serway1 (1)

Alternating Current Circuits 251

= 1 = 1 = 2.65 Ω
ωC 1.00 ×10−3
P33.16 ( )XC s)
2π (60.0 CV

∆vC (t ) = ∆Vmax sinω t, to be zero at t = 0

iC = ∆Vmax sin(ω t +φ) = 2 (120 V ) sin ⎡ 2π 60 s−1 + 90.0°⎦⎥⎤ = (64.0 A)sin(120° + 90.0°)
XC ⎢⎣ 180 s−1
2.65 Ω

= −32.0 A

Section 33.5 The RLC Series Circuit

)(P33.17 (a) XL = ω L = 2π (50.0) 400 × 10−3 = 126 Ω

=1 = 1 = 719 Ω
ωC 4.43 × 10−6
)(XC
2π (50.0)

Z = )(R2 + XL − XC 2 = 5002 + (126 − 719)2 = 776 Ω FIG. P33.17

( )∆Vmax = ImaxZ = 250 × 10−3 (776) = 194 V

(b) φ = tan−1 ⎛ XL − XC ⎞ = tan−1 ⎛ 126 − 719⎞ = −49.9° . Thus, the
⎝ R ⎠ ⎝ 500 ⎠

current leads the voltage .

P33.18 ( )( )ω L = 1 → ω = 1 = 1 = 1.75 × 104 rad s
ω C LC 57.0 × 10−6 57.0 × 10−6
f = ω = 2.79 kHz

2π

P33.19 (a) ( )( )XL = ω L = 2π 50.0 s−1 250 × 10−3 H = 78.5 Ω
(b)
(c) 1 2.00 × 10−6 F ⎦⎤−1 = 1.59 kΩ
(d) ( )( )XC ωC
(e) = = ⎣⎡2π 50.0 s−1

Z = )(R2 + XL − XC 2 = 1.52 kΩ

I max = ∆Vmax = 210 V Ω = 138 mA
Z 1.52 × 103

φ = tan−1 ⎡ XL − XC ⎤ = tan−1 (−10.1) = −84.3°
⎣⎢ R ⎦⎥

252 Chapter 33

P33.20 (a) Z = )(R2 + XL − XC 2 = 68.02 + (16.0 − 101)2 = 109 Ω

(b) XL = ω L = (100)(0.160) = 16.0 Ω
(c)
= 1 = 1 = 101 Ω
ωC 99.0 × 10−6
)(XC
(100)

I max = ∆Vmax = 40.0 V = 0.367 A
Z 109 Ω

tanφ = XL − XC = 16.0 −101 = −1.25:
R 68.0

φ = −0.896 rad = −51.3°

Imax = 0.367 A ω = 100 rad s φ = −0.896 rad = −51.3°

P33.21 XL = 2π f L = 2π (60.0)(0.460) = 173 Ω

= 1 = 1 = 126 Ω
2π f C 21.0 ×10−6
( )XC
2π (60.0)

(a) tanφ = XL − XC = 173 Ω − 126 Ω = 0.314
R 150 Ω

φ = 0.304 rad = 17.4°

(b) Since XL > XC , f is positive; so voltage leads the current .

P33.22 For the source-capacitor circuit, the rms source voltage is ∆Vs = (25.1 mA) XC . For the circuit
with resistor, ∆Vs = (15.7 mA) R2 + XC2 = (25.1 mA) XC . This gives R = 1.247XC . For the cir-
cuit with ideal inductor, ∆Vs = (68.2 mA) XL − XC = (25.1 mA) XC . So XL − XC = 0.368 0 XC .

Now for the full circuit

)(∆Vs = I R2 + XL − XC 2
(25.1 mA) XC = I (1.247XC )2 + (0.368XC )2

I = 19.3 mA

XC = 1 = 1 = 1.33×108 Ω
2π f C 20.0 × 10−12
( )P33.23
2π (60.0 Hz) F

( ) ( )Z = 50.0 × 103 Ω 2 + 1.33 × 108 Ω 2 ≈ 1.33 × 108 Ω

I rms = ∆Vrms = 5 000 V = 3.77 × 10−5 A
Z 1.33 × 108 Ω

( ) ( )( )∆Vrms body = Irms Rbody = 3.77 × 10−5 A 50.0 × 103 Ω = 1.88 V

Alternating Current Circuits 253

= 1 = 1 = 49.0 Ω
ωC 65.0 × 10−6
P33.24 )(XC
2π (50.0)

)(XL = ω L = 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

= ∆Vmax = 150 = FIG. P33.24
Z 41.0
I max 3.66 A

(a) ∆VR = Imax R = (3.66)(40) = 146 V

(b) ∆VL = Imax XL = (3.66)(58.1) = 212.5 = 212 V

(c) ∆VC = Imax XC = (3.66)(49.0) = 179.1 V = 179 V

(d) ∆VL − ∆VC = 212.5 − 179.1 = 33.4 V

P33.25 R = 300 Ω

XL = ω L = 2π ⎛ 500 s−1 ⎞ ( 0.200 H) = 200 Ω XL = 200 Ω
⎝ π ⎠
{XL – XC = 109 Ω
=1 ⎡⎣⎢2π ⎛ 500 ⎞ ⎤−1 Z
ωC ⎝ π ⎠ ⎦⎥ XC = 90.9 Ω f
( )XC −1 R = 300 Ω
= s 11.0 × 10−6 F = 90.9 Ω

Z = R2 + ( XL − )XC 2 = 319 Ω and

φ = tan−1 ⎛ XL − XC ⎞ = 20.0°
⎝ R ⎠

FIG. P33.25

*P33.26 Let Xc represent the initial capacitive reactance. Moving the plates to half their original separation

doubles the capacitance and cuts XC = 1 in half. For the current to double, the total impedance
ωC

must be cut in half: Zi = 2Z f , )(R2 + XL − XC 2 = 2 R2 + ⎛ XL − XC ⎞2 . With XL = R, algebra
⎝ 2 ⎠

then gives

R2 + (R − XC )2 = ⎛ R2 + ⎛ R − XC ⎞2⎞
4 ⎝⎜ ⎝ 2 ⎠ ⎠⎟

2R2 − 2RXC + XC2 = 8R2 − 4RXC + XC2

XC = 3R

254 Chapter 33

Section 33.6 Power in an AC Circuit

P33.27 ω = 1 000 rad s, R = 400 Ω, C = 5.00 × 10−6 F, L = 0.500 H
∆Vmax = 100 V, ωL = 500 Ω,
⎝⎜⎛ω1C ⎞ = 200 Ω
⎟
⎠

Z= ⎛ 1 ⎞2 4002 + 3002 = 500 Ω
R2 + ⎜ωL − ⎟=
⎝ ωC⎠

I max = ∆Vmax = 100 = 0.200 A
Z 500

The average power dissipated in the circuit is P = I 2 R = ⎛ I 2 ⎞ R.
rms ⎝⎜ max ⎟⎠

P = (0.200 A)2 (400 Ω) = 8.00 W 2

2

P33.28 Z = R2 + ( XL )− XC 2 or ( XL − XC ) = Z 2 − R2

( XL − XC ) = (75.0 Ω)2 − (45.0 Ω)2 = 60.0 Ω

φ = tan−1 ⎛ XL − XC ⎞ = tan−1 ⎛ 60.0 Ω⎞ = 53.1°
⎝ R ⎠ ⎝ 45.0 Ω⎠

I rms = ∆Vrms = 210 V = 2.80 A
Z 75.0 Ω

P ( )= ∆Vrms Irms cosφ = (210 V)(2.80 A)cos(53.1°) = 353 W

P33.29 (a) P = ( )Irms ∆Vrms cosφ = (9.00)180 cos(−37.0°) = 1.29 × 103 W

P = I 2 R so 1.29 × 103 = (9.00)2 R and R = 16.0 Ω
rms

(b) tanφ = XL − XC becomes tan(−37.0°) = XL − XC : so XL − XC = −12.0 Ω

R 16

P33.30 XL = ω L = 2π (60.0 s)(0.025 0 H) = 9.42 Ω

Z = )(R2 + XL − XC 2 = (20.0)2 + (9.42)2 Ω = 22.1 Ω

(a) I rms = ∆Vrms = 120 V = 5.43 A
Z 22.1 Ω

(b) φ = tan−1 ⎛ 9.42 ⎞ = 25.2° so power factor = cosφ = 0.905
⎝ 20.0 ⎠

(c) We require φ = 0. Thus, XL = XC : 1
60.0 s−1
)(9.42 Ω = C
2π

and C = 281 µF

( ) ( ) ( )(d) 2
Pb = Pd or ∆Vrms b Irms b cosφb = ∆Vrms d

R

( )∆Vrms d = ( ) ( )R ∆Vrms b Irms b cosφb = (20.0 Ω)(120 V)(5.43 A)(0.905) = 109 V

Alternating Current Circuits 255

*P33.31 Consider a two-wire transmission line taking in power P R1

I rms = P . Then power loss = I R2 = P. R1
rms line
∆Vrms 100 ∆Vrms FIG. P33.31 RL

Thus, ⎛ P ⎞2 (2R1 ) = P or R1 = ( )∆Vrms 2
⎜⎝ ⎟⎠
∆Vrms 100 200P

)(R1 = ρ = ∆Vrms 2 or A = π (2r)2 = 200ρP
A
200P 4 ( )∆Vrms 2

and the diameter is 2r = 800ρP
π (∆V )2

(a) 2r = 800(1.7 × 10−8 Ωm) 20 000 W (18 000 m) = 39.5 V ⋅ m/∆V

π (∆V )2

(b) The diameter is inversely proportional to the potential difference.

(c) 2r = 39.5 V⋅m/1 500 V = 2.63 cm

(d) ∆V = 39.5 V⋅m/0.003 m = 13.2 kV

*P33.32 (a) XL = ωL = 2p (60/s) 0.1 H = 37.7 Ω
(b) Z = (1002 + 37.72)1/2 = 107 Ω
(c) power factor = cosf = 100/107 = 0.936

The power factor cannot in practice be made 1.00. If the inductor were removed or if the
generator were replaced with a battery, so that either L = 0 or f = 0, the power factor would
be 1, but we would not have a magnetic buzzer.

We want resonance, with f = 0. We insert a capacitor in series with

XL = XC so 37.7 Ω = 1 s/2p C 60 and C = 70.4 mF

P33.33 One-half the time, the left side of the generator is positive, the top
diode conducts, and the bottom diode switches off. The power supply
sees resistance R 2R
R R

⎡ 1 + 1 ⎤−1 = R and the power is ( ∆Vrms )2 .
⎢⎣ 2R 2R ⎥⎦
R ~

The other half of the time the right side of the generator is positive, ∆V
the upper diode is an open circuit, and the lower diode has zero
resistance. The equivalent resistance is then FIG. P33.33

P = ∆Vrms 2 = 4 ∆Vrms 2

Req 7R
( ) ( )Req= R + ⎡1 + 1 ⎤−1 = 7R
⎢⎣ 3R R ⎦⎥ 4 and

The overall time average power is: ) ) )( ( (⎣⎡ ∆Vrms 2 R⎦⎤ + ⎡⎣4 ∆Vrms 2 7R⎦⎤ = 11 ∆Vrms 2

2 14R

256 Chapter 33

Section 33.7 Resonance in a Series RLC Circuit

P33.34 (a) f= 1
2π LC

= 1 = 1 A⎛ C ⎞ = 6.33 × 10−13 F
4π 2 f 1010 s Vs ⎝ As⎠
( )C 2L 2 400 × 10−12
4π 2

(b) C = κ ∈0 A = κ ∈0 2
dd

⎛ Cd ⎞1 2 ⎛ 6.33 × 10−13 F × 10−3 mm ⎞ 1 2
⎜⎝ κ ∈0 ⎠⎟ ⎝⎜ 1 × 8.85 × 10−12 F ⎠⎟
= = = 8.46 × 10−3 m

(c) XL = 2π f L = 2π ×1010 s × 400 ×10−12 Vs A = 25.1 Ω

P33.35 ( )ω0 = 2π 99.7 × 106 = 6.26 × 108 rad s = 1
LC

= 1 = 1 =
6.26 × 108 2 1.40 × 10−6
( ) ( )Cω2L 1.82 pF
0

P33.36 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 1 = 3.56 kHz .
2π LC

(b) At resonance, I max = ∆Vmax = 5.00 A
R

(c) From Equation 33.38, Q = ω0 L = 22.4
R

(d) ∆VL, max = XL Imax = ω0 LImax = 2.24 kV

P33.37 The resonance frequency is ω0 = 1 . Thus, if ω = 2ω0 ,
LC

XL = ω L = ⎛ 2 ⎟⎞⎠ L = 2 L and XC = 1 = LC = 1 L
⎝⎜ LC C ωC 2C 2 C

Z= )(R2 + XL − XC 2 = R2 + 2.25 ⎛⎝ L ⎞ so I rms = ∆Vrms = ∆Vrms
C ⎠ Z
R2 + 2.25(L C )

and the energy delivered in one period is E = P ∆t:

( )∆Vrms 2 R ⎛ 2π ⎞ ( )∆Vrms 2 RC 4π ( )∆Vrms 2 RC LC
⎝ ω ⎠
R2 + 2.25(L C ) R2C + 2.25L 4R2C + 9.00L
( )E= = π LC =

With the values specified for this circuit, this gives:

( ) ( )4π (50.0 V)2 (10.0 Ω) 100 × 10−6 F 3 2 10.0 × 10−3 H 1 2
) )( (E = 4(10.0 Ω)2 100 × 10−6 F + 9.00 10.0 × 10−3 H = 242 mJ

Alternating Current Circuits 257

P33.38 The resonance frequency is ω0 = 1. Thus, if ω = 2ω0 ,
LC

XL =ω L = ⎛ 2 ⎞ L = 2 L and XC = 1 = LC = 1 L
⎝⎜ LC ⎠⎟ C ωC 2C 2 C

Then Z = )(R2 + XL − XC 2 = R2 + 2.25 ⎝⎛ L ⎞ so I rms = ∆Vrms = ∆Vrms
C ⎠ Z
R2 + 2.25(L C )

and the energy delivered in one period is

)(∆Vrms 2 R ⎛ 2π ⎞ ( )∆Vrms 2 RC )(4π ∆Vrms 2 RC LC

R2 + 2.25(L C ) ⎝ ω ⎠ R2C + 2.25L 4R2C + 9.00L
( )E = P ∆t = = π LC =

1= 1 = 251 rad s
LC 160 × 10−3 H 99.0 × 10−6 F
P33.39 For the circuit of Problem 20, ( )( )ω0 =

)(Q = ω0 L = (251 rad s) 160 × 10−3 H = 0.591
R 68.0 Ω

For the circuit of Problem 21, Q = ω0L = L = 1 L= 1 460 × 10−3 H = 0.987
R R LC R C 150 Ω 21.0 × 10−6 F

The circuit of Problem 21 has a sharper resonance.

