electromagnetic waves - university of california, san diego · chapter 24 663 q24.15 welding...

661

Electromagnetic WavesCHAPTER OUTLINE 24.1 Displacement Current

and the Generalized Ampère’s Law

24.2 Maxwell’s Equations 24.3 Electromagnetic Waves 24.4 Hertz’s Discoveries 24.5 Energy Carried by

Electromagnetic Waves 24.6 Momentum and Radiation

Pressure 24.7 The Spectrum of

Electromagnetic Waves 24.8 Polarization 24.9 Context ConnectionThe

Special Properties of Laser Light

ANSWERS TO QUESTIONS Q24.1 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 1010

62×

×− m

3 10 m s s8 ~ . To go halfway around the Earth 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.

Q24.2 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.

Q24.3 Acceleration of electric charge. Q24.4 A wire connected to the terminals of a battery does not radiate electromagnetic waves. The battery

establishes an electric field, which produces current in the wire. The current in the wire creates a magnetic field. Both fields are constant in time, so no electromagnetic induction or “magneto-electric induction” happens. Neither field creates a new cycle of the other field. No wave propagation occurs.

Q24.5 No. Static electricity is just that: static. Without acceleration of the charge, there can be no

electromagnetic wave. Q24.6 Sound

The world of sound extends to the top of the atmosphere and stops there; sound requires a material medium. Sound propagates by a chain reaction of density and pressure disturbances recreating each other. Sound in air moves at hundreds of meters per second. Audible sound has frequencies over a range of three decades (ten octaves) from 20 Hz to 20 kHz. Audible sound has wavelengths of ordinary size (1.7 cm to 17 m). Sound waves are longitudinal.

Light The universe of light fills the whole universe. Light moves through materials, but faster in a vacuum. Light propagates by a chain reaction of electric and magnetic fields recreating each other. Light in air moves at hundreds of millions of meters per second. Visible light has frequencies over a range of less than one octave, from 430 to 750 Terahertz. Visible light has wavelengths of very small size (400 nm to 700 nm). Light waves are transverse.

continued on next page

662 Electromagnetic Waves

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.

Q24.7 When the capacitor is fully discharged, the current in the circuit is a maximum. The inductance of

the coil is making the current continue to flow. At this time the magnetic field of the coil contains all the energy that was originally stored in the charged capacitor. The current has just finished discharging the capacitor and is proceeding to charge it up again with the opposite polarity.

Q24.8 The Poynting vector

rS describes the energy flow associated with an electromagnetic wave. The

direction of rS is along the direction of propagation and the magnitude of

rS is the rate at which

electromagnetic energy crosses a unit surface area perpendicular to the direction of rS.

Q24.9 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 polarization is determined by your surroundings, so antennas need to be adjusted to align with the polarization of the wave.

Q24.10 You become part of the receiving antenna! You are a big sack of salt water. Your contribution usually

increases the gain of the antenna by a few tenths of a dB, enough to noticeably improve reception. Q24.11 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 vibrations of the electrons in the rod but at frequencies that imitate those of the carrier.

Q24.12 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.

Q24.13 Lightbulbs 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.

Q24.14 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.

Chapter 24 663

Q24.15 Welding produces ultraviolet light, along with high intensity visible and infrared. Q24.16 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.

Q24.17 Stimulated emission coerces atoms to emit photons along a specific axis, rather than in the random

directions of spontaneously emitted photons. The photons that are emitted through stimulation can be made to accumulate over time. The fraction allowed to escape constitutes the intense, collimated, and coherent laser beam. If this process relied solely on spontaneous emission, the emitted photons would not exit the laser tube or crystal in the same direction. Neither would they be coherent with one another.

Q24.18 Photons carry momentum. Recalling what we learned in Chapter 8, 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.

SOLUTIONS TO PROBLEMS Section 24.1 Displacement Current and the Generalized Ampère’s Law P24.1 We use the extended form of Ampère’s law, Equation 24.7. Since

no moving charges are present, I = 0 and we have

r r

lBz ⋅ = ∈ddd t

Eµ0 0Φ

.

