24
Electromagnetic Waves
CHAPTER OUTLINE
24.1
Displacement Current and the Generalized Form of Ampère’s Law
24.2
Maxwell’s Equations and Hertz’s Discoveries
24.3
Electromagnetic Waves
24.4
Energy Carried by Electromagnetic Waves
24.5
Momentum and Radiation Pressure
24.6
The Spectrum of Electromagnetic Waves
24.7
Polarization of Light Waves
24.8
Context Connection: The Special Properties of Laser Light
* An asterisk indicates an item new to this edition.
ANSWERS TO OBJECTIVE QUESTIONS
*OQ24.1
Answers (c) and (d). The relationship between frequency,
wavelength, and the speed of a wave is f λ = v. In a vacuum, all
electromagnetic waves travel at the same speed: v = c.
Electromagnetic waves, consisting of oscillating electric and
magnetic fields, are transverse waves.
*OQ24.2
Answer (d). The peak values of the electric and magnetic field
components of an electromagnetic wave are related by Emax Bmax = c,
where c is the speed of light in vacuum. Thus,
Emax = cBmax = ( 3.00 × 108 m/s ) ( 1.50 × 10−7 T ) = 45.0 N/C
*OQ24.3
Answer (e). Accelerating charge, changing electric field, or changing
magnetic field can be the source of a radiated electromagnetic wave.
*OQ24.4
Answer (b). From Malus’s law, the intensity of the light transmitted
through a polarizer (analyzer) having its transmission axis oriented
383
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
384
Electromagnetic Waves
at angle 45° to the plane of polarization of the incident polarized
light is I = Imax cos2 45° = Imax/2. Therefore, the intensity passing
through the second polarizer having its transmission axis oriented at
angle θ = 90° – 45° = 45° is I = (Imax/2) cos2 45° = Imax/4.
*OQ24.5
Answer (b). Frequency, wavelength, and the speed of light are
related:
fλ = c
*OQ24.6
→
c 3.00 × 108 m/s
λ= =
= 0.122 m = 12.2 cm
f
2.45 × 109 Hz
(i) Answer (a). According to f = 1 2π LC , to make f half as large,
the capacitance should be made four times larger.
(ii) Answer (b). According to f λ = c, if frequency is halved,
wavelength is doubled.
*OQ24.7
(i) Answer (c). (ii) Answer (c). (iii) Answer (c). (iv) Answer (b). (v)
Answer (b). The same amount of energy passes through concentric
spheres of increasing area as the wave travels outward from its
source, so the amplitude and the intensity, which is proportional to
the square of the amplitude, decrease.
*O24.8
(i) through (v) have the same answer (c). The same amount of energy
passes through equal areas parallel to the yz plane as the wave
travels in the +x direction, so the amplitude and the intensity, which
is proportional to the square of the amplitude, do not change.
*OQ24.9
(i) The ranking is c > b > d > e > a. Gamma rays have the shortest
wavelength.
(ii) The ranking is a > e > d > b > c. According to f λ = c, as
wavelength decreases, frequency increases.
(iii) The ranking is a = b = c = d = e. All electromagnetic waves
travel at the speed of light c in vacuum, which is assumed here.
*OQ24.10 (i) Answer (b). Electric and magnetic fields both carry the same
energy, so their amplitudes are proportional to each other.
(ii) Answer (a). The intensity is proportional to the square of the
amplitude.
*OQ24.11 (i) Answer (c). Both the light intensity and the gravitational force
follow inverse-square laws.
(ii) Answer (a). The smaller grain presents less face area and feels a
smaller force due to light pressure.
*OQ24.12 Answer (d). An electromagnetic wave travels in the direction of the
  
Poynting vector: S = E × B µ0 .
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
385
ANSWERS TO CONCEPTUAL QUESTIONS
*CQ24.1
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.
*CQ24.2
Sound
Light
1) Sound is a longitudinal wave.
1) Light is a transverse wave.
2) Sound requires a material
medium.
2) Light does not require a
material medium.
3) Sound in air moves at hundreds
of meters per second.
3) Light in air moves at hundreds
of millions of meters per
second.
4) The speed of sound through a
medium, depending the material
of the medium, can be faster or
slower than that in air.
4) The speed of light through
materials is less than in
vacuum.
5) Sound propagates by a chain
reaction of density and pressure
disturbances recreating each
other.
5) Light propagates by a chain
reaction of electric and
magnetic fields recreating each
other.
6) Audible sound has frequencies
over a range of three decades (ten
octaves) from 20 Hz to 20 kHz.
6) Visible light has frequencies
over a range of less than one
octave, from 430 to 700 THz
(THz = Terahertz = 1012 Hz).
7) Audible sound has wavelengths
of ordinary size (1.7 cm to 17 m).
7) Visible light has wavelengths of
very small size (400 nm to
750 nm).
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
386
Electromagnetic Waves
*CQ24.3
Maxwell included a term in Ampère’s law to account for the
contributions to the magnetic field by changing electric fields, by
treating those changing electric fields as “displacement currents.”
*CQ24.4
(a) The electric and magnetic fields of the light wave oscillate in
time at each point in space, like sports fans in a grandstand when the
crowd does “the wave.”
(b)
The wave transports energy.
*CQ24.5
No. Radio waves travel at a finite speed, the speed of light. Radio
waves 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
5 × 106 m
transit time of
~ 10−2 s.
8
3 × 10 m/s
*CQ24.6
Electromagnetic waves carry momentum. Recalling what we learned
in Chapter 8, the impulse imparted by a particle that bounces
elastically off a wall is twice that imparted by an object that sticks to
a wall. Similarly, the impulse, and hence the pressure exerted by a
wave reflecting from a surface, must be twice that exerted by a wave
that is absorbed.
*CQ24.7
The changing magnetic field of the solenoid induces eddy currents in
the conducting core. This is accompanied by I2 R conversion by
heating of electrically-transmitted energy into internal energy in the
conductor.
*CQ24.8
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.

The Poynting vector S describes the energy flow associated with an

electromagnetic wave. The direction of S is along the direction of

propagation and the magnitude of S is the rate at which
electromagnetic energy crosses a unit surface area perpendicular to

the direction of S.
*CQ24.9
*CQ24.10
Consider a typical metal rod antenna for a car radio. Charges in the
rod respond to 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
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
387
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.
*CQ24.11
An infrared photograph records the infrared light reflected, but also
emitted by a person’s face. When a person blushes or exercises or
becomes excited, warmer areas glow brighter in the infrared. A
person’s nostrils and the openings of the ear canals are bright;
brighter still are just the pupils of the eyes.
*CQ24.12
The entire room and its contents have a soft glow. Incandescent light
bulbs shine brightly in the infrared, but fluorescent lights do not. The
top of a computer monitor glows brighter than the screen, which
glows faintly. Windowpanes appear dark if they are cool, and a
patch of wall where sunlight falls glows brighter than where the
sunlight does not fall. Heating resistors or warm air outlets shine,
and the air near to them has a faint glow, but cold air outlets are
dark, and the nearby air has no glow.
SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 24.1
P24.1
Displacement Current and the
Generalized Form of Ampère’s Law
(a)
We use the right-hand rule for both
 real and displacement
currents. Thus, the direction of B is counterclockwise, and the

direction of B at P is upwards.
(b)
We use the extended form of Ampère’s law, Equation 24.7. Since
no moving charges are present, I = 0 and we have
 
dΦE
B
∫ ⋅ d  = µ0 ∈0 d t
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.
ANS. FIG. P24.1
We know the magnetic field lines are
circles about the axis. Therefore, as we
travel around such a magnetic field circle, the magnetic field
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
388
Electromagnetic Waves
remains constant in magnitude. Setting aside until later the

 
determination of the direction of B, we integrate ∫ B ⋅ d  around
the circle at R = 0.150 m to obtain 2 π RB.
Differentiating the expression ΦE = AE, we have
dΦE ⎛ π d 2 ⎞ dE
=⎜
dt
⎝ 4 ⎟⎠ dt
2⎞
⎛
 