Section 33.8 The Transformer and Power Transmission

P33.40 (a) ∆V2,rms = 1 (120 V) = 9.23 V
(b)
13
(c)
∆V1,rms I1,rms = ∆V2,rms I2,rms

(120 V)(0.350 A) = (9.23 V) I2,rms

I 2 , rms = 42.0 W = 4.55 A for a transformer with no energy loss.
9.23 V

P = 42.0 W from part (b).

= N2 ⎛ 2 000⎞
Nmax ⎝ 350 ⎠

1
( ) )(∆Vout (170 V) =
P33.41 ∆Vin max = 971 V

)( ∆Vout = (971 V) = 687 V

rms 2

( ) ( )∆V2,rms = N2 (2 200)(80)
P33.42 (a) N1 ∆V1,rms N2 = = 1 600 windings
(b) 110

( ) ( )I1,rms ∆V1,rms = I2,rms ∆V2,rms I1,rms = (1.50) (2 200) = 30.0 A

110

( ) ( )(c) 0.950I1,rms ∆V1,rms = I2,rms ∆V2,rms I1,rms = (1.20)(2 200) = 25.3 A
110 (0.950)

258 Chapter 33

( )( )R = 4.50 × 10−4 Ω M P 5.00 × 106 W
P33.43 (a) 6.44 × 105 m = 290 Ω and I rms = = 5.00 × 105 V = 10.0 A
(b) ∆Vrms

Ploss = I 2 R = (10.0 A)2 (290 Ω) = 29.0 kW
rms

Ploss = 2.90 × 104 = 5.80 × 10−3
P 5.00 × 106

(c) It is impossible to transmit so much power at such low voltage. Maximum power transfer
occurs 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.

Section 33.9 Rectifiers and Filters

P33.44 (a) Input power = 8 W

Useful output power = I ∆V = 0.3 A(9 V) = 2.7 W

efficiency = useful output = 2.7 W = 0.34 = 34%
total input 8 W

(b) Total input power = Total output power
8 W = 2.7 W + wasted power

wasted power = 5.3 W

(c) E =P ∆t =8 W (6)(31 d) ⎛ 86 400 s⎞ ⎛ 1 J ⎞ = 1.29 × 108 J ⎛ $0.135 ⎞ = $4.8
⎝ 1d ⎠ ⎝ 1 Ws⎠ ⎜⎝ 3.6 × 106 J ⎠⎟

P33.45 (a) The input voltage is ∆Vin = IZ = I R2 + XC2 = I R2 + ⎛ 1 ⎞2 . The output voltage is
⎝⎜ ωC ⎠⎟

∆Vout = IR. The gain ratio is ∆Vout = IR = R.
∆Vin I
R2 + (1 ωC)2 R2 + (1 ωC)2

(b) As ω → 0, 1 → ∞ and ∆Vout → 0 .
ωC ∆Vin

As ω → ∞, 1 → 0 and ∆Vout → R = 1 .
ωC ∆Vin R

(c) 1 = R

2 R2 + (1 ωC)2

R2 + ω 1 2 = 4R2 ω 2C 2 = 1 ω = 2π f = 1 f= 1
2C 3R2 3RC 2π 3RC

Alternating Current Circuits 259

P33.46 (a) The input voltage is ∆Vin = IZ = I R2 + XC2 = I R2 + (1 ωC)2 . The output voltage is

∆Vout = IXC = I. The gain ratio is ∆Vout = I ωC = 1 ωC
ωC ∆Vin .
I R2 + (1 ωC )2
R2 + (1 ωC )2

(b) As ω → 0, 1 → ∞ and R becomes negligible in comparison. Then ∆Vout → 1 ωC = 1 .
ωC ∆Vin 1 ωC

As ω → ∞, 1 → 0 and ∆Vout → 0 .
ωC ∆Vin

(c) 1 = 1 ωC R2 + ⎛ 1 ⎞2 = 4 R2ω 2C2 = 3 ω = 2π f = 3
⎜⎝ ωC ⎟⎠ ω2C2 RC
2 R2 + (1 ωC)2

f= 3
2π RC

P33.47 For this RC high-pass filter, ∆Vout = R .
∆Vin R2 + XC2

(a) When ∆Vout = 0.500,
∆Vin

then 0.500 Ω = 0.500 or XC = 0.866 Ω.

(0.500 Ω)2 + XC2

If this occurs at f = 300 Hz, the capacitance is

C= 1 = 2π (300 1 Ω)
2π f XC
Hz ) ( 0.866

= 6.13 × 10−4 F = 613 µF

(b) With this capacitance and a frequency of 600 Hz,

= 1 = 0.433 Ω
6.13 × 10−4
)(XC
2π (600 Hz ) F

∆Vout = R = 0.500 Ω = 0.756

∆Vin R2 + XC2 (0.500 Ω)2 + (0.433 Ω)2 FIG. P33.47

260 Chapter 33

P33.48 For the filter circuit, ∆Vout = XC .
∆Vin R2 + XC2

(a) At f = 600 Hz, = 1 = 1 = 3.32 ×104 Ω
2π f 8.00 ×10−9
( )XC
C 2π (600 Hz) F

and ∆Vout = 3.32 × 104 Ω ≈ 1.00

( )∆Vin (90.0 Ω)2 + 3.32 × 104 Ω 2

(b) At f = 600 kHz, = 1 = 1 = 33.2 Ω
and 2π f 600 ×103 Hz
( )( )XC
C 2π 8.00 ×10−9 F

∆Vout = 33.2 Ω = 0.346

∆Vin (90.0 Ω)2 + (33.2 Ω)2

P33.49 ∆vout = R

( )∆vin
R2 + XL − XC 2

(a) At 200 Hz: 1 = (8.00 Ω)2 (8.00 Ω)2 1 400π C ]2
4
+ [400π L −

⎢⎣⎡8 1 ⎤2 FIG. P33.49(a)
000π ⎦⎥
At 4 000 Hz: (8.00 Ω)2 + 000π L − 8 C = 4 (8.00 Ω)2

At the low frequency, XL − XC < 0. This reduces to 400π L − 1 = −13.9 Ω [1]
400π C

For the high frequency half-voltage point, 8 000π L − 1 = +13.9 Ω [2]
8 000π C

Solving Equations (1) and (2) simultaneously gives C = 54.6 µF and L = 580 µH

(b) When XL = XC , ∆vout = ⎛ ∆vout ⎞ = 1.00
∆vin ⎜⎝ ∆vin ⎟⎠
max

XL = XC requires = 1 = 1 = 894 Hz
LC 5.80 × 10−4 H 5.46 × 10−5 F
( )( )(c) f0 2π 2π

(d) At 200 Hz, ∆vout = R = 1 and XC > XL , R ∆Vout
∆vin Z 2 ∆Vout
XL – XC f or f
XL – XC
so the phasor diagram is as shown: Z ∆Vin

φ = −cos−1 ⎛ R ⎞ = −cos−1 ⎛ 1 ⎞ so Z ∆Vin
⎜⎝ Z ⎠⎟ ⎜⎝ 2 ⎟⎠ or
ff

∆vout leads ∆vin by 60.0° . R

At f0 , XL = XC so FIG. P33.49(d)

∆vout and ∆vin have a phase difference of 0° .

At 4 000 Hz, ∆vout =R=1 and XL − XC >0
∆vin Z2

Thus, φ = cos−1 ⎛ 1⎞ = 60.0°
⎝ 2⎠

or ∆vout lags ∆vin by 60.0°

continued on next page

Alternating Current Circuits 261

(e) At 200 Hz and at 4 kHz,

∆vout,rms 2 = 2 = (1 2) ⎡⎣(1 2) ∆vin,max ⎦⎤ 2
R
( ) ( )P = R
( )1 2 ∆vin,rms = (10.0 V)2 = 1.56 W
8(8.00 Ω)
R

∆vout,rms 2 = 2 = (1 )2 ⎣⎡∆vin,max ⎤⎦ 2
R
( ) ( )At f0 , P = R (10.0 V)2
∆vin,rms = 2(8.00 Ω) = 6.25 W
R

( )(f ) We take Q = ω0 L = 2π f0 L = 2π (894 Hz) 5.80 ×10−4 H = 0.408
RR 8.00 Ω

Additional Problems

P33.50 The equation for ∆v(t) during the first period (using
y = mx + b) is:

∆v (t ) = (2 ∆Vmax )t − ∆Vmax

T

⎤ =1 T (∆Vmax )2 T ⎡2 ⎤ 2
⎦avg T 0 0 ⎢⎣T 1⎥⎦
T
∫ ∫⎡⎣(∆v
)2 ⎡⎣∆v (t )⎦⎤2 dt = t − dt FIG. P33.50

2 t =T 2 2
) ) )( ( (⎣⎡(∆v)2 ⎦⎤avg =
∆Vmax ⎛T⎞ [2t T − 1]3 = ∆Vmax ⎡⎣(+1)3 − (−1)3 ⎦⎤ = ∆Vmax
T ⎝ 2⎠ 6 3
3 t=0

( )∆Vrms = ⎡⎣(∆v)2 ⎤⎦avg = ∆Vmax 2 = ∆Vmax
33

*P33.51 (a) Z 2 = R2 + (XL – XC)2 760 2 = 4002 + (700 − XC)2 417 600 = (700 − XC)2
There are two values for the square root. We can have 646.2 = 700 − XC or −646.2 = 700 − XC.
XC can be 53.8 Ω or it can be 1.35 kΩ.

(b) If we were below resonance, with inductive reactance 700 Ω and capacitive reactance
1.35 kΩ, raising the frequency would increase the power. We must be above resonance,

with inductive reactance 700 Ω and capacitive reactance 53.8 Ω.

(c) 7602 = 2002 + (700 − XC)2 537 600 = (700 − XC)2 Here +733 = 700 − XC has no solution
so we must have −733.2 = 700 − XC and XC = 1.43 kΩ .

262 Chapter 33

P33.52 The angular frequency is ω = 2π 60 s = 377 s. When S is open, R, L, and C are in series with the

source:

R2 + (XL − )XC 2 = ⎛ ∆Vs ⎞ 2 = ⎛ 20 V ⎞ 2 = 1.194 × 104 Ω2 (1)
⎝ I⎠ ⎝ 0.183 A⎠

R
When S is in position 1, a parallel combination of two R’s presents equivalent resistance , in
2
series with L and C:

⎛ R⎞2 + (XL − )XC 2 = ⎛ 20 V ⎞ 2 = 4.504 × 103 Ω2 (2)
⎝ 2⎠ ⎝ 0.298 A⎠

When S is in position 2, the current by passes the inductor. R and C are in series with the source:

R2 + XC2 = ⎛ 20 V ⎞ 2 = 2.131 × 104 Ω2 (3)
⎝ 0.137 A⎠

Take equation (1) minus equation (2):

3 R2 = 7.440 × 103 Ω2 R = 99.6 Ω
4

Only the positive root is physical. We have shown than only one resistance value is possible.

Now equation (3) gives

= ⎡⎣2.131×104 − ( )2 ⎤1 2 Ω = 106.7 Ω= 1
⎦ ωC
XC 99.6 Only the positive root is physical and only

one capacitance is possible.

C = (ω XC )−1 = ⎡⎣(377 s)106.7 Ω⎦⎤ −1 = 2.49 × 10−5 F = C

Now equation (1) gives

− XC = ±⎣⎡1.194 ×104 − (99.6)2 ⎤1 2 Ω = ±44.99 Ω
⎦
XL

XL = 106.7 Ω + 44.99 Ω = 61.74 Ω or 151.7 Ω = ω L
L = XL = 0.164 H or 0.402 H = L

ω
Two values for self-inductance are possible.

ω0 = 1= 1 = 2 000 s−1
LC
(0.050 0 H) 5.00 × 10−6 F
( )P33.53

so the operating angular frequency of the circuit is

ω = ω0 = 1 000 s−1
2

2 Rω 2
() )(Using Equation 33.37, P
= ∆Vrms ω2 − ω 2 2
R2ω 2 + L2 0

P = (400)2 (8.00)(1 000)2 = 56.7 W FIG. P33.53
(8.00)2 (1 000)2 + (0.050 0)2 ⎣⎡(1.00 − 4.00) × 106 ⎤⎦ 2

Alternating Current Circuits 263

*P33.54 (a) At the resonance frequency XL and XC are equal. The certain frequency must be higher
(b) than the resonance frequency for the inductive reactance to be the greater.

It is possible to determine the values for L and C, because we have three independent
equations in the three unknowns L, C, and the unknown angular frequency w.
The equations are
2 0002 = 1/LC 12 = w L and 8 = 1/w C

(c) We eliminate w = 12/L to have 8 w C = 1 = 8(12/L)C = 96C/L so L = 96C

Then 4 000 000 = 1/96 C 2 so C = 51.0 mF and L = 4.90 mH

*P33.55 The lowest-frequency standing-wave state is NAN. The distance between the clamps we

represent as d = dNN = λ . The speed of transverse waves on the string is v = fλ = T = f 2d.
2 µ

The magnetic force on the wire oscillates at 60 Hz, so the wire will oscillate in resonance at 60 Hz.

