 1 – cos θ = λ 0λ C

= 0.001600.00243 → θ = 70.0°

40.28 We may write down four equations, not independent, in the three unknowns λ 0 , ′λ , and vusing the conservation laws:

hcλ 0

= hc′λ+ γmec

2 − mec2 (Energy conservation)

hλ 0

= γmev cos 20.0° (momentum in x-direction)

0 = h

′λ− γmev sin 20.0° (momentum in y-direction)

and Compton's equation

′λ − λ 0 = hmec

1 − cos 90.0°( ).

It is easiest to ignore the energy equation and, using the two momentum equations, write

h / λ 0

h / ′λ= γ mev cos 20.0°

γ mev sin 20.0° or λ 0 = ′λ tan 20.0°

Then, the Compton equation becomes ′λ − ′λ tan 20.0°= 0.00243 nm,

′λ = 0.00243 nm1 − tan 20.0°

= 0.00382 nm = 3.82 pm

Chapter 40 Solutions 11

40.29 (a) Conservation of momentum in the x direction gives: pγ = ′pγ cos θ + pe cos φ

or since θ = φ,

hλ 0

= pe + h′λ

cos θ [1]

Conservation of momentum in the y direction gives: 0 = ′pγ sin θ − pe sin θ ,

which (neglecting the trivial solution θ = 0) gives: pe = ′pγ = h

′λ [2]

Substituting [2] into [1] gives:

hλ 0

= 2h′λ

cos θ , or ′λ = 2λ 0 cos θ [3]

Then the Compton equation is

′λ − λ 0 = hmec

(1 − cos θ)

giving 2λ 0 cos θ − λ 0 = h

mec(1 − cos θ)

or 2 cos θ − 1 = hc

λ 0

1mec

2 (1 − cos θ)

Since Eγ =

hcλ 0

, this may be written as: 2 cos θ − 1 =

Eγ

mec2

(1 − cos θ)

which reduces to:

2 +Eγ

mec2

cos θ = 1 +

Eγ

mec2

or cos θ =

mec2 + Eγ

2mec2 + Eγ

= 0.511 MeV + 0.880 MeV1.02 MeV + 0.880 MeV

= 0.732 so that θ = φ = 43.0°

(b) Using Equation (3):

′Eγ = hc′λ

= hcλ 0 2 cos θ( ) =

Eγ

2 cos θ= 0.880 MeV

2 cos 43.0°= 0.602 MeV = 602 keV

Then,

′pγ =′Eγ

c= 0.602 MeV c = 3.21× 10−22 kg ⋅ m s

From energy conservation: Ke = Eγ − ′Eγ = 0.880 MeV − 0.602 MeV = 0.278 MeV = 278 keV

12 Chapter 40 Solutions

40.30 The energy of the incident photon is E0 = pγc = hc λ 0 .

(a) Conserving momentum in the x direction gives

pγ = pe cos φ + ′pγ cos θ , or since φ = θ ,

c= pe + ′pγ( )cos θ [1]

Conserving momentum in the y direction (with φ = θ ) yields

0 = ′pγ sin θ − pe sin θ , or pe = ′pγ = h

′λ[2]

Substituting Equation [2] into Equation [1] gives

c= h

′λ+ h

′λ

cos θ , or

′λ = 2hc

E0cos θ [3]

By the Compton equation,

′λ − λ 0 = hmec

1 − cos θ( ),

2hcE0

cos θ − 2hcE0

= hmec

1 − cos θ( )

which reduces to 2mec

2 + E0( )cos θ = mec2 + E0

Thus, φ = θ = cos−1 mec

2 + E0

2mec2 + E0

(b) From Equation [3],

′λ = 2hcE0

cos θ = 2hcE0

mec2 + E0

2mec2 + E0

Therefore,

′Eγ = hc′λ

= hc

2hc E0( ) mec2 + E0( ) 2mec

2 + E0( ) =

22mec

2 + E0

mec2 + E0

and

′pγ =′Eγ

2c2mec

2 + E0

mec2 + E0

22mec

2 + E0

mec2 + E0

or Ke = E0

22mec

2 + 2E0 − 2mec2 − E0

mec2 + E0

E02

2 mec2 + E0( )

Finally, from Equation (2), pe = ′pγ =

2c2mec

2 + E0

mec2 + E0

Chapter 40 Solutions 13

40.31 (a) Thanks to Compton we have four equations in the unknowns φ, v, and ′λ :

hcλ 0

= hc′λ+ γ mec

2 − mec2 (energy conservation) [1]

hλ 0

= hλ '

cos 2φ + γmev cos φ (momentum in x direction) [2]

0 = h

′λsin 2φ − γ mev sin φ (momentum in y direction) [3]

′λ − λ 0 = h

mec(1 − cos 2φ) (Compton equation) [4]

Using sin 2φ = 2 sin φ cos φ in Equation [3] gives γ mev = 2h

′λcos φ.

Substituting this into Equation [2] and using cos 2φ = 2 cos2 φ − 1 yields

hλ 0

= h′λ(2 cos2φ − 1) + 2h

′λ cos2φ = h

′λ(4 cos2φ − 1),

or λ ' = 4λ 0 cos2φ − λ 0 [5]

Substituting the last result into the Compton equation gives

4λ 0 cos2φ − 2λ 0 = h

mec1 − 2 cos2φ − 1( )[ ] = 2

hcmec

2 1 − cos2 φ( ) .

With the substitution λ 0 = hc E0 , this reduces to

cos2 φ = mec

2 + E0

2mec2 + E0

= 1 + x2 + x

where x ≡ E0

mec2 .