µ0 ∈0 d 2 dE
⎜ π d ⎟ dE
Thus, ∫ B ⋅ d  = 2π RB = µ0 ∈0 ⎜
→ B=
⎟
⎜⎝ 4 ⎟⎠ dt
8 R dt
Substituting numerical values,
( 4π × 10 T ⋅ m/A )( 8.85 × 10
B=
−7
−12
C2 /N ⋅ m 2 ) ( 0.100 m )
8 ( 0.150 m )
2
× ( 20.0 V/m ⋅ s )
= 1.85 × 10−18 T
*P24.2
dQ dt
dΦE d
I
= ( EA ) =
=
dt
dt
∈0
∈0
(a)
dE
I
0.200 A
=
=
dt ∈0 A ( 8.85 × 10−12 C2 /N ⋅ m 2 ) ⎡π ( 10.0 × 10−2 m ) ⎤
⎣
⎦
= 7.19 × 1011 V m ⋅ s
(b)
∫ B ⋅ ds = ∈0 µ0
dΦE
:
dt
2π rB = ∈0 µ0
⎤
d⎡ Q
⋅ π r2 ⎥
⎢
dt ⎢⎣ ∈ 0 A
⎥⎦
µ Ir µ0 ( 0.200 A ) ( 5.00 × 10 m )
B= 0 =
= 2.00 × 10−7 T
2
−2
2A
2 ⎡π ( 10.0 × 10 m ) ⎤
⎣
⎦
−2
*P24.3
The electric field in the space between the plates is E =
The flux of this field is
(a)
σ
Q
=
.
∈0
∈0 A
 
⎛ Q ⎞
Q
ΦE = E ⋅ A = ⎜
A cos 0 =
.
⎟
∈0
⎝ ∈0 A ⎠
The rate of change of flux is
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
389
dΦE dQ dt
I
0.100 A
=
=
=
dt
∈0
∈0 8.85 × 10−12 C2 /N ⋅ m 2
= 11.3 × 109 V ⋅ m/s
(b)
The displacement current is defined as
dΦE
dt
= (8.85 × 10 -12 C2 / N ⋅ m 2 )(1.13 × 1010 N ⋅ m 2 / C ⋅ s)
I d = ∈0
= 0.100 A
Section 24.2
P24.4
Maxwell’s Equations and Hertz’s Discoveries
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 =
P24.5
1
( 2π f0 ) L
2
=
1
2π LC
.
1
⎡⎣ 2π ( 6.30 × 106 Hz ) ⎤⎦ ( 1.05 × 10−6 H )
2
= 608 pF .



 
F = ma = qE + qv × B so
 e   
a = ⎡⎣E + v × B ⎤⎦ , where
m
 
v×B=
î
ĵ
k̂
200
0
0
0.200 0.300 0.400
= − ( 200 m/s ) ( 0.400 T ) ĵ + ( 200 m/s ) ( 0.300 T ) k̂
= − ( 80.0 T ⋅ m/s ) ĵ + ( 60.0 T ⋅ m/s ) k̂
Then
 1.60 × 10−19 C
a=
1.67 × 10−27 kg
× ⎡⎣( 50.0 V/m ) ĵ − ( 80.0 T ⋅ m/s ) ĵ + ( 60.0 T ⋅ m/s ) k̂ ⎤⎦
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
390
Electromagnetic Waves
= 9.58 × 107 ⎡⎣ −30.0 ĵ + 60.0k̂ ⎤⎦ m/s 2

a = 2.87 × 109 ⎡⎣ − ĵ + 2 k̂ ⎤⎦ m/s 2 =
P24.6
(a)
(−2.87 ĵ + 5.75k̂ ) × 10 m/s
9
The very long rod creates the same electric
field that it would if stationary. We apply
Gauss’s law to a cylinder, concentric with
the rod, of radius r = 20.0 cm and length :
  qinside
∫ E ⋅ dA = ∈
0
E ( 2π r ) cos 0° =

E=
2
ANS. FIG. P24.6
λ
∈0
λ
radially outward
2π ∈0 r
35.0 × 10−9 C/m
=
ĵ
2π ( 8.85 × 10−12 C2 /N ⋅ m 2 ) ( 0.200 m )
= 3.15 × 103 ĵ N/C
(b)
The charge in motion constitutes a current of (35.0 × 10–9 C/m)
×(15.0 × 106 m/s) = 0.525 A. This current creates a magnetic field.
 µI
B= 0
2π r
( 4π × 10 T ⋅ m/A )( 0.525 A ) k̂ =
=
−7
2π ( 0.200 m )
(c)
5.25k̂ × 10−7 T


 
The Lorentz force on the electron is F = qE + qv × B.

F = ( −1.60 × 10−19 C ) 3.15 × 103 ĵ N/C
(
(
)
+ ( −1.60 × 10−19 C ) 240 × 106 î m/s
(
)
× 5.25 × 10−7 k̂ T
)

F = 5.04 × 10−16 − ĵ N + 2.02 × 10−17 + ĵ N
( )
( )
( )
= 4.83 − ĵ × 10−16 N
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
P24.7
1
f =
(b)
Q = Cε = ( 1.00 × 10−6 F ) ( 12.0 V ) = 12.0 µC
(c)
1 2 1 2
Cε = LI max
2
2
2π LC
I max = ε
=
1
(a)
= 503 Hz
C
1.00 × 10−6 F
= ( 12.0 V )
= 37.9 mA
L
0.100 H
(d) At all times U =
*P24.8
( 0.100 H ) (1.00 × 10−6 F )
2π
391
1 2 1
2
Cε = ( 1.00 × 10−6 F ) ( 12.0 V ) = 72.0 µ J .
2
2



 
F = ma = qE + qv × B so
 −e   
⎡E + v × B ⎤⎦ where
a=
m⎣
î
ĵ
k̂
 
v × B = 10.0 0
= − ( 4.00 T ⋅ m/s ) ĵ
0
0
0 0.400
Then
−19
 ( −1.60 × 10 C )
⎡( 2.50 V/m ) î + ( 5.00 V/m ) ĵ − ( 4.00 T ⋅ m/s ) ĵ ⎤
a=
⎦
9.11 × 10−31 kg ⎣
= ( −1.76 × 1011 ) ⎡⎣ 2.50î + 1.00 ĵ ⎤⎦ m/s 2

a=
( −4.39î − 1.76 ĵ) × 10 m/s
11
2
Section 24.3
Electromagnetic Waves
P24.9
v=
1
1
=
c = 0.750c = 2.25 × 108 m/s
κµ0 ∈0
1.78
P24.10
(a)
Since the light from this star travels at 3.00 × 108 m/s, the last bit
of light will hit the Earth in
6.44 × 1018 m
= 2.15 × 1010 s = 680 years
8
3.00 × 10 m s
Therefore, it will disappear from the sky in the year
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
392
Electromagnetic Waves
2 012 + 680 = 2 692 C.E.
The star is 680 light-years away.
P24.11
Δx 1.496 × 1011 m
=
= 499 s = 8.31 min
v
3.00 × 108 m s
(b)
Δt =
(c)
8
Δx 2 ( 3.84 × 10 m )
Δt =
=
= 2.56 s
v
3.00 × 108 m s
(d)
Δt =
Δx 2π ( 6.37 × 10 m )
=
= 0.133 s
v
3.00 × 108 m s
(e)
Δt =
Δx
10.0 × 103 m
=
= 3.33 × 10−5 s
8
v
3.00 × 10 m s
6
In the fundamental mode, there is a single loop in the standing wave
between the plates. Therefore, the distance between the plates is equal
to half a wavelength.
λ = 2L = 2 ( 2.00 m ) = 4.00 m
c 3.00 × 108 m/s
= 7.50 × 107 Hz = 75.0 MHz .
Thus, f = =
λ
4.00 m
P24.12
E
220 V/m
= c or
= 3.00 × 108 m/s, so
B
B
B = 7.33 × 10−7 T = 733 nT
P24.13
(a)
f λ = c or f (50.0 m) = 3.00 × 108 m/s
so
(b)
f = 6.00 × 106 Hz = 6.00 MHz .
Emax
22.0 V/m
= c or
= 3.00 × 108 m/s
Bmax
Bmax

so Bmax = −73.3k̂ nT .
(c)
k=
and
2π
2π
=
= 0.126 m −1
λ 50.0 m
ω = 2π f = 2π ( 6.00 × 106 s −1 ) = 3.77 × 107 rad/s.
Then,
 