In order to evaluate the integral, we make use of the symmetry

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

rB, we integrate r r

lB ⋅z d around the circle

rBr

ErE = 0 here

FIG. P24.1

at R = 0 15. m

to obtain 2π RB .

Differentiating the expression ΦE AE=

we have dd t

d dEd t

EΦ=FHGIKJ

π 2

4.

Thus, r r

lBz ⋅ = = ∈F

HGGG

I

KJJJd RB

d dEd t

240 0

2

π µ π.



Solving for B gives BR

d dEd t

=∈ F

HGGG

I

KJJJ

µπ

π0 02

2 4.

Substituting numerical values, B =× × ⋅− −4 10 8 85 10 0 10 20

2 0 15 4

7 12 2π π

π

H m F m m V m s

m

e je j a f b ga fa f

. .

.

B = × −1 85 10 18. T .

In Figure 24.1, 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

rB is counterclockwise.

Thus, the direction of rB at P is upwards.

Section 24.2 Maxwell’s Equations

P24.2 (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 l :

r rE A⋅ =

∈

°=∈

z dq

E rll

inside

0

02 0π λb gcos

rv

rE

rB

FIG. P24.2

rE 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.

rB =

µπ0

2Ir

=× ⋅

= ×−

−4 10 0 525

2 0 25 25 10

77

ππ T m A A

m T

e ja fa f

.

.$ . $k k

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

r r rE v B

r

r

F j i k

F j j j

= − × × + − × × × × ⋅⋅

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

Chapter 24 665

*P24.3 r r r r rF a E v B= = + ×m q q

r r r ra E v B= + ×e

m where

r rv B

i j kj k× = = − +

$ $ $

. . .. $ . $200 0 0

0 200 0 300 0 400200 0 400 200 0 300a f a f

r

r

a j j k j k

a j k j k

= ××

− + = × − +

= × − + = − × ×

−

−1 60 101 67 10

50 0 80 0 60 0 9 58 10 30 0 60 0

2 87 10 2 2 87 10

19

277

9 9

.

.. $ . $ . $ . . $ . $

. $ $ . $ $ m s +5.75 10 m s2 9 2e j

Section 24.3 Electromagnetic Waves P24.4 (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..

××

= × = m3.00 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) ∆ ∆t

xv

= = ××

= =1 496 10499 8 31

11..

m3 10 m s

s min8

(c) ∆ ∆t

xv

= =×

×=

2 3 84 10

3 102 56

8

8

..

m

m s s

e j

(d) ∆ ∆t

xv

= =×

×=

2 6 37 10

3 100 133

6

8

π ..

m

m s s

e j

(e) ∆ ∆t

xv

= = ××

= × −10 103 33 10

35 m

3 10 m s s8 .

P24.5 v c c=∈

= = = ×1 11 78

0 750 2 25 100 0

8

κµ .. . m s

P24.6 EB

c=

or 220

3 00 108

B= ×.

so B = × =−7 33 10 7337. T nT .


P24.7 (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=

or 22 0

3 00 108..

maxB= ×

so rB 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 j

r rB B k= − = − − ×max cos . cos . . $kx t x tωb g e j73 3 0 126 3 77 107 nT .

P24.8 (a) BEc

= =×

= × =−1003 00 10

3 33 10 0 33387 V m

m s T T

.. . µ

(b) λ π π µ= =×

=−2 2

1 00 100 6287 1k ..

m m

(c) fc= =

××

= ×−λ3 00 106 28 10

4 77 108

714.

..

m s m

Hz

P24.9 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, because k

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.

Chapter 24 667

P24.10 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= =

×= × =

λ3 00 10

4 007 50 10 75 0

87.

.. .

m s m

Hz MHz .

P24.11 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

*P24.12 The orbital speed of the Earth is as described by F ma∑ = : Gm m

rm v

rS E E2

2

=

vGm

rS= =

× ⋅ ×

×= ×

−6 67 10 1 99 10

1 496 102 98 10

11 30

114

. .

..

N m kg kg

m m s

2 2e je j.