T = (60 s)2 4d2 )(T = 274 kg m ⋅s2 d2

0.019 kg m

Any values of T and d related according to this expression will work, including
if d = 0.200 m T = 10.9 N . We did not need to use the value of the current and magnetic

field. If we assume the subsection of wire in the field is 2 cm wide, we can find the rms value of
the magnetic force:

FB = I B sinθ = (9 A)(0.02 m)(0.015 3T )sin 90° = 2.75 mN

So a small force can produce an oscillation of noticeable amplitude if internal friction is small.

*P33.56 φ = tan−1 ⎛ ω L −1/ ωC⎞ changes from −90° for w = 0 to 0 f
⎝ R⎠ 90°

at the resonance frequency to +90° as w goes to infinity. −90°

The slope of the graph is df/dw:

dφ = 1 1 ⎛ L − 1 (−1) 1 ⎞ At resonance w0 w
dω − 1/ωC R ⎝⎜ C ω2 ⎠⎟
1 + ⎛ ω L R ⎞2
⎜⎝ ⎟⎠

we have w L = 1/w0C and LC = 1/w02. FIG. P33.56
0

Substituting, the slope at the resonance point is

dφ = 1 1 ⎛ L + 1 LC⎞⎠ = 2L = 2Q
dω 1+ 02 R ⎝ C R ω0
ω0

P33.57 (a) When ωL is very large, the bottom branch carries negligible current. Also, 1 will be
ωC

negligible compared to 200 Ω and 45.0 V = 225 mA flows in the power supply and the
200 Ω

top branch.

(b) Now 1 → ∞ and ωL → 0 so the generator and bottom branch carry 450 mA .
ωC

264 Chapter 33

P33.58 (a) With both switches closed, the current goes only through
generator and resistor.

i (t ) = ∆Vmax cosω t

R

(b) )(P = 1 ∆Vmax 2

2R

i(t) = ∆Vmax cos ⎣⎢⎡ω t −1 ⎛ ωL ⎞ ⎤ FIG. P33.58
R2 + ω 2 L2 ⎝ R ⎠ ⎥⎦
(c) + tan

(d) For 0 = φ = tan−1 ⎛ ω0 L − (1 ω0C )⎞
⎜⎝
R ⎠⎟

We require ω0 L = 1 , so C = 1
ω0C
ω 2 L
0

(e) At this resonance frequency, Z = R

(f ) U = 1 C (∆VC )2 = 1 CI 2 XC2
2 2

( ) ( )Umax 2 ∆Vmax 2 L
1 1C ∆Vmax 1 2R2
= 2 CI 2 XC2 = 2 R2 ω02C 2 =
max

( )(g) 2
1 1L ∆Vmax
Umax = 2 LI 2 = 2 R2
max

(h) Now ω = 2ω0 = 2.
LC

So φ = tan−1 ⎛ ω L − (1 ωC)⎞ = tan−1 ⎛ 2 L C − (1 2) L C⎞ = tan−1 ⎛ 3 L⎞ .
⎝⎜ ⎝⎜ ⎟⎠ ⎝⎜ 2R C ⎠⎟
R ⎟⎠ R

(i) Now ωL = 1 1 ω = 1 = ω0
2 ωC 2LC 2

P33.59 (a) I R,rms = ∆Vrms = 100 V = 1.25 A IR ∆V
(b) R 80.0 Ω f
I
The total current will lag the applied voltage as seen in the phasor IL
diagram at the right. FIG. P33.59

= ∆Vrms = 100 V = 1.33 A
XL 2π 60.0 s−1
( )IL,rms (0.200 H)

Thus, the phase angle is: φ = tan−1 ⎛ I L , rms ⎞ = tan−1 ⎛ 1.33 A ⎞ = 46.7° .
⎜⎜⎝ I R,rms ⎟⎟⎠ ⎜⎝ 1.25 A ⎟⎠

Alternating Current Circuits 265

P33.60 Suppose each of the 20 000 people uses an average power of 500 W. (This means 12 kWh per day,

or $36 per 30 days at 10c| per kWh.) Suppose the transmission line is at 20 kV. Then

I rms = P = (20 000)(500 W) ~103 A

∆Vrms 20 000 V

If the transmission line had been at 200 kV, the current would be only ~102 A .

P33.61 R = 200 Ω, L = 663 mH, C = 26.5 µF, ω = 377 s−1, ∆Vmax = 50.0 V

ω L = 250 Ω, ⎜⎝⎛ω1C ⎞ = 100 Ω, Z = R2 + ( XL − )XC 2 = 250 Ω
⎟
⎠

(a) I max = ∆Vmax = 50.0 V = 0.200 A
Z 250 Ω

φ = tan−1 ⎛ XL − XC ⎞ = 36.8° (∆V leads I)
⎝ R ⎠

(b) ∆VR,max = Imax R = 40.0 V at φ = 0°

(c) ∆VC , max = I max = 20.0 V at φ = −90.0° (I leads ∆V)
ωC

(d) ∆VL,max = ImaxωL = 50.0 V at φ = +90.0° (∆V leads I )

P33.62 L = 2.00 H, C = 10.0 × 10−6 F, R = 10.0 Ω, ∆v (t) = (100 sinω t)

(a) The resonant frequency ω0 produces the maximum current and thus the maximum power
delivery to the resistor.

1= 1 = 224 rad s
LC
(2.00) 10.0 × 10−6
)(ω0 =

(b) P = ( )∆Vmax 2 = (100)2 = 500 W
2(10.0)
2R

∆Vrms ∆Vrms ∆Vrms
Z R2 + ω L − 1 ω C 2 R
( ) ( )( )(c)I rms= = and I rms max =

2 2

R= 1 R
2
( ) ( )or
( )I2 R = 1 I2 R ∆Vrms ∆Vrms
rms 2 rms Z2 R2
max

⎛ 1 ⎞ 2
⎜⎝ ωC ⎠⎟
This occurs where Z 2 = 2R2: R2 + ω L − = 2R2

ω 4 L2C 2 − 2Lω 2C − R2ω 2C 2 + 1 = 0 ( )or L2C2ω 4 − 2LC + R2C2 ω 2 + 1 = 0

( ) ( ) ( )⎡⎣(2.00)2 2 ⎤ 2 ⎦⎤ω 2 + 1 = 0
10.0 × 10−6 ⎦ ω 4 − ⎡⎣2(2.00) 10.0 × 10−6 + (10.0)2 10.0 × 10−6

Solving this quadratic equation, we find that ω 2 = 51130, or 48 894

ω1 = 48 894 = 221 rad s and ω2 = 51130 = 226 rad s

266 Chapter 33 N1 = ∆V1 .
N2 ∆V2
P33.63 (a) From Equation 33.41,
Z1 = ∆V1 and the output impedance Z2 = ∆V2
Let input impedance I1 I2
so that
N1 = Z1I1 But from Eq. 33.42, I1 = ∆V2 = N2
N2 Z2I2 I2 ∆V1 N1

So, combining with the previous result we have N1 = Z1 .
N2 Z2

(b) N1 = Z1 = 8 000 = 31.6
N2 Z2 8.00

⎛ ∆Vrms ⎞ 2 (120 V)2 ⎛⎜⎝ ω 1 ⎞ 2
⎝ Z ⎠ ωC ⎟⎠
P33.64 P = I2 R = R, so 250 W = Z 2 (40.0 Ω): Z= R2 + L −
rms

(120)2 (40.0) and
( )250 =
(40.0)2 + ⎣⎡2π f (0.185) − ⎣⎡1 2π f 65.0 ×10−6 ⎤⎦ ⎤ 2
⎦

576 000 f 2
1.162 4 f 2 − 2 448.5
( )250 = + 2
1 600 f 2

1 = 1 600 f 2 + 1.3511 2 304 f2 f 2 + 5 995 300 so 1.3511 f 4 − 6 396.3 f 2 + 5 995 300 = 0
f4 −5 692.3

f 2 = 6 396.3 ± (6 396.3)2 − 4(1.3511)(5 995 300) = 3 446.5 or 1 287.4

2(1.3511)

f = 58.7 Hz or 35.9 Hz There are two answers because we could be above resonance or

below resonance.

P33.65 IR = ∆Vrms ; = ∆Vrms ∆Vrms
R ωL ω C −1
IL ; )(IC =

(a) Irms = I 2 + (IC − IL )2 = ∆Vrms ⎛ 1 ⎞ + ⎜⎝⎛ωC − 1 ⎞2
R ⎝⎜ R2 ⎟⎠ ωL ⎠⎟

(b) tanφ = IC − IL = ∆Vrms ⎡ 1 − 1 ⎤⎛ 1 ⎞
IR ⎢ XC XL ⎥ ∆Vrms R ⎠⎟
⎣ ⎦ ⎜⎝

tan φ = ⎡ 1 − 1 ⎤
R⎢ XC XL ⎥
⎦
⎣

FIG. P33.65

Alternating Current Circuits 267

*P33.66 An RLC series circuit, containing a 35.0-Ω resistor, a 205-mH inductor,
a capacitor, and a power supply with rms voltage 200 V and frequency
100 Hz, carries rms current 4.00 A. Find the capacitance of the capacitor.

We solve for C FIG. P33.66
2500 = 352 + (129 − 1/628C)2 1275 = (129 − 1/628C)2

There are two possibilities: 35.7 = 129 − 1/628C and −35.7 = 129 − 1/628C
1/628C = 93.1 or 1/628C = 164.5

C = either 17.1 m F or 9.67 mF

P33.67 (a) XL = XC = 1 884 Ω when f = 2 000 Hz f (Hz) XL (Ω) XC (Ω) Z (Ω)
300 283 12 600 1 2300
L = XL = 1 884 Ω = 0.150 H and 600 565 6 280 5 720
2π f 4 000π rad s 800 754 4 710 3 960

C = (2π 1 = (4 000π 1 Ω) = 42.2 nF 1 000 942 3 770 2 830
1 500 1 410 2 510 1100
f ) XC rad s)(1 884 2 000 1 880 1 880 40
3 000 2 830 1 260 1 570
1 4 000 3 770 942 2 830
4.22 ×10−8 6 000 5 650 628 5 020
( )XL = 2π f (0.150 H) 10 000 9 420 377 9 040
XC = (2π f ) F

Z = (40.0 Ω)2 + ( XL − XC )2

(b) Impedance, Ω

FIG. P33.67(b)

268 Chapter 33

P33.68 ω0 = 1 = 1.00 × 106 rad s ω ωL (Ω) 1 (Ω) Z (Ω) P = I 2R (W)
LC ω0
0.9990 999.0 ωC 2.24 0.19984
For each angular frequency, we find 0.9991 999.1 1001.0 2.06 0.23569
0.9993 999.3 1000.9 1.72 0.33768
Z = R2 + (ωL −1 ωC)2 0.9995 999.5 1000.7 1.41 0.49987
0.9997 999.7 1000.5 1.17 0.73524
then I = 1.00 V 0.9999 999.9 1000.3 1.02 0.96153
Z 1.0000 1000 1000.1 1.00 1.00000
1.0001 1000.1 1000.0 1.02 0.96154
and P = I 2 (1.00 Ω) 1.0003 1000.3 1.17 0.73535
1.0005 1000.5 999.9 1.41 0.50012
The full width at half maximum is: 1.0007 1000.7 999.7 1.72 0.33799
1.0009 1000.9 999.5 2.06 0.23601
∆f = ∆ω = (1.000 5 − 0.999 5)ω0 1.0010 1001 999.3 2.24 0.20016
999.1
2π 2π 999.0
∆f = 1.00 × 103 s−1 = 159 Hz

2π

while

1.00 Ω
1.00 × 10−3 H
)(R = = 159 Hz

2π L 2π

1.0 0.998 1 1.002 1.004
w/w 0
0.8

I 2R 0.6
(W) 0.4

0.2

0.0
0.996

FIG. P33.68

*P33.69 (a) We can use sin A + sin B = 2 sin ⎛ A + B⎞ cos ⎛ A − B⎞ to find the sum of the two sine
⎝ 2 2⎠ ⎝ 2 2⎠

functions to be

E1 + E2 = (24.0 cm)sin(4.5t + 35.0°)cos35.0°
E1 + E2 = (19.7 cm)sin(4.5t + 35.0°)

Thus, the total wave has amplitude 19.7 cm and has a constant phase difference of

35.0° from the first wave.