For x = 0.700 MeV

0.511 MeV= 1.37, this gives

φ = cos−1 1 + x

2 + x= 33.0°

(b) From Equation [5],

′λ = λ 0 4 cos2φ − 1( ) = λ 0 41 + x2 + x

− 1

= λ 02 + 3x2 + x

Then, Equation [1] becomes

hcλ 0

= hcλ 0

2 + x2 + 3x

+ γ mec

2 − mec2 or

mec2 − E0

mec2

2 + x2 + 3x

+ 1 = γ .

Thus, γ = 1 + x − x

2 + x2 + 3x

, and with x = 1.37 we get γ = 1.614.

Therefore,

= 1 − γ −2 = 1 − 0.384 = 0.785 or v = 0.785 c .

14 Chapter 40 Solutions

40.32

′λ − λ =h

mec(1 − cos θ)

′′λ − ′λ =

hmec

1 − cos(π − θ)[ ]

′′λ − λ =

hmec

− hmec

cos(π − θ) +h

mec− h

meccos θ

Now cos (π – θ) = – cos θ, so

′′λ − λ = 2h

mec= 0.00486 nm

40.33 (a) K = 1

2 mev2 = 1

2 9.11× 10−31 kg( ) 1.40 × 106 m s( )2= 8.93 × 10−19 J = 5.58 eV

E0 = hc

λ 0= 1240 eV ⋅ nm

0.800 nm= 1550 eV

′E = E0 − K , and

′λ = hc′E

= 1240 eV ⋅ nm1550 eV − 5.58 eV

= 0.803 nm

∆λ = ′λ − λ 0 = 0.00288 nm = 2.88 pm

(b) ∆λ = λ C 1 − cos θ( ) ⇒ cos θ = 1 − ∆λ

λ C= 1 − 0.00288 nm

0.00243 nm= − 0.189, so θ = 101°

*40.34 Maximum energy loss appears as maximum increase in wavelength, which occurs forscattering angle 180°. Then ∆λ = 1 − cos 180°( ) h / mc( ) = 2h / mc where m is the mass of thetarget particle. The fractional energy loss is

E0 − ′EE0

=hc λ 0 − hc ′λ

hc λ 0=

′λ − λ 0

′λ= ∆λ

λ 0 + ∆λ= 2h mc

λ 0 + 2h mc

Further, λ 0 = hc E0 , so

E0 − ′EE0

= 2h mchc E0 + 2h mc

= 2E0

mc2 + 2E0.

(a) For scattering from a free electron, mc2 = 0.511 MeV, so

E0 − ′EE0

= 2 0.511 MeV( )0.511 MeV + 2 0.511 MeV( ) = 0.667

(b) For scattering from a free proton, mc2 = 938 MeV, and

E0 − ′EE0

= 2 0.511 MeV( )938 MeV + 2 0.511 MeV( ) = 0.00109

Chapter 40 Solutions 15

40.35 Start with Balmer's equation,

1λ

= RH1

22 − 1n 2

, or

λ = (4n 2 / RH )

(n 2 − 4).

Substituting RH = 1.0973732 × 107 m−1, we obtain

λ = (3.645 × 10− 7 m)n 2

n 2 − 4= 364.5n 2

n 2 − 4nm, where n = 3, 4, 5, . . .

40.36 (a) Using

1λ

= RH1

nf2 − 1

ni2

, for nf = 2, and ni ≥ 3, we get:

λ = 4n2

RH n2 − 4( ) = 4n2

2.00 × 107 m−1( ) n2 − 4( ) = 200.0( )n2

n2 − 4nm

This says that 200 nm ≤ λ ≤ 360 nm, which is ultraviolet .

(b) Using n ≥ 3 ,

λ = 4n2

RH n2 − 4( ) = 4n2

0.500 × 107 m− 1( ) n2 − 4( ) = (800.0)n2

n2 − 4nm

This says that 800 nm ≤ λ ≤ 1440 nm, which is in the infrared .

40.37 (a) Lyman series:

1λ

= R 1 − 1n2

n = 2, 3, 4, . . .

1λ

= 194.96 × 10−9 = (1.097 × 107 ) 1 − 1

n = 5

(b) Paschen series:

1λ

= R1

32 − 1n2

n = 4, 5, 6, . . .

The shortest wavelength for this series corresponds to n = ∞ for ionization

1λ

= 1.097 × 107 19

− 1n2

For n = ∞, this gives λ = 820 nm

This is larger than 94.96 nm, so this wave length cannot be associated with the Paschen series

Brackett series:

1λ

= R1

42 − 1n2

n = 5, 6, 7, . . .

1λ

= 1.097 × 107 116

− 1n2

n = ∞ for ionization λ min = 1458 nm

Once again this wavelength cannot be associated with the Brackett series

16 Chapter 40 Solutions

40.38 (a) λ min = hc

Emax

Lyman (nf = 1): λ min = hc

E1= 1240 eV ⋅ nm

13.6 eV= 91.2 nm (Ultraviolet)

Balmer (nf = 2): λ min = hc

E2= 1240 eV ⋅ nm

14( )13.6 eV

= 365 nm (UV)

Paschen (nf = 3): λ min = . . . = 32(91.2 nm) = 821 nm (Infrared)

Bracket (nf = 4): λ min = . . . = 42(91.2 nm) = 1460 nm (IR)

(b) Emax = hc

λ min

Lyman: Emax = 13.6 eV = E1( )Balmer: Emax = 3.40 eV = E2( )Paschen: Emax = 1.51 eV = E3( )Brackett: Emax = 0.850 eV = E4( )

40.39 Liquid O2 λ abs = 1269 nm

E = hc

λ= 1.2398 × 10−6

1.269 × 10−6 = 0.977 eV for each molecule.

For two molecules, λ = hc

2E= 634 n