B = Bmax cos ( kx − ω t ) = −73.3k̂ cos ( 0.126x − 3.77 × 107 t ) nT
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
P24.14
393
E = Emax cos ( kx − ω t )
∂E
∂2E
= −Emax sin ( kx − ω t ) ( k ) →
= −Emax cos ( kx − ω t ) ( k 2 )
2
∂x
∂x
∂E
∂2E
2
= −Emax sin ( kx − ω t ) ( −ω ) →
= −Emax cos ( kx − ω t ) ( −ω )
2
∂t
∂t
We must show:
∂2E
∂2E
=
µ
∈
0
0
∂x 2
∂t 2
( )
That is, − k 2 Emax cos ( kx − ω t ) = − µ0 ∈0 ( −ω ) Emax cos ( kx − ω t ) .
2
2
k2 ⎛ 1 ⎞
1
But this is true, because 2 = ⎜
= 2 = µ0 ∈0 .
⎟
ω
c
⎝ fλ⎠
The proof for the wave of the magnetic field follows precisely the same
steps.
P24.15
dA to A = 6 cm ± 5% =
λ
2
λ = 12 cm ± 5%
v = λ f = ( 0.12 m ± 5% ) ( 2.45 × 109 s −1 ) = 2.9 × 108 m/s ± 5%
P24.16
P24.17
The wave is of the form Ey = Emax sin ( kx − ω t ) .
Emax
100 V/m
=
= 3.33 × 10−7 T = 0.333 µT
c
3.00 × 108 m/s
(a)
Bmax =
(b)
λ=
2π
2π
=
= 0.628 µm
k
1.00 × 107 m −1
(c)
f =
c 3.00 × 108 m/s
=
= 4.77 × 1014 Hz
λ
6.28 × 10−7 m
For the light as observed, λ = 650 nm and λ ′ = 520 nm. From Equation
24.23,
f′ =
1+ v c
1+ v c c
c
=
f =
λ′
1− v c
1− v c λ
1+ v c λ
=
1 − v c λ′
v ⎛λ⎞ ⎛
v⎞
1+ = ⎜ ⎟ ⎜1− ⎟
c ⎝ λ′ ⎠ ⎝
c⎠
2
→
Thus,
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
394
Electromagnetic Waves
2
2
v⎡ ⎛ λ⎞ ⎤ ⎛ λ⎞
1
+
=
⎢ ⎜ ⎟ ⎥ ⎜ ⎟ −1
c ⎣ ⎝ λ′ ⎠ ⎦ ⎝ λ′ ⎠
⎛λ⎞
⎜⎝ ⎟⎠ − 1
λ′
2
v
=
c
⎛λ⎞
1+ ⎜ ⎟
⎝ λ′ ⎠
2
=
2
⎛ 650 nm ⎞
⎜⎝
⎟ −1
520 nm ⎠
⎛ 650 nm ⎞
1+ ⎜
⎝ 520 nm ⎟⎠
2
= 0.220
v = 0.220c = 6.59 × 107 m/s
P24.18
(a)
The mirror is approaching the source. Let fm be the frequency as
seen by the mirror. Thus,
c+v
c−v
fm = f
After reflection, the mirror acts as a source, approaching the
receiver. If f' is the frequency of the reflected wave,
f ′ = fm
c+v
c−v
Combining gives
(b)
f′ =
c+v
f .
c−v
Using the above result, the beat frequency is
f beat = f ′ − f = f ′ =
c+v
⎛c+v
⎞
f − f = f⎜
− 1⎟
⎝c−v
⎠
c−v
2v 2v
⎛ c + v − (c − v)⎞
⎛ 2v ⎞
f beat = f ⎜
= f⎜
≈ f
=
⎟
⎟
⎝ c − v⎠
⎝
⎠
c−v
c
c f
f beat =
(c)
(d)
2v
λ
The wavelength is λ =
v=
c 3.00 × 108 m/s
=
= 0.030 0 m:
f
10.0 × 109 Hz
f beat =
2v ( 2 ) ( 30.0 m/s )
=
= 2 000 Hz = 2.00 kHz
λ
( 0.030 0 m )
f beat λ
2
so
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
Δv =
395
Δf beat λ ( 5.0 Hz ) ( 0.030 0 m )
=
2
2
= 0.075 0 m/s ≈ 0.17 mi/h
P24.19
The orbital speed of the Earth is as described by ∑ F = ma:
GmSmE mE v 2
=
r2
r
6.67 × 10−11 N⋅ m 2 / kg 2 ) ( 1.99 × 1030 kg )
(
GmS
v=
=
r
1.496 × 1011 m
= 2.98 × 10 4 m/s
The maximum frequency received by the extraterrestrials is
f max
′ = f
1+ v c
= ( 57.0 × 106 Hz )
1− v c
×
1 + ( 2.98 × 10 4 m/s ) ( 3.00 × 108 m/s )
1 − ( 2.98 × 10 4 m/s ) ( 3.00 × 108 m/s )
= 57.005 66 × 106 Hz
The minimum frequency received is
f min
′ = f
1+ v c
= ( 57.0 × 106 Hz )
1− v c
×
1 − ( 2.98 × 10 4 m/s ) ( 3.00 × 108 m/s )
1 + ( 2.98 × 10 4 m/s ) ( 3.00 × 108 m/s )
= 56.994 34 × 106 Hz
The difference, which lets them figure out the speed of our planet, is
( 57.005 66 − 56.994 34) × 106 Hz = 1.13 × 104 Hz
P24.20
(a)
When the source moves away from an observer, the observed
frequency is
⎛ c + v⎞
f′ = f ⎜
⎝ c − v ⎟⎠
12
⎛ c − vS ⎞
= f⎜
⎝ c + v ⎟⎠
12
S
where v = vsource = −vS because the source is moving away from
the observer. When vs  c, the binomial expansion gives
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
396
Electromagnetic Waves
⎛ c − vS ⎞
⎜⎝ c + v ⎟⎠
S
So,
12
12
−1 2
⎡ ⎛ v ⎞⎤ ⎡ ⎛ v ⎞⎤
= ⎢1 − ⎜ S ⎟ ⎥ ⎢1 + ⎜ S ⎟ ⎥
⎣ ⎝ c ⎠⎦ ⎣ ⎝ c ⎠⎦
v ⎞⎛
v ⎞ ⎛
v ⎞
⎛
≈ ⎜1− S ⎟ ⎜1− S ⎟ ≈ ⎜1− S ⎟
⎝
2c ⎠ ⎝
2c ⎠ ⎝
c⎠
v ⎞
⎛
f ′ ≈ f ⎜1− S ⎟ .
⎝
c⎠
The observed wavelength is found from c = λ ′ f ′ = λ f :
λ′ =
λf
λf
λ
≈
=
f′
f ( 1 − vS c ) 1 − vS c
⎛
⎞
⎛ v c ⎞
1
Δλ = λ ′ − λ = λ ⎜
− 1⎟ = λ ⎜ S
⎝ 1 − vS c
⎠
⎝ 1 − vS c ⎟⎠
Since 1 −
(b)
P24.21
(a)
vS
≈ 1,
c
Δλ vS
≈
λ
c
⎛ Δλ ⎞
⎛ 20.0 nm ⎞
vS = c ⎜
= c⎜
= 0.050 4c
⎟
⎝ 397 nm ⎟⎠
⎝ λ ⎠
Consider the raindrops moving toward the station, at speed v.
They receive radio waves with the Doppler-enhanced frequency
c+v
f′ = f
, where f = 2.85 GHz. These raindrops reflect the
c−v
waves at frequency f ′. The waves are received by the station
with another upward Doppler shift in frequency to
f ′′ = f ′
c+v
c+v c+v
= f
c−v
c−v c−v
2.85 × 109 Hz + 254 Hz = 2.85 × 109 Hz
1 + 8.91 × 10−8 =
c+v
c−v
( )
c+v
c−v
c + 8.91 × 10−8 c − v − 8.91 × 10−8 v = c + v
8.91 × 10−8 c = 2.000 000 089v
v = ( 4.46 × 10−8 ) ( 3.00 × 108 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.
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
(b)
Radio waves travel to the rain and back again in 180 µs, so the
1
one-way distance is ( 3.00 × 108 m s ) ( 180 × 10−6 s ) = 27 000 m.
2
The diameter of the vortex is
π rad
s = rθ = 27 000 m ( 1° )
= 471 m. Its radius is therefore
180°
1
( 471 m ) = 236 m and the angular speed of the rain is
2
v 13.4 m s
ω= =
= 0.056 7 rad s . A Doppler weather radar
r
236 m
computer performs a calculation like this to detect a “tornado
vortex signature.”
(
*P24.22
397
)
The amplitudes of the electric and magnetic fields are in the correct
ratio so that Emax/Bmax = c. The ratio of ω to k, however, must also equal
the speed of light:
ω
3.00 × 1015 s −1
=
= 3.33 × 108 m/s
6
−1
k
9.00 × 10 m
This value is higher than the speed of light in a vacuum, so the wave as
described is impossible.
Section 24.4
P24.23
P24.24
Energy Carried by Electromagnetic Waves
r = ( 5.00 mi ) ( 1 609 m/mi ) = 8.04 × 103 m
S=
P
250 × 103 W
=
= 307 µ W/m 2
4π r 2 4π ( 8.04 × 103 W )2
(a)
 