(b) In units of cm, the resultant phasor is y wt

( ) ( )yR = y1 + y2 = 12.0ˆi + 12.0 cos 70.0°ˆi +12.0 sin 70.0ˆj yR y2
= 16.1ˆi +11.3ˆj
α 70.0°
yR = (16.1)2 + (11.3)2 at tan−1 ⎜⎝⎛ 11.3 ⎠⎟⎞ = 19.7 cm at 35.0°
16.1 y1 x

The answers are identical. FIG. P33.69(b)

continued on next page

Alternating Current Circuits 269

(c) yR = 12.0 cos 70.0°ˆi +12.0 sin 70.0°ˆj y
+15.5 cos 80.0°ˆi −15.5 sin 80.0°ˆj
+17.0 cos160°ˆi +17.0 sin160°ˆj y1 kx - w t

yR = −9.18ˆi +1.83ˆj = 9.36 cm at 169° yR y2
y3
The wave function of the total wave is x

yR = (9.36 cm)sin(15x − 4.5t + 169°). FIG. P33.69(c)

ANSWERS TO EVEN PROBLEMS

P33.2 (a)193 Ω (b) 144 Ω

P33.4 (a) 25.3 rad/s (b) 0.114 s

P33.6 3.38 W

P33.8 3.14 A

P33.10 3.80 J

P33.12 (a) greater than 41.3 Hz (b) less than 87.5 Ω

P33.14 2C (∆Vrms )

P33.16 –32.0 A

P33.18 2.79 kHz

P33.20 (a) 109 Ω (b) 0.367 A (c) Imax = 0.367 A, ω = 100 rad s , φ = −0.896 rad
P33.22 19.3 mA

P33.24 (a) 146 V (b) 212 V (c) 179 V (d) 33.4 V

P33.26 Cutting the plate separation in half doubles the capacitance and cuts in half the capacitive

reactance to X / 2. The new impedance must be half as large as the old impedance for the new
C

current to be doubled. For the new impedance we then have

(R2 + [R − X / 2]2)1 / 2 = 0.5(R2 + [R − XC]2 )1 / 2. Solving yields X = 3R.
C C

P33.28 353 W

P33.30 (a) 5.43 A (b) 0.905 (c) 281 µF (d) 109 V

P33.32 (a) 0.936 (b) Not in practice. If the inductor were removed or if the generator were replaced
with a battery, so that either L = 0 or f = 0, the power factor would be 1, but we would not have
a magnetic buzzer. (c) 70.4 mF

P33.34 (a) 633 fF (b) 8.46 mm (c) 25.1 Ω

270 Chapter 33

P33.36 (a) 3.56 kHz (b) 5.00 A (c) 22.4 (d) 2.24 kV

P33.38 )(4π ∆Vrms 2 RC LC (c) 42.0 W
P33.40
4R2C + 9L

(a) 9.23 V (b) 4.55 A

P33.42 (a) 1 600 turns (b) 30.0 A (c) 25.3 A

P33.44 (a) 0.34 (b) 5.3 W (c) $4.8

P33.46 (a) See the solution. (b) 1; 0 (c) 3
P33.48 (a) 1.00 (b) 0.346 2π RC

P33.50 See the solution.

P33.52 Only one value for R and only one value for C are possible. Two values for L are possible.
R = 99.6 Ω, C = 24.9 mF, and L = 164 mH or 402 mH

P33.54 (a) Higher. At the resonance frequency XL = XC. As the frequency increases, XL goes up and
XC goes down. (b) It is. We have three independent equations in the three unknowns L, C, and
the certain f. (c) L = 4.90 mH and C = 51.0 m F

P33.56 See the solution. (b) P = ( )∆Vmax 2 (c) i(t ) = ∆Vmax cos ⎢⎡⎣ω t + tan −1 ⎛ ωL ⎞ ⎤
P33.58 R2 + ω 2 L2 ⎝ R ⎠ ⎦⎥
(a) i (t ) = ∆Vmax cosω t 2R

R

∆Vmax 2 L ∆Vmax 2 L
2R2 2R2
) )(d)= 1 ((f) ((g) tan−1 ⎛ 3 L⎞ (i) 1
C (e) Z = R (h) ⎜⎝ 2R C ⎠⎟ 2LC
ω 2 L
0

P33.60 ~103 A

P33.62 (a) 224 rad s (b) 500 W (c) 221 rad s and 226 rad s

P33.64 The frequency could be either 58.7 Hz or 35.9 Hz. We can be either above or below resonance.

P33.66 An RLC series circuit, containing a 35.0-Ω resistor, a 205-mH inductor, a capacitor, and a
power supply with rms voltage 200 V and frequency 100 Hz, carries rms current 4.00 A. Find the
capacitance of the capacitor. Answer: It could be either 17.1 mF or 9.67 mF.

P33.68 See the solution.

34

Electromagnetic Waves

CHAPTER OUTLINE ANSWERS TO QUESTIONS

34.1 Displacement Current and the *Q34.1 Maxwell included a term in Ampère’s law to account
General Form of Ampère’s Law for the contributions to the magnetic field by changing
electric fields, by treating those changing electric fields
34.2 Maxwell’s Equations and Hertz’s as “displacement currents.”
Discoveries
*Q34.2 No, they do not. Specifically, Gauss’s law in magnetism
34.3 Plane Electromagnetic Waves prohibits magnetic monopoles. If magnetic monopoles
34.4 Energy Carried by existed, then the magnetic field lines would not have to be
closed loops, but could begin or terminate on a magnetic
Electromagnetic Waves monopole, as they can in Gauss’s law in electrostatics.
34.5 Momentum and Radiation

Pressure
34.6 Production of Electromagnetic

Waves by an Antenna
34.7 The Spectrum of Electromagnetic

Waves

Q34.3 Radio waves move at the speed of light. They can

travel around the curved surface of the Earth, bouncing

between the ground and the ionosphere, which has an

altitude that is small when compared to the radius of the

Earth. The distance across the lower forty-eight states is

approximately 5 000 km, requiring a transit time of

5 × 106 m ~ 10−2 s . To go halfway around the Earth
3 × 108 m s

takes only 0.07 s. In other words, a speech can be heard

on the other side of the world before it is heard at the

back of a large room.

Q34.4 Energy moves. No matter moves. You could say that electric and magnetic fields move, but it is
nicer to say that the fields at one point stay at that point and oscillate. The fields vary in time, like
sports fans in the grandstand when the crowd does the wave. The fields constitute the medium for
the wave, and energy moves.

Q34.5 The changing magnetic field of the solenoid induces eddy currents in the conducting core. This
is accompanied by I 2R conversion of electrically-transmitted energy into internal energy in the

conductor.

*Q34.6 (i) According to f = (2π )−1 (LC)−1/2, to make f half as large, the capacitance should be made four
times larger. Answer (a).

(ii) Answer (b).

*Q34.7 Answer (e). Accelerating charge, changing electric field, or changing magnetic field can be the
source of a radiated electromagnetic wave.

*Q34.8 (i) Answer (c). (ii) Answer (c). (iii) Answer (c). (iv) Answer (b). (v) Answer (b).

*Q34.9 (i) through (v) have the same answer (c).

271

272 Chapter 34

Q34.10 Sound Light

The world of sound extends to the top of The universe of light fills the whole universe.
the atmosphere and stops there; sound requires Light moves through materials, but faster in a
a material medium. Sound propagates by a vacuum. Light propagates by a chain reac-
chain reaction of density and pressure dis- tion of electric and magnetic fields recreating
turbances recreating each other. Sound in each other. Light in air moves at hundreds of
air moves at hundreds of meters per second. millions of meters per second. Visible light
Audible sound has frequencies over a range has frequencies over a range of less than one
of three decades (ten octaves) from 20 Hz to octave, from 430 to 750 Terahertz. Visible
20 kHz. Audible sound has wavelengths of light has wavelengths of very small size
ordinary size (1.7 cm to 17 m). Sound waves (400 nm to 700 nm). Light waves are
are longitudinal. transverse.

Sound and light can both be reflected, refracted, or absorbed to produce internal energy. Both
have amplitude and frequency set by the source, speed set by the medium, and wavelength set by
both source and medium. Sound and light both exhibit the Doppler effect, standing waves, beats,
interference, diffraction, and resonance. Both can be focused to make images. Both are described
by wave functions satisfying wave equations. Both carry energy. If the source is small, their
intensities both follow an inverse-square law. Both are waves.

Q34.11 The Poynting vector S describes the energy flow associated with an electromagnetic wave. The
direction of S is along the direction of propagation and the magnitude of S is the rate at which
electromagnetic energy crosses a unit surface area perpendicular to the direction of S .

*Q34.12 (i) Answer (b). Electric and magnetic fields must both carry the same energy, so their amplitudes
are proportional to each other. (ii) Answer (a). The intensity is proportional to the square of the
amplitude.

*Q34.13 (i) Answer (c). Both the light intensity and the gravitational force follow inverse-square laws.

(ii) Answer (a). The smaller grain presents less face area and feels a smaller force due to light
pressure.

Q34.14 Photons carry momentum. Recalling what we learned in Chapter 9, the impulse imparted to a
particle that bounces elastically is twice that imparted to an object that sticks to a massive wall.
Similarly, the impulse, and hence the pressure exerted by a photon reflecting from a surface must
be twice that exerted by a photon that is absorbed.

Q34.15 Different stations have transmitting antennas at different locations. For best reception align your
rabbit ears perpendicular to the straight-line path from your TV to the transmitting antenna. The
transmitted signals are also polarized. The polarization direction of the wave can be changed by
reflection from surfaces—including the atmosphere—and through Kerr rotation—a change in
polarization axis when passing through an organic substance. In your home, the plane of polar-
ization is determined by your surroundings, so antennas need to be adjusted to align with the
polarization of the wave.

Q34.16 Consider a typical metal rod antenna for a car radio. The rod detects the electric field portion of
the carrier wave. Variations in the amplitude of the incoming radio wave cause the electrons in
the rod to vibrate with amplitudes emulating those of the carrier wave. Likewise, for frequency
modulation, the variations of the frequency of the carrier wave cause constant-amplitude vibra-
tions of the electrons in the rod but at frequencies that imitate those of the carrier.

Electromagnetic Waves 273

*Q34.17 (i) Gamma rays have the shortest wavelength. The ranking is a < g < e < f < b < c < d.
(ii) Gamma rays have the highest frequency: d < c < b < f < e < g < a.
(iii) All electromagnetic waves have the same physical nature. a = b = c = d = e = f = g.

Q34.18 The frequency of EM waves in a microwave oven, typically 2.45 GHz, is chosen to be in a
band of frequencies absorbed by water molecules. The plastic and the glass contain no water
molecules. Plastic and glass have very different absorption frequencies from water, so they
may not absorb any significant microwave energy and remain cool to the touch.

Q34.19 People of all the world’s races have skin the same color in the infrared. When you blush or
exercise or get excited, you stand out like a beacon in an infrared group picture. The brightest
portions of your face show where you radiate the most. Your nostrils and the openings of your ear
canals are bright; brighter still are just the pupils of your eyes.

Q34.20 Light bulbs and the toaster shine brightly in the infrared. Somewhat fainter are the back of the
refrigerator and the back of the television set, while the TV screen is dark. The pipes under the sink
show the same weak sheen as the walls until you turn on the faucets. Then the pipe on the right
turns very black while that on the left develops a rich glow that quickly runs up along its length.
The food on your plate shines; so does human skin, the same color for all races. Clothing is dark as
a rule, but your bottom glows like a monkey’s rump when you get up from a chair, and you leave
behind a patch of the same blush on the chair seat. Your face shows you are lit from within, like a
jack-o-lantern: your nostrils and the openings of your ear canals are bright; brighter still are just
the pupils of your eyes.

Q34.21 12.2-cm waves have a frequency of 2.46 GHz. If the Q value of the phone is low (namely if it is
cheap), and your microwave oven is not well shielded (namely, if it is also cheap), the phone can
likely pick up interference from the oven. If the phone is well constructed and has a high Q value,
then there should be no interference at all.

SOLUTIONS TO PROBLEMS

Section 34.1 Displacement Current and the General Form of Ampère’s Law

*P34.1 (a) dΦE = dQ dt = I = (0.100 A) = 11.3 × 109 V ⋅ m s
dt ∈0 ∈0
8.85 × 10−12 C2 N ⋅ m2

(b) Id = ∈0 dΦE = I = 0.100 A
dt

*P34.2 dΦE = d (EA) = dQ dt = I
dt dt ∈0 ∈0

(a) dE = I = 7.19 × 1011 V m ⋅ s
dt ∈0 A

∫(b) B ⋅ ds = ∈0 µ0 dΦE so 2π rB = ∈0 µ0 d ⎡Q ⋅ π r 2 ⎤
dt dt ⎢ ⎥
⎣ ∈0 A ⎦

( )B
= µ0 Ir = µ0 (0.200) 5.00 × 10−2 = 2.00 × 10−7 T
2A 2π (0.100)2

274 Chapter 34

*P34.3 We use the extended form of Ampère’s law, Equation 34.7.
Since no moving charges are present, I = 0 and we have

∫ B⋅d = µ0 ∈0 dΦE .
dt

In order to evaluate the integral, we make use of the symmetry FIG. P34.3
of the situation. Symmetry requires that no particular direction
from the center can be any different from any other direction.
Therefore, there must be circular symmetry about the central axis.
We know the magnetic field lines are circles about the axis.
Therefore, as we travel around such a magnetic field circle, the
magnetic field remains constant in magnitude. Setting aside until

later the determination of the direction of B, we integrate ∫ B ⋅ dl

around the circle

at R = 0.15 m

to obtain 2π RB

Differentiating the expression ΦE = AE
we have
dΦE ⎛ πd 2 ⎞ dE
dt ⎝⎜ 4 ⎠⎟ dt
=

⎛ π d 2 ⎞ dE
⎜ ⎟ dt
Thus, ∫ B⋅d = 2π RB = µ0 ∈0 ⎜ ⎟
Solving for B gives ⎝⎜ 4 ⎟⎠

µ0 ∈0 ⎛ π d 2 ⎞ dE
2π R ⎜ ⎟ dt
B = ⎜ ⎟
⎝⎜ 4 ⎠⎟

( ) ( )Substituting numerical values, B = 4π ×10−7 H m 8.85 ×10−12 F m ⎡⎣π (0.10 m)2 ⎤⎦(20 V m ⋅ s)

2π (0.15 m) (4)

B = 1.85 × 10−18 T

In Figure P34.3, the direction of the increase of the electric field is out the plane of the
paper. By the right-hand rule, this implies that the direction of B is counterclockwise.

Thus, the direction of B at P is upwards .

Electromagnetic Waves 275

Section 34.2 Maxwell’s Equations and Hertz’s Discoveries

P34.4 (a) The rod creates the same electric field that it would if stationary.
We apply Gauss’s law to a cylinder of radius r = 20 cm and length :

∫ E ⋅ dA = qinside

∈0

E (2π rl) cos 0° = λl FIG. P34.4

∈0

(( ) )E =
λ r radially outward = 2π 35 × 10 −9 C m N⋅m 2 ) ˆj = 3.15 × 103 ˆj N C
2π ∈0 8.85 × 10−12 C2
(0.2 m

) )( ((b) The charge in motion constitutes a current of 35 × 10−9 C m 15 × 106 m s = 0.525 A.

