Electromagnetic Waves
CHAPTER OUTLINE
24.1
24.2
24.3
24.4
24.5
24.6
24.7
24.8
24.9
Displacement Current
and the Generalized
Ampère’s Law
Maxwell’s Equations
Electromagnetic Waves
Hertz’s Discoveries
Energy Carried by
Electromagnetic Waves
Momentum and Radiation
Pressure
The Spectrum of
Electromagnetic Waves
Polarization
Context ConnectionThe
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 × 10 6 m
~ 10 −2 s . To go halfway around the Earth takes only
8
3 × 10 m s
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
Light
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.
continued on next page
661
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.
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
Q24.8
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.
r
The Poynting
vector
S
describes the energy flow associated with an electromagnetic
wave. The
r
r
direction of S is along the direction of propagation and the magnitude of S is the rate
r at which
electromagnetic energy crosses a unit surface area perpendicular to the direction of S.
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-olantern: 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
z
r r
dΦ E
.
B ⋅ dl = µ 0 ∈0
dt
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. rSetting aside
until
the determination of the direction of B, we integrate
r later
r
B⋅ dl around the circle
z
at
R = 0.15 m
to obtain
2π RB .
Differentiating the expression
Φ E = AE
we have
π d 2 dE
dΦ E
.
=
4
dt
dt
Thus,
z
continued on next page
F
GH
I
JK
F π d 2 I dE
r r
J
B ⋅ dl = 2π RB = µ 0 ∈0 G
GH 4 JK d t .
r
E = 0 here
r
E
FIG. P24.1
r
B
664
Electromagnetic Waves
µ 0 ∈0 FG π d 2 IJ dE
.
2π R GH 4 JK d t
Solving for B gives
B=
Substituting numerical values,
e4π × 10
B=
−7
j a
2π a0.15 mfa 4f
je
f b20 V m ⋅ sg
H m 8.85 × 10 −12 F m π 0.10 m
2
B = 1.85 × 10 −18 T .
In Figure 24.1, the direction of the increase of the electric field is out
r the plane of the
B
is counterclockwise.
paper. By the right-hand
rule,
this
implies
that
the
direction
of
r
Thus, the direction of B 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 :
z
r
v
r r q
E ⋅ dA = inside
∈0
b g
E 2π rl cos 0° =
r
E=
(b)
r
B
r
E
λl
∈0
FIG. P24.2
e35 × 10 C mj N ⋅ m $j =
2π e8.85 × 10
C ja0.2 mf
−9
λ
2π ∈0 r
radially outward =
2
−12
2
e
3.15 × 10 3 $j N C .
je
j
The charge in motion constitutes a current of 35 × 10 −9 C m 15 × 10 6 m s = 0.525 A . This
current creates a magnetic field.
e4π × 10
=
r µ I
B= 0
2π r
(c)
ja
2π a0. 2 mf
−7
T ⋅ m A 0.525 A
f k$ =
5.25 × 10 −7 k$ T
r
r r
The Lorentz force on the electron is F = qE + qv × B
j FGH
r
N ⋅s
F = −1.6 × 10 −19 C 3.15 × 10 3 $j N C + −1.6 × 10 −19 C 240 × 10 6 i$ m s × 5.25 × 10 −7 k$
C⋅m
r
−16
−
17
−
16
− $j N + 2.02 × 10
+ $j N = 4.83 × 10
− $j N
F = 5.04 × 10
e
je
e j
j e
e j
je
e j
IJ
K
Chapter 24
*P24.3
r
r
r
r r
F = ma = qE + qv × B
$i
$j
k$
r e r r r
r r
a=
E + v × B where v × B = 200
0
0 = −200 0.400 $j + 200 0.300 k$
m
0.200 0.300 0.400
a
f
a
f
r 1.60 × 10 −19
50.0 $j − 80.0 $j + 60.0k$ = 9.58 × 10 7 −30.0 $j + 60.0k$
a=
1.67 × 10 −27
r
a = 2.87 × 10 9 − $j + 2k$ m s 2 = −2.87 × 10 9 $j +5.75 × 10 9 k$ m s 2
e
j
Section 24.3 Electromagnetic Waves
P24.4
Since the light from this star travels at 3.00 × 10 8 m s
(a)
the last bit of light will hit the Earth in
6.44 × 10 18 m
= 2.15 × 10 10 s = 680 years .
3.00 × 10 8 m s
Therefore, it will disappear from the sky in the year 2 004 + 680 = 2.68 × 10 3 C. E. .
The star is 680 light-years away.
(b)
∆t =
∆x 1.496 × 10 11 m
=
= 499 s = 8.31 min
v
3 × 10 8 m s
(c)
∆t =
8
∆x 2 3.84 × 10 m
=
= 2.56 s
v
3 × 10 8 m s
(d)
∆t =
6
∆x 2π 6.37 × 10 m
=
= 0.133 s
v
3 × 10 8 m s
(e)
∆t =
∆x 10 × 10 3 m
=
= 3.33 × 10 −5 s
v
3 × 10 8 m s
P24.5
v=
P24.6
E
=c
B
or
1
κµ 0 ∈0
e
j
e
=
1
1.78
j
c = 0.750 c = 2.25 × 10 8 m s
220
= 3.00 × 10 8
B
so B = 7.33 × 10 −7 T = 733 nT .