E ⋅ B = 80.0î + 32.0 ĵ − 64.0k̂ ( N/C ) ⋅ 0.200î + 0.080 0 ĵ + 0.290k̂ µT
(
)
(
)
 
E ⋅ B = ( 16.0 + 2.56 − 18.56 ) µT ⋅ N/C2 = 0
(b)
 1  
S=
E×B
µ0
(
)
1
⎛
⎞⎡
=⎜
80.0î + 32.0 ĵ − 64.0k̂ N/C ⎤
−7
⎦
⎝ 4π × 10 T ⋅ m/A ⎟⎠ ⎣
(
)
× ⎡ 0.200î + 0.080 0 ĵ + 0.290k̂ µT ⎤
⎣
⎦
(
)
6.40k̂ − 23.2 ĵ − 6.40k̂ + 9.28î − 12.8 ĵ + 5.12 î × 10

S=
4π × 10−7
−6
W/m 2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
398
Electromagnetic Waves

S=
(11.5î − 28.6 ĵ) W/m
2
= 30.9 W/m 2 at – 68.1° from the + x axis
P24.25
In time interval ∆t, sunlight travels distance ∆x = c∆t. The intensity of
the sunlight passing into a volume ∆V = A∆x in time ∆t is
S= I =
U
U
Uc
=
=
= uc
AΔt A Δx c V
Energy
1 000 W/m 2
I
=u= =
= 3.33 µ J/m 3
8
Unit Volume
c 3.00 × 10 m/s
P24.26
(a)
We assume that the starlight moves through space without any of
it being absorbed. The radial distance is
20 ly = 20c ( 1 yr ) = 20 ( 3.00 × 108 m/s ) ( 3.16 × 107 s )
= 1.89 × 1017 m
I=
P
4.00 × 1028 W
−8
=
W/m 2
2 = 8.88 × 10
2
17
4π r
4π ( 1.89 × 10 m )
= 88.8 nW/m 2
(b)
The Earth presents the projected target area of a flat circle:
2
P = IA = ( 8.88 × 10−8 W/m 2 ) ⎡π ( 6.37 × 106 m ) ⎤
⎣
⎦
= 1.13 × 107 W= 11.3 MW
P24.27
2
Emax
Savg =
=
4π r 2 2 µ0 c
Pavg
Emax = 2 µ0 cSavg = µ0 c
=
Pavg
2π r 2
( 4 π × 10 T ⋅ m/A )( 3.00 × 10 m/s )
−7
8
4.00 × 103 W
2 π [ 4.00 ( 1 609 m )]
2
= 0.076 1 V/m
The maximum emf (amplitude) induced in a length L of wire is
ΔVmax = Emax L = ( 76.1 mV m ) ( 0.650 m ) = 49.5 mV
2
( 3.00 × 10 V/m )
Emax
I=
=
2 µ0 c 2 ( 4π × 10−7 T ⋅ m/A ) ( 3.00 × 108 m/s )
6
P24.28
(a)
2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
399
I = 1.19 × 1010 W/m 2
(b)
⎛ 5.00 × 10−3 m ⎞
P = IA = Iπ r = ( 1.19 × 10 W/m ) π ⎜
⎟⎠
2
⎝
2
10
2
2
= 2.34 × 105 W
P24.29
Power output = (power input)(efficiency).
Power input =
Thus,
Power out 1.00 × 106 W
=
= 3.33 × 106 W
eff
0.300
P
3.33 × 106 W
and A = =
= 3.33 × 103 m 2 .
3
2
I 1.00 × 10 W/m
P24.30
Power = SA =
2
Emax
4π r 2 )
(
2 µ0c
Solving for r,
r=
(100 W) ( 4π × 10−7 T ⋅ m/A ) ( 3.00 × 108 m/s )
P µ0c
=
2
2π Emax
2π (15.0 V m)2
= 5.16 m
P24.31
(a)
P = I 2 R = 150 W
A = 2π rL = 2π ( 0.900 × 10−3 m ) ( 0.080 0 m ) = 4.52 × 10−4 m 2
(b)
S=
P
150 W
=
= 332 kW m 2
−4
2
A 4.52 × 10 m
B=
−7
µ0 I ( 1.00 A ) ( 4π × 10 T ⋅ m/A )
=
= 222 µT
2π r
2π ( 0.900 × 10−3 m )
E=
ΔV IR
150 V
=
=
= 1.88 kV m
Δx
L 0.080 0 m
Note that these values yield S =
(points radially inward)
EB
= 332 kW m 2 , in agreement
µ0
with the result from part (a).
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
400
Electromagnetic Waves
Section 24.5
P24.32
(a)
Momentum and Radiation Pressure
The radiation pressure is
P=
2S 2I
=
c
c
The force on area A is
F = PA =
(b)
2 ( 1 370 W/m 2 )
(6.00 × 10 m ) = 5.48 N
3.00 × 10 m/s
5
2
8
The acceleration is:
F
5.48 N
=
= 9.13 × 10−4 m/s 2
m 6 000 kg
a=
= 913 µm/s 2 away from the Sun
(c)
It will arrive at time t, where d =
2d
=
a
t=
P24.33
(a)
1 2
at or,
2
2 ( 3.84 × 108 m )
( 9.13 × 10 m/s )
−4
2
= 9.17 × 105 s = 10.6 days
2
P
Emax
, and r = 1.00 × 10–3 m:
I=
=
2
πr
2 µ0c
Emax =
2 µ0 cP
π r2
=
2 ⎡⎣ 4π × 10−7 T ⋅ m/A ⎤⎦ ( 3.00 × 108 m/s ) ( 15.0 × 10−3 W )
π ( 1.00 × 10−3 m )
2
= 1.90 × 108 J = 1.90 kN/C
(b)
The beam carries power P. The amount of energy ∆E in the length
of a beam of length  is the amount of power that passes a point
in time interval Δt = /c:
P=
or
(c)
ΔE ΔE
=
Δt  c
P 15.0 × 10−3 W
ΔE =
=
(1.00 m ) = 50.0 pJ .
c
3.00 × 108 m/s
From Equation 24.31 and our result in part (b), the momentum
and energy carried a light beam are related by
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
401
TER ΔE
50.0 × 10−12 J
p=
=
=
= 1.67 × 10−19 kg ⋅ m/s
8
c
c
3.00 × 10 m/s
*P24.34
2
P
Emax
=
π r 2 2 µ0c
(b)
The beam carries power P. The amount of energy ∆E in the length
of a beam of length  is the amount of power that passes a point
in time interval Δt = /c:
P=
(c)
→ Emax =
2 µ0 cP
π r2
(a)
I=
ΔE ΔE
P
=
→ ΔE =
Δt  c
c
From Equation 24.31 and our result in part (b), the momentum
and energy carried a light beam are related by
p=
TER ΔE P
=
= 2
c
c
c
Section 24.6
The Spectrum of Electromagnetic Waves
c 3.00 × 108 m s
=
= 5.45 × 1014 Hz
λ
5.50 × 10−7 m
P24.35
f =
P24.36
The time interval for the radio signal to travel 100 km is:
Δtr =
100 × 103 m
= 3.33 × 10−4 s
3.00 × 108 m/s
The sound wave travels 3.00 m across the room in:
Δts =
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
Δt = 8.75 × 10−3 s − 3.33 × 10−4 s = 8.41 × 10−3 s
P24.37
(a)
f λ = c gives
( 5.00 × 10 Hz ) λ = 3.00 × 10 m/s:
19
8
λ = 6.00 × 10−12 m = 6.00 pm
(b)
f λ = c gives
( 4.00 × 10 Hz ) λ = 3.00 × 10 m/s:
9
8
λ = 0.075 0 m = 7.50 cm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
402
P24.38
Electromagnetic Waves
From the electromagnetic spectrum chart and accompanying text
discussion, the following identifications are made:
c
f
Frequency, f
Wavelength, λ =
2 Hz = 2 × 100 Hz
150 Mm
Radio
2 KHz = 2 × 103 Hz
150 km
Radio
2 MHz = 2 × 106 Hz
150 m
Radio
2 GHz = 2 × 109 Hz
15 cm
Microwave
2 THz = 2 × 1012 Hz
150 µm
Infrared
2 PHz = 2 × 1015 Hz
150 nm
Ultraviolet
2 EHz = 2 × 1018 Hz
150 pm
X-ray
2 ZHz = 2 × 1021 Hz
150 fm
Gamma ray
2 YHz = 2 × 1024 Hz
150 am
Gamma ray
Wavelength, λ
Frequency, f =
2 km = 2 × 103 m
1.5 × 105 Hz
Radio
2 m = 2 × 100 m
1.5 × 108 Hz
Radio
2 mm = 2 × 10–3 m
1.5 × 1011 Hz
Microwave
2 µm = 2 × 10–6 m
1.5 × 1014 Hz
Infrared
2 nm = 2 × 10–9 m
1.5 × 1017 Hz
Ultraviolet/X-ray
2 pm = 2 × 10–12 m
1.5 × 1020 Hz
X-ray/Gamma ray
2 fm = 2 × 10–18 m
1.5 × 1023 Hz
Gamma ray
2 am = 2 × 10–18 m
1.5 × 1026 Hz
Gamma ray
c
λ
Classification
Classification
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
P24.39
For the proton,
∑ F = ma:
403
mv 2
q vB sin 90.0° =
.
R
The period and frequency of the proton’s circular motion are therefore:
T=
2π ( 1.67 × 10−27 kg )
2π R 2π m
=
=
= 1.87 × 10−7 s
−19
v
qB
(1.60 × 10 C)( 0.350 T )
and f = 5.34 × 106 Hz.
The charge will radiate at this same frequency, with
c 3.00 × 108 m s
λ= =
= 56.2 m
f
5.34 × 106 Hz
P24.40
c 3 × 108 m/s
=
~ 108 Hz
λ
1.7 m
(a)
f =
(b)
1 000 pages, 500 sheets, is about 3 cm thick so one sheet is about
6 × 10–5 m thick.
f =
P24.41
radio wave
3.00 × 108 m/s
~ 1013 Hz
6 × 10−5 m
infrared
fλ = c:
Channel 3: f = 60.0 MHz to 66.0 MHz.
(a)
Channel 4: f = 66.0 MHz to 72.0 MHz, λ = 4.17 m to 4.55 m .
72.0–76.0 MHz is reserved for non-TV purposes.
Channel 5: f = 76.0 MHz to 82.0 MHz.
(b)
Channel 6: f = 82.0 MHz to 88.0 MHz, λ = 3.41 m to 3.66 m .
88.0–174 MHz is reserved for non-TV purposes.
Channel 7: f = 174 MHz to 180 MHz.
(c)
Channel 8: f = 180 MHz to 186 MHz, λ = 1.61 m to 1.67 m .
P24.42
λ=
c 3.00 × 108 m s
=
= 11.0 m
f 27.33 × 106 Hz
P24.43
(a)
λ=
c 3.00 × 108 m s
=
= 261 m so
f 1 150 × 103 s −1
180 m
= 0.690 wavelengths
261 m
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
404
Electromagnetic Waves
(b)
c 3.00 × 108 m s
λ= =
= 3.06 m so
f
98.1 × 106 s −1
180 m
= 58.9 wavelengths
3.06 m
P24.44
Time to reach object
=
1
1
( total time of flight ) = ( 4.00 × 10−4 s ) = 2.00 × 10−4 s
2
2
Thus,
d = vt = ( 3.00 × 108 m s ) ( 2.00 × 10−4 s ) = 6.00 × 10 4 m = 60.0 km
Section 24.7
P24.45
I = I max cos 2 θ
(a)
(b)
(c)
P24.46
Polarization of Light Waves
I
I max
I
I max
I
I max
⇒ θ = cos −1
I
I max
=
1
1
⇒ θ = cos −1
= 54.7°
3.00
3.00
=
1
1
⇒ θ = cos −1
= 63.4°
5.00
5.00
=
1
1
⇒ θ = cos −1
= 71.6°
10.0
10.0
For incident unpolarized light of intensity Imax, the average value of the
cosine-squared function is one-half, so the intensity after transmission
1
by the first disk is I = 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
where the angle between the first and second disk is θ = ω t .
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
405
ANS. FIG. P24.46
Using trigonometric identities cos 2 θ =
and cos 2 ( 90° − θ ) = sin 2 θ =
we have I =
I=
1
(1 + cos 2θ )
2
1
(1 − cos 2θ ) ,
2
1
⎡ ( 1 + cos 2θ ) ⎤ ⎡ ( 1 − cos 2θ ) ⎤
I max ⎢
⎥⎢
⎥
2
2
2
⎣
⎦⎣
⎦
1
1
⎛ 1⎞
I max ( 1 − cos 2 2θ ) = I max ⎜ ⎟ ( 1 − cos 4θ )
⎝ 2⎠
8
8
Since θ = ω t, the intensity of the emerging beam is given by
I=
P24.47
The intensity of unpolarized light passing through the first polarizing
3
filter is reduced by 1/2. The second transmits cos 2 30.0° = .
4
I
I max
P24.48
1
I max ( 1 − cos 4ω t )
16
P=
( ΔV )2
R
=
1 3 3
× = = 0.375
2 4 8
or P ∝ ( ΔV )2
ΔV = ( − ) Ey ⋅ Δy = Ey ⋅  cos θ
∆y
θ
l
receiving
antenna
ΔV ∝ cos θ so P ∝ cos θ
2
(a)
θ = 15.0°:
ANS. FIG. P24.48
P = Pmax cos 2 ( 15.0° ) = 0.933Pmax = 93.3%
P24.49
(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
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.
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
406
Electromagnetic Waves
The first sheet passes intensity I max cos 2 θ ,
(I
The second sheet passes
max
cos 2 θ ) cos 2 θ = I max cos 4 θ ,
and the nth sheet lets through I max cos 2n θ ≥ 0.90I max , where θ =
45°
.
n
Try different integers to find n; for example,
⎛ 45° ⎞
cos 2 × 5 ⎜
= 0.885,
⎝ 5 ⎟⎠
P24.50
(a)
So
n= 6 .
(b)
θ=
45°
= 7.50°
6
I
=
I max
⎛ 45° ⎞
cos 2 ×6 ⎜
= 0.902,
⎝ 6 ⎟⎠
⎛ 45° ⎞
cos 2 ×7 ⎜
= 0.915
⎝ 7 ⎟⎠
1
1
cos 2 45.0° ) ( cos 2 45.0° ) =
(
2
8
Section 24.8
P24.51
Context Connection: The Special Properties of
Laser Light
(a)
Bmax =
(b)
I=
Emax
7.00 × 105 N/C
: Bmax =
= 2.33 mT
c
3.00 × 108 m/s
2
Emax
:
2 µ0c
(7.00 × 10 V/m )
I=
= 6.50 × 10 W m
2 ( 4π × 10 T ⋅ m/A ) ( 3.00 × 10 m/s )
5
2
8
−7
2
8
= 650 MW m 2
(c)
P24.52
I=
2⎤
P
⎡π
: P = IA = ( 6.50 × 108 W/m 2 ) ⎢ ( 1.00 × 10−3 m ) ⎥ = 511 W
A
⎣4
⎦
The electron in the E3* state drops to the E2 state, emitting a photon of
energy hf. The process conserves energy:
E3 * = E2 + hf →
hf = E3 * − E2
The photon energy is