The maximum frequency received by the extraterrestrials is

f fv cv cobs source Hz

m s m s

m s m s Hz=

+−

= ×+ × ×

− × ×= ×

11

57 0 101 2 98 10 3 00 10

1 2 98 10 3 00 1057 005 66 106

4 8

4 86.

. .

. ..e j e j e j

e j e j.

The minimum frequency received is

f fv cv cobs source Hz

m s m s

m s m s Hz=

−+

= ×− × ×

+ × ×= ×

11

57 0 101 2 98 10 3 00 10

1 2 98 10 3 00 1056 994 34 106

4 8

4 86.

. .

. ..e j e j e j

e j e j.

The difference, which lets them figure out the speed of our planet, is

57 005 66 56 994 34 10 1 13 106 4. . .− × = ×b g Hz Hz .

*P24.13 (a) Let fc be the frequency as seen by the car. Thus, f fc vc vc =+−source

and, if f is the frequency of the reflected wave, f fc vc vc= +−

.

Combining gives f fc vc v

=+−sourcea fa f .

(b) Using the above result, f c v f c v− = +a f a fsource

which gives f f c f f v f v− = + ≈source source sourceb g b g 2 .

The beat frequency is then f f ff v

cv

beat sourcesource= − = =2 2

λ.



(c) fbeat

9 m s 10 Hz

m s

m s

m Hz= 2.00 kHz=

×

×= =

2 30 0 10 0

3 00 10

2 30 0

0 030 02 0008

a fb ge j a fb gb g

. .

.

.

.

λ = =××

=cfsource

m s Hz

cm3 00 1010 0 10

3 008

9

..

.

(d) vf

= beatλ2

so ∆∆

vf

= = = ≈beat Hz m m s mi h

λ2

5 0 030 0

20 075 0 0 2

a fb g.. .

P24.14 (a) When the source moves away from an observer, the observed frequency is

f fc vc v

s

sobs source=

−+FHGIKJ

1 2

where v vs = source .

When v cs << , the binomial expansion gives

c vc v

vc

vc

vc

vc

vc

s

s

s s s s s−+FHGIKJ = − FHG

IKJ

LNM

OQP + FHG

IKJ

LNM

OQP ≈ −FHG

IKJ −FHG

IKJ ≈ −FHG

IKJ

−1 2 1 2 1 2

1 1 12

12

1 .

So, f fvcs

obs source≈ −FHGIKJ1 .

The observed wavelength is found from c f f= =λ λobs obs source :

λ λ λ λ

λ λ λ λ λ

obssource

obs

source

source

obs

= ≈−

=−

= − =−

−FHG

IKJ = −FHG

IKJ

ff

ff v c v c

v cv c

v c

s s

s

s

s

1 1

11

11

b g∆ .

Since 1 1− ≈vcs ,

∆λλ

≈v

csource .

(b) v c c csource nm

397 nm= FHGIKJ =FHG

IKJ =

∆λλ

20 00 050 4

..

*P24.15 For the light as observed

fc v c

v cf

v cv c

c

v cv c

v cv c

vc

vc

vc

v c

obsobs

sourcesource

source

obs

nm520 nm

m s

= =+−

=+−

+−

= =+−

= =

+ = − = =

= = ×

λ λ

λλ

11

11

11

650 11

1 25 1 562

1 1 562 1 5620 5622 562

0 220

0 220 6 59 10

2

7

. .

. ...

.

. .

Chapter 24 669

*P24.16 (a) Consider the raindrops moving toward the station, at speed v. They receive radio waves

with the Doppler-enhanced frequency ′ = +−

f fc vc v

where f = 2 85. GHz. These raindrops

reflect the waves at frequency ′f . The waves are received by the station with another upward Doppler shift in frequency to

′′ = ′+−

= +−

+−

× + = × +−FHGIKJ

+ × = +−

+ × − − × = +

× =

= × × =

−

− −

−

−

f fc vc v

fc vc v

c vc v

c vc v

c vc v

c c v v c v

c v

v

2 85 10 254 2 85 10

1 8 91 10

8 91 10 8 91 10

8 91 10 2 000 000 089

4 46 10 3 10 13 4

9 9

8

8 8

8

8 8

. .