665
666
P24.7
Electromagnetic Waves
(a)
fλ = c
a
f = 6.00 × 10 6 Hz = 6.00 MHz .
so
(b)
(c)
f
f 50.0 m = 3.00 × 10 8 m s
or
E
=c
B
or
22.0
= 3.00 × 10 8
Bmax
so
r
Bmax = −73.3k$ nT .
k=
2π
λ
=
2π
= 0.126 m −1
50.0
e
j
ω = 2π f = 2π 6.00 × 10 6 s −1 = 3.77 × 10 7 rad s
and
r r
B = Bmax cos kx − ω t = −73.3 cos 0.126 x − 3.77 × 10 7 t k$ nT .
b
P24.8
P24.9
g
e
j
(a)
B=
100 V m
E
=
= 3.33 × 10 −7 T = 0.333 µT
c 3.00 × 10 8 m s
(b)
λ=
2π
2π
=
= 0.628 µm
k
1.00 × 10 7 m −1
(c)
f=
c
λ
=
b
3.00 × 10 8 m s
6. 28 × 10
E = Emax cos kx − ω t
−7
m
= 4.77 × 10 14 Hz
g
ga f
ga f
ge j
ga f
b
b
∂E
= −Emax sin kx − ω t k
∂x
∂E
= −Emax sin kx − ω t −ω
∂t
∂2E
= −Emax cos kx − ω t k 2
∂x 2
∂2E
= −Emax cos kx − ω t −ω
∂t 2
b
b
2
We must show:
∂E
∂2E
=
∈
µ
.
0
0
∂x 2
∂t 2
That is,
− k 2 Emax cos kx − ω t = − µ 0 ∈0 −ω Emax cos kx − ω t .
But this is true, because
b
e j
k2
ω
2
F1I
=G J
H fλ K
2
=
g
a f
2
1
= µ 0 ∈0 .
c2
The proof for the wave of magnetic field follows precisely the same steps.
b
g
Chapter 24
P24.10
667
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.
a
f
λ = 2L = 2 2.00 m = 4.00 m
Thus,
P24.11
f=
c
λ
d A to A = 6 cm ± 5% =
λ = 12 cm ± 5%
a
=
3.00 × 10 8 m s
= 7.50 × 10 7 Hz = 75.0 MHz .
4.00 m
λ
2
fe
j
v = λ f = 0.12 m ± 5% 2.45 × 10 9 s −1 = 2.9 × 10 8 m s ± 5%
*P24.12
The orbital speed of the Earth is as described by
GmS
v=
=
r
e6.67 × 10
−11
∑ F = ma :
je
N ⋅ m 2 kg 2 1.99 × 10 30 kg
1.496 × 10
11
m
GmS m E
r2
j = 2.98 × 10
4
=
mE v 2
r
m s.
The maximum frequency received by the extraterrestrials is
f obs = fsource
1+v c
= 57.0 × 10 6 Hz
1−v c
e
j 1 − ee2.98 × 10
j e3.00 × 10
m sj e3.00 × 10
1 + 2.98 × 10 4 m s
4
8
8
j = 57.005 66 × 10
m sj
6
Hz .
j = 56.994 34 × 10
m sj
6
Hz .
ms
The minimum frequency received is
f obs = fsource
1−v c
= 57.0 × 10 6 Hz
1+v c
e
j 1 + ee2.98 × 10
j e3.00 × 10
m sj e3.00 × 10
1 − 2.98 × 10 4 m s
4
8
8
ms
The difference, which lets them figure out the speed of our planet, is
b57.005 66 − 56.994 34g × 10
*P24.13
(a)
6
Hz = 1.13 × 10 4 Hz .
Let f c be the frequency as seen by the car. Thus,
f c = f source
and, if f is the frequency of the reflected wave,
f = fc
c+v
.
c−v
which gives
a c + vf .
a c − vf
f a c − vf = f
a c + vf
b f − f gc = b f + f gv ≈ 2 f
The beat frequency is then
f beat = f − fsource =
Combining gives
(b)
c+v
c−v
Using the above result,
continued on next page
f = fsource
source
source
source
source v .
2 f source v
2v
=
.
λ
c
668
Electromagnetic Waves
(c)
f beat
λ=
P24.14
a2fb30.0 m sge10.0 × 10
=
9
Hz
8
3.00 × 10 m s
c
fsource
f beat λ
2
=
3.00 × 10 8 m s
10.0 × 10 9 Hz
2.00 kHz
= 3.00 cm
a
fb
g
5 Hz 0.030 0 m
∆f beat λ
=
= 0.075 0 m s ≈ 0.2 mi h
2
2
(d)
v=
(a)
When the source moves away from an observer, the observed frequency is
so
∆v =
j = a2fb30.0 m sg = 2 000 Hz =
b0.030 0 mg
f obs = fsource
FG c − v IJ
Hc+v K
12
s
where v s = v source .
s
When v s << c , the binomial expansion gives
FG c − v IJ
Hc+v K
s
s
12
L F v IO L F v IO
= M1 − G J P M1 + G J P
N H c KQ N H c KQ
FG1 − v IJ .
H cK
12
s
FG
H
−1 2
≈ 1−
s
f obs ≈ fsource
So,
s
The observed wavelength is found from c = λ obs fobs = λ fsource :
λ obs =
λ fsource
λ fsource
λ
≈
=
f obs
1 − vs c
fsource 1 − v s c
∆λ = λ obs − λ = λ
Since 1 −
(b)
*P24.15
vs
≈1,
c
v source = c
∆λ
λ
≈
F 1
GH 1 − v
b
s
v source
.
c
FG ∆λ IJ = cFG 20.0 nmIJ =
H λ K H 397 nm K
0.050 4c
For the light as observed
fobs =
1+v c
1+v c
c
c
fsource =
=
λ obs
1−v c
1 − v c λ source
1 + v c λ source 650 nm
=
=
520 nm
λ obs
1− v c
v
v
= 1.562 − 1.562
c
c
v = 0.220 c = 6.59 × 10 7 m s
1+
1+v c
= 1.25 2 = 1.562
1−v c
v 0.562
=
= 0.220
c 2.562
g
c
I F v c I.
JK GH 1 − v c JK
−1 = λ
s
s
vs
2c
IJ FG 1 − v IJ ≈ FG1 − v IJ .
