34

CHAPTER OUTLINE

34.1 Maxwell’s Equations and Hertz’s Discoveries34.2 Plane Electromagnetic Waves34.3 Energy Carried by Electromagnetic Waves34.4 Momentum and Radiation Pressure34.5 Production of Electromagnetic Waves by an Antenna34.6 The Spectrum of

Electromagnetic Waves

Electromagnetic Waves

ANSWERS TO QUESTIONS

Q34.1 Radio waves move at the speed of light. They can travel aroundthe curved surface of the Earth, bouncing between the groundand the ionosphere, which has an altitude that is small whencompared to the radius of the Earth. The distance across thelower forty-eight states is approximately 5 000 km, requiring a

transit time of 5 10

106

2××

− m3 10 m s

s8 ~ . To go halfway around the

Earth takes only 0.07 s. In other words, a speech can be heardon the other side of the world before it is heard at the back of alarge room.

Q34.2 The Sun’s angular speed in our sky is our rate of rotation,36024

15°= °

hh . In 8.3 minutes it moves west by

θ ω= = ° FHGIKJ = °t 15

18 3 2 1h

h60 min

minb g a f. . . This is about four

times the angular diameter of the Sun.

Q34.3 Energy moves. No matter moves. You could say that electric and magnetic fields move, but it is nicerto say that the fields at one point stay at that point and oscillate. The fields vary in time, like sportsfans in the grandstand when the crowd does the wave. The fields constitute the medium for thewave, and energy moves.

Q34.4 No. If a single wire carries DC current, it does not emit electromagnetic waves. In this case, there is aconstant magnetic field around the wire. Alternately, if the cable is a coaxial cable, it ideally does notemit electromagnetic waves even while carrying AC current.

Q34.5 Acceleration of electric charge.

Q34.6 The changing magnetic field of the solenoid induces eddy currents in the conducting core. This isaccompanied by I R2 conversion of electrically-transmitted energy into internal energy in theconductor.

Q34.7 A wire connected to the terminals of a battery does not radiate electromagnetic waves. The batteryestablishes an electric field, which produces current in the wire. The current in the wire creates amagnetic field. Both fields are constant in time, so no electromagnetic induction or “magneto-electricinduction” happens. Neither field creates a new cycle of the other field. No wave propagationoccurs.

291

292 Electromagnetic Waves

Q34.8 No. Static electricity is just that: static. Without acceleration of the charge, there can be noelectromagnetic wave.

Q34.9 Sound

The world of sound extends to the top of theatmosphere and stops there; sound requires amaterial medium. Sound propagates by a chainreaction of density and pressure disturbancesrecreating each other. Sound in air moves athundreds of meters per second. Audible soundhas frequencies over a range of three decades(ten octaves) from 20 Hz to 20 kHz. Audiblesound has wavelengths of ordinary size (1.7 cmto 17 m). Sound waves are longitudinal.

Light

The universe of light fills the whole universe.Light moves through materials, but faster in avacuum. Light propagates by a chain reaction ofelectric and magnetic fields recreating eachother. Light in air moves at hundreds of millionsof meters per second. Visible light hasfrequencies over a range of less than one octave,from 430 to 750 Terahertz. Visible light haswavelengths of very small size (400 nm to700 nm). Light waves are transverse.

Sound and light can both be reflected, refracted, or absorbed to produce internal energy. Both haveamplitude and frequency set by the source, speed set by the medium, and wavelength set by bothsource 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 bywave functions satisfying wave equations. Both carry energy. If the source is small, their intensitiesboth follow an inverse-square law. Both are waves.

Q34.10 The Poynting vector S describes the energy flow associated with an electromagnetic wave. Thedirection of S is along the direction of propagation and the magnitude of S is the rate at whichelectromagnetic energy crosses a unit surface area perpendicular to the direction of S.

Q34.11 Photons carry momentum. Recalling what we learned in Chapter 9, the impulse imparted to aparticle 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 betwice that exerted by a photon that is absorbed.

Q34.12 Different stations have transmitting antennas at different locations. For best reception align yourrabbit ears perpendicular to the straight-line path from your TV to the transmitting antenna. Thetransmitted signals are also polarized. The polarization direction of the wave can be changed byreflection from surfaces—including the atmosphere—and through Kerr rotation—a change inpolarization axis when passing through an organic substance. In your home, the plane ofpolarization is determined by your surroundings, so antennas need to be adjusted to align with thepolarization of the wave.

Q34.13 You become part of the receiving antenna! You are a big sack of salt water. Your contribution usuallyincreases the gain of the antenna by a few tenths of a dB, enough to noticeably improve reception.

Q34.14 On the TV set, each side of the dipole antenna is a 1/4 of the wavelength of the VHF radio wave. Theelectric field of the wave moves free charges in the antenna in electrical resonance, giving maximumcurrent in the center of the antenna, where the cable connects it to the receiver.

Q34.15 The loop antenna is essentially a solenoid. As the UHF radio wave varies the magnetic field insidethe loop, an AC emf is induced in the loop as described by Faraday’s and Lenz’s laws. This signal isthen carried down a cable to the UHF receiving circuit in the TV. An excellent reference for antennasand all things radio is the ARRL Handbook.

Chapter 34 293

Q34.16 The voltage induced in the loop antenna is proportional to the rate of change of the magnetic field inthe wave. A wave of higher frequency induces a larger emf in direct proportion. The instantaneousvoltage between the ends of a dipole antenna is the distance between the ends multiplied by theelectric field of the wave. It does not depend on the frequency of the wave.

Q34.17 The radiation resistance of a broadcast antenna is the equivalent resistance that would take the samepower that the antenna radiates, and convert it into internal energy.

Q34.18 Consider a typical metal rod antenna for a car radio. The rod detects the electric field portion of thecarrier wave. Variations in the amplitude of the incoming radio wave cause the electrons in the rodto 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 vibrations of theelectrons in the rod but at frequencies that imitate those of the carrier.