.

. .

. .

. . .

Hz Hz Hz

m s m se je j

The same calculation with 254 Hz replaced by –254 Hz applies to the receding raindrops and

given the same velocity magnitude. Thus the velocities are 13.4 m/s toward the station and 13.4 m/s away from the station.

(b) Radio waves travel to the rain and back again in 180 sµ , so the one-way distance is

12

3 10 180 10 27 0008 6× × =− m s s me je j . The diameter of the vortex is

s r= = °°

FHGIKJ =θ π

27 000 471 m 1 rad

180 ma f . So its radius is

12

471 236 m ma f = and the angular

speed of the rain is ω = = =vr

13 4236

0 056 7.

.m sm

rad s . A Doppler weather radar computer

performs a calculation like this to detect a “tornado vortex signature.”

Section 24.4 Hertz’s Discoveries P24.17 This radio is a radiotelephone on a ship, according to frequency assignments made by international

treaties, laws, and decisions of the National Telecommunications and Information Administration.

The resonance frequency is fLC0

12

=π

.

Thus, Cf L

= =× ×

=−

1

2

1

2 6 30 10 1 05 10608

02 6 2 6π πb g e j e j. . Hz H

pF .

P24.18 fLC

= 12π

: Lf C

= =×

=−

1

2

1

2 120 8 00 100 2202 2 6π πb g a f e j.. H


P24.19 (a) fLC

= =×

=−

12

1

2 0 100 1 00 10503

6π π . . H F Hz

a fe j

(b) Q C= = × =−ε µ1 00 10 12 0 12 06. . . F V Ce ja f

(c) 12

12

2 2C LIε = max

ICLmax .= = × =

−ε 12 37 9

6

V1.00 10 F

0.100 H mA

(d) At all times U C= = × =−12

12

1 00 10 12 0 72 02 6 2ε µ. . . F V Je ja f .

FIG. P24.19

Section 24.5 Energy Carried by Electromagnetic Waves

P24.20 S IUAt

UcV

uc= = = = Energy

Unit Volume W m

m s J m

23= = =

×=u

Ic

1 0003 00 10

3 338.. µ

P24.21 r = = ×5 00 1 609 8 04 103. . mi m mi ma fb g

Sr

= = ×

×=P

4250 10

3072

3

2π πµ W

4 8.04 10 W W m

3

2

e j

P24.22 Srav

2 W

4 4.00 1 609 m W m= = ×

×=P

44 00 10

7 682

3

2π πµ.

.b g

E 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)

P24.23 Power output = (power input)(efficiency).

Thus, Power inputPower out

eff W

0.300 W= = × = ×1 00 10

3 33 106

6..

and AI

= = ××

= ×P 3 33 103 33 10

63.

. W

1.00 10 W m m3 2

2 .

Chapter 24 671

P24.24 (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 10

5 102 34 1010

3 25. . W m

m2

W2e jπ

P24.25 (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 g

SA

= =P332 kW m2 (points radially inward)

(b) BIr

= =×

=−

µπ

µπ

µ0 032

1 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

P24.26 (a) r rE 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

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

(b) r r rS E B

i 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

rS

k 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

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

P24.27 Power = =SAE

crmax

2

0

2

24

µπe j

Solving for r , rc

Ec

= = =P µπ

µπ

022

1002 15 0

5 16max

(( .

. W)

V m) m0

2


*P24.28 (a) We assume that the starlight moves through space without any of it being absorbed. The radial distance is

20 20 1 20 3 10 3 16 10 1 89 10

44 10

8 88 10

8 7 17

2

28

28

ly yr m s s m

W

4 1.89 10 m W m

17

2

= = × × = ×

= = ×

×= × −

c

Ir

b g e je j

e j

. .

. .Pπ π

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

P = = × × = ×−IA 8 88 10 6 37 10 1 13 108 6 2 7. . . . W m m W2e j e jπ

Section 24.6 Momentum and Radiation Pressure

P24.29 Ir

Ec

= =Pπ 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

P24.30 (a) The radiation pressure is 2 1 370

3 00 109 13 108

6 W m

m s N m

2

22e j

..