K H 2c K H c K
s
s
Chapter 24
*P24.16
669
Consider the raindrops moving toward the station, at speed v. They receive radio waves
c+v
with the Doppler-enhanced frequency f ′ = f
where f = 2.85 GHz . These raindrops
c−v
reflect the waves at frequency f ′ . The waves are received by the station with another
upward Doppler shift in frequency to
(a)
c+v
c+v
=f
c−v
c−v
f ′′ = f ′
c+v
c−v
2.85 × 10 9 Hz + 254 Hz = 2.85 × 10 9 Hz
FG c + v IJ
H c − vK
c+v
c−v
c + 8.91 × 10 −8 c − v − 8.91 × 10 −8 v = c + v
1 + 8.91 × 10 −8 =
8.91 × 10 −8 c = 2.000 000 089 v
e
je
j
v = 4.46 × 10 −8 3 × 10 8 m s = 13.4 m s .
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.
Radio waves travel to the rain and back again in 180 µs , so the one-way distance is
1
3 × 10 8 m s 180 × 10 −6 s = 27 000 m . The diameter of the vortex is
2
π rad
1
s = rθ = 27 000 m 1°
= 471 m . So its radius is
471 m = 236 m and the angular
180°
2
v 13.4 m s
= 0.056 7 rad s . A Doppler weather radar computer
speed of the rain is ω = =
r
236 m
performs a calculation like this to detect a “tornado vortex signature.”
(b)
e
je
j
IJ
K
a fFGH
a
f
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
f0 =
Thus,
C=
1
2π LC
1
b 2π f g L
0
P24.18
f=
1
2π LC
:
L=
.
1
b2π f g C
2
=
2
=
1
e
6
2π 6.30 × 10 Hz
1
a f e8.00 × 10 j
2π 120
2
−6
2
j e
1.05 × 10 −6 H
= 0.220 H
j
= 608 pF .
670
P24.19
Electromagnetic Waves
1
1
(a)
f=
(b)
Q = Cε = 1.00 × 10 −6 F 12.0 V = 12.0 µC
(c)
1
1 2
Cε 2 = LI max
2
2
2π LC
=
2π
a
fe
0.100 H 1.00 × 10 −6 F
ja
e
f
FIG. P24.19
C
1.00 × 10 −6 F
= 12 V
= 37.9 mA
L
0.100 H
I max = ε
(d)
j
= 503 Hz
U=
At all times
ja
1
1
Cε 2 = 1.00 × 10 −6 F 12.0 V
2
2
e
f
2
= 72.0 µJ .
Section 24.5 Energy Carried by Electromagnetic Waves
Energy
1 000 W m 2
I
=u= =
= 3.33 µJ m3
Unit Volume
c 3.00 × 10 8 m s
U Uc
=
= uc
At V
P24.20
S=I=
P24.21
r = 5.00 mi 1 609 m mi = 8.04 × 10 3 m
a
S=
P24.22
fb
g
250 × 10 3 W
P
=
4π r 2 4π 8.04 × 10 3 W
Sav =
e
j
2
= 307 µW m 2
4.00 × 10 3 W
P
=
2
4π r
4π 4.00 × 1 609 m
b
g
2
= 7.68 µW m 2
Emax = 2 µ 0 cSav = 0.076 1 V m
b
ga
f
b
g
∆Vmax = Emax L = 76.1 mV m 0.650 m = 49.5 mV amplitude
P24.23
or 35.0 mV (rms)
Power output = (power input)(efficiency).
Thus,
Power input =
and
A=
P
I
=
Power out 1.00 × 10 6 W
=
= 3.33 × 10 6 W
eff
0.300
3.33 × 10 6 W
= 3.33 × 10 3 m 2 .
1.00 × 10 3 W m 2
Chapter 24
P24.24
(a)
e
j
2
2
3 × 10 6 V m
Emax
=
I=
2 µ 0 c 2 4π × 10 −7 T ⋅ m A 3 × 10 8 m s
e
je
FG J IJ FG C IJ FG T ⋅ C ⋅ mIJ FG N ⋅ mIJ
j H V ⋅CK H A ⋅sKH N ⋅s KH J K
2
I = 1.19 × 10 10 W m 2
P24.25
e
(b)
P = IA = 1.19 × 10
(a)
P = I 2 R = 150 W
10
F 5 × 10
W m jπ G
H 2
−3
2
jb
e
m
I
JK
2
= 2.34 × 10 5 W
g
A = 2π rL = 2π 0.900 × 10 −3 m 0.080 0 m = 4.52 × 10 −4 m 2
S=
(b)
P
A
= 332 kW m 2
(a)
a f
B=
µ 0 1.00
µ0I
=
= 222 µT
2π r 2π 0.900 × 10 −3
E=
∆V IR
150 V
=
=
= 1.88 kV m
∆x
L 0.080 0 m
e
Note:
P24.26
(points radially inward)
S=
EB
µ0
j
= 332 kW m 2
r r
E • B = 80.0 i$ + 32.0 $j − 64.0k$ N C • 0.200 i$ + 0.080 0 $j + 0.290k$ µT
e
jb g e
r r
E • B = a16.0 + 2.56 − 18.56f N ⋅ s C ⋅ m =
2
(b)
e
2
j
j
0
e
j
80.0 $i + 32.0 $j − 64.0k$ N C × 0.200 $i + 0.080 0 $j + 0.290k$ µT
r
1 r r
S=
E×B=
µ0
4π × 10 −7 T ⋅ m A
e
j
−6
2
r 6.40k$ − 23.2 $j − 6.40k$ + 9.28 i$ − 12.8 $j + 5.12 i$ × 10 W m
S=
4π × 10 −7
r
S = 11.5 i$ − 28.6 $j W m 2 = 30.9 W m 2 at −68.2° from the +x axis.