Q34.19 The frequency of EM waves in a microwave oven, typically 2.45 GHz, is chosen to be in a band offrequencies 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 anysignificant microwave energy and remain cool to the touch.

Q34.20 People of all the world’s races have skin the same color in the infrared. When you blush or exerciseor get excited, you stand out like a beacon in an infrared group picture. The brightest portions ofyour face show where you radiate the most. Your nostrils and the openings of your ear canals arebright; brighter still are just the pupils of your eyes.

Q34.21 Light bulbs and the toaster shine brightly in the infrared. Somewhat fainter are the back of therefrigerator and the back of the television set, while the TV screen is dark. The pipes under the sinkshow the same weak sheen as the walls until you turn on the faucets. Then the pipe on the rightturns very black while that on the left develops a rich glow that quickly runs up along its length. Thefood 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 apatch 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 pupilsof your eyes.

Q34.22 Welding produces ultraviolet light, along with high intensity visible and infrared.

Q34.23 12.2 cm waves have a frequency of 2.46 GHz. If the Q value of the phone is low (namely if it ischeap), and your microwave oven is not well shielded (namely, if it is also cheap), the phone canlikely 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.

294 Electromagnetic Waves

SOLUTIONS TO PROBLEMS

Section 34.1 Maxwell’s Equations and Hertz’s Discoveries

*P34.1 (a) The rod creates the same electric field that it would ifstationary. We apply Gauss’s law to a cylinder of radiusr = 20 cm and length :

E A⋅ =∈

°=∈

z dq

E rll

inside

0

02 0π

λb gcosFIG. P34.1

E j j=∈

=× ⋅

×= ×

−

−

λπ π2

35 10

2 8 85 10 0 23 15 10

0

9

123

r radially outward

C m N m

C m N C

2

2

e je ja f. .

. .

(b) The charge in motion constitutes a current of 35 10 15 10 0 5259 6× × =− C m m s Ae je j . . This

current creates a magnetic field.

B =µπ0

2Ir

=× ⋅

= ×−

−4 10 0 525

2 0 25 25 10

77

π

π

T m A A

m T

e ja fa f

.

..k k

(c) The Lorentz force on the electron is F q q= + ×E v B

F

F

= − × × + − × × × ×⋅⋅

FHG

IKJ

= × − + × + = × −

− − −

− − −

1 6 10 3 15 10 1 6 10 240 10 5 25 10

5 04 10 2 02 10 4 83 10

19 3 19 6 7

16 17 16

. . . .

. . .

C N C C m s N sC m

N N N

e je j e je j

e j e j e j

j i k

j j j

Section 34.2 Plane Electromagnetic Waves

P34.2 (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 10

2 15 10 68018

10..

××

= × = m

3.00 10 m s s years8 .

Therefore, it will disappear from the sky in the year 2 004 680 2 68 103+ = ×. C.E. .

The star is 680 light-years away.

(b) ∆∆

tx

v= =

××

= =1 496 10

499 8 3111.

. m

3 10 m s s min8

(c) ∆∆

tx

v= =

×

×=

2 3 84 10

3 102 56

8

8

..

m

m s s

e j

(d) ∆∆

tx

v= =

×

×=

2 6 37 10

3 100 133

6

8

π ..

m

m s s

e j

(e) ∆∆

tx

v= =

××

= × −10 103 33 10

35 m

3 10 m s s8 .

Chapter 34 295

P34.3 v c c=∈

= = = ×1 1

1 780 750 2 25 10

0 0

8

κµ .. . m s

P34.4EB

c=

or 220

3 00 108

B= ×.

so B = × =−7 33 10 7337. T nT .

P34.5 (a) f cλ =

or f 50 0 3 00 108. . m m sa f = ×

so f = × =6 00 10 6 006. . Hz MHz .

(b)EB

c=

or22 0

3 00 108..

maxB= ×

so B kmax .= −73 3 nT .

(c) k = = = −2 250 0

0 126 1πλ

π.

. m

and ω π π= = × = ×−2 2 6 00 10 3 77 106 1 7f . . s rad se jB B k= − = − − ×max cos . cos . .kx t x tωb g e j73 3 0 126 3 77 107 nT .

P34.6 ω π π= = × = ×− −2 6 00 10 1 88 109 1 10 1f . . s s

kc

= = =×

×= =

−−2 6 00 10

3 00 1020 0 62 8

9 1

81π

λω π

π..

. . s

m s m B

Ecmax .

.= =×

=300

3 00 101 008

V m m s

Tµ

E x t= − ×300 62 8 1 88 1010 V mb g e jcos . . B x t= − ×1 00 62 8 1 88 1010. cos . . Tµb g e j

P34.7 (a) BEc

= =×

= × =−1003 00 10

3 33 10 0 33387 V m

m s T T

.. . µ

(b) λπ π

µ= =×

=−2 2

1 00 100 6287 1k ..

m m

(c) fc

= =×

×= ×−λ

3 00 106 28 10

4 77 108

714.

..

m s m

Hz

296 Electromagnetic Waves

P34.8 E E kx t= −max cos ωb g∂∂

= − −

∂∂

= − − −

∂∂

= − −

∂∂

= − − −

Ex

E kx t k

Et

E kx t

Ex

E kx t k

Et

E kx t

max

max

max

max

sin

sin

cos

cos

ω

ω ω

ω

ω ω

b ga f

b ga f

b ge j

b ga f

2

22

2

22

We must show:∂∂

= ∈∂∂

Ex

Et2 0 0

2

2µ .

That is, − − = − ∈ − −k E kx t E kx t20 0

2e j b g a f b gmax maxcos cosω µ ω ω .

But this is true, becausek

f c

2

2

2

2 0 01 1

ω λµ=

FHGIKJ = = ∈ .