×= × − .

Multiplying by the total area, A = ×6 00 105. m2 gives: F = 5 48. N .

(b) The acceleration is: aFm

= = = × −5 489 13 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

9 13 109 17 10 10 6

8

45

.

.. .

m

m s s days

2

e je j

.

Chapter 24 673

Section 24.7 The Spectrum of Electromagnetic Waves

P24.31 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 newsroom by a

total time difference of

∆t = × − × = ×− − −8 75 10 3 33 10 8 41 103 4 3. . .s s s . P24.32 From the electromagnetic spectrum chart and accompanying text discussion, the following

identifications are made:

Frequency, f Wavelength, λ = cf

Classification

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

Wavelength, λ Frequency, fc=λ

Classification

2 2 103 km m= × 1 5 105. × Hz Radio 2 2 100 m m= × 1 5 108. × Hz Radio 2 2 10 3 mm m= × − 1 5 1011. × Hz Microwave 2 2 10 6 m mµ = × − 1 5 1014. × Hz Infrared

2 2 10 9 nm m= × − 1 5 1017. × Hz Ultraviolet/X-ray 2 2 10 12 pm m= × − 1 5 1020. × Hz X-ray/Gamma ray

2 2 10 15 fm m= × − 1 5 1023. × Hz Gamma ray 2 2 10 18 am m= × − 1 5 1026. × Hz Gamma ray

P24.33 fc= =

××

= ×−λ3 00 105 50 10

5 45 108

714.

..

m s m

Hz


P24.34 (a) fc= =

×λ

3 101 7

108

8 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 sm

Hz infrared

P24.35 (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

*P24.36 λ = =××

=cf

3 00 1027 33 10

11 08

6

..

. m s Hz

m

P24.37 For the proton, ΣF ma= : q vBmv

Rsin . .90 0

2

°=

The period and frequency of the proton’s circular motion are therefore:

TR

vm

qB= = =

×

×= ×

−

−−2 2 2 1 67 10

1 60 10 0 3501 87 10

27

197π π π .

. ..

kg

C Ts e j

e ja f

f = ×5 34 106. . Hz

The charge will radiate at this same frequency,

with λ = =××

=cf

3 00 1056 2

8.. .

m s5.34 10 Hz

m6

P24.38 (a) Channel 4: fmin = 66 MHz λmax .= 4 55 m

fmax = 72 MHz λmin .= 4 17 m

(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

P24.39 (a) λ = =××

=−cf

3 00 101 150

2618

1. m s

10 s m3 so

1800 690

m261 m

wavelengths= .

(b) λ = =××

=−cf

3 00 1098 1 10

3 068

6 1.

..

m s s

m so 180

58 9 m

3.06 m wavelengths= .

Chapter 24 675

*P24.40 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 .

Section 24.8 Polarization

P24.41 I I= max cos2 θ ⇒ θ = −cosmax

1 II

(a) I

Imax .= 1

3 00 ⇒ θ = = °−cos

..1 1

3 0054 7

(b) I

Imax .= 1

5 00 ⇒ θ = = °−cos

..1 1

5 0063 4

(c) I

Imax .= 1

10 0 ⇒ θ = = °−cos

..1 1

10 071 6

P24.42 The average value of the cosine-squared function is one-half, so the first polarizer transmits 12

the

light. The second transmits cos .2 30 034

°= .

I I If i i= × =12

34

38

P24.43 For incident unpolarized light of intensity Imax :

After transmitting 1st disk: I I= 12 max .

After transmitting 2nd disk: I I= 12

2max cos θ .

After transmitting 3rd disk: I I= °−12

902 2max cos cosθ θa f .

where the angle between the first and second disk is θ ω= t .

Using trigonometric identities cos cos2 12

1 2θ θ= +a f

FIG. P24.43

and cos sin cos2 29012

1 2°− = = −θ θ θa f a f



we have I I=+LNM

OQP

−LNM

OQP

12

1 22

1 22max

cos cosθ θa f a f

I I I= − = FHGIKJ −1

81 2

18

12

1 42max maxcos cosθ θe j a f .