This current creates a magnetic field.

B = µ0I ( )= 4π ×10−7 T ⋅ m A (0.525 A) kˆ = 5.25 ×10−7 kˆ T
2π r
2π (0.2 m)

(c) The Lorentz force on the electron is F = qE + qv × B

( )( )F = −1.6 ×10−19 C 3.15 ×103 ˆj N C

( )( )+ −1.6 ×10−19 C 240 ×106 ˆi m s

× ⎛ 5.25 × 10−7 kˆ N⋅s ⎞
⎝⎜ C⋅m ⎠⎟

( ) ( ) ( )F = 5.04 ×10−16 −ˆj N + 2.02 ×10−17 +ˆj N = 4.83×10−16 −ˆj N

*P34.5 F = ma = qE + qv × B

e ˆi ˆj kˆ
m 0
a = ⎣⎡E + v × B⎤⎦ where v × B = 200 0.300 0 = −200 (0.400) ˆj + 200 (0.300) kˆ

0.200 0.400

a = 1.60 × 10−19 ⎡⎣50.0ˆj − 80.0ˆj + 60.0kˆ ⎦⎤ = 9.58 × 107 ⎡⎣−30.0ˆj + 60.0kˆ ⎤⎦
1.67 × 10−27

( )a = 2.87 × 109 ⎣⎡−ˆj + 2kˆ ⎤⎦ m s2 = −2.87 × 109 ˆj + 5.75 × 109kˆ m s2

−e ˆi ˆj kˆ
m
*P34.6 F = ma = qE + qv × B so a = ⎡⎣E + v × B⎦⎤ where v × B = 10.0 0 0 = −4.00ˆj

0 0 0.400

( ) ( )a =
−1.60 × 10−19 ⎣⎡2.50ˆi + 5.00ˆj − 4.00ˆj⎦⎤ = −1.76 × 1011 ⎡⎣2.50ˆi + 1.00ˆj⎤⎦
9.11 × 10−31

( )a = −4.39 × 1011 ˆi − 1.76 × 1011 ˆj m s2

276 Chapter 34

Section 34.3 Plane Electromagnetic Waves

P34.7 (a) Since the light from this star travels at 3.00 ×108 m s

the last bit of light will hit the Earth in 6.44 × 1018 m = 2.15 × 1010 s = 680 years.
3.00 × 108 m s

Therefore, it will disappear from the sky in the year 2 007 + 680 = 2.69 × 103 A.D.
The star is 680 light-years away.

(b) ∆t = ∆x = 1.496 × 1011 m = 499 s = 8.31 min
v 3 × 108 m s

)((c)
∆t = ∆x = 2 3.84 × 108 m = 2.56 s
v 3 × 108 m s

)((d)
∆t = ∆x = 2π 6.37 × 106 m = 0.133 s
v 3 × 108 m s

(e) ∆t = ∆x = 10 × 103 m = 3.33 × 10−5 s
v 3 × 108 m s

P34.8 v = 1 = 1 c = 0.750c = 2.25 × 108 m s
κµ0 ∈0 1.78

P34.9 (a) f λ = c or f (50.0 m) = 3.00 × 108 m s

so f = 6.00 × 106 Hz = 6.00 MHz

(b) E = c or 22.0 = 3.00 × 108
B Bmax
so Bmax = −73.3kˆ nT

(c) k = 2π = 2π = 0.126 m−1
λ 50.0

( )and ω = 2π f = 2π 6.00 × 106 s−1 = 3.77 × 107 rad s
( )B = Bmax cos(kx −ω t) = −73.3cos 0.126x − 3.77 ×107 t kˆ nT

P34.10 E = c or 220 = 3.00 × 108
BB
so B = 7.33 × 10−7 T = 733 nT

Electromagnetic Waves 277

P34.11 (a) B = E = 100 V m s = 3.33 × 10−7 T = 0.333 µT
c 3.00 × 108 m

(b) λ = 2π = 2π m−1 = 0.628 µm
k 1.00 × 107

(c) f = c = 3.00 × 108 m s = 4.77 × 1014 Hz
λ 6.28 × 10−7 m

P34.12 E = Emax cos (kx − ω t )

∂E = −Emax sin (kx −ωt)(k)
∂x

∂E = −Emax sin (kx − ωt) (−ω)
∂t

( )∂2E k2

∂x 2
= −Emax cos (kx − ωt)

∂2 E = −Emax cos (kx − ωt) (−ω)2
∂t 2

We must show: ∂E = µ0 ∈0 ∂2E
∂x 2 ∂t 2

That is, ( )− k2 Emax cos (kx − ω t ) = −µ0 ∈0 (−ω )2 Emax cos (kx − ωt )

But this is true, because k2 = ⎛ 1 ⎞2 = 1 = µ0 ∈0
ω2 ⎜⎝ f λ ⎟⎠ c2

The proof for the wave of magnetic field follows precisely the same steps.

P34.13 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 = 3.00 ×108 m s = 7.50 ×107 Hz = 75.0 MHz
λ 4.00 m

P34.14 dA to A = 6 cm ± 5% = λ
2

λ = 12 cm ± 5%

( )v = λ f = (0.12 m ± 5%) 2.45 × 109 s−1 = 2.9 × 108 m s ± 5%

278 Chapter 34

Section 34.4 Energy Carried by Electromagnetic Waves

P34.15 S = I = U = Uc = uc Energy = u = I = 1 000 W m2 = 3.33 µJ m3
At V Unit Volume c 3.00 × 108 m s

P34.16 Sav = P = 4π 4.00 × 103 W = 7.68 µW m2

4π r2 (4.00 × 1 609 m)2

Emax = 2µ0cSav = 0.076 1 V m

∆Vmax = Emax L = (76.1 mV m)(0.650 m) = 49.5 mV (amplitude) or 35.0 mV (rms)
P34.17 r = (5.00 mi)(1 609 m mi) = 8.04 × 103 m

= P = 250 × 103 W = 307 µW m2
8.04 × 103 W
4π r2
( )S 4π 2

P = energy = 600 kWh = 600 × 103 (J s)h ⎝⎛⎜ 1d ⎟⎞⎠ = 6.75 W m2
∆t ⋅area 24 h
area d)(13 m)(9.5 30 d 123.5 m2
( )*P34.18 (a) (30 m)

(b) The car uses gasoline at the rate (55 mi h) ⎛ gal ⎞ . Its rate of energy conversion is
⎝ 25 mi ⎠

P = 44 × 106 J kg ⎛ 2.54 kg ⎞ (55 mi h)⎛⎜⎝ gal ⎟⎠⎞ ⎛ 3 1h ⎞ = 6.83 × 104 W . Its power-
⎝⎜ 1 gal ⎟⎠ 25 mi ⎜⎝ 600 s ⎟⎠

per-footprint-area is P = 6.83×104 W = 6.64 ×103 W m2 .

area 2.10 m (4.90 m)

(c) For an automobile of typical weight and power to run on sunlight, it would have to carry
a solar panel huge compared to its own size. Rather than running a conventional car, it is
much more natural to use solar energy for agriculture, forestry, lighting, space heating,
drying, water purification, water heating, and small appliances.

P34.19 Power output = (power input)(efficiency).

Thus, Power input = Power out = 1.00 × 106 W = 3.33 × 106 W
eff 0.300

and A = P = 3.33×106 W = 3.33×103 m2

I 1.00 ×103 W m2

P34.20 I = B2 c = P
max
4π r2
2µ0

( ( ) ) (( ))Bmax =
⎛ P 2 ⎞ ⎛ 2µ0 ⎞ = 10.0 ×103 (2) 4π ×10−7 = 5.16 ×10−10 T
⎝⎜ ⎟⎠ ⎜⎝ c ⎠⎟
4πr 4π 5.00 ×103 2 3.00 ×108

Since the magnetic field of the Earth is approximately 5 × 10−5 T, the Earth’s field is some
100 000 times stronger.

Electromagnetic Waves 279

P34.21 (a) P = I 2R = 150 W

( )A = 2π rL = 2π 0.900 × 10−3 m (0.080 0 m) = 4.52 × 10−4 m2

S = P = 332 kW m2 (points radially inward)

A

B = µ0I = µ0 (1.00) = 222 µT
2πr 2π
0.900 × 10−3
( )(b)

E = ∆V = IR = 150 V = 1.88 kV m
∆x L 0.080 0 m

Note: S = EB = 332 kW m2
µ0

( ( )( ) )P34.22
(a) I= E2 = 3 × 106 V m 2 ⎛ J ⎞2 ⎛ C ⎞ ⎛ T⋅C⋅m⎞ ⎛ N⋅m⎞
max 4π × 10−7 T ⋅ m A 3 × 108 ⎝ V⋅C⎠ ⎝ A⋅s⎠ ⎝ N⋅s ⎠ ⎝ J ⎠

2µ0c 2 m s

I = 1.19 × 1010 W m2

⎛ 5 × 10−3 m ⎞ 2
⎜⎝ 2 ⎠⎟
( )(b)
P = IA = 1.19 × 1010 W m2 π = 2.34 × 105 W

( ) ( )P34.23 (a) E ⋅ B = 80.0ˆi + 32.0ˆj − 64.0kˆ (N C) ⋅ 0.200ˆi + 0.080 0ˆj + 0.290kˆ µT

E ⋅ B = (16.0 + 2.56 − 18.56) N2 ⋅ s C2 ⋅ m = 0

( ) ( )(b)
S = 1 E × B = ⎣⎡ 80.0ˆi + 32.0ˆj − 64.0kˆ N C⎤⎦ × ⎡⎣ 0.200ˆi + 0.080 0ˆj + 0.290kˆ µT⎦⎤
µ0 4π × 10−7 T ⋅ m A

( )6.40kˆ − 23.2ˆj − 6.40kˆ + 9.28ˆi − 12.8ˆj + 5.12ˆi × 10−6 W m2

S = 4π × 10−7

( )S = 11.5ˆi − 28.6ˆj W m2 = 30.9 W m2 at −68.2° from the +x axis

P34.24 The energy put into the water in each container by electromagnetic radiation can be written as

eP ∆t = eIA∆t where e is the percentage absorption efficiency. This energy has the same effect as

heat in raising the temperature of the water:

eIA∆t = mc∆T = ρVc∆T

∆T = eI 2∆t = eI ∆t
ρ 3c ρc

where is the edge dimension of the container and c the specific heat of water. For the small

container,

)(0.7 25 × 103 W m2 480 s =
)(∆T = kg ⋅°C 33.4°C
103 kg m3 (0.06 m) 4 186 J

For the larger,

0.91(25 J )s ⋅ m2 480 s
)∆T = (0.12 m2 4 186 J °C = 21.7°C

280 Chapter 34

P34.25 (a) Bmax = Emax : Bmax = 7.00 × 105 N C = 2.33 mT
c 3.00 × 108 m s

(b) I = E2 : )(7.00 × 105 2 = 650 MW m2
max ) )( (I = 2 4π × 10−7 3.00 × 108

2µ0c

(c) I = P : ( ) ( )P = IA = ⎡π 2 ⎤
6.50 ×108 W m2 ⎣⎢ 4 1.00 ×10−3 m ⎥⎦ = 510 W
A

) )( (P34.26 (a) E = cB = 3.00 × 108 m s 1.80 × 10−6 T = 540 V m

( )(b)
uav = B2 = 1.80 × 10−6 2 2.58 µJ m3
µ0 4π × 10−7 =

( )( )(c) Sav = cuav = 3.00 × 108 2.58 × 10−6 = 773 W m2

*P34.27 (a) We assume that the starlight moves through space without any of it being absorbed. The
radial distance is

( )( )20 ly = 20c (1 yr) = 20 3×108 m s 3.156 ×107 s = 1.89 ×1017 m

=P = 4 ×1028 W = 8.88 ×10−8 W m2
1.89 ×1017 m
4πr2
( )I 4π 2

(b) The Earth presents the projected target area of a flat circle:

( ) ( )P = IA = 8.88 ×10−8 W m2 π 6.37 ×106 m 2 = 1.13×107 W

Section 34.5 Momentum and Radiation Pressure

( )*P34.28 (a)
The radiation pressure is 2 1 370 W m2 = 9.13 × 10−6 N m2 .
3.00 × 108 m s2

Multiplying by the total area, A = 6.00 × 105 m2 gives: F = 5.48 N .

(b) The acceleration is: a = F = 5.48 N = 9.13 × 10−4 m s2
m 6 000 kg

(c) It will arrive at time t where d = 1 at 2
2

( ( ))or
t = 2d = 2 3.84 ×108 m
a 9.13×10−4 m s2 = 9.17 ×105 s = 10.6 days

P34.29 For complete absorption, P = S = 25.0 = 83.3 nPa .
c 3.00 × 108

Electromagnetic Waves 281

*P34.30 (a) The magnitude of the momentum transferred to the assumed totally reflecting surface in
time t is p = 2TER /c = 2SAt/c. Then the vector momentum is

p = 2SAt/c = 2(6 ˆi W/m2 )(40 ×10−4 m2 )(1 s)/(3×108 m/s)

= 1.60 × 10−10 ˆi kg ⋅ m/s each second

(b) The pressure on the assumed totally reflecting surface is P = 2S/c. Then the force is
PAˆi = 2SAˆi/c = 2(6 W/m2)(40 × 10–4 m2)(1 s)/(3 × 108 m/s) = 1.60 × 10−10 ˆi N

(c) The answers are the same. Force is the time rate of momentum transfer.

P34.31 I= P = E2
max

πr2 2µ0c

(a) Emax = P (2µ0c) = 1.90 kN C

πr2

(b) 15 × 10−3 J s (1.00 m) = 50.0 pJ
3.00 × 108
ms

(c) p = U = 5 × 10−11 = 1.67 × 10−19 kg ⋅ m s
c 3.00 × 108

*P34.32 (a) The light pressure on the absorbing Earth is P = S = I .
cc

( )The force is F = PA = I π R2 = (1370 W/m2 )π (6.37 × 106 m)2 = 5.82 × 108 N away
c 3.00 × 108 m s
from the Sun.