hf = E3 * − E2 = ( 20.66 − 18.70 ) eV = 1.96 eV =
hc
λ
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
hc 1 240 eV ⋅ nm
=
= 633 nm
E
1.96 eV
λ=
P24.53
(a)
407
The energy of the pulse is spread over the area of a circle of
radius R = 15.0 µm:
P ΔE Δt 3.00 × 10−3 J 1.00 × 10−9 s
I= =
=
A
π R2
⎡π ( 15.0 × 10−6 m )2 ⎤
⎣
⎦
= 4.24 × 1015 W m 2
(b)
The absorbed energy falls within the area of a circle of radius r =
0.300 nm:
r2
⎛ ΔE Δt ⎞
2
ΔE = IΔtA = ⎜
Δt
π
r
=
ΔE
⎝ π R 2 ⎟⎠
R2
( 0.300 × 10 m ) = 1.20 × 10
= ( 3.00 × 10 J )
(15.0 × 10 m )
−9
2
−6
2
−3
P24.54
J
(a)
 = vt = ( 3.00 × 108 m s ) ( 14.0 × 10−12 s ) = 4.20 mm
(b)
E=
−34
8
hc ( 6.626 × 10 J ⋅ s ) ( 3.00 × 10 m/s )
=
= 2.86 × 10−19 J
λ
694.3 × 10−9 m
N=
3.00 J
= 1.05 × 1019 photons
−19
2.86 × 10 J
(c)
V = ( 4.20 mm ) ⎡⎣π ( 3.00 mm )2 ⎤⎦ = 119 mm 3
n=
P24.55
−12
(a)
1.05 × 1019 photons
= 8.82 × 1016 mm −3
119 mm 3
E = hf = 0.117 eV:
f =
⎛ 1.60 × 10−19 J ⎞
E
0.117 eV
=
⎟⎠
h 6.626 × 10−34 J ⋅ s ⎜⎝
eV
= 2.83 × 1013 s −1 = 28.3 × 1012 s −1 = 28.3 THz
c 3.00 × 108 m/s
=
= 10.6 µm
f
2.83 × 1013 s −1
(b)
λ=
(c)
infrared
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
408
P24.56
Electromagnetic Waves
(a)
The distance between nodes is
λ
, so we require solutions to
2
N
λ , where N is an integer and λ is in the
2
required range. The midpoint of the range is 632.809 10 nm,
2 ( 35.124 103 × 10−2 m )
= 1 110 101.07. So we try
giving N trial =
632.809 1 × 109 m
N = 1 110 101, 1 110 102, 1 110 100, 1 110 103, and so on:
35.124 103 cm =
2 ( 35.124 103 × 10−2 m )
λ1 =
= 632.809 14 nm
1 110 101
2 ( 35.124 103 × 10−2 m )
λ2 =
= 632.808 57 nm
1 110 102
2 ( 35.124 103 × 10−2 m )
λ3 =
= 632.809 71 nm
1 110 100
2 ( 35.124 103 × 10−2 m )
λtrial =
= 632.808 00 nm, outside the range.
1 110 103
Thus the laser light has just three wavelength components.
(b)
1
3
m0 v 2 = kT. We use the periodic table.
2
2
v=
3kT
=
m0
3 ( 1.38 × 10−23 J K ) ( 393 K ) ⎛
⎞
1u
−27
⎜
⎝ 1.66 × 10 kg ⎟⎠
20.18 u
= 697 m s
(c)
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
c−v
3 × 108 − 697
= ( 632.809 1 nm )
= 632.807 63 nm
c+v
3 × 108 + 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.
P24.57
Elaser = P Δt = ( 1.00 × 106 W ) ( 10.0 × 10−9 s ) = 0.010 0 J
−34
8
hc ( 6.626 × 10 J ⋅ s ) ( 3.00 × 10 m/s )
E = hf =
=
= 2.86 × 10−19 J
λ
694.3 × 10−9 m
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
P24.58
N=
Elaser
0.010 0 J
=
= 3.49 × 1016 photons
−19
E
2.86 × 10 J
(a)
The equilibrium ratio is
409
−E k T
N4 * N ge 3 B
=
= e − (E3 −E2 ) kBT = e − ΔE kBT
N3
N g e −E2 kBT
where temperature T = 27.0 °C + 293.15 = 320.2 K, and energy
difference (from Figure P24.58) is
ΔE = E4* − E3 = 20.66 eV − 18.70 eV = 1.96 eV
N4 *
− ( 1.96 eV )( 1.602 ×10−19 J/eV ) ( 1.381×10−23 J K )( 320.2 K )
= e − ΔE kBT = e
N3
= 1.45 × 10−31
(b)
Now, we require
N4 *
= e − ΔE kBT = 1.02
N3
where ΔE = E4* − E3 = 1.96 eV.
Thus,
ln ( 1.02 ) = −
(1.96 eV ) (1.602 × 10−19 J/eV )
(1.381 × 10
−23
J K )T
T = −1.15 × 106 K
(c)
The population inversion requires the temperature be negative.
Because ΔE = E4* − E3 > 0, and in any real equilibrium state T > 0,
the ratio N 4 * N 3 = e − ΔE kBT < 1. Thus, a population inversion
cannot happen in thermal equilibrium.
Additional Problems
P24.59
(a)
2
P = SA: P = ( 1 370 W/m 2 ) ⎡ 4π ( 1.496 × 1011 m ) ⎤ = 3.85 × 1026 W
⎣
⎦
(b)
S=
2
Emax
so
2 µ0c
Emax = 2 µ0 cS = 2 ( 4π × 10−7 T ⋅ m/A ) ( 3.00 × 108 m s ) ( 1 370 )
= 1.02 kV m
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
410
Electromagnetic Waves
(c)
2
cBmax
S=
2 µ0
Bmax =
so
2 µ 0S
=
c
2 ( 4π × 10−7 T ⋅ m/A ) ( 1 370 W m 2 )
3.00 × 108 m s
= 3.39 µT
P24.60
(a)
The power incident on the mirror is:
PI = IA = ( 1 370 W m 2 ) ⎡⎣π ( 100 m )2 ⎤⎦ = 4.30 × 107 W .
The power reflected through the atmosphere is
PR = 0.746 ( 4.30 × 107 W ) = 3.21 × 107 W .
PR
3.21 × 107 W
=
= 0.639 W m 2
A π ( 4.00 × 103 m )2
(b)
S=
(c)
Noon sunshine in St. Petersburg produces this power-per-area on
a horizontal surface:
PN
= 0.746 ( 1 370 W m 2 ) sin 7.00° = 125 W m 2 .
A
The radiation intensity received from the mirror is
⎛ 0.639 W m 2 ⎞
⎜⎝ 125 W m 2 ⎟⎠ 100% = 0.513% of that from the noon Sun in
January.
P24.61
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 effective target area you
fill in the Sun’s light is
A = ( 1.7 m ) ( 0.3 m ) cos 30° = 0.4 m 2
Now I =
P
ΔE
=
, so
A AΔt
ΔE = IAΔt = ( 0.5 ) ⎡⎣( 0.6 ) ( 1 370 W/m 2 ) ⎤⎦ ( 0.4 m 2 ) ( 3 600 s ) ~ 106 J
P24.62
(a)
ε = − dΦB = − d ( BA cosθ )
dt
dt
ε = −A d ( Bmax cos ω t cosθ ) = ABmaxω ( sin ω t cosθ )
dt
ε (t ) = 2π fBmax A sin 2π f t cosθ → ε (t ) = 2π 2 r 2 f Bmax cosθ sin 2π f t
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
ε max = 2π 2 r 2 f Bmax cosθ ,
Thus,
P24.63
411
(b)
where θ is the angle between the magnetic field and the normal to
the loop.


