Download Ch 22 Solutions

ELECTROMAGNETIC WAVES
22
Responses to Questions
1.
If the direction of travel for the EM wave is north and the electric field oscillates in an east–west plane,
then the magnetic field must oscillate up and down. For an EM wave, the direction of travel, the
electric field, and the magnetic field must all be perpendicular to each other.
2.
No, sound is not an electromagnetic wave. Sound is a longitudinal mechanical (pressure) wave, which
requires a medium in which to travel. The medium can be a gas, a liquid, or a solid. EM waves do not
need a medium in which to travel.
3.
EM waves can travel through a perfect vacuum. The energy is carried in the oscillating electric and
magnetic fields, and no medium is required to travel. Sound waves cannot travel through a perfect
vacuum. A medium is needed to carry the energy of a mechanical wave such as sound, and there is no
medium in a perfect vacuum.
4.
No. Electromagnetic waves travel at a very large but finite speed. When you flip on a light switch, it
takes a very small amount of time for the electrical signal to travel along the wires from the switch to
the lightbulb.
5.
The wavelengths of radio and TV signals are much longer than those of visible light. Radio waves are
on the order of 3 m–30,000 m. TV waves are on the order of 0.3 m–3 m. Visible waves are on the
order of 10−7 m.
6.
It is not necessary to make the lead-in wires to your speakers the exact same length. Since electrical
signals in the wires travels at nearly the speed of light, the difference in time between the signals
getting to the different speakers will be too small for your ears to detect.
7.
Wavelength of 103 km: sub-radio waves (or very long radio waves; for example, ELF waves for
submarine communication fall into this category). Wavelength of 1 km: radio waves. Wavelength
of 1 m: TV signals and microwaves. Wavelength of 1 cm: microwaves and satellite TV signals.
Wavelength of 1 mm: microwaves and infrared waves. Wavelength of 1 μ m: infrared waves.
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22-1
22-2
Chapter 22
8.
Yes, radio waves can have the same frequencies as sound waves. These 20- to 20,000-Hz EM waves
would have extremely long wavelengths when compared to the sound waves, because of their high
speed. A 5000-Hz sound wave has a wavelength of about 70 mm, while a 5000-Hz EM wave has a
wavelength of about 60 km.
9.
The receiver antenna should also be vertical for obtaining the best reception. The oscillating carrier
electric field is up and down, so a vertical antenna would “pick up” that signal better. That is because
the electrons in the metal antenna would be forced to oscillate up and down along the entire length
of the vertical antenna, creating a stronger signal.
10.
Diffraction effects (the bending of waves around the edge of an object) are evident only when the size
of the wavelength of the wave is on the order of the size of the object (or larger). AM waves have
wavelengths that are on the order of 300 m long, while FM waves have wavelengths on the order of
3 m long. Buildings and hills are much larger than FM waves, so FM waves will not diffract around
the buildings and hills. Thus the FM signal will not be received behind the hills or buildings. On the
other hand, these objects are smaller than AM waves, so the AM waves will diffract around them
easily. The AM signal can be received behind the objects.
11.
Cordless phones utilize EM waves when sending information back and forth between the handset (the
part you hold up to your ear/mouth) and its base (which is sitting in your house, physically connected
to the wire phone lines that lead outside to the phone company’s network). These EM waves are
usually very weak—you can’t walk very far away from the base before you lose the signal. Cell
phones utilize EM waves when sending information back and forth between the phone and the nearest
tower in your geographical area (which could be miles away from your location). These EM waves
need to be much stronger than cordless phone waves (or the cell phone electronics need to be more
sensitive) because of the larger distances involved.
12.
Transmitting Morse code by flashing a flashlight on and off creates an AM wave. The amplitude of the
carrier wave is increasing/decreasing every time you turn the flashlight on and off. The frequency of
the carrier wave is visible light, which is on the order of 1014 – 1015 Hz.
Responses to MisConceptual Questions
1.