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

P34.9 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 = = =2 2 2 00 4 00L . . m ma f

Thus, fc

= =×

= × =λ

3 00 104 00

7 50 10 75 08

7..

. . m s

m Hz MHz .

P34.10 dA to A cm= ± =6 5%2λ

λ

λ

= ±

= = ± × = × ±−

12 5%

0 12 5% 2 45 10 2 9 10 5%9 1 8

cm

m s m sv f . . .a fe j

Section 34.3 Energy Carried by Electromagnetic Waves

P34.11 S IUAt

UcV

uc= = = =Energy

Unit Volume W m

m s J m

23= = =

×=u

Ic

1 0003 00 10

3 338.. µ

P34.12 Srav

2 W

4 4.00 1 609 m W m= =

×

×=

P4

4 00 107 682

3

2π πµ

..

b gE cSmax .= =2 0 076 10µ av V m

∆V E Lmax max . . .= = =76 1 0 650 49 5 mV m m mV amplitudeb ga f b g or 35.0 mV (rms)

Chapter 34 297

P34.13 r = = ×5 00 1 609 8 04 103. . mi m mi ma fb gS

r= =

×

×=

P4

250 103072

3

2π πµ

W

4 8.04 10 W W m

3

2

e j

P34.14 I = =100

7 962 W

4 1.00 m W m2

π a f .

uIc

= = × =−2 65 10 26 58. . J m nJ m3 3

(a) u uE = =12

13 3. nJ m3

(b) u uB = =12

13 3. nJ m3

(c) I = 7 96. W m2

P34.15 Power output = (power input)(efficiency).

Thus, Power inputPower out

eff W

0.300 W= =

×= ×

1 00 103 33 10

66.

.

and AI

= =×

×= ×

P 3 33 103 33 10

63.

. W

1.00 10 W m m3 2

2 .

P34.16 IB c

r= =max

2

022 4µ π

P

Br cmax

.

. ..=

FHGIKJFHGIKJ =

× ×

× ×= ×

−−P

42 10 0 10 2 4 10

4 5 00 10 3 00 105 16 102

03 7

3 2 8

10

πµ π

π

e ja fe je j e j

T

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

P34.17 (a) P = =I R2 150 W

A rL= = × = ×− −2 2 0 900 10 0 080 0 4 52 103 4π π . . . m m m2e jb gS

A= =P

332 kW m2 (points radially inward)

(b) BIr

= =×

=−

µπ

µ

πµ0 0

321 00

2 0 900 10222

.

.

a fe j

T

EVx

IRL

= = = =∆∆

1501 88

V0.080 0 m

kV m.

Note: SEB

= =µ0

332 kW m2

298 Electromagnetic Waves

*P34.18 (a) IE

c= =

×

× ⋅ × ⋅FHGIKJ ⋅FHGIKJ

⋅ ⋅⋅

FHG

IKJ

⋅FHGIKJ−

max2

0

6 2

7 8

2

2

3 10

2 4 10 3 10µ π

V m

T m A m s

JV C

CA s

T C mN s

N mJ

e je je j

I = ×1 19 1010. W m2

(b) P = = ××FHG

IKJ = ×

−

IA 1 19 105 10

2 34 10103 2

5. . W m m

2 W2e jπ

P34.19 (a) E B i j k i j k⋅ = + − ⋅ + +80 0 32 0 64 0 0 200 0 080 0 0 290. . . . . .e jb g e jN C Tµ

E B⋅ = + − ⋅ ⋅ =16 0 2 56 18 56 0. . .a f N s C m2 2

(b) S E Bi j k i j k

= × =+ − × + +

× ⋅−1 80 0 32 0 64 0 0 200 0 080 0 0 290

4 1007µ

µ

π

. . . . . .e j e j N C T

T m A

Sk j k i j i

=− − + − + ×

×

−

−

6 40 23 2 6 40 9 28 12 8 5 12 10

4 10

6

7

. . . . . .e j W m2

π

S i j= − =11 5 28 6 30 9. . .e j W m W m2 2 at − °68 2. from the +x axis.

*P34.20 The energy put into the water in each container by electromagnetic radiation can be written ase t eIA tP ∆ ∆= where e is the percentage absorption efficiency. This energy has the same effect as heatin raising the temperature of the water:eIA t mc T Vc T

TeI t

ceI t

c

∆ ∆ ∆

∆∆ ∆

= =

= =

ρ

ρ ρ

2

3

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

∆T =×

⋅°= °

0 7 25 10 480

0 06 4 18633 4

3.

..

W m s

10 kg m m J kg CC

2

3 3

e je ja f .

For the larger,

∆T =⋅

°= °

0 91 25 480

0 12 4 18621 7

.

..

J s m s

m J CC

2

2

e je j

.

P34.21 We call the current Irms and the intensity I. The power radiated at this frequency is

P = = =0 010 00 010 0

1 312

..

.b gb g b g∆

∆V I

VRrms rms

rms W .

If it is isotropic, the intensity one meter away is

IA

Sc

B

BI

c

= = = = =

= =× ⋅

×=

−

P 1 310 104

2

2 2 4 10 0 104

3 00 1029 5

20

2

07

8

..

.

..

max

max

W

4 1.00 m W m

T m A W m

m s nT

2av

2

π µ

µ π

a fe je j

Chapter 34 299

P34.22 (a) efficiencyuseful power output

total power input W

1 400 W= × = FHG

IKJ × =100%

700100% 50 0%.

(b) SAav

2 W0.068 3 m m

W m= = = ×P 700

0 038 12 69 105

b gb g..

Sav2 kW m toward the oven chamber= 269

(c) SE

cav = max2

02µ

Emax . . .

.

= × ⋅ × × = ×

=

−2 4 10 3 00 10 2 69 10 1 42 10

14 2

7 8 5 4π T m A m s W m V m

kV m

2e je je j

P34.23 (a) BE

cmaxmax= : Bmax

..