P24.27
e
j
Power = SA =
Emax 2
4π r 2
2µ 0 c
Solving for r ,
e
r=
j
P µ0c
(100 W) µ 0 c
=
= 5.16 m
2
2π Emax
2π (15.0 V m) 2
671
672
*P24.28
Electromagnetic Waves
(a)
We assume that the starlight moves through space without any of it being absorbed. The
radial distance is
b g e
je
j
20 ly = 20 c 1 yr = 20 3 × 10 8 m s 3.16 × 10 7 s = 1.89 × 10 17 m
I=
(b)
P
4 × 10 28 W
=
2
4π r
4π 1.89 × 10 17 m
e
j
2
= 8.88 × 10 −8 W m 2 .
The Earth presents the projected target area of a flat circle:
e
je
j
2
P = IA = 8.88 × 10 −8 W m 2 π 6.37 × 10 6 m
= 1.13 × 10 7 W .
Section 24.6 Momentum and Radiation Pressure
P24.29
I=
(a)
(b)
(c)
P24.30
(a)
P
π r2
=
2
Emax
2µ 0 c
Emax =
b
P 2µ 0 c
πr
2
15 × 10 −3 J s
3.00 × 10 8 m s
p=
g=
1.90 kN C
a1.00 mf =
50.0 pJ
U 5 × 10 −11
=
= 1.67 × 10 −19 kg ⋅ m s
c 3.00 × 10 8
e
2 1 370 W m 2
The radiation pressure is
8
3.00 × 10 m s
j = 9.13 × 10
2
−6
N m2 .
Multiplying by the total area, A = 6.00 × 10 5 m 2 gives: F = 5.48 N .
(b)
The acceleration is:
a=
5.48 N
F
=
= 9.13 × 10 −4 m s 2 .
m 6 000 kg
(c)
It will arrive at time t where
d=
1 2
at
2
or
t=
2d
=
a
e
j
2 3.84 × 10 8 m
e9.13 × 10
−4
m s2
j
= 9.17 × 10 5 s = 10.6 days .
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 × 10 3 m
= 3.33 × 10 −4 s .
3.00 × 10 8 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
total time difference of
∆t = 8.75 × 10 −3 s − 3.33 × 10 −4 s = 8.41 × 10 −3 s .
P24.32
From the electromagnetic spectrum chart and accompanying text discussion, the following
identifications are made:
Frequency, f
2
2
2
2
2
2
2
2
2
Hz = 2 × 10 0 Hz
kHz = 2 × 10 3 Hz
MHz = 2 × 10 6 Hz
GHz = 2 × 10 9 Hz
THz = 2 × 10 12 Hz
PHz = 2 × 10 15 Hz
EHz = 2 × 10 18 Hz
ZHz = 2 × 10 21 Hz
YHz = 2 × 10 24 Hz
Wavelength, λ
150 Mm
150 km
150 m
15 cm
Classification
Radio
Radio
Radio
Microwave
Infrared
Ultraviolet
X-ray
Gamma ray
Gamma ray
150 µm
150 nm
150 pm
150 fm
150 am
Frequency, f =
c
f
c
λ
Classification
km = 2 × 10 3 m
m = 2 × 10 0 m
mm = 2 × 10 −3 m
µm = 2 × 10 −6 m
1.5 × 10 5 Hz
1.5 × 10 8 Hz
1.5 × 10 11 Hz
1.5 × 10 14 Hz
Radio
Radio
Microwave
Infrared
2 nm = 2 × 10 −9 m
2 pm = 2 × 10 −12 m
1.5 × 10 17 Hz
1.5 × 10 20 Hz
Ultraviolet/X-ray
X-ray/Gamma ray
2 fm = 2 × 10 −15 m
2 am = 2 × 10 −18 m
1.5 × 10 23 Hz
1.5 × 10 26 Hz
Gamma ray
Gamma ray
2
2
2
2
P24.33
Wavelength, λ =
f=
c
λ
=
3.00 × 10 8 m s
5.50 × 10 −7 m
= 5.45 × 10 14 Hz
674
P24.34
Electromagnetic Waves
(a)
f=
f=
*P24.36
P24.37
λ
=
3 × 10 8 m s
~ 10 8 Hz
1.7 m
radio wave
1 000 pages, 500 sheets, is about 3 cm thick so one sheet is about 6 × 10 −5 m thick.
(b)
P24.35
c
3.00 × 10 8 m s
6 × 10 −5 m
~ 10 13 Hz
infrared
(a)
fλ = c
gives
e5.00 × 10
19
(b)
fλ = c
gives
e4.00 × 10
9
λ=
j
λ = 6.00 × 10 −12 m = 6.00 pm
j
λ = 0.075 m = 7.50 cm
Hz λ = 3.00 × 10 8 m s :
Hz λ = 3.00 × 10 8 m s :
c 3.00 × 10 8 m s
=
= 11.0 m
f 27.33 × 10 6 Hz
For the proton, ΣF = ma :
q vB sin 90.0° =
mv 2
.
R
The period and frequency of the proton’s circular motion are therefore:
T=
e
j
2π 1.67 × 10 −27 kg
2π R 2π m
=
=
= 1.87 × 10 −7 s
v
qB
1.60 × 10 −19 C 0.350 T
e
ja
f
f = 5.34 × 10 6 Hz .
The charge will radiate at this same frequency,
λ=
with
P24.38
(a)
(b)
(c)
P24.39
Channel 4:
Channel 6:
Channel 8:
3.00 × 10 8 m s
c
=
= 56.2 m .
f
5.34 × 10 6 Hz
fmin = 66 MHz
λ max = 4.55 m
fmax = 72 MHz
λ min = 4.17 m
fmin = 82 MHz
λ max = 3.66 m
fmax = 88 MHz
λ min = 3.41 m
fmin = 180 MHz
λ max = 1.67 m
fmax = 186 MHz
λ min = 1.61 m
(a)
λ=
c 3.00 × 10 8 m s
=
= 261 m
f 1 150 × 10 3 s −1
so
180 m
= 0.690 wavelengths
261 m
(b)
λ=
8
c 3.00 × 10 m s
=
= 3.06 m
f
98.1 × 10 6 s −1
so
180 m
= 58.9 wavelengths
3.06 m
Chapter 24
*P24.40
Time to reach object =
b
g e
675
1
1
total time of flight = 4.00 × 10 −4 s = 2.00 × 10 −4 s .