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.
(a)
Bmax =
(b)
( 0.200 × 10−6 V/m )
E2
Savg = max =
2 µ0 c 2 ( 4π × 10−7 T ⋅ m/A ) ( 3.00 × 108 m/s )
Emax 0.200 × 10−6 V/m
=
= 6.67 × 10−16 T
8
c
3.00 × 10 m/s
2
= 5.31 × 10−17 W/m 2
2
(c)
⎛ d⎞
⎛ 20.0 m ⎞
P = Savg A = Savgπ ⎜ ⎟ = ( 5.31 × 10−17 W/m 2 ) π ⎜
⎝ 2⎠
⎝ 2 ⎟⎠
2
= 1.67 × 10−14 W
(d)
⎛ 5.31 × 10−17 W/m 2 ⎞ ⎛ 20.0 m ⎞
⎛ Savg ⎞
F = PA = ⎜
A=⎜
π⎜
⎟
⎝ c ⎟⎠
⎝ 3.00 × 108 m/s ⎟⎠ ⎝ 2 ⎠
2
= 5.56 × 10−23 N
(approximately the weight of 3 000 hydrogen atoms!)
*P24.64
(a)
ANS. FIG. P24.64
(b)
uE =
1
1
2
∈0 E 2 = ∈0 Emax
cos 2 (kx)
2
2
(c)
uB =
1 2
1 2
B =
Bmax cos 2 (kx)
2 µ0
2 µ0
(d) Note that
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
412
Electromagnetic Waves
2
2
1 Emax
1
Emax
2
uB =
cos (kx) =
cos 2 (kx)
2
2 µ0 c
2 µ0 ( 1 µ0 ∈0 )
1
2
= ∈0 Emax
cos 2 (kx) = uE
2
2
Therefore, u = uE + uB = ∈0 Emax
cos 2 (kx) .
(e)
λ
Eλ = ∫0 uA dx
λ
λ
⎡1 1
⎤
2
2
Eλ = ∫0 ∈0 Emax
cos 2 (kx)A dx = ∫0 ∈0 Emax
A ⎢ + cos(2kx) ⎥ A dx
⎣2 2
⎦
=
1
∈ E2
λ
λ
2
∈0 Emax
A x 0 + 0 max A sin(2kx) 0
2
4k
=
1
∈ E2
2
∈0 Emax
Aλ + 0 max A [ sin(4π ) − sin(0)]
2
4k
=
1
2
∈0 Emax
λA
2
(f)
2
Eλ 1 ∈0 Emax
λA 1
1
2
2
P=
=
= ∈0 Emax
λ f ) A = ∈0 cEmax
A
(
T 2 (1 f )
2
2
(g)
2
P 21 ∈0 cEmax
A 1
2
I= =
= ∈0 cEmax
A
A
2
(h)
From part (g), we have
1
µ ∈ cE 2
cE 2
1 cE 2
E2
2
∈0 cEmax
= 0 0 max = ( µ0 ∈0 ) max = 2 max = max
2
µ0
2
2 µ0
c 2 µ0
2 µ0c
2
Emax
The result in part (g) agrees with I =
in Equation 24.26.
2 µ0c
*P24.65
Fgrav =
GMSm ⎛ GMS ⎞ ⎡ ⎛ 4 3 ⎞ ⎤
= ⎜ 2 ⎟ ⎢ρ ⎜ π r ⎟ ⎥
⎝ R ⎠⎣ ⎝3
⎠⎦
R2
where MS = mass of Sun, r = radius of particle, and R = distance from
Sun to particle.
The particle absorbs all the radiation so, by Equation 24.32,
Frad = PA =
Frad = Fgrav
S 2
πr
c
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
413
S 2 ⎛ GMS ⎞ ⎡ ⎛ 4 3 ⎞ ⎤
π r = ⎜ 2 ⎟ ⎢ρ ⎜ π r ⎟ ⎥
⎝ R ⎠ ⎣ ⎝3
⎠⎦
c
3SR 2
r=
4cGMS ρ
Suppressing units,
r=
3 ( 214 ) ( 3.75 × 1011 )
2
4 ( 3.00 × 108 ) ( 6.67 × 10−11 ) ( 1.991 × 1030 ) ( 1 500 )
= 3.78 × 10−7 m = 378 nm
*P24.66
Fgrav =
GMSm ⎛ GMS ⎞ ⎡ ⎛ 4 3 ⎞ ⎤
= ⎜ 2 ⎟ ⎢ρ ⎜ π r ⎟ ⎥
⎝ R ⎠⎣ ⎝3
⎠⎦
R2
where MS = mass of Sun, r = radius of particle, and R = distance from
Sun to particle.
The particle absorbs all the radiation so, by Equation 24.32,
Frad = PA =
S 2
πr
c
Frad = Fgrav
S 2 ⎛ GMS ⎞ ⎡ ⎛ 4 3 ⎞ ⎤
π r = ⎜ 2 ⎟ ⎢ρ ⎜ π r ⎟ ⎥
⎝ R ⎠⎣ ⎝3
⎠⎦
c
3SR 2
r=
4cGMS ρ
P24.67
The mirror intercepts power
2
P = I1 A1 = ( 1.00 × 103 W/m 2 ) ⎡⎣π ( 0.500 m ) ⎤⎦ = 785 W .
(a)
In the image, I 2 =
(b)
I2 =
P
:
A2
I2 =
785 W
π ( 0.020 0 m )
2
= 625 kW/m 2
2
Emax
, so
2 µ0c
Emax = 2 µ0 cI 2
= 2 ( 4π × 10−7 T ⋅ m/A ) ( 3.00 × 108 m/s ) ( 6.25 × 105 W/m 2 )
= 21.7 kN C
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
414
Electromagnetic Waves
Emax
= 72.4 µT
c
(c)
Bmax =
(d)
0.400 ( P Δt ) = mcΔT
0.400 ( 785 W ) Δt = ( 1.00 kg ) ( 4 186 J/kg ⋅ °C ) ( 100°C − 20.0°C )
Δt =
P24.68
3.35 × 105 J
= 1.07 × 103 s = 17.8 min
314 W
We use the relationship between energy density and electric field
magnitude that we studied previously for a static field. The energy
density can be written as
2
u = 21 ∈0 Emax
P24.69
2u
=
∈0
2 ( 4.00 × 10 –14 N ⋅ m 2 )
so
Emax =
(a)
A hemisphere is half a sphere:
8.85 × 10 –12 C2/N ⋅ m 2
= 95.1 mV/m .
⎡1⎛ 4
⎞⎤
m = ρV = ρ ⎢ ⎜ π r 3 ⎟ ⎥ = 5.50 + 4(0.800) kg = 8.70 kg
⎠⎦
⎣2 ⎝ 3
⎛ 6m ⎞
r=⎜
⎝ ρ 4π ⎟⎠
13
⎛
6 ( 8.7 kg ) ⎞
=⎜
⎟
3
⎝ ( 990 kg/m ) 4π ⎠
13
= 0.161 m
1
2
4π r 2 = 2π ( 0.161 m ) = 0.163 m 2
2
(b)
A=
(c)
P = eσ AT 4 and T = 31.0 + 273.0 = 304 K:
P = 0.970 ( 5.67 × 10−8 W/m 2 ⋅ K 4 ) ( 0.163 m 2 ) ( 304 K )
4
= 76.8 W
(d)
P
= eσ T 4
A
4
I = 0.970 ( 5.67 × 10−8 W m 2 ⋅ K 4 ) ( 304 K )
I=
2
= 470 W m
(e)
I=
2
Emax
2 µ0c
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
415
Emax = ( 2 µ0 cI )
12
= ⎡⎣ 2 ( 4π × 10−7 Tm/A ) ( 3.00 × 108 m/s ) ( 470 W/m 2 ) ⎤⎦
12
= 595 N/C
595 N/C
= 1.98 µT
3 × 108 m/s
(f)
Emax = cBmax → Bmax =
(g)
⎛ 6m ⎞
Each kitten has radius rk = ⎜
⎝ ρ 4π ⎟⎠
13
⎡ 6 ( 0.800 ) ⎤
=⎢
⎥
⎣ 990 × 4π ⎦
13
= 0.072 8 m
and radiating area 2π ( 0.072 8 m ) = 0.033 3 m 2 . The mother cat
2
⎡ 6 ( 5.50 ) ⎤
has area 2π ⎢
⎥
⎣ 990 × 4π ⎦
23
= 0.120 m 2 . The total glowing area is
0.120 m 2 + 4 ( 0.033 3 m ) = 0.254 m and has power output
2
2
P = IA = ( 470 W/m 2 ) ( 0.254 m 2 ) = 119 W .
P24.70
We take R to be the planet’s distance from its star, and r to be the
radius of the planet.
(a)
The effective area of the planet over which it absorbs light is its
projection onto a plane perpendicular to the light from its sun.
The projected area of a planet of radius r is π r2, so the planet
absorbs light over area π r2 .
(b)
The planet radiates over its entire surface area, 4π r2.
(c)
At steady-state, Pin = Pout: eI in π r 2 = eσ 4π r 2 T 4
(
)
(
)
⎛ 6.00 × 1023 W ⎞
e⎜
π r 2 ) = eσ ( 4π r 2 ) T 4 , so that
(
2
⎟
4π R
⎝
⎠
6.00 × 1023 W = 16πσ R 2T 4
R=
=
P24.71
(a)
6.00 × 1023 W
16πσ T 4
6.00 × 1023 W
9
4 = 4.77 × 10 m
−8
2
4
16π ( 5.67 × 10 W/m ⋅ K ) ( 310 K )
At steady state, Pin = Pout and the power radiated out is
Pout = eσ AT 4 .
(
)
(
)
Thus, 0.900 1 000 W/m 2 A = 0.700 5.67 × 10−8 W/m 2 ⋅ K 4 AT 4 ,
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
416
Electromagnetic Waves
⎡
⎤
900 W/m 2
or T = ⎢
−8
2
4 ⎥
⎢⎣ 0.700 ( 5.67 × 10 W/m ⋅ K ) ⎥⎦
(b)
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,
0.900 ( 1 000 W/m 2 ) A sin 50.0° = 0.700 ( 5.67 × 10−8 W/m 2 ⋅ K 4 ) AT 4
⎡
⎤
900 W/m 2 ) sin 50.0°
(
or T = ⎢
−8
2
4 ⎥
⎢⎣ 0.700 ( 5.67 × 10 W/m ⋅ K ) ⎥⎦
P24.72
14
c 3.00 × 108 m s
=
= 1.50 cm
f
20.0 × 109 s −1
(a)
λ=
(b)
U = P ( Δt ) = ( 25.0 × 103 J s ) ( 1.00 × 10−9 s )
= 25.0 × 10−6 J = 25.0 µ J
(c)
= 363 K = 90.0°C
uav =
=
ANS. FIG. P24.72
U
U
=
V (π r 2 ) 
U
25.0 × 10−6 J
=
(π r 2 ) c ( Δt ) π ( 0.060 0 m )2 ( 3.00 × 108 m s )(1.00 × 10−9 s )
uav = 7.37 × 10−3 J m 3 = 7.37 mJ m 3
(d)
Emax =
2uav
=
∈0
2 ( 7.37 × 10−3 J m 3 )
= 4.08 × 10 4 V m
−12
2
2
8.85 × 10 C N ⋅ m
= 40.8 kV m
Bmax =
(e)
Emax 4.08 × 10 4 V m
=
= 1.36 × 10−4 T = 136 µT
c
3.00 × 108 m s
F = PA =
()
S
A = uav A = ( 7.37 × 10−3 J m 3 ) π ( 0.060 0 m )2
c
= 8.33 × 10−5 N = 83.3 µN
P24.73
Emax
175 V/m
=
= 5.83 × 10−7 T = 583 nT
8
c
3.00 × 10 m/s
(a)
Bmax =
(b)
k=
(c)
ω = kc = ( 419 m −1 ) ( 3.00 × 108 m/s ) = 1.26 × 1011 s −1
2π
2π
–1
=
= 419 m
λ 0.015 0 m
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
(d)
(e)
(f)
(g)
417