.=×

×=

7 00 103 00 10

2 335

8

N C m s

mT

(b) IE

c= max

2

02µ: I =

×

× ×=

−

7 00 10

2 4 10 3 00 10650

5 2

7 8

.

.

e je je jπ

MW m2

(c) IA

=P

: P = = × ×LNM

OQP =

−IA 6 50 104

1 00 10 5108 3 2. . W m m W2e j e jπ

P34.24 (a) I =×

×=

−

−

10 0 104 97

3

3 2

..

e je j

W

0.800 10 m kW m2

π

(b) uIcav

23 J m s

m s J m= =

× ⋅

×=

4 97 103 00 10

16 63

8

..

. µ

P34.25 (a) E cB= = × × =−3 00 10 1 80 10 5408 6. . m s T V me je j

(b) uB

av3 J m= =

×

×=

−

−

2

0

6 2

7

1 80 10

4 102 58

µ πµ

..

e j

(c) S cuav av2 W m= = × × =−3 00 10 2 58 10 7738 6. .e je j

(d) This is 77 3%. of the intensity in Example 34.5 . It may be cloudy, or the Sun may be setting.

300 Electromagnetic Waves

Section 34.4 Momentum and Radiation Pressure

P34.26 The pressure P upon the mirror is PSc

=2 av

where A is the cross-sectional area of the beam and SAav =P

.

The force on the mirror is then F PAc A

Ac

= = FHGIKJ =

2 2P P.

Therefore, F =×

×= ×

−−

2 100 10

3 106 67 10

3

810e j

e j. N .

P34.27 For complete absorption, PSc

= =×

=25 0

3 00 1083 38

..

. nPa .

P34.28 (a) The radiation pressure is2 1 340

3 00 108 93 108

6 W m

m s N m

2

22e j

..

×= × − .

Multiplying by the total area, A = ×6 00 105. m2 gives: F = 5 36. N .

(b) The acceleration is: aFm

= = = × −5 368 93 10 4..

N6 000 kg

m s2 .

(c) It will arrive at time t where d at=12

2

or tda

= =×

×= × =

−

2 2 3 84 10

8 93 109 27 10 10 7

8

45

.

.. .

m

m s s days

2

e je j

.

P34.29 Ir

Ec

= =P

π µ2

2

02max

(a) Ec

rmax .= =P 2

1 9002

µπb g

kN C

(b)15 10

3 00 101 00 50 0

3

8

×

×=

− J s m s

m pJ.

. .a f

(c) pUc

= =××

= × ⋅−

−5 103 00 10

1 67 1011

819

.. kg m s

Chapter 34 301

P34.30 (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:

IrE

S

E

=P

4 2π

and IrMS

M

=P

4 2π.

Thus, I IrrM E

E

M=FHGIKJ =

××

FHG

IKJ =

2 11 2

1 3401 496 10

577 W m m

2.28 10 m W m2

112e j .

.

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

PM M MI R= = ×LNM

OQP = ×π π2 6 2 16577 3 37 10 2 06 10e j e j e j W m m W2 . . .

(c) If Mars behaves as a perfect absorber, it feels pressure PS

cIc

M M= =

and force F PAIc

Rc

MM

M= = = =××

= ×π 216

72 06 106 87 10e j P ..

W3.00 10 m s

N8 .

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

FGM M

rgS M

M

= =× ⋅ × ×

×= ×

−

2

11 30 23

11 221

6 67 10 1 991 10 6 42 10

2 28 101 64 10

. . .

..

N m kg kg kg

m N

2 2e je je je j

which is ~1013 times stronger than the repulsive force of part (c).

P34.31 (a) The total energy absorbed by the surface is

U I At= FHGIKJ = LNM

OQP × =

12

12

750 0 500 1 00 60 0 11 3 W m m s kJ2 2e j e ja f. . . . .

(b) The total energy incident on the surface in this time is 2 22 5U = . kJ , with U = 11 3. kJ beingabsorbed and U = 11 3. kJ being reflected. The total momentum transferred to the surface is

p

pUc

Uc

Uc

= +

= FHGIKJ +FHGIKJ = =

×

×= × ⋅−

momentum from absorption momentum from reflection

J

m s kg m s

b g a fe j2 3 3 11 3 10

3 00 101 13 10

3

84

.

..

302 Electromagnetic Waves

*P34.32 The radiation pressure on the disk is PSc

Ic

FA

Fr

= = = =π 2 .

Then Fr Ic

=π 2

.

Take torques about the hinge: τ∑ = 0

H H mgrr Ircx y0 0 02

a f a f+ − + =sinθπ

θπ π

= =×

FHG

IKJ

= = °

− −

−

sin sin.

sin . .

12

12 7

1

0 4 10 11

0 071 2 4 09

r Imgc

m W s s

0.024 kg m 9.8 m 3 10 m

kg m W s

2

2 8

2

3

a fb g a fe j

mgPA

HxHy

r

θ

FIG. P34.32

Section 34.5 Production of Electromagnetic Waves by an Antenna

P34.33 λ = =cf

536 m so h = =λ4

134 m

λ = =cf

188 m so h = =λ4

46 9. m

P34.34 P =∆V

Ra f2

or P ∝ ∆Va f2

∆ ∆V E y Ey y= − ⋅ = ⋅a f cosθ

∆V ∝ cosθ so P ∝ cos2 θ

(a) θ = °15 0. : P P P= ° = =max maxcos . . .2 15 0 0 933 93 3%a f

(b) θ = °45 0. : P P P= ° = =max maxcos . . .2 45 0 0 500 50 0%a f

(c) θ = °90 0. : P P= ° =max cos .2 90 0 0a f

θ

receiving antenna∆y

FIG. P34.34

Chapter 34 303

P34.35 (a) Constructive interference occurs whend ncosθ λ= for some integer n.

cosθλ λ

λ= =

FHGIKJ =n

dn n

22

n = ± ±0 1 2, , , …

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

(b) Destructive interference occurs when

dn

cosθ λ=+F

HGIKJ

2 12

: cosθ = +2 1n

∴ = ± = ° °−weak signal @ θ cos ,1 1 0 180a f

FIG. P34.35

P34.36 For the proton, F ma∑ = yields qvBmv

Rsin .90 0

2

°= .