(b) The attractive gravitational force exerted on Earth by the Sun is

( ( )( ) )( )Fg
= GM S M M = 6.67 × 10−11 N ⋅ m2 kg2 1.991 × 1030 kg 5.98 × 1024 kg
rM2 1.496 × 1011 m 2

= 3.55 × 1022 N

which is 6.10 × 1013 times stronger and in the opposite direction compared to the

repulsive force in part (a).

282 Chapter 34

*P34.33 (a) If PS is the total power radiated by the Sun, and rE and rM are the radii of the orbits of the

planets Earth and Mars, then the intensities of the solar radiation at these planets are:

IE = PS

4π rE2

and IM = PS
Thus,
4π rM2

⎛ rE ⎞ 2 ⎛ 1.496 × 1011 m⎞2
⎜⎝ rM ⎠⎟ ⎝⎜ 2.28 × 1011 m ⎟⎠
( )IM = = m2 = 590 W m2
IE 1 370 W

(b) Mars intercepts the power falling on its circular face:

( ) ( ) ( )PM = IM ⎡⎣π 2⎤
π RM2 = 590 W m2 3.37 × 106 m ⎦ = 2.10 × 1016 W

(c) If Mars behaves as a perfect absorber, it feels pressure P = SM = IM
cc

( )F = PA = IM = PM 2.10 × 1016 W
c 3.00 × 108 m s
and force π RM2 c = = 7.01 × 107 N

(d) The attractive gravitational force exerted on Mars by the Sun is

( ( )( ) )( )Fg
= GM S M M = 6.67 × 10−11 N ⋅ m2 kg2 1.991 × 1030 kg 6.42 × 1023 kg
rM2 2.28 × 1011 m 2

= 1.64 × 1021 N
which is ~1013 times stronger than the repulsive force of part (c).

(e) The relationship between the gravitational force and the light-pressure force is similar at
very different distances because both forces follow inverse-square laws. The force ratios are
not identical for the two planets because of their different radii and densities.

P34.34 The radiation pressure on the disk is P = S = I = F = F . Hx Hy
c c A πr2
Then F = π r2I . PA
c mg

Take torques about the hinge: ∑ τ = 0 r

Hx (0) + Hy (0) − mgr sinθ + π r2 Ir = 0 q
c
FIG. P34.34
π (0.4 m)2 107 W s2 ⎛ 1 kg m2 ⎞
⎜⎝ 1 W s3 ⎟⎠
(0.024 kg) m2 (9.8 m) 3 ×
( )θ = sin−1 π r2I= sin−1 s m
mgc 108

= sin−1 0.0712 = 4.09°

Section 34.6 Production of Electromagnetic Waves by an Antenna

P34.35 λ = c = 536 m so h = λ = 134 m
f 4

λ = c = 188 m so h = λ = 46.9 m
f 4

Electromagnetic Waves 283

P34.36 P = (∆V )2 or P ∝ (∆V )2

R receiving
antenna
∆V = (−) Ey ⋅ ∆y = Ey ⋅ cosθ ∆y
q

∆V ∝ cosθ so P ∝ cos2 θ

(a) θ = 15.0°: P P= max cos2 (15.0°) = 0.933Pmax = 93.3% FIG. P34.36

(b) θ = 45.0°: P P= max cos2 (45.0°) = 0.500Pmax = 50.0%

(c) θ = 90.0°: P P= max cos2 (90.0°) = 0

P34.37 (a) Constructive interference occurs when
d cosθ = nλ for some integer n.

cosθ = n λ = ⎛ λ ⎞ = 2n
d n⎜ λ2 ⎟
⎠
⎝

n = 0, ±1, ±2, …

∴ strong signal @ θ = cos−1 0 = 90 , 270

(b) Destructive interference occurs when

d cosθ = ⎛ 2n + 1⎞ λ : cosθ = 2n + 1
⎝ 2 ⎠

∴ weak signal @ θ = cos−1 (±1) = 0 , 180

FIG. P34.37

P34.38 For the proton, ΣF = ma yields qvB sin 90.0 = mv2
R
The period of the proton’s circular motion is therefore:
The frequency of the proton’s motion is T = 2π R = 2π m
The charge will radiate electromagnetic waves at this frequency, with v qB

f=1
T

λ = c = cT = 2π mc
f qB

P34.39 (a) The magnetic field B = 1 µ0 J max cos ( kx − ω t ) kˆ applies for x> 0, since it describes a
2

wave moving in the ˆi direction. The electric field direction must satisfy S = 1 E × B as
µ0

ˆi = ˆj × kˆ so the direction of the electric field is ˆj when the cosine is positive. For its

magnitude we have E = cB, so altogether we have E = 1 µ0 cJ max cos ( kx − ω t ) ˆj .
2

continued on next page

284 Chapter 34

(b) S = 1 E× B = 1 1 µ02cJm2 ax cos2 (kx − ω t ) ˆi
µ0 µ0 4

S = 1 µ0 cJ 2 cos2 ( kx −ω t ) ˆi
4 max

(c) The intensity is the magnitude of the Poynting vector averaged over one or more cycles.

The average of the cosine-squared function is 1, so I = 1 µ0 cJ 2 .
2 8 max

( )(d)
Jmax = 8I = 8 570 W m2 = 3.48 A m
µ0c ms
4π × 10−7 (Tm A)3 × 108

Section 34.7 The Spectrum of Electromagnetic Waves

P34.40 From the electromagnetic spectrum chart and accompanying text discussion, the following
identifications are made:

Frequency, f Wavelength, λ = c Classification
f
2 Hz = 2 × 100 Hz Radio
2 kHz = 2 × 103 Hz 150 Mm Radio
2 MHz = 2 × 106 Hz 150 km Radio
2 GHz = 2 × 109 Hz 150 m Microwave
2 THz = 2 × 1012 Hz 15 cm Infrared
2 PHz = 2 × 1015 Hz 150 mm Ultraviolet
2 EHz = 2 × 1018 Hz 150 nm X-ray
2 ZHz = 2 × 1021 Hz 150 pm Gamma ray
2 YHz = 2 × 1024 Hz 150 fm Gamma ray
150 am

Wavelengh, l Frequency, f = c Classification
λ
2 km = 2 × 103 m Radio
2 m = 2 × 100 m 1.5 × 105 Hz Radio
2 mm = 2 × 10−3 m 1.5 × 108 Hz Microwave
2 mm = 2 × 10−6 m 1.5 × 1011 Hz Infrared
2 nm = 2 × 10−9 m 1.5 × 1014 Hz Ultraviolet or X-ray
2 pm = 2 × 10−12 m 1.5 × 1017 Hz X-ray or Gamma ray
2 fm = 2 × 10−15 m 1.5 × 1020 Hz Gamma ray
2 am = 2 × 10−18 m 1.5 × 1023 Hz Gamma ray
1.5 × 1026 Hz

Electromagnetic Waves 285

( )f λ = c gives
P34.41 (a) 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

P34.42 (a) f = c = 3 × 108 m s ~108 Hz radio wave
λ 1.7 m

(b) 1 000 pages, 500 sheets, is about 3 cm thick so one sheet is about 6 × 10−5 m thick.

f = 3.00 × 108 m s ~1013 Hz infrared
6 × 10−5 m

*P34.43 (a) Channel 4: fmin = 66 MHz λmax = 4.55 m

f = 72 MHz λmin = 4.17 m

max

(b) Channel 6: fmin = 82 MHz λmax = 3.66 m
fmax = 88 MHz λmin = 3.41 m

(c) Channel 8: fmin = 180 MHz λmax = 1.67 m
fmax = 186 MHz λmin = 1.61 m

P34.44 The time for the radio signal to travel 100 km is: ∆tr = 100 × 103 m = 3.33 × 10−4 s
3.00 × 108 m s

The sound wave travels 3.00 m across the room in: ∆ts = 3.00 m = 8.75 × 10−3 s
343 m s

Therefore, listeners 100 km away will receive the news before the people in the newsroom by a

P34.45 total time difference of ∆t = 8.75 × 10−3 s − 3.33 × 10−4 s = 8.41 × 10−3 s.
The wavelength of an ELF wave of frequency 75.0 Hz is λ = c = 3.00 × 108 m s = 4.00 × 106 m.

f 75.0 Hz
The length of a quarter-wavelength antenna would be L = 1.00 × 106 m = 1.00 × 103 km

or L = (1 000 km) ⎛ 0.621 mi ⎞ = 621 mi
⎝⎜ 1.00 km ⎠⎟

Thus, while the project may be theoretically possible, it is not very practical.

286 Chapter 34

Additional Problems
P34.46 ω = 2π f = 6.00π × 109 s−1 = 1.88 × 1010 s−1

k = 2π = ω = 6.00π × 109 s−1 = 20.0π = 62.8 m −1 Bmax = E = 300 V m s = 1.00 µT
λ c 3.00 × 108 m s c 3.00 × 108 m

E = (300 V )(m) cos 62.8x − 1.88 × 1010 t B = (1.00 µT) cos(62.8x − 1.88 × 1010 t )

( ) ( )P = 1 370 W m2 2
*P34.47 (a) P = SA: ⎣⎡4π 1.496 × 1011 m ⎤ = 3.85 × 1026 W
⎦

( )( )(b)
S = cBm2 ax so Bmax = 2µ0S = 2 4π × 10−7 N A2 1 370 W m2 = 3.39 µT
2µ0 c 3.00 × 108 m s

E2
max
( )( )S =
2µ0c so Emax = 2µ0cS = 2 4π × 10−7 3.00 × 108 (1 370) = 1.02 kV m

*P34.48 Suppose you cover a 1.7 m × 0.3 m section of beach blanket. Suppose the elevation angle of the
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 = U ( ) ( )U = IAt = 1 370 W m2 ⎡⎣(0.6)(0.5) 0.4 m2 ⎤⎦(3 600 s) ~106 J

A At (d) u
*P34.49 (a) y

0x
l 2l

(a) Ey

(b) uE = 1 ∈0 E 2 = 1 ∈0 E2 cos2 (kx)
2 2 max

(c) uB = 1 B2 = 1 B2 cos2 (kx) = 1 E2 cos2 (kx) = µ0 ∈0 E2 cos2 (kx) = uE
2µ0 2µ0 max 2µ0 max 2µ0 max
c2

(d) u = uE + uB = ∈0 E2 cos2 (kx)
max

∫ ∫(e) λ λ
[ ]Eλ = 0 0
∈ 0 E2 cos2 (kx)Adx = ∈ 0 E2 1 + 1 cos(2kx) Adx
max max 2 2

λ ∈ E2 A λ ∈0 E2 A
0 max max
[ ]=
1 ∈ E2 A x + 0 sin(2 kx ) = 1 ∈ E2 Aλ + sin(4π ) − sin(0)
2 max 2 max
0 4k 0 0 4k

= 1 ∈0 E2 Aλ
2 max

(f) I = Eλ = ∈0 E2 Aλ = 1 ∈0cEm2 ax
AT max 2

2AT

This result agrees with equation 34.24, I = E2 = cEm2 ax = cEm2 axµ0 ∈0 .
max 2µ0c2 2µ0

2µ0c

Electromagnetic Waves 287

P34.50 (a) Fgrav = GM S m = ⎛ GM S ⎞ ⎣⎡⎢ρ ⎛ 4 π r 3 ⎞ ⎤
R2 ⎝ R2 ⎠ ⎝ 3 ⎠ ⎦⎥

where M S = mass of Sun, r = radius of particle, and R = distance from Sun to particle.

Since Frad = Sπ r2 ,
c

Frad = ⎛ 1⎞ ⎛ 3SR2 ⎞ ∝1
Fgrav ⎝ r ⎠ ⎜⎝ 4cGM S ρ ⎠⎟ r

(b) From the result found in part (a), when Fgrav = Frad,
we have r = 3SR2
4cGM S ρ

) )( (3 214 W m2 3.75 × 1011 m 2
) ) ) )( ( ( (r =

4 6.67 × 10−11 N ⋅ m2 kg2 1.991 × 1030 kg 1 500 kg m3 3.00 × 108 m s
= 3.78 × 10−7 m

P34.51 (a) Bmax = Emax = 6.67 × 10−16 T
c

(b) Sav = E2 = 5.31 × 10−17 W m2
max

2µ0c

(c) P = SavA = 1.67 ×10−14 W

(d) F = PA = ⎛ Sav ⎞ A = 5.56 × 10−23 N (approximately the
⎝ c ⎠

weight of 3 000 hydrogen atoms!) FIG. P34.51

*P34.52 (a) In E = q/4p ∈0r2, the net flux is q/∈0, so
E = Φ rˆ /4p r2 = (487 N ⋅ m2/C) rˆ /4p r2 = (38.8/r2 ) rˆ N ⋅ m2/C

(b) The radiated intensity is I = P /4p r2 = E2max/2m0c. Then

Emax = (P m 0c /2p )1/2/r

= [(25 N ⋅ m/s)(4p × 10–7 T ⋅ m/2p A)(3 × 108 m/s)(1 N ⋅ s/1 T ⋅ C ⋅ m)(1 A⋅ s/1 C)]1/2/r

= (38.7/r) N ⋅ m/C

(c) For 3 × 106 N/C = (38.7 N ⋅ m/C)/r we find r = 12.9 µm , but the expression in part (b)
does not apply if this point is inside the source.

(d) In the radiated wave, the field amplitude is inversely proportional to distance. As the distance
doubles, the amplitude is cut in half and the intensity is reduced by a factor of 4. In the static
case, the field is inversely proportional to the square of distance. As the distance doubles, the
field is reduced by a factor of 4. The intensity of radiated energy is everywhere zero.