2
2
e
je
j
j
Thus, d = vt = 3.00 × 10 8 m s 2.00 × 10 −4 s = 6.00 × 10 4 m = 60.0 km .
Section 24.8 Polarization
P24.41
I = I max cos 2 θ
(a)
(b)
(c)
P24.42
I
I max
I
I max
I
I max
⇒
θ = cos −1
I
I max
=
1
3.00
⇒
θ = cos −1
1
= 54.7°
3.00
=
1
5.00
⇒
θ = cos −1
1
= 63.4°
5.00
=
1
10.0
⇒
θ = cos −1
1
= 71.6°
10.0
The average value of the cosine-squared function is one-half, so the first polarizer transmits
light. The second transmits cos 2 30.0° =
If =
P24.43
3
.
4
1 3
3
× Ii =
Ii
2 4
8
For incident unpolarized light of intensity I max :
After transmitting 1st disk:
I=
1
I max .
2
After transmitting 2nd disk:
I=
1
I max cos 2 θ .
2
After transmitting 3rd disk:
I=
1
I max cos 2 θ cos 2 90°−θ .
2
a
f
FIG. P24.43
where the angle between the first and second disk is θ = ω t .
Using trigonometric identities cos 2 θ =
and
continued on next page
a
a
1
1 + cos 2θ
2
f
f
cos 2 90°−θ = sin 2 θ =
a
1
1 − cos 2θ
2
f
1
the
2
676
Electromagnetic Waves
we have
LM a
f OPLM a1 − cos 2θ f OP
QN 2 Q
N
e1 − cos 2θ j = 18 I FGH 12 IJK a1 − cos 4θ f .
I=
1 + cos 2θ
1
I max
2
2
I=
1
I max
8
2
max
Since θ = ω t , the intensity of the emerging beam is given by I =
* P24.44
P=
a∆V f
2
a f
or P ∝ ∆V
R
b
g
1
I max 1 − cos 4ω t .
16
2
∆y
af
∆V = − Ey ⋅ ∆y = E y ⋅ l cos θ
l
θ
receiving
antenna
∆V ∝ cos θ so P ∝ cos 2 θ
P24.45
P24.46
a
f
a
f
a
f
(a)
θ = 15.0° : P = Pmax cos 2 15.0° = 0.933Pmax = 93.3%
(b)
θ = 45.0° : P = Pmax cos 2 45.0° = 0.500Pmax = 50.0%
(c)
θ = 90.0° : P = Pmax cos 2 90.0° = 0
FIG. P24.44
1
1
I
= cos 2 45.0° cos 2 45.0° =
8
I max 2
e
je
j
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
I max cos 2 θ .
The second sheet passes
I max cos 4 θ
and the nth sheet lets through
I max cos 2n θ ≥ 0.90 I max
Try different integers to find
cos 2 × 5
(a)
So n = 6
(b)
θ = 7.50°
FG 45° IJ = 0.885
H5K
where θ =
cos 2 × 6
45°
.
n
FG 45° IJ = 0.902 .
H6K
Chapter 24
Section 24.9 Context Connection
The Special Properties of Laser Light
P24.47
a
λ=
P24.48
(a)
(b)
(c)
Bmax =
f
E3* − E2 = 20.66 − 18.70 eV = 1.96 eV =
The photon energy is
Emax
:
c
e6.626 × 10
je
−34
J ⋅ s 3.00 × 10 8 m s
e
j
1.96 1.60 × 10 −19 J
7.00 × 10 5 N C
Bmax =
3.00 × 10 8 m s
hc
j=
λ
633 nm .
= 2.33 mT
e7.00 × 10 j
I=
= 650 MW m
2e 4π × 10 je3.00 × 10 j
P
Lπ
O
I= :
P = IA = e6.50 × 10 W m jM e1.00 × 10 mj P =
A
N4
Q
e3.00 × 10 Jj
I=
= 4.24 × 10 W m
O
L
π
×
×
1
.
00
10
s
15
.
0
10
m
e
jMN e
j PQ
5 2
E2
I = max :
2µ 0 c
2
−7
8
8
2
−3
−3
P24.49
(a)
15
−9
−6
0.600 × 10 m
(b)
e3.00 × 10 Jj e 30.0 × 10 m j = 1.20 × 10 J = 7.50 MeV
e
j
F 1.60 × 10 C I FG 1 J IJ = 2.82 × 10 s
0.117 eV
E
f= =
G e JK H 1 V ⋅ C K
h 6.63 × 10
J⋅s H
2
−9
−3
−6
P24.50
2
−12
−19
13
−34
λ=
*P24.51
2
2
c 3.00 × 10 8 m s
=
= 10.6 µm , infrared
f 2.82 × 10 13 s −1
e
je
j
E = P ∆t = 1.00 × 10 6 W 1.00 × 10 −8 s = 0.010 0 J
Eγ = hf =
N=
hc
λ
e6.626 × 10 je3.00 × 10 j J = 2.86 × 10
−34
=
694.3 × 10
8
−9
0.010 0
E
=
= 3.49 × 10 16 photons
Eγ 2.86 × 10 −19
−19
J
−1
2
510 W
677
678
P24.52
Electromagnetic Waves
(a)
e3.00 × 10
(b)
E=
hc
(a)
j
3.00 J
= 1.05 × 10 19 photons
2.86 × 10 −19 J
a
f a
f
V = 4.20 mm π 3.00 mm
n=
*P24.53
je
m s 14.0 × 10 −12 s = 4.20 mm
= 2.86 × 10 −19 J
λ
N=
(c)
8
2
= 119 mm3
1.05 × 10 19
= 8.82 × 10 16 mm −3
119
λ
N
, so we require solutions to 35.124 103 cm = λ where N is
2
2
an integer and λ is in the required range. The midpoint of the range is 632.809 10 nm, giving
2 35.124 103 × 10 −2 m
= 1 110 101.07 . So we try N = 1 110 101, 1 110 102, 1 110 100,
N trial =
632.809 1 × 10 9 m
1 110 103, and so on:
The distance between nodes is
e
λ1 =
e
j=
632.809 14 nm
j=
632.808 57 nm
j=
632.809 71 nm
2 35.124 103 × 10 −2 m
1 110 101
e
2 35.124 103 × 10 −2 m
λ2 =
λ3 =
j
1 110 102
e
2 35.124 103 × 10 −2 m
1 110 100
λ trial =
e
j = 632.808 00 nm , outside the range. Thus the laser light has just
2 35.124 103 × 10 −2 m
1 110 103
three wavelength components.