The period of the proton’s circular motion is therefore: TR

vm

qB= =

2 2π π.

The frequency of the proton’s motion is fT

=1

.

The charge will radiate electromagnetic waves at this frequency, with λπ

= = =cf

cTmc

qB2

.

*P34.37 (a) The magnetic field B k= −12 0µ ωJ kx tmax cosb g applies for x > 0 , since it describes a wave

moving in the i direction. The electric field direction must satisfy S E B= ×1

0µ as i j k= × so

the direction of the electric field is j when the cosine is positive. For its magnitude we have

E cB= , so altogether we have E j= −12 0µ ωcJ kx tmax cosb g .

(b) S E B i= × = −1 1 1

40 002 2 2

µ µµ ωcJ kx tmax cos b g

S i= −14 0

2 2µ ωcJ kx tmax cos b g

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

average of the cosine-squared function is 12

, so I cJ=18 0

2µ max .

(d) JIcmax .= =

× ×=−

8 8 570

4 10 3 103 48

07 8µ π

W m

Tm A m s A m

2e jb g

304 Electromagnetic Waves

Section 34.6 The Spectrum of Electromagnetic Waves

P34.38 From the electromagnetic spectrum chart and accompanying text discussion, the followingidentifications are made:

Frequency, f Wavelength, λ =cf

Classification

2 2 100 Hz Hz= × 150 Mm Radio2 2 103 kHz Hz= × 150 km Radio2 2 106 MHz Hz= × 150 m Radio2 2 109 GHz Hz= × 15 cm Microwave2 2 1012 THz Hz= × 150 µm Infrared2 2 1015 PHz Hz= × 150 nm Ultraviolet2 2 1018 EHz Hz= × 150 pm X-ray2 2 1021 ZHz Hz= × 150 fm Gamma ray2 2 1024 YHz Hz= × 150 am Gamma ray

Wavelength, λ Frequency, fc

=λ

Classification

2 2 103 km m= × 1 5 105. × Hz Radio2 2 100 m m= × 1 5 108. × Hz Radio2 2 10 3 mm m= × − 1 5 1011. × Hz Microwave2 2 10 6 m mµ = × − 1 5 1014. × Hz Infrared2 2 10 9 nm m= × − 1 5 1017. × Hz Ultraviolet/X-ray2 2 10 12 pm m= × − 1 5 1020. × Hz X-ray/Gamma ray2 2 10 15 fm m= × − 1 5 1023. × Hz Gamma ray2 2 10 18 am m= × − 1 5 1026. × Hz Gamma ray

P34.39 fc

= =×

×= ×−λ

3 00 105 50 10

5 45 108

714.

..

m s m

Hz

P34.40 (a) fc

= =×

λ3 10

1 710

88 m s

m Hz radio wave

.~

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

f =×

× −

3 00 106 10

108

513.

~ m s m

Hz infrared

P34.41 (a) f cλ = gives 5 00 10 3 00 1019 8. .× = × Hz m se jλ : λ = × =−6 00 10 6 0012. . m pm

(b) f cλ = gives 4 00 10 3 00 109 8. .× = × Hz m se jλ : λ = =0 075 7 50. . m cm

Chapter 34 305

P34.42 (a) λ = =×

×=−

cf

3 00 101 150

2618

1

. m s10 s

m3 so180

0 690 m

261 m wavelengths= .

(b) λ = =×

×=−

cf

3 00 1098 1 10

3 068

6 1

..

. m s s

m so180

58 9 m

3.06 m wavelengths= .

P34.43 Time to reach object = = × = ×− −12

12

4 00 10 2 00 104 4total time of flight s sb g e j. . .

Thus, d vt= = × × = × =−3 00 10 2 00 10 6 00 10 60 08 4 4. . . . m s s m kme je j .

P34.44 The time for the radio signal to travel 100 km is: ∆tr =××

= × −100 103 33 10

34 m

3.00 10 m s s8 . .

The sound wave travels 3.00 m across the room in: ∆ts = = × −3 008 75 10 3..

m343 m s

s .

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

total time difference of

∆t = × − × = ×− − −8 75 10 3 33 10 8 41 103 4 3. . . s s s .

P34.45 The wavelength of an ELF wave of frequency 75.0 Hz is λ = =×

= ×cf

3 00 1075 0

4 00 108

6..

. m s

Hz m .

The length of a quarter-wavelength antenna would be L = × = ×1 00 10 1 00 106 3. . m km

or L = FHG

IKJ =1 000

0 621621 km

mi1.00 km

mib g ..

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

P34.46 (a) For the AM band, λmaxmin

.= =

×

×=

cf

3 00 10540 10

5568

3

m s Hz

m

λminmax

.= =

×

×=

cf

3 00 101 600 10

1878

3

m s Hz

m .

(b) For the FM band, λmaxmin

..

.= =×

×=

cf

3 00 1088 0 10

3 418

6

m s Hz

m

λminmax

..= =

×

×=

cf

3 00 10108 10

2 788

6

m s Hz

m .

306 Electromagnetic Waves

Additional Problems

P34.47 (a) P = SA : P = ×LNM

OQP = ×1 340 4 1 496 10 3 77 1011 2 26 W m m W2e j e jπ . .

(b) ScB

= max2

02µso B

Scmax .