288 Chapter 34

P34.53 u = 1 ∈0 E2 Emax = 2u = 95.1 mV m
2 max ∈0

P34.54 The area over which we model the antenna as radiating is the lateral surface of a cylinder,

( )A = 2πr = 2π 4.00 ×10−2 m (0.100 m) = 2.51×10−2 m2

(a) The intensity is then: S = P = 0.600 W = 23.9 W m2 .
2.51 × 10−2 m2
A

(b) The standard is

( )0.570 ⎛ 1.00 × 10−3 W ⎞ ⎛ 1.00 × 104 cm2 ⎞ m2
mW cm2 = 0.570 mW cm 2 ⎝⎜ 1.00 mW ⎟⎠ ⎝⎜ 1.00 m2 ⎠⎟ = 5.70 W

While it is on, the telephone is over the standard by 23.9 W m2 = 4.19 times .
5.70 W m2

P34.55 (a) Bmax = Emax = 175 V m s = 5.83 × 10−7 T
c 3.00 × 108 m

k = 2π = 2π m = 419 rad m ω = kc = 1.26 × 1011 rad s
λ 0.015 0

Since S is along x, and E is along y, B must be in the z direction . (That is, S ∝ E × B. )

(b) Sav = Emax Bmax = 40.6 W m2 ( )Sav = 40.6 W m2 ˆi
2µ0

(c) Pr = 2S = 2.71 × 10−7 N m2
c

∑ ( )( )(d) a = F = PA = 2.71 × 10−7 N m2 0.750 m2 = 4.06 × 10−7 m s2
mm 0.500 kg

a = (406 nm )s2 ˆi

*P34.56 Of the intensity S = 1 370 W m2
the 38.0% that is reflected exerts a pressure
P1 = 2Sr = 2 (0.380) S
The absorbed light exerts pressure c
c

P2 = Sa = 0.620S
c c

Altogether the pressure at the subsolar point on Earth is

( )(a)
Ptotal = P1 + P2 = 1.38S = 1.38 1 370 W m2 = 6.30 × 10−6 Pa
c 3.00 × 108 m s

(b) Pa = 1.01 × 105 N m2 = 1.60 × 1010 times smaller than atmospheric pressure
Ptotal 6.30 × 10−6 N m2

Electromagnetic Waves 289

P34.57 (a) P = F = I F = IA =P = 100 J s s = 3.33 × 10−7 N = (110 kg) a
Ac c 3.00 × 108 m
c
a = 3.03 × 10−9 m s2 and
x = 1 at 2
2

t = 2x = 8.12 × 104 s = 22.6 h
a

(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 = 36.0 = 0.327 m s t = 30.6 s
110

( )P34.58 The mirror intercepts power P = I1A1 = 1.00 × 103 W m2 ⎣⎡π (0.500 m)2 ⎦⎤ = 785 W .
In the image,

(a) I2 = P : I2 = 785 W = 625 kW m2

A2 π (0.020 0 m)2

(b) I2 = E2 so ( )( )( )Emax = 2µ0cI2 = 2 4π × 10−7 3.00 × 108 6.25 × 105
max
= 21.7 kN C
2µ0c

Bmax = Emax = 72.4 µT
c

(c) 0.400 P ∆t = mc∆T

0.400(785 W) ∆t = (1.00 kg)(4 186 J kg ⋅ C)(100 C − 20.0 C)

∆t = 3.35 × 105 J = 1.07 × 103 s = 17.8 min
314 W

P34.59 Think of light going up and being absorbed by the bead which presents a face area π rb2

The light pressure is P = S = I
.
cc

(a) F = Iπ rb2 = mg = ρ 4 π rb3 g and I = 4ρ gc ⎛ 3m ⎞1 3 = 8.32 × 107 W m2
c 3 3 ⎝⎜ 4π ρ ⎟⎠

( ) ( )(b) P = IA = 8.32 × 107 W m2 π 2.00 × 10−3 m 2 = 1.05 kW

290 Chapter 34

P34.60 Think of light going up and being absorbed by the bead, which presents face area π rb2 .

If we take the bead to be perfectly absorbing, the light pressure is P = Sav = I = F .
c cA

(a) F = Fg

so I = Fc = Fg c = mgc
A A π rb2

From the definition of density, ρ = m = (4 m
V
3)π rb3

so 1 = ⎛ (4 3)π ρ⎞1 3
Substituting for rb , rb ⎟⎠
⎝⎜ m
(b) P = IA
mgc ⎛ 4π ρ ⎞ 2 3 ⎛ 4ρ ⎞ 2 3 ⎛ m⎞1 3 4ρgc ⎛ 3m ⎞1 3
π ⎝ 3m ⎠ ⎝ 3⎠ ⎝ π⎠ 3 ⎜⎝ 4π ρ ⎠⎟
I = = gc =

P= 4π r2ρ gc ⎛ 3m ⎞ 1 3
3 ⎜⎝ 4πρ ⎠⎟

( ( ) )P34.61
(a) On the right side of the equation, C2 m s2 2 = N ⋅ m2 ⋅ C2 ⋅ m2 ⋅ s3 = N⋅m = J = W.
C2 ⋅ s4 ⋅ m3 s s
C2 N ⋅ m2 (m s)3

( )a = qE =
m
(b) F = ma = qE or 1.60 × 10−19 C (100 N C) = 1.76 × 1013 m s2

9.11 × 10−31 kg

1.60 × 10−19 2 1.76 × 1013 2
6π 8.85 × 10−12 3.00 × 108 3
( ( ) ()( ))The
radiated power is then: P = q2a2 =
6π ∈0 c3

= 1.75 × 10−27 W

(c) F = mac = ⎛ v2 ⎞ = qvB so v = qBr
m ⎝⎜ r ⎠⎟ m

( )The proton accelerates at a = v2
( )r
= q2B2r = 1.60 × 10−19 2 (0.350)2 (0.500)
m2
1.67 × 10−27 2

= 5.62 × 1014 m s2

1.60 × 10−19 2 5.62 × 1014 2
6π 8.85 × 10−12 3.00 × 108 3
( ( ) ()( ))The
proton then radiates P = q2a2 = = 1.80 × 10−24 W
6π ∈0 c3

Electromagnetic Waves 291

P34.62 f = 90.0 MHz, Emax = 2.00 × 10−3 V m = 200 mV m

(a) λ = c = 3.33 m
f

T = 1 = 1.11 × 10−8 s = 11.1 ns
f

Bmax = Emax = 6.67 × 10−12 T= 6.67 pT
c

(b) E = (2.00 mV m) cos 2π ⎛x m −t ⎞ ˆj
⎝ 3.33 11.1 ns ⎠

B = (6.67 pT) kˆ cos 2π ⎛ x m − t ns ⎞
⎝ 3.33 11.1 ⎠

( ( )( ) )(c)
I = E2 = 2 4π 2.00 × 10−3 2 = 5.31 × 10−9 W m2
max × 10−7 3.00 × 108

2µ0c

(d) I = cuav so uav = 1.77 × 10−17 J m3

( )(e)
P = 2I = (2) 5.31 × 10−9 = 3.54 × 10−17 Pa
c
3.00 × 108

P34.63 (a) ε = − dΦB = − d (BA cosθ ) ε = −A d ( Bmax cos ω t cosθ ) = ABmaxω (sin ω t cosθ )
dt
dt dt
ε (t ) = 2π 2r2 fBmax cosθ sin 2π f t
ε (t ) = 2π fBmax A sin 2π f t cosθ
εmax = 2π 2r2 f Bmax cosθ
Thus,

where θ is the angle between the magnetic field and the normal to the loop.

(b) If E is vertical, B is horizontal, so the plane of the loop should be vertical

and the plane should contain the line of sight of the transmitter .

292 Chapter 34

P34.64 (a) m = ρV = ρ 1 4 πr3
23

⎛ ⎞1 3 ⎛ 6 (8.7 kg) ⎞1 3
⎝⎜ ⎠⎟ ⎝⎜ ⎠⎟
= 6m = 990 kg m3 = 0.161 m
( )r ρ4π 4π

(b) A = 1 4πr2 = 2π (0.161 m)2 = 0.163 m2

2

( )(c) I = eσ T 4 = 0.970 5.67 × 10−8 W m2 ⋅ K4 (304 K)4 = 470 W m2

( )(d) P = IA = 470 W m2 0.163 m2 = 76.8 W

(e) I = E2
max

2µ0c

( )( )( )( )Emax = 2µ0cI 1 2 = ⎡⎣ 8π × 10−7 Tm A 3 × 108 m s 470 W m2 ⎤⎦1 2 = 595 N C

(f) Emax = cBmax

Bmax = 595 N C = 1.98 µT
3 × 108 m s

(g) The sleeping cats are uncharged and nonmagnetic. They carry no macroscopic current.
They are a source of infrared radiation. They glow not by visible-light emission but by
infrared emission.

⎛ 6(0.8) ⎞ 13
⎝⎜ ⎠⎟
(h) Each kitten has radius rk = 990 × 4π = 0.072 8 m and radiating area

⎛ 6(5.5) ⎞ 23
⎝⎜ ⎠⎟
2π (0.072 8 m )2 = 0.033 3 m2. Eliza has area 2π 990 × 4π = 0.120 m. The

( )total glowing area is 0.120 m2 + 4 0.033 3 m2 = 0.254 m2 and has power output

( )P = IA = 470 W m2 0.254 m2 = 119 W

P34.65 (a) At steady state, Pin P= out and the power radiated out is Pout = eσ AT 4.

Thus, ( ) ( )0.900 1 000 W m2 A = 0.700 5.67 × 10−8 W m2 ⋅ K4 AT 4

= ⎡ 900 W m2 ⎤1 4 388 K = 115°C
5.67 × 10−8 W ⎥=
( )or T ⎢ ⎦⎥
⎢⎣ 0.700 m2 ⋅K4

(b) The box of horizontal area A presents projected area A sin 50.0° perpendicular to the
sunlight. Then by the same reasoning,

( ) ( )0.900 1 000 W m2 A sin 50.0° = 0.700 5.67 × 10−8 W m2 ⋅ K4 AT 4

= ( )⎡ 900 W m2 sin 50.0° ⎤1 4 = = 90.0°C
( )or T ⎢ ⎥ 363 K
⎣⎢ 0.700 5.67 × 10−8 W m2 ⋅ K4 ⎦⎥

Electromagnetic Waves 293

P34.66 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 4π r2.

At steady-state, Pin = Pout: ) )( (eIin π r2 = eσ 4π r2 T 4

⎛ 6.00 × 1023 W⎞
⎜⎝ 4π R2 ⎠⎟
( ) ( )e
πr2 = eσ 4π r2 T 4 so that 6.00 × 1023 W = 16πσ R2T 4

6.00 × 1023 W 6.00 × 1023 W 4.77 × 109 m = 4.77 Gm
16πσ T 4 5.67 × 10−8 W m2 ⋅ K4
( )R = = (310 K)4 =
16π

ANSWERS TO EVEN PROBLEMS

P34.2 (a) 7.19 × 1011 V m ⋅ s (b) 200 nT
P34.4 (a) 3.15ˆj kN C (b) 525 nTkˆ (c) −483ˆj aN
P34.6
( )−4.39ˆi − 1.76ˆj 1011 m s2

P34.8 2.25 × 108 m/s

P34.10 733 nT

P34.12 See the solution.

P34.14 2.9 × 108 m s ± 5%

P34.16 49.5 mV

P34.18 (a) 6.75 W/m2 (b) 6.64 kW/m2 (c) A powerful automobile running on sunlight would have to
carry on its roof a solar panel huge compared to the size of the car. Agriculture and forestry for
food and fuels, space heating of large and small buildings, water heating, and heating for drying
and many other processes are clear present and potential applications of solar energy.

P34.20 516 pT, ~105 times stronger than the Earth’s field

P34.22 (a) 11.9 GW m2 (b) 234 kW

P34.24 33.4°C for the smaller container and 21.7°C for the larger

P34.26 (a) 540 V/m (b) 2.58 ␮ J/m3 (c) 773 W/m2

P34.28 (a) 5.48 N away from the Sun (b) 913 µm/s2 away from the Sun (c) 10.6 d

P34.30 (a) 1.60 × 10−10 ˆi kg⋅m/s each second (b) 1.60 × 10−10 ˆi N (c) The answers are the same.
Force is the time rate of momentum transfer.

P34.32 (a) 582 MN away from the Sun (b) The gravitational force is 6.10 × 1013 times stronger and in
the opposite direction.

294 Chapter 34

P34.34 4.09°

P34.36 (a) 93.3% (b) 50.0% (c) 0

P34.38 2π mpc
P34.40 eB

radio, radio, radio, radio or microwave, infrared, ultraviolet, x-ray, γ -ray, γ -ray; radio, radio,
microwave, infrared, ultraviolet or x-ray, x- or γ -ray, γ -ray, γ -ray

P34.42 (a) ~108 Hz radio wave (b) ~1013 Hz infrared light

P34.44 The radio audience gets the news 8.41 ms sooner.

) )P34.46 E = (300 V (m) cos 62.8x − 1.88 × 1010 t B = (1.00 µT) cos(62.8x − 1.88 × 1010 t

P34.48 ~106 J

P34.50 (a) See the solution. (b) 378 nm

P34.52 (a) E = (38.8/r2) rˆ N ⋅ m2/C (b) E = (38.7/r) (W⋅T ⋅ m2/A ⋅ s)1/ 2 = (38.7/r) N ⋅ m/C
max
(c) 12.9 µm, but the expression in part (b) does not apply if this point is inside the source.

(d) In the radiated wave, the field amplitude is inversely proportional to distance. As the distance

doubles, the amplitude is cut in half and the intensity is reduced by a factor of 4. In the static

case, the field is inversely proportional to the square of distance. As the distance doubles, the field

is reduced by a factor of 4. The intensity of radiated energy is everywhere zero in the static case.