(a, b) All electromagnetic waves have the same velocity in a vacuum. The velocity is the product of
the wavelength and frequency. Since X-rays and radio waves have different wavelengths but the
same speed, they will also have different frequencies.
2.
(c)
3.
(a, b, c, e, f, g, h) All electromagnetic waves travel at the speed of light. The only listed wave that is
not an electromagnetic wave is (d ) ultrasonic wave, which is a sound wave and would travel at
the speed of sound.
4.
(e)
A common misconception is that electromagnetic waves travel at different speeds. They do not,
as they all travel at the speed of light.
5.
(c)
In empty space, X-ray radiation and a radio wave both travel at the speed of light. X-ray
radiation has a much smaller wavelength than a radio wave. Since wave speed is equal to the
product of the wavelength and the frequency, and both waves travel at the same speed, the X-ray
Visible light has a wavelength on the order of 10−7 to 10−8 m. This is over a thousand times
larger than the size of an atom.
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Electromagnetic Waves
22-3
radiation has a higher frequency than a radio wave. The amplitude depends upon the intensity of
the wave, so either the X-ray radiation or the radio wave could have the greater amplitude.
6.
(b)
Frequency is the measure of the number of oscillations made per second. The rate at which the
electrons oscillate will be equal to the oscillation frequency of the radiation emitted.
7.
(d)
A common misconception is to think that the radiation decreases linearly with distance, such that
the intensity would decrease by a factor of two. However, the radiation intensity decreases as the
square of the distance. Doubling the distance will decrease the radiation intensity to one-fourth
the initial intensity.
8.
(a)
The direction of the magnetic field is perpendicular to the direction the wave is traveling and
perpendicular to the electric field, so only north and south are possible answers. The right-hand
rule can be used to determine which direction is correct by pointing fingers in the direction of the
electric field (west) and bending them in the direction of the magnetic field (north), resulting in
the thumb pointing in the direction of the wave (down).
9.
(c)
A misconception that can arise is thinking that the wave intensity is proportional to just one of
the field amplitudes. However, the intensity is proportional to the product of the amplitudes, with
the field amplitudes proportional to each other. If the intensity doubles, then each field (electric
and magnetic) must increase by the same factor, 2.
10.
(d)
A common misconception is that every color surface will experience the same radiation pressure.
However, the pressure is related to the change in momentum of the light. With a black object, all
of the light is absorbed, so the radiation pressure is proportional to the incident momentum. With
a colored object, some of the light is reflected, thus increasing the change in momentum of the
light and therefore increasing the radiation pressure as compared to the black object. For a white
object, the maximum amount of light is reflected off the surface, creating the greatest change in
momentum of the light and the greatest radiation pressure.
11.
(c)
A common misconception is that the digital broadcast uses a different frequency than the analog
broadcast. This is incorrect—the carrier wave frequency is the same. Since the antenna is
designed for operation at the broadcast frequency, the antenna will work just as well with a
digital signal.
Solutions to Problems
1.
The electric field between the plates is given by Eq. 17–4a, E = V /d , where d is the distance between
the plates. We use that to find the rate of change of the electric field.
E=
2.
V
d
→
⎞
ΔE 1 Δ V ⎛
1
5 V/m
=
=⎜
⎟ (120 V/s) = 1.1× 10
Δt d Δt ⎝ 0.0011 m ⎠
s
The displacement current is shown in Section 22–1 to be I D = ε 0
ΔΦE
, and the flux is given by
Δt
ΦE = EA, where A is the area of the capacitor plates.
ID = ε0
ΔΦE
ΔE
V ⎞
⎛
−8
= ε0 A
= (8.85 × 10−12 C2 /N ⋅ m 2 )(0.058 m) 2 ⎜ 1.6 × 106
⎟ = 4.8 × 10 A
m ⋅s ⎠
Δt
Δt
⎝
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22-4
3.
Chapter 22
The current in the wires must also be the displacement current in the capacitor. Use the displacement
current to find the rate at which the electric field is changing.
ID = ε0
4.
ΔΦE