.= =×

×=

−2 2 4 10 1 340

3 00 103 350

7

8µ π

µ N A W m

m s T

2 2e je j

SE

c= max

2

02µso E cSmax . .= = × × =−2 2 4 10 3 00 10 1 340 1 010

7 8µ πe je jb g kV m

P34.48 Suppose you cover a 1.7 m-by-0.3 m section of beach blanket. Suppose the elevation angle of the Sunis 60°. Then the target area you fill in the Sun’s field of view is

1 7 0 3 30 0 4. . cos . m m m2a fa f °= .

Now IA

UAt

= =P

U IAt= = 1 340 0 6 0 5 0 4 3 600 106 W m m s J2 2e j a fa fe j b g. . . ~ .

P34.49 (a) ε θ= − = −d

dtddt

BABΦcosa f ε ω θ ω ω θ= − =A

ddt

B t AB tmax maxcos cos sin cosb g b g

ε π π θt fB A f ta f = 2 2max sin cos ε π θ πt r fB f ta f = 2 22 2max cos sin

Thus, ε π θmax max cos= 2 2 2r f B

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 .

P34.50 (a) FGM m

RGM

RrS S

grav = = FHGIKJFHGIKJ

LNM

OQP2 2

343

ρ π

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

Since FS r

crad =π 2

,

FF r

SRcGM r

rad

Sgrav= FHGIKJFHG

IKJ ∝

1 34

12

ρ.

(b) From the result found in part (a), when F Fradgrav = ,

we have rSR

cGMS=

34

2

ρ

r =×

× ⋅ × ×

= ×

−

−

3 214 3 75 10

4 6 67 10 1 991 10 1 500 3 00 10

3 78 10

11 2

11 30 8

7

W m m

N m kg kg kg m m s

m

2

2 2 3

e je je je je je j

.

. . .

.

Chapter 34 307

P34.51 (a) BE

cmaxmax .= = × −6 67 10 16 T

(b) SE

cav2 W m= = × −max .

2

0

17

25 31 10

µ

(c) P = = × −S Aav W1 67 10 14.

(d) F PAS

cA= = FHGIKJ = × −av N5 56 10 23. (≅ the weight of

3 000 H atoms!) FIG. P34.51

P34.52 (a) The power incident on the mirror is: PI IA= = = ×1 340 100 4 21 102 7 W m m W2e j a fπ . .

The power reflected through the atmosphere is PR = × = ×0 746 4 21 10 3 14 107 7. . . W We j .

(b) SAR= =

×

×=

P 3 14 100 625

7

2.

. W

4.00 10 m W m

3

2

π e j

(c) Noon sunshine in Saint Petersburg produces this power-per-area on a horizontal surface:

PN

A= °=0 746 1 340 7 00 122. sin . W m W m2 2e j .

The radiation intensity received from the mirror is

0 625100% 0 513%

..

W m122 W m

2

2

FHG

IKJ = of that from the noon Sun in January.

P34.53 u E= ∈12 0

2max E

umax .=

∈=

295 1

0 mV m

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

A r= = × = ×− −2 2 4 00 10 0 100 2 51 102 2π π . . . m m m2e ja f .

(a) The intensity is then: SA

= =×

=−P 0 600

23 92.

. W

2.51 10 m W m2

2 .

(b) The standard is:

0 570 0 5701 00 10 1 00 10

1 005 70

3 4

. .. .

.. mW cm mW cm

W1.00 mW

cm m

W m2 22

22=

×FHG

IKJ

×FHG

IKJ =

−

e j .

While it is on, the telephone is over the standard by 23 95 70

4 19.

..

W m W m

times2

2 = .

308 Electromagnetic Waves

P34.55 (a) BE

cmaxmax

..= =

×= × −175

3 00 105 83 108

7 V m m s

T

k = = =2 2

0 015 0419

πλ

π. m

rad m ω = = ×kc 1 26 1011. rad s

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

(b) SE B

av = =max max .2

40 60µ

W m2 S iav = 40 6. W m2e j

(c) PScr = = × −2

2 71 10 7. N m2

(d) aF

mPAm

= = =×

= ×∑ −−

2 71 10 0 750

0 5004 06 10

77

. .

..

N m m

kg m s

2 22e je j

a i= 406 nm s2e j

P34.56 Of the intensity S = 1 340 W m2

the 38.0% that is reflected exerts a pressure PSc

Sc

r1

2 2 0 380= =

.a f.

The absorbed light exerts pressure PSc

Sc

a2

0 620= =

..

Altogether the pressure at the subsolar point on Earth is

(a) P P PS

ctotal

2 W m

m s Pa= + = =

×= × −

1 2 861 38 1 38 1 340

3 00 106 16 10

. .

..

e j

(b)P

Pa

total

2

2

N m N m

times smaller than atmospheric pressure=×

×= ×−

1 01 106 16 10

1 64 105

610.

..

P34.57 (a) PFA

Ic

= = FIAc c

a= = =×

= × =−P 1003 00 10

3 33 10 11087 J s

m s N kg

.. b g

a = × −3 03 10 9. m s2 and x at=12

2

txa

= = × =2

8 12 10 22 64. . s h

(b) 0 107 3 00 12 0 107 36 0 3 00= − − = − ⋅ + kg kg m s kg kg m s kgb g b gb g b g b gv v v v. . . .

v = =36 0110

0 327.

. m s t = 30 6. s

Chapter 34 309

P34.58 The mirror intercepts power P = = × =I A1 13 21 00 10 0 500 785. . W m m W2e j a fπ .

In the image,

(a) IA2

2=P

: I2 2785

625= = W

0.020 0 m kW m2

π b g

(b) IE

c2

2

02= max

µ so E cImax . . .= = × × × =−2 2 4 10 3 00 10 6 25 10 21 70 2

7 8 5µ πe je je j kN C

BE

cmaxmax .= = 72 4 Tµ

(c) 0 400. P∆ ∆t mc T=

0 400 785 1 00 4 186 100 20 0

3 35 101 07 10 17 8

53

. . .