P34.54 (a) 23.9 W m2 (b) 4.19 times the standard

P34.56 (a) 6.30 µPa (b) 1.60 × 1010 times less than atmospheric pressure

P34.58 (a) 625 kW m2 (b) 21.7 kN C and 72.4 µT (c) 17.8 min

P34.60 (a) ⎛ 16mρ2 ⎞ 1 3 gc (b) ⎛ 16π 2mρ2 ⎞ 1 3 r 2 gc
⎝⎜ 9π ⎟⎠ ⎝⎜ 9 ⎠⎟

P34.62 (a) 3.33 m, 11.1 ns, 6.67 pT (b) E = (2.00 mV m) cos 2π ⎛ x m − t ns ⎞ ˆj ;
⎜⎝ 3.33 11.1 ⎟⎠

B = (6.67 pT) kˆ cos 2π ⎛ x m − t ns ⎞ (c) 5.31 nW m2 (d) 1.77 × 10−17 J m3
⎝ 3.33 11.1 ⎠

(e) 3.54 × 10−17 Pa

P34.64 (a) 16.1 cm (b) 0.163 m2 (c) 470 W m2 (d) 76.8 W (e) 595 N/C (f) 1.98 µ T

(g) The cats are nonmagnetic and carry no macroscopic charge or current. Oscillating charges
within molecules make them emit infrared radiation. (h) 119 W

P34.66 p r2; 4p r2 where r is the radius of the planet; 4.77 Gm

35

The Nature of Light and the
Laws of Geometric Optics

CHAPTER OUTLINE ANSWERS TO QUESTIONS

35.1 The Nature of Light Q35.1 Light travels through a vacuum at a speed of 300 000 km
35.2 Measurements of the Speed per second. Thus, an image we see from a distant star
or galaxy must have been generated some time ago. For
of Light example, the star Altair is 16 light-years away; if we
35.3 The Ray Approximation in look at an image of Altair today, we know only what
was happening 16 years ago. This may not initially seem
Geometric Optics significant, but astronomers who look at other galaxies
35.4 The Wave Under Reflection can gain an idea of what galaxies looked like when they
35.5 The Wave Under Refraction were significantly younger. Thus, it actually makes sense
35.6 Huygens’s Principle to speak of “looking backward in time.”
35.7 Dispersion
35.8 Total Internal Reflection

*Q35.2 104 m/(3 × 108 m/s) is 33 µs. Answer (c).

*Q35.3 We consider the quantity l /d. The smaller it is, the better the ray approximation works. In (a) it is
like 0.34 m/1 m ≈ 0.3. In (b) we can have 0.7 µm/2 mm ≈ 0.000 3. In (c), 0.4 µm/2 mm ≈ 0.000 2.
In (d), 300 m/1 m ≈ 300. In (e) 1 nm/1 mm ≈ 0.000 001. The ranking is then e, c, b, a, d.

Q35.4 With a vertical shop window, streetlights and his own reflection
can impede the window shopper’s clear view of the display.
The tilted shop window can put these reflections out of the way.
Windows of airport control towers are also tilted like this, as are
automobile windshields.

FIG. Q35.4

295

296 Chapter 35

Q35.5 We assume that you and the child are

always standing close together. For a

flat wall to make an echo of a sound

that you make, you must be stand-

ing along a normal to the wall. You

must be on the order of 100 m away,

to make the transit time sufficiently

long that you can hear the echo sepa-

rately from the original sound. Your (a)
sound must be loud enough so that

you can hear it even at this consider-

able range. In the picture, the dashed

rectangle represents an area in which

you can be standing. The arrows

represent rays of sound.

Now suppose two vertical

perpendicular walls form an inside

corner that you can see. Some of

the sound you radiate horizontally

will be headed generally toward the

corner. It will reflect from both walls

with high efficiency to reverse in

direction and come back to you. You

can stand anywhere reasonably far

away to hear a retroreflected echo of

sound you produce.

If the two walls are not perpen-

dicular, the inside corner will not

produce retroreflection. You will

generally hear no echo of your shout

or clap. (b)
If two perpendicular walls have

a reasonably narrow gap between FIG. Q35.5

them at the corner, you can still hear

a clear echo. It is not the corner line itself that retroreflects the sound, but the perpendicular walls

on both sides of the corner. Diagram (b) applies also in this case.

Q35.6 The stealth fighter is designed so that adjacent panels are not joined at right angles, to prevent any
retroreflection of radar signals. This means that radar signals directed at the fighter will not be
channeled back toward the detector by reflection. Just as with sound, radar signals can be treated
as diverging rays, so that any ray that is by chance reflected back to the detector will be too weak
in intensity to distinguish from background noise. This author is still waiting for the automotive
industry to utilize this technology.

*Q35.7 Snell originally stated his law in terms of cosecants. From v = c/n and sinθ = 1/cscθ and λ = c/nf
with c and f constant between media, we conclude that a, b, and c are all correct statements.

Q35.8 An echo is an example of the reflection of sound. Hearing the noise of a distant highway on a
cold morning, when you cannot hear it after the ground warms up, is an example of acoustical
refraction. You can use a rubber inner tube inflated with helium as an acoustical lens to concen-
trate sound in the way a lens can focus light. At your next party, see if you can experimentally
find the approximate focal point!

The Nature of Light and the Laws of Geometric Optics 297

*Q35.9 (a) Yes. (b) No. (c) Yes. (d) No. If the light moves into a medium of higher refractive index, its
wavelength decreases. The frequency remains constant. The speed diminishes by a factor equal
to the index of refraction. If its angle of incidence is 0°, it will continue in the same direction.

Q35.10 If a laser beam enters a sugar solution with a concentration gradient (density and index of refrac-
tion increasing with depth) then the laser beam will be progressively bent downward (toward the
normal) as it passes into regions of greater index of refraction.

*Q35.11 (a) Yes. It must be traveling in the medium in which it moves slower, water, to undergo total
internal reflection.

(b) Yes. It must be traveling in the medium in which it moves slower, air, to undergo total internal
reflection.

Q35.12 Diamond has higher index of refraction than glass and consequently a smaller critical angle
for total internal reflection. A brilliant-cut diamond is shaped to admit light from above, reflect
it totally at the converging facets on the underside of the jewel, and let the light escape only at
the top. Glass will have less light internally reflected.

Q35.13 Highly silvered mirrors reflect about 98% of the incident light. With a 2-mirror periscope,
that results in approximately a 4% decrease in intensity of light as the light passes through the
periscope. This may not seem like much, but in low-light conditions, that lost light may mean
the difference between being able to distinguish an enemy armada or an iceberg from the sky
beyond. Using prisms results in total internal reflection, meaning that 100% of the incident light
is reflected through the periscope. That is the “total” in total internal reflection.

*Q35.14 The light with the greater change in speed will have the larger deviation. Since the glass has a
higher index than the surrounding air, A travels slower in the glass.

Q35.15 Immediately around the dark shadow of my head, I see a halo brighter than the rest of the dewy
grass. It is called the heiligenschein. Cellini believed that it was a miraculous sign of divine
favor pertaining to him alone. Apparently none of the people to whom he showed it told him
that they could see halos around their own shadows but not around Cellini’s. Thoreau knew that
each person had his own halo. He did not draw any ray diagrams but assumed that it was entirely
natural. Between Cellini’s time and Thoreau’s, the Enlightenment and Newton’s explanation
of the rainbow had happened. Today the effect is easy to see whenever your shadow falls on a
retroreflecting traffic sign, license plate, or road stripe. When a bicyclist’s shadow falls on a paint
stripe marking the edge of the road, her halo races along with her. It is a shame that few people
are sufficiently curious observers of the natural world to have noticed the phenomenon.

Q35.16 At the altitude of the plane the surface of the Earth need not block off the lower half of the
rainbow. Thus, the full circle can be seen. You can see such a rainbow by climbing on a steplad-
der above a garden sprinkler in the middle of a sunny day. Set the sprinkler for fine mist. Do not
let the slippery children fall from the ladder.

*Q35.17 Light from the lamps along the edges of the sheet enters the plastic. Then it is totally internally
reflected by the front and back faces of the plastic, wherever the plastic has an interface with air.
If the refractive index of the grease is intermediate between 1.55 and 1.00, some of this light can
leave the plastic into the grease and leave the grease into the air. The surface of the grease is
rough, so the grease can send out light in all directions. The customer sees the grease shining
against a black background. The spotlight method of producing the same effect is much less
efficient. With it, much of the light from the spotlight is absorbed by the blackboard. The
refractive index of the grease must be less than 1.55. Perhaps the best choice would be

1.55 × 1.00 = 1.24.

298 Chapter 35

*Q35.18 Answer (c). We want a big difference between indices of refraction to have total internal reflec-
tion under the widest range of conditions.

Q35.19 A mirage occurs when light changes direction as it moves between batches of air having different
indices of refraction because they have different densities at different temperatures. When the sun
makes a blacktop road hot, an apparent wet spot is bright due to refraction of light from the bright
sky. The light, originally headed a little below the horizontal, always bends up as it first enters
and then leaves sequentially hotter, lower-density, lower-index layers of air closer to the road
surface.

SOLUTIONS TO PROBLEMS

Section 35.1 The Nature of Light

Section 35.2 Measurements of the Speed of Light

*P35.1 The Moon’s radius is 1.74 × 106 m and the Earth’s radius is 6.37 × 106 m. The total distance
traveled by the light is:

( )d = 2 3.84 × 108 m − 1.74 × 106 m − 6.37 × 106 m = 7.52 × 108 m

This takes 2.51 s, so v = 7.52 × 108 m = 2.995 × 108 m s = 299.5 Mm s . The sizes of the
2.51 s

objects need to be taken into account. Otherwise the answer would be too large by 2%.

P35.2 ∆ x = ct ; c = ∆x = 2(1.50 × 108 km)(1 000 m km) = 2.27 × 108 m s = 227 Mm s
t
(22.0 min)(60.0 s min)

P35.3 The experiment is most convincing if the wheel turns fast enough to pass outgoing light through
one notch and returning light through the next: t = 2

c

( ( ) )θ
= ω t = ω ⎛ 2 ⎞ so ω = cθ = 2.998 × 108 ⎡⎣2π (720)⎤⎦ = 114 rad s
⎝⎜ c ⎠⎟
2 2 11.45 × 103

The returning light would be blocked by a tooth at one-half the angular speed, giving another data
point.

The Nature of Light and the Laws of Geometric Optics 299

Section 35.3 The Ray Approximation in Geometric Optics

Section 35.4 The Wave Under Reflection

Section 35.5 The Wave Under Refraction

P35.4 (a) Let AB be the originally horizontal ceiling, BC its originally vertical A B

normal, AD the new ceiling, and DE its normal. Then angle BAD = φ. D

By definition DE is perpendicular to AD and BC is perpendicular to AB.

Then the angle between DE extended and BC is φ because angles are EC
equal when their sides are perpendicular, right side to right side and

left side to left side. FIG. P35.4 (a)

(b) Now CBE = φ is the angle of incidence of the vertical light beam. A D
Its angle of reflection is also φ. The angle between the vertical inci-

dent beam and the reflected beam is 2φ.

E

FIG. P35.4 (b)

(c) tan 2φ = 1.40 cm = 0.001 94 φ = 0.055 7°
720 cm

P35.5 (a) From geometry, 1.25 m = d sin 40.0° Mirror 2

so d = 1.94 m Mirror 2 Light 50°
beam i2 = 50°
(b) 50.0° above the horizontal 40.0°
or parallel to the incident ray. 40° d
1.25 m P 40° i1 = 40°

Mirror 1 50° 50°
1.25 m Mirror 1

FIG. P35.5

P35.6 (a) Method One:
The incident ray makes angle α = 90° − θ1
with the first mirror. In the picture, the law of reflection

implies that

θ1 = θ1′

FIG. P35.6

Then β = 90° − θ1′ = 90 − θ1 = α

In the triangle made by the mirrors and the ray passing between them,

Further, β + 90° + γ = 180°
and γ = 90° − β
δ = 90° − γ = β = α
∈= δ = α

Thus the final ray makes the same angle with the first mirror as did the incident ray. Its
direction is opposite to the incident ray.

Method Two:
The vector velocity of the incident light has a component vy perpendicular to the first mirror
and a component vx perpendicular to the second. The vy component is reversed upon the first

continued on next page

300 Chapter 35

reflection, which leaves vx unchanged. The second reflection reverses vx and leaves vy
unchanged. The doubly reflected ray then has velocity opposite to the incident ray.

(b) The incident ray has velocity vxˆi + vyˆj + vzkˆ . Each reflection reverses one component
and leaves the other two unchanged. After all the reflections, the light has velocity
−vxˆi − vyˆj − vzkˆ , opposite to the incident ray.

P35.7 Let d represent the perpendicular distance
from the person to the mirror. The distance
between lamp and person measured parallel d tan f q 2d tan q
to the mirror can be written in two ways: fq d tan q
2d tanθ + d tanθ = d tan φ. The condition on the
distance traveled by the light 2d
dd
is 2d = 2d + d . We have the two
cosφ cosθ cosθ FIG. P35.7

equations 3 tanθ = tanφ and 2 cosθ = 3cosφ.
To eliminate φ we write

9 sin2 θ = sin2 φ 4 cos2 θ = 9 cos2 φ
cos2 θ cos2 φ

( )9 cos2 φ sin2 θ = cos2 θ 1 − cos2 φ

4 cos2 θ sin2 θ = cos2 θ ⎛⎝1 − 4 cos2 θ ⎞
9 ⎠

( )4 sin2 θ = 1 − 4 1 − sin2 θ 36 sin2 θ = 9 − 4 + 4 sin2 θ
9

sin2 θ = 5 θ = 23.3°
32

*P35.8 The excess time the second pulse spends in the ice is 6.20 m/[(3.00 × 108 m/s)/1.309] = 27.1 ns

P35.9 Using Snell’s law, sin θ 2 = n1 sin θ1
n2

θ2 = 25.5°

λ2 = λ1 = 442 nm
n1
FIG. P35.9