Since θ ω= t , the intensity of the emerging beam is given by I I t= −116

1 4max cos ωb g .

*P24.44 P =

∆VRa f2

or P ∝ ∆Va f2

∆ ∆V E y Ey y= − ⋅ = ⋅a f lcosθ

∆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 l

receiving antenna∆y

FIG. P24.44

P24.45 I

Imaxcos . cos .= ° ° =1

245 0 45 0

18

2 2e je j

P24.46 Let the first sheet have its axis at angle θ to the original plane of polarization, and let each further

sheet have its axis turned by the same angle.

The first sheet passes intensity Imax cos2 θ .

The second sheet passes Imax cos4 θ

and the nth sheet lets through I Inmax maxcos .2 0 90θ ≥ where θ = °45

n.

Try different integers to find cos .2 5 455

0 885× °FHGIKJ = cos .2 6 45

60 902× °F

HGIKJ = .

(a) So n = 6

(b) θ = °7 50.

Chapter 24 677

Section 24.9 Context ConnectionThe Special Properties of Laser Light

P24.47 The photon energy is E Ehc

3 2 20 66 18 70 1 96* . . .− = − = =a f eV eVλ

λ =× ⋅ ×

×=

−

−

6 626 10 3 00 10

1 96 1 60 10633

34 8

19

. .

. .

J s m s

J nm

e je je j

.

P24.48 (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

P24.49 (a) I =×

× ×LNM

OQP= ×

−

− −

3 00 10

1 00 10 15 0 104 24 10

3

9 6 215

.

. ..

J

s m W m2e j

e j e jπ

(b) 3 00 100 600 10

30 0 101 20 10 7 503

9 2

6 212.

.

.. .×

×

×= × =−

−

−

− J m

m J MeVe j e j

e j

P24.50 fEh e

= =× ⋅

×FHG

IKJ ⋅FHG

IKJ = ×−

−−0 117

6 63 101 60 10

2 82 1034

1913 1.

..

. eV

J s C 1 J

1 V C s

λ µ= =××

=−cf

3 00 102 82 10

10 68

13 1

..

. m s s

m , infrared

*P24.51 E t= = × × =−P ∆ 1 00 10 1 00 10 0 010 06 8. . . W s Je je j

E hfhc

γ λ= = =

× ×

×= ×

−

−−

6 626 10 3 00 10

694 3 102 86 10

34 8

919

. .

..

e je j J J

NE

E= =

×= ×−

γ

0 010 02 86 10

3 49 101916.

.. photons


P24.52 (a) 3 00 10 14 0 10 4 208 12. . .× × =− m s s mme je j

(b) Ehc= = × −

λ2 86 10 19. J

N =×

= ×−3 00

2 86 101 05 1019

19..

.J

J photons

(c) V = =4 20 3 00 1192. . mm mm mm3a f a fπ

n = × = × −1 05 10119

8 82 1019

16 3.. mm

*P24.53 (a) The distance between nodes is λ2

, so we require solutions to 35 124 1032

. cm = N λ where N is

an integer and λ is in the required range. The midpoint of the range is 632.809 10 nm, giving

Ntrial

m

m=

×

×=

−2 35 124 103 10

632 809 1 101 110 101 07

2

9

.

..

e j. So we try N = 1 110 101, 1 110 102, 1 110 100,

1 110 103, and so on:

λ

λ

λ

1

2

2

2

3

2

2 35 124 103 10

1 110 101632 809 14

2 35 124 103 10

1 110 102632 808 57

2 35 124 103 10

1 110 100632 809 71

=×

=

=×

=

=×

=

−

−

−

..

..

..

m nm

m nm

m nm

e j

e j

e j

λ trial

m nm=

×=

−2 35 124 103 10

1 110 103632 808 00

2..

e j, outside the range. Thus the laser light has just

three wavelength components.