.. .

W kg J kg C C C

J314 W

s min

a f b gb ga f∆

∆

t

t

= ⋅° ° − °

=×

= × =

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

The light pressure is PSc

Ic

= = .

(a) FI r

cmg r gb

b= = =π

ρ π2

343

and Igc m

=FHGIKJ = ×

43

34

8 32 101 3

7ρπ ρ

. W m2

(b) P = = × × =−IA 8 32 10 2 00 10 1 057 3 2. . . W m m kW2e j e jπ

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 PS

cIc

FA

= = =av .

(a) F Fg=

so IF cA

F c

Amgc

rg

b

= = =π 2 .

From the definition of density, ρπ

= =mV

mrb4 3 3b g

so1 4 3

1 3

r mb=FHG

IKJ

b gπ ρ.

Substituting for rb , Imgc

mgc

m gc m= FHGIKJ = FHG

IKJFHGIKJ =

FHGIKJπ

π ρ ρπ

ρπ ρ

43

43

43

34

2 3 2 3 1 3 1 3

.

(b) P = IA P =FHGIKJ

43

34

2 1 3π ρ

πρr gc m

310 Electromagnetic Waves

P34.61 (a) λ = =×

×=−

cf

3 00 1020 0 10

1 508

9 1

..

. m s s

cm

(b) U t= = × × = ×

=

− −P ∆a f e je j25 0 10 1 00 10 25 0 10

25 0

3 9 6. . .

.

J s s J

Jµ FIG. P34.61

(c) uUV

U

r

U

r c tav

J

0.060 0 m m s s= = = =

×

× ×

−

−π π π2 2

6

2 8 9

25 0 10

3 00 10 1 00 10e j e j a f b g e je j∆

.

. .

uav3 3 J m mJ m= × =−7 37 10 7 373. .

(d) Eu

max

.

.. .=

∈=

×

× ⋅= × =

−

−2 2 7 37 10

8 85 104 08 10 40 8

0

3

124av

3

2 2

J m

C N m V m kV m

e j

BE

cmaxmax .

..= =

×

×= × =−4 08 10

3 00 101 36 10 136

4

84 V m

m s T Tµ

(e) F PASc

A u A= = FHGIKJ = = × = × =− −

av3 J m m N N7 37 10 0 060 0 8 33 10 83 33 2 5. . . .e j b gπ µ

P34.62 (a) On the right side of the equation, C m s

C N m m s

N m C m sC s m

N ms

Js

W2 2

2 2

2 2 2 3

2 4 3

e je jb g

2

3⋅

=⋅ ⋅ ⋅ ⋅

⋅ ⋅=

⋅= = .

(b) F ma qE= = or aqEm

= =×

×= ×

−

−

1 60 10 100

9 11 101 76 10

19

3113

.

..

C N C

kg m s2e jb g

.

The radiated power is then: P =∈

=× ×

× ×= ×

−

−

−q ac

2 2

03

19 2 13 2

12 8 327

6

1 60 10 1 76 10

6 8 85 10 3 00 101 75 10

π π

. .

. ..

e j e je je j

W .

(c) F ma mvr

qvBc= =FHGIKJ =

2

so vqBrm

= .

The proton accelerates at avr

q B rm

= = =×

×= ×

−

−

2 2 2

2

19 2 2

27 214

1 60 10 0 350 0 500

1 67 105 62 10

. . .

..

e j a f a fe j

m s2 .

The proton then radiates Pq a

c=

∈=

× ×

× ×= ×

−

−

−2 2

03

19 2 14 2

12 8 324

6

1 60 10 5 62 10

6 8 85 10 3 00 101 80 10

π π

. .

. ..

e j e je je j

W .

P34.63 PSc Ac r c

= = = =×

= × −Power W

2 0.050 0 m m m s Pa

P2

60 0

1 00 3 00 106 37 10

87

π π.

. ..b ga fe j

Chapter 34 311

P34.64 F PASAc

A A

c c= = = =

P Pb g, τ = FHG

IKJ =F

c2 2P

, and τ κθ= .

Therefore, θκ

= =×

× ×= ×

−

−−P

2

3 00 10 0 060 0

2 3 00 10 1 00 103 00 10

3

8 112

c

. .

. ..

e jb ge je j

deg .

P34.65 The light intensity is I SE

c= =av

2

02µ.

The light pressure is PSc

Ec

E= = = ∈2

02 0

2

212µ

.

For the asteroid, PA ma= and aE Am

=∈0

2

2.

P34.66 f = 90 0. MHz , Emax .= × =−2 00 10 2003 V m mV m

(a) λ = =cf

3 33. m

Tf

BE

c

=1= × =

= = × =

−

−

1 11 10 11 1

6 67 10 6 67

8

12

. .

. .maxmax

s ns

T pT

(b) E j= −FHG

IKJ2 00 2

3 33 11 1. cos

. . mV m

m nsb g π

x tB k= −F

HGIKJ6 67 2

3 33 11 1. cos

. . pT

m nsb g π

x t

(c) IE

c= =

×

× ×= ×

−

−−max

.

..

2

0

3 2

7 89

2

2 00 10

2 4 10 3 00 105 31 10

µ π

e je je j

W m2

(d) I cu= av so uav3 J m= × −1 77 10 17.

(e) PI

c= =

×

×= ×

−−2 2 5 31 10

3 00 103 54 10

9

817

a fe j.

.. Pa

312 Electromagnetic Waves

*P34.67 (a) m V r= =ρ ρ π12

43

3

rm

=FHGIKJ =FHGG

IKJJ =

64

6 8 7

990 40 161

1 31 3

ρ π π

..

kg

kg m m

3

b ge j

(b) A r= = =12

4 2 0 161 0 1632 2π π . . m m2a f

(c) I e T= = × ⋅ =−σ 4 8 40 970 5 67 10 304 470. . W m K K W m2 4 2e ja f

(d) P = = =IA 470 0 163 76 8 W m m W2 2e j . .