(b)
1
3
m 0 v 2 = kT . We use the periodic table.
2
2
v=
(c)
ja
e
3 1.38 × 10 −23 J K 393 K
3 kT
=
20.18 u
m0
fF 1 u
GH 1.66 × 10
−27
I=
kg JK
697 m s
For a neon atom moving toward one mirror at the rms speed as it emits, the Doppler shift is
described by
f′= f
c+v
c
c
=
=
c − v λ′ λ
c+v
c−v
λ′ = λ
3 × 10 8 − 697
c−v
= 632.809 1 nm
= 632.807 63 nm .
c+v
3 × 10 8 + 697
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
−E
P24.54
(a)
3
N3 N g e
=
N 2 N g e − E2
b k ⋅300 K g
b k ⋅300 K g
B
B
=e
b
− E3 − E2
679
g b k ⋅300 K g = e − hc λ b k ⋅300 K g
B
B
where λ is the wavelength of light radiated in the 3 → 2 transition.
N3
− 6.63 × 10 −34 J⋅s je 3 ×10 8 m s j
=e e
N2
N3
= e −75.9 = 1.07 × 10 −33
N2
(b)
Nu
− E −E
= e b u lg
Nl
e632.8 ×10
−9
ja
je
m 1.38 ×10 −23 J K 300 K
f
k BT
where the subscript u refers to an upper energy state and the subscript l to a lower energy
state.
Since Eu − El = Ephoton =
hc
λ
Nu
= e − hc λkBT .
Nl
1.02 = e − hc λkBT
Thus, we require
a f
or
ln 1.02 = −
T=−
e6.63 × 10
e632.8 × 10
−34
−9
je
J ⋅ s 3 × 10 8 m s
je
m 1.38 × 10
−23
j
j
JKT
2.28 × 10 4
= −1.15 × 10 6 K .
ln 1.02
a f
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 Eu − El > 0 , and in any real equilibrium state T > 0 ,
e
b
g
− Eu − El k BT
<1
and
Nu < N l .
Thus, a population inversion cannot happen in thermal equilibrium.
Additional Problems
P24.55
(a)
P = SA :
(b)
cB 2
S = max
2µ 0
S=
2
Emax
2µ 0 c
j OPQ =
jLMN e
e
2
P = 1 370 W m 2 4π 1.496 × 10 11 m
2 µ 0S
=
c
e
3.85 × 10 26 W
je
2 4π × 10 −7 N A 2 1 370 W m 2
j=
3.39 µT
so
Bmax =
so
Emax = 2 µ 0 cS = 2 4π × 10 −7 3.00 × 10 8 1 370 = 1.02 kV m
8
3.00 × 10 m s
e
je
jb
g
680
P24.56
Electromagnetic Waves
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
a1.7 mfa0.3 mf cos 30° = 0.4 m .
U = IAt = e1 370 W m j a0.6 fa0.5fe0.4 m j b3 600 sg ~ 10
2
Now I =
P24.57
(a)
P
A
=
U
At
2
ε = − dΦ B
dt
=−
a
d
BA cos θ
dt
f
ε atf = 2π fBmax A sin 2π f t cos θ
2
6
J .
ε = − A d bBmax cos ω t cos θ g = ABmaxω bsin ω t cos θ g
dt
ε atf = 2π 2 r 2 fBmax cos θ sin 2π f t
ε max = 2π 2 r 2 f Bmax cos θ
Thus,
where θ is the angle between the magnetic field and the normal to the loop.
(b)
r
r
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 .
*P24.58
(a)
j a
e
f
The power incident on the mirror is: PI = IA = 1 370 W m 2 π 100 m
2
e
= 4.30 × 10 7 W .
j
The power reflected through the atmosphere is PR = 0.746 4.30 × 10 7 W = 3.21 × 10 7 W .
PR
3.21 × 10 7 W
= 0.639 W m 2
(b)
S=
(c)
Noon sunshine in St. Petersburg produces this power-per-area on a horizontal surface:
PN
A
A
=
e
3
j
π 4.00 × 10 m
e
2
j
= 0.746 1 370 W m 2 sin 7.00° = 125 W m 2 .
The radiation intensity received from the mirror is
F 0.639 W m I 100% =
GH 125 W m JK
2
2
0.513% of that from the noon Sun in January.
Chapter 24
P24.59
Emax
= 6.67 × 10 −16 T
c
(a)
Bmax =
(b)
Sav =
(c)
P = Sav A = 1.67 × 10 −14 W
(d)
F = PA =
2
Emax
= 5.31 × 10 −17 W m 2
2µ 0 c
FG S IJ A =
HcK
av
5.56 × 10 −23 N (≈ the weight of
FIG. P24.59
3 000 H atoms!)