(b) 12

320

2m v kT= . We use the periodic table.

vkT

m= =

×

×

FHG

IKJ =

−

−3 3 1 38 10 393

20 181

6970

23

27

.

.

J K K

u u

1.66 10 kg m s

e ja f

(c) For a neon atom moving toward one mirror at the rms speed as it emits, the Doppler shift is

described by

′ = +

−=

′= +

−

′ = −+

= × −× +

=

f fc vc v

c c c vc v

c vc v

λ λ

λ λ 632 809 13 10 6973 10 697

632 807 638

8. . . nm nm

This is outside the given range. Many atoms are moving faster than the rms speed, so we should expect still more Doppler broadening of the resonance amplification peak.

Chapter 24 679

P24.54 (a) NN

N e

N ee eg

E k

gE k

E E k hc kB

B

B B3

2

300

300300 300

3

2

3 2= = =− ⋅

− ⋅− − ⋅ − ⋅

K

K K K

b g

b gb g b g b gλ

where λ is the wavelength of light radiated in the 3 2→ transition.

NN

e

NN

e

3

2

6 63 10 3 10 632 8 10 1.38 10 300

3

2

75 9 33

34 8 9 23

1 07 10

=

= = ×

− × ⋅ × × ×

− −

− − −. .

. .

J s m s m J K Ke je j e je ja f

(b) NN

eu E E k Tu B

l

l= − −b g

where the subscript u refers to an upper energy state and the subscript l to a lower energy state.

Since E E Ehc

u − = =l photon λ

NN

eu hc k TB

l

= − λ .

Thus, we require 1 02. = −e hc k TBλ

or ln ..

. .1 02

6 63 10 3 10

632 8 10 1 38 10

34 8

9 23a f e je j

e je j= −

× ⋅ ×

× ×

−

− −

J s m s

m J K T

T = − × = − ×2 28 101 02

1 15 104

6.ln .

.a f K .

A negative-temperature state is not achieved by cooling the system below 0 K, but by heating it above T = ∞ , for as T →∞ the populations of upper and lower states approach equality.

(c) Because E Eu − >l 0 , and in any real equilibrium state T > 0 ,

e E E k Tu B− − <lb g 1 and N Nu < l . Thus, a population inversion cannot happen in thermal equilibrium.

Additional Problems

P24.55 (a) P = SA : P = ×LNM

OQP = ×1 370 4 1 496 10 3 85 1011 2 26 W m m W2e j e jπ . .

(b) ScB

= max2

02µ so B

Scmax .

.= =×

×=

−2 2 4 10 1 370

3 00 103 390

7

8µ π

µ N A W m

m s T

2 2e je j

SE

c= max

2

02µ so E cSmax . .= = × × =−2 2 4 10 3 00 10 1 370 1 020

7 8µ πe je jb g kV m


P24.56 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 0 3 30 0 4. . cos . m m m2a fa f °= .

Now IA

UAt

= =P U IAt= = 1 370 0 6 0 5 0 4 3 600 106 W m m s J2 2e j a fa fe j b g. . . ~ .

P24.57 (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

rE is vertical,

rB is horizontal, so the plane of the loop should be vertical

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

*P24.58 (a) The power incident on the mirror is: PI IA= = = ×1 370 100 4 30 102 7 W m m W2e j a fπ . .

The power reflected through the atmosphere is PR = × = ×0 746 4 30 10 3 21 107 7. . . W We j .

(b) SAR= = ×

×=

P 3 21 100 639

7

2.

. W

4.00 10 m W m

3

2

πe j

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

PN

A= °=0 746 1 370 7 00 125. sin . W m W m2 2e j .

The radiation intensity received from the mirror is

0 639

100% 0 513%.

. W m

125 W m

2

2

FHG

IKJ = of that from the noon Sun in January.

Chapter 24 681

P24.59 (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. P24.59

P24.60 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 πl . . . 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= ×F

HGIKJ

×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 = .