(e) IE

c= max

2

02µ

E cImax = = × × =−2 8 10 3 10 470 59501 2 7 8

1 2µ πb g e je je j Tm A m s W m N C2

(f) E cBmax max=

Bmax .=×

=595

3 101 988

N C m s

Tµ

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

(h) Each kitten has radius rk = ×FHG

IKJ =

6 0 8990 4

0 072 81 3

..

a fπ

m and radiating area

2 0 072 8 0 033 32π . . m m2b g = . Eliza has area 26 5 5

990 40 120

2 3

ππ

..

a f×

FHG

IKJ = m2 . The total glowing

area is 0 120 4 0 033 3 0 254. . . m m m2 2 2+ =e j and has power output

P = = =IA 470 0 254 119 W m m W2 2e j . .

P34.68 (a) At steady state, P Pin out= and the power radiated out is Pout = e ATσ 4 ..

Thus, 0 900 1 000 0 700 5 67 10 8 4. . . W m W m K2 2 4e j e jA AT= × ⋅−

or T =× ⋅

L

NMM

O

QPP = = °

−

900

0 700 5 67 10388 115

8

1 4

W m

W m K K C

2

2 4. .e j.

(b) The box of horizontal area A, presents projected area A sin .50 0° perpendicular to thesunlight. Then by the same reasoning,

0 900 1 000 50 0 0 700 5 67 10 8 4. sin . . . W m W m K2 2 4e j e jA AT°= × ⋅−

or T =°

× ⋅

L

NMM

O

QPP = = °

−

900 50 0

0 700 5 67 10363 90 0

8

1 4 W m

W m K K C

2

2 4

e je j

sin .

. .. .

Chapter 34 313

P34.69 We take R to be the planet’s distance from its star. The planet, of radius r, presents a

projected area π r 2 perpendicular to the starlight. It radiates over area 4 2π r .

At steady-state, P Pin out= : eI r e r Tin π σ π2 2 44e j e j=

eR

r e r T6 00 10

423

2 2 4. ×FHG

IKJ =

W4 2π

π σ πe j e j so that 6 00 10 1623 2 4. × = W πσR T

RT

=×

=×

× ⋅= × =

−

6 00 10 6 00 10

3104 77 10 4 77

23

4

23

8 49. .

. . W

16 W

16 5.67 10 W m K K m Gm

2 4πσ π e ja f .

ANSWERS TO EVEN PROBLEMS

P34.2 (a) 2 68 103. × AD ; (b) 8 31. min ; (c) 2 56. s ; P34.30 (a) 577 W m2 ; (b) 2 06 1016. × W;(d) 0 133. s; (e) 33 3. sµ (c) 68 7. MN ; (d) The gravitational force is

~1013 times stronger and in the oppositedirection.P34.4 733 nT

P34.6 E x t= − ×300 62 8 1 88 1010 V mb g e jcos . . ;

B x t= − ×1 00 62 8 1 88 1010. cos . . Tµb g e jP34.32 4.09°

P34.34 (a) 93 3%. ; (b) 50 0%. ; (c) 0

P34.8 see the solutionP34.36

2π m c

eBp

P34.10 2 9 10 5%8. × ± m sP34.38 see the solution

P34.12 49 5. mVP34.40 (a) ~108 Hz radio wave;

P34.14 (a) 13 3. nJ m3 ; (b) 13 3. nJ m3 ; (b) ~1013 Hz infrared light(c) 7 96. W m2

P34.42 (a) 0 690. wavelengths ;P34.16 516 pT, ~105 times stronger than the

Earth’s field(b) 58 9. wavelengths

P34.44 The radio audience gets the news 8 41. mssooner.P34.18 (a) 11 9. GW m2 ; (b) 234 kW

P34.46 (a) 187 m to 556 m; (b) 2 78. m to 3 41. mP34.20 33.4°C for the smaller container and 21.7°Cfor the larger

P34.48 ~106 JP34.22 (a) 50 0%. ;

P34.50 (a) see the solution; (b) 378 nm(b) 269 kW m toward the oven chamber2 ;(c) 14 2. kV m

P34.52 (a) 31.4 MW; (b) 0 625. W m2 ; (c) 0.513%

P34.24 (a) 4 97. kW m2 ; (b) 16 6. J m3µP34.54 (a) 23 9. W m2 ; (b) 4 19. times the standard

P34.26 667 pNP34.56 (a) 6 16. Paµ ; (b) 1 64 1010. × times less than

atmospheric pressureP34.28 (a) 5 36. N; (b) 893 m s2µ ; (c) 10 7. days

314 Electromagnetic Waves

P34.58 (a) 625 kW m2 ; P34.66 (a) 3 33. m, 11 1. ns , 6 67. pT ;

(b) E j= −FHG

IKJ2 00 2

3 33 11 1. cos

. .mV m

m nsb g π

x t;

(b) 21 7. kN C and 72 4. Tµ ; (c) 17 8. min

P34.60 (a) 16

9

2 1 3m

gcρπ

FHG

IKJ ; (b)

169

2 2 1 32π ρm

r gcFHG

IKJ B k= −F

HGIKJ6 67 2

3 33 11 1. cos

. . pT

m nsb g π

x t;

(c) 5 31. nW m2 ; (d) 1 77 10 17. × − J m3 ;(e) 3 54 10 17. × − PaP34.62 (a) see the solution;

(b) 17 6. Tm s2 , 1 75 10 27. × − W ;P34.68 (a) 388 K; (b) 363 K(c) 1 80 10 24. × − W

P34.64 3 00 10 2. × − deg