   
S ∝ E × B, S is in the x direction, and E vibrates in the y direction

(xy plane), so B must vibrate in the z direction, thus

B vibrates in the xz plane .
−7
Emax Bmax ( 175 V/m ) ( 5.83 × 10 T )
Savg =
=
= 40.6 W/m 2
−7
2 µ0
2 ( 4π × 10 T ⋅ m/A )

→ Savg = 40.6î W/m 2
2
2S 2 ( 40.6 W m )
Pr =
=
= 2.71 × 10−7 N m 2 = 271 nPa
8
c
3.00 × 10 m/s
∑ F = PA = ( 2.71 × 10
a=
m
m
−7
N/m 2 ) ( 0.750 m 2 )
0.500 kg
= 4.07 × 10−7 m/s 2

a = 407 î nm/s 2
P24.74
(a)
On the right side of the equation,
C2 ( m/s 2 )
2
( C / N ⋅ m )( m/s )
2
(b)
F = ma = qE
2
3 =
N ⋅ m 2 ⋅ C2 ⋅ m 2 ⋅ s 3 N ⋅ m J
=
= =W
C2 ⋅ s 4 ⋅ m 3
s
s
or
−19
qE ( 1.60 × 10 C ) ( 100 N/C )
a=
=
=
m
9.11 × 10−31 kg
(c)
1.76 × 1013 m/s 2
The radiated power is then (suppressing units):
1.60 × 10−19 ) ( 1.76 × 1013 )
(
q2 a2
P=
=
6π ∈0 c 3 6π ( 8.85 × 10−12 ) ( 3.00 × 108 )3
2
2
= 1.75 × 10−27 W
(d)
⎛ v2 ⎞
F = mac = m ⎜ ⎟ = qvB
⎝ r ⎠
so
v=
qBr
.
m
The proton accelerates at (suppressing units)
−19
v 2 q 2 B2 r ( 1.60 × 10 ) ( 0.350 ) ( 0.500 )
a=
=
=
2
r
m2
(1.67 × 10−27 )
2
2
= 5.62 × 1014 m/s 2
The proton then radiates (suppressing units)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
418
Electromagnetic Waves
1.60 × 10−19 ) ( 5.62 × 1014 )
(
q2 a2
P=
=
= 1.80 × 10−24 W
6π ∈0 c 3 6π ( 8.85 × 10−12 ) ( 3.00 × 108 )3
2
P24.75
(a)
P=
F I
IA P
100 J/s
= F=
= =
= 3.33 × 10−7 N = ( 110 kg ) a
A c
c
c 3.00 × 108 m/s
a = 3.03 × 10−9 m/s 2 and x =
t=
(b)
2
2x
=
a
1 2
at
2
2 ( 10.0 m )
= 8.12 × 10 4 s = 22.6 h
−9
2
3.03 × 10 m s
Let A = astronaut, SP = spacecraft, and F = flashlight. The
astronaut-flashlight system is isolated. With respect to the
spacecraft,
0 = MvA,SP + mvF,SP
(
0 = MvA,SP + m vF,A + vA,SP
)
0 = ( 107 kg ) v + ( 3.00 kg ) ( −12.0 m s + v )
0 = ( 107 kg ) v − 36.0 kg ⋅ m s + ( 3.00 kg ) v
= ( 110 kg ) v − 36.0 kg ⋅ m s
36.0 kg ⋅ m s
= 0.327 m s
110 kg
d
10.0 m
t= =
= 30.6 s
v 0.327 m s
v=
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 24
419
ANSWERS TO EVEN-NUMBERED PROBLEMS
P24.2
(a) 7.19 × 1011 V m ⋅ s ; (b) 2.00 × 10–7 T
P24.4
608 pF
P24.6
(a) 3.15 × 103 ĵ N C ; (b) 5.25k̂ × 10−7 T ; (c) 4.83 − ĵ × 10−16 N
P24.8
( −4.39î − 1.76 ĵ) × 10 m/s
P24.10
(a) 2 692 C.E.; (b) 499 s; (c) 2.56 s; (d) 0.133 s; (e) 3.33 × 10–5 s
P24.12
733 nT
P24.14
See P24.14 for full explanation.
P24.16
(a) 0.333 µT; (b) 0.628 µm; (c) 4.77 × 1014 Hz
P24.18
(a) See P24.18(a) for full explanation; (b) See P24.18(b) for full
explanation; (c) 2.00 kHz; (d) 0.075 0 m/s ≈ 0.17 mi/h
P24.20
(a) See P24.20(a) for full explanation; (b) 0.050 4c
P24.22
The ratio of ω to k is higher than the speed of light in a vacuum, so the
wave as described is impossible.
P24.24
(a) 0; (b) 11.5î − 28.6 ĵ W/m 2
P24.26
(a) 88.8 nW/m2; (b) 11.3 MW
P24.28
(a) 1.19 × 1010 W/m2; (b) 2.34 × 105 W
P24.30
5.16 m
P24.32
(a) 5.48 N; (b) 913 µm/s2 away from the Sun; (c) 10.6 days
P24.34
(a)
P24.36
Listeners 100 km away will receive the news before the people in the
newsroom.
P24.38
See table in P24.38 for full description.
P24.40
(a) ~108 Hz radio wave; (b) ~1013 Hz infrared
P24.42
11.0 m
P24.44
60.0 km
P24.46
1
I max ( 1 − cos 4ω t )
16
( )
11
(
2
)
2 µ0 cP
P
P
; (b)
; (c) 2
2
πr
c
c
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
420
Electromagnetic Waves
P24.48
(a) 93.3%; (b) 50.0%; (c) 0
P24.50
1/8
P24.52
633 nm
P24.54
(a) 4.20 mm; (b) 1.05 × 1019 photons; (c) 8.82 × 1016 mm–3
P24.56
(a) λ1 = 632.809 14 nm, λ2 = 632.808 57 nm, λ3 = 632.809 71 nm, three;
(b) 697 m/s (c) See P24.56(c) for full description.
P24.58
(a) 1.45 × 10–31; (b) –1.15 × 106 K; (c) The population inversion requires
the temperature be negative. Thus, a population inversion cannot
happen in thermal equilibrium.
P24.60
(a) 3.21 × 107 W; (b) 0.639 W/m2; 0.513% of that from the noon Sun in
January
P24.62
(a) ε max = 2π 2 r 2 f Bmax cos θ ; (b) The plane of the loop should be vertical
and the plane should contain the line of sight of the transmitter.
P24.64
(a) See ANS. FIG. P24.64; (b)
P24.66
3SR 2
4cGMS ρ
P24.68
95.1 mV/m
P24.70
(a) The projected area of a planet of radius r is π r2, which radiates over
an area 4π r2; (b) 4.77 × 109 m
P24.72
(a) 1.50 cm; (b) 25.0 μJ; (c) 7.37 mJ/m3; (d) Emax = 40.8 kV/m,
Bmax = 136 μT; (e) 83.3 μN
P24.74
(a) See P24.74 for full description; (b) 1.76 × 1013 m/s2;
(c) 1.75 × 10−27 W; (d) 1.80 × 10−24 W
1 2
1
2
Bmax cos 2 (kx) ;
∈0 Emax
cos 2 (kx); (c)
2 µ0
2
1
1
1
2
2
2
2
cos 2 (kx); (e) ∈0 Emax
(d) ∈0 Emax
λ A; (f) ∈0 cEmax
A; (g) ∈0 cEmax
;
2
2
2
E2
(h) The result in part (g) agrees with I = max in Equation 24.26.
2 µ0c
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.