P24.61 u E= ∈12 0

2max E

umax .=

∈=2

95 10

mV m

P24.62 (a) BE

cmaxmax

..= =

×= × −175

3 00 105 83 108

7 V m m s

T

k = = =2 20 015 0

419πλ

π. m

rad m ω = = ×kc 1 26 1011. rad s

Since rS is along x, and

rE is along y,

rB must be in the directionz . (That is

r r rS E B∝ × .)

(b) SE B

av = =max max .2

40 60µ

W m2 rS 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

ra i= 406 nm s2e j$


P24.63 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

= ⋅° ° − °

= × = × =

*P24.64 (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. P24.64

(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 jl ∆.

. .

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 08 103 00 10

1 36 10 1364

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

Chapter 24 683

P24.65 (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 P =∈

=× ×

× ×= ×

−

−

−q ac

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 .

*P24.66 (a) m V r= =ρ ρ π12

43

3

rm=FHGIKJ =FHGG

IKJJ =6

4

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. They are a source of infrared radiation. They glow not by visible-light emission but by infrared emission.

(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 . .

P24.67 (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 the

sunlight. 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 .

. .. .

*P24.68 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 = W

4 2ππ 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 πe ja f .

Chapter 24 685

P24.69 (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

= = × =28 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

ANSWERS TO EVEN PROBLEMS P24.2 (a) 3 15. $j kN C ; (b) 525 nT $k ; (c) −483$j aN P24.4 (a) 2 68 103. × A.D.; (b) 499 s; (c) 2 56. s; (d) 0 133. s; (e) 33 3. sµ P24.6 733 nT P24.8 (a) 0 333. Tµ ; (b) 0 628. mµ ; (c) 477 THz P24.10 75.0 MHz P24.12 11.3 kHz P24.14 0 050 4. c P24.16 (a) 13.4 m/s toward the station and 13.4 m/s

away from the station. (b) 0.056 7 rad/s P24.18 0.220 H P24.20 3 33. J m3µ P24.22 49.5 mV P24.24 (a) 11 9. GW m2 ; (b) 234 kW

P24.26 (a) r rE B⋅ = 0 ; (b) 11 5 28 6. $ . $i j−e j W m2

P24.28 (a) 88 8. nW m2 ; (b) 11.3 MW P24.30 (a) 5.48 N; (b) 913 m s2µ ; (c) 10.6 d

P24.32 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

P24.34 (a) ~108 Hz radio wave; (b) ~1013 Hz infrared light P24.36 11.0 m P24.38 (a) 4.17 m to 4.55 m; (b) 3.41 m to 3.66 m;

(c) 1.61 m to 1.67 m P24.40 60.0 km

P24.42 38

P24.44 (a) 93.3%; (b) 50.0%; (c) 0 P24.46 (a) 6; (b) 7.50° P24.48 (a) 2.33 mT; (b) 650 MW m2 ; (c) 510 W P24.50 28.2 THz, 10 6. mµ , infrared P24.52 (a) 4.20 mm; (b) 1 05 1019. × photons; (c) 8 82 1016. × mm3 P24.54 (a) 1 07 10 33. × − ; (b) − ×1 15 106. K ; (c) no real T is below 0 K


P24.56 ~106 J P24.58 (a) 32.1 MW; (b) 0 639. W m2 ; (c) 0.513% P24.60 (a) 23 9. W m2 ; (b) It is 4.19 times the standard. P24.62 (a)

rB kmax

$= 583 nT , k = 419 rad m, ω = 12 6. Trad s ;

(b) Sav2 W m= 40 6. ; (c) 271 nPa;

(d) 406 nm s2

P24.64 (a) 1.50 cm; (b) 25 0. Jµ ; (c) 7 37. m J m3 ; (d) 40.8 kV/m, 136 Tµ ; (e) 83 3. Nµ P24.66 (a) 16.1 cm; (b) 0 163. m2 ; (c) 470 W m2 ;

(d) 76.8 W; (e) 595 N/C; (f) 1 98. Tµ ; (g) The cats are nonmagnetic and carry no

macroscopic charge or current. Oscillating charges within molecules make them emit infrared radiation.

(h) 119 W P24.68 4.77 Gm

electromagnetic waves - university of california, san diego · chapter 24 663 q24.15 welding...

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