P24.60
681
The area over which we model the antenna as radiating is the lateral surface of a cylinder,
ja
e
f
A = 2π rl = 2π 4.00 × 10 −2 m 0.100 m = 2.51 × 10 −2 m 2 .
(a)
The intensity is then: S =
(b)
The standard is:
P
A
=
e
0.600 W
= 23.9 W m 2 .
2.51 × 10 −2 m 2
0.570 mW cm 2 = 0.570 mW cm 2
jFGH 1.001.00× 10mWW IJK FGH 1.001×.0010m cm IJK = 5.70 W m
−3
4
While it is on, the telephone is over the standard by
P24.61
u=
P24.62
(a)
1
2
∈0 Emax
2
Emax =
Bmax =
23.9 W m 2
5.70 W m 2
2
.
= 4.19 times .
2u
= 95.1 mV m
∈0
175 V m
Emax
=
= 5.83 × 10 −7 T
8
c
3.00 × 10 m s
2π
= 419 rad m
ω = kc = 1.26 × 10 11 rad s
0.015 0 m
r
r
r
r r r
Since S is along x, and E is along y, B must be in the z direction . (That is S ∝ E × B .)
k=
2π
2
2
λ
=
r
S av =
Emax Bmax
= 40.6 W m 2
2µ 0
(b)
S av =
(c)
Pr =
(d)
∑ F = PA = e2.71 × 10
a=
e40.6 W m ji$
2
2S
= 2.71 × 10 −7 N m 2
c
m
m
−7
je
N m 2 0.750 m 2
0.500 kg
j = 4.06 × 10
−7
m s2
r
a=
e406 nm s ji$
2
682
P24.63
Electromagnetic Waves
j a
e
f
P = I 1 A1 = 1.00 × 10 3 W m 2 π 0.500 m
The mirror intercepts power
2
= 785 W .
In the image,
(a)
I2 =
(b)
I2 =
P
A2
I2 =
:
2
Emax
so
2µ 0 c
b
g
π 0.020 0 m
= 625 kW m 2
2
e
je
je
Emax
= 72.4 µT
c
0.400P∆t = mc∆T
a
f b
gb
ga
0.400 785 W ∆t = 1.00 kg 4 186 J kg⋅° C 100° C − 20.0° C
∆t =
*P24.64
f
5
3.35 × 10 J
= 1.07 × 10 3 s = 17.8 min
314 W
8
c 3.00 × 10 m s
=
= 1.50 cm
f
20.0 × 10 9 s −1
(a)
λ=
(b)
U = P ∆t = 25.0 × 10 3 J s 1.00 × 10 −9 s = 25.0 × 10 −6 J
a f e
je
j
FIG. P24.64
= 25.0 µJ
(c)
j
Emax = 2 µ 0 cI 2 = 2 4π × 10 −7 3.00 × 10 8 6.25 × 10 5 = 21.7 kN C
Bmax =
(c)
785 W
uav =
U
U
U
=
=
=
2
2
V
π r l π r c ∆t π 0.060 0 m
e j e ja f b
25.0 × 10 −6 J
g e3.00 × 10
2
8
je
m s 1.00 × 10 −9 s
j
uav = 7.37 × 10 −3 J m3 = 7.37 mJ m 3
(d)
Emax =
Bmax =
(e)
e
j
2 7.37 × 10 −3 J m3
2uav
=
= 4.08 × 10 4 V m = 40.8 kV m
∈0
8.85 × 10 −12 C 2 N ⋅ m 2
Emax 4.08 × 10 4 V m
=
= 1.36 × 10 −4 T = 136 µT
c
3.00 × 10 8 m s
F = PA =
FG S IJ A = u
H cK
av A
e
jb
g
= 7.37 × 10 −3 J m3 π 0.060 0 m
2
= 8.33 × 10 −5 N = 83.3 µN
Chapter 24
P24.65
(a)
On the right side of the equation,
(b)
F = ma = qE or
(c)
F = ma c = m
F v I = qvB so
GH r JK
2
2
2
jb g
N ⋅ m2 m s
3
N ⋅ m2 ⋅ C 2 ⋅ m2 ⋅ s 3 N ⋅ m J
=
= = W.
s
s
C 2 ⋅ s 4 ⋅ m3
=
jb
e
P=
v=
eC
j
g
−19
C 100 N C
qE 1.60 × 10
=
= 1.76 × 10 13 m s 2 .
m
9.11 × 10 −31 kg
a=
The radiated power is then:
e
C2 m s2
683
q2 a2
6π ∈0 c 3
e1.60 × 10 j e1.76 × 10 j
=
6π e8.85 × 10 je3.00 × 10 j
−19 2
13 2
−12
8 3
= 1.75 × 10 −27 W .
qBr
.
m
ja
e
2
fa
j
f
2
1.60 × 10 −19 0.350 0.500
v 2 q 2B2r
=
=
The proton accelerates at a =
= 5.62 × 10 14 m s 2 .
−27 2
r
m2
1.67 × 10
e
The proton then radiates P =
*P24.66
(a)
m = ρV = ρ
F 6m IJ
r =G
H ρ 4π K
q2 a2
6π ∈0 c 3
e1.60 × 10 j e5.62 × 10 j
=
6π e8.85 × 10 je3.00 × 10 j
13
F 6b8.7 kg g I
=G
GH e990 kg m j4π JJK
−12
8 3
= 1.80 × 10 −24 W .
3
a
f
1
4π r 2 = 2π 0.161 m
2
2
13
= 0.161 m
= 0.163 m 2
A=
(c)
I = eσ T 4 = 0.970 5.67 × 10 −8 W m 2 ⋅ K 4 304 K
(d)
P = IA = 470 W m 2 0.163 m 2 = 76.8 W
(e)
I=
ja
e
e
f
4
= 470 W m 2
j
2
Emax
2µ 0 c
b
Emax = 2 µ 0 cI
g
12
e
Bmax =
je
je
= 8π × 10 −7 Tm A 3 × 10 8 m s 470 W m 2
Emax = cBmax
continued on next page
14 2
14 3
πr
23
(b)
(f)
−19 2
595 N C
3 × 10 8 m s
= 1.98 µT
j
12
= 595 N C
684
Electromagnetic Waves
(g)
(h)
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.
Each kitten has radius rk
b
g
2π 0.072 8 m
2
F 6a0.8f IJ
=G
H 990 × 4π K
13
= 0.072 8 m and radiating area
= 0.033 3 m 2 . Eliza has area 2π
e
j
FG 6a5.5f IJ
H 990 × 4π K
23
= 0.120 m 2 . The total glowing
area is 0.120 m 2 + 4 0.033 3 m 2 = 0.254 m 2 and has power output
e
2
j
2
P = IA = 470 W m 0.254 m = 119 W .
P24.67
(a)
(b)
At steady state, Pin = Pout and the power radiated out is Pout = eσAT 4 .
.
e
j
e
j
Thus,
0.900 1 000 W m 2 A = 0.700 5.67 × 10 −8 W m 2 ⋅ K 4 AT 4
or
L
900 W m
T=M
MN 0.700e5.67 × 10 W m
2
−8
2
⋅K4
OP
j PQ
14
= 388 K = 115° C .
The box of horizontal area A, presents projected area A sin 50.0° perpendicular to the
sunlight. Then by the same reasoning,
e
j
e
j
0.900 1 000 W m 2 A sin 50.0° = 0.700 5.67 × 10 −8 W m 2 ⋅ K 4 AT 4
L e900 W m j sin 50.0° OP
T=M
MN 0.700e5.67 × 10 W m ⋅ K j PQ
2
or
*P24.68
2
−8
14
4
= 363 K = 90.0° C .
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 4π r 2 .
e j
F 6.00 × 10
GH 4π R
e
R=
23
2
W
e
j
eI in π r 2 = eσ 4π r 2 T 4
At steady-state, Pin = Pout :
I π r = eσ 4π r T
JK e j e j
2
2
4
so that 6.00 × 10 23 W = 16πσR 2 T 4
6.00 × 10 23 W
6.00 × 10 23 W
=
16πσ T 4
16π 5.67 × 10 −8 W m 2 ⋅ K 4 310 K
e
ja
f
4
= 4.77 × 10 9 m = 4.77 Gm .
Chapter 24
P24.69
(a)
P=
F I
=
A c
a = 3.03 × 10 −9 m s 2 and
100 J s
IA P
= =
= 3.33 × 10 −7 N = 110 kg a
c
c 3.00 × 10 8 m s
x=
1 2
at
2
b
g
2x
= 8.12 × 10 4 s = 22.6 h
a
t=
(b)
F=
685
b
g b
gb
g b
g
b
g
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
110
t = 30.6 s
ANSWERS TO EVEN PROBLEMS
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) ~ 10 8 Hz radio wave;
(b) ~ 10 13 Hz infrared light
75.0 MHz
P24.36
11.0 m
P24.12
11.3 kHz
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
P24.14
0.050 4c
P24.40
60.0 km
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
P24.42
3
8
P24.18
0.220 H
P24.44
(a) 93.3%; (b) 50.0%; (c) 0
P24.20
3.33 µ J m3
P24.46
(a) 6; (b) 7.50°
P24.22
49.5 mV
P24.48
(a) 2.33 mT; (b) 650 MW m 2 ; (c) 510 W
P24.24
(a) 11.9 GW m 2 ; (b) 234 kW
P24.50
28.2 THz, 10.6 µ m , infrared
P24.26
r r
(a) E ⋅ B = 0 ; (b) 11.5 $i − 28.6 $j W m 2
P24.52
(a) 4.20 mm; (b) 1.05 × 10 19 photons;
(c) 8.82 × 10 16 mm3
P24.28
(a) 88.8 nW m 2 ; (b) 11.3 MW
P24.54
P24.30
(a) 5.48 N; (b) 913 µ m s 2 ; (c) 10.6 d
(a) 1.07 × 10 −33 ; (b) −1.15 × 10 6 K ;
(c) no real T is below 0 K
P24.2
(a) 3.15 $j kN C ; (b) 525 nTk$ ; (c) −483 $j aN
P24.4
(a) 2.68 × 10 3 A.D.; (b) 499 s; (c) 2.56 s;
(d) 0.133 s; (e) 33.3 µ s
P24.6
733 nT
P24.8
(a) 0.333 µ T ; (b) 0.628 µ m ; (c) 477 THz
P24.10
e
j
686
Electromagnetic Waves
P24.56
~ 10 6 J
P24.64
P24.58
(a) 32.1 MW; (b) 0.639 W m 2 ;
(c) 0.513%
(a) 1.50 cm; (b) 25.0 µ J ; (c) 7.37 m J m 3 ;
(d) 40.8 kV/m, 136 µ T ; (e) 83.3 µ N
P24.66
(a) 16.1 cm; (b) 0.163 m 2 ; (c) 470 W m 2 ;
(d) 76.8 W; (e) 595 N/C; (f) 1.98 µ T ;
(g) The cats are nonmagnetic and carry no
macroscopic charge or current. Oscillating
charges within molecules make them emit
infrared radiation.
(h) 119 W
P24.68
4.77 Gm
P24.60
P24.62
(a) 23.9 W m 2 ;
(b) It is 4.19 times the standard.
r
(a) Bmax = 583 nTk$ , k = 419 rad m,
ω = 12.6 Trad s ;
(b) Sav = 40.6 W m 2 ; (c) 271 nPa;
(d) 406 nm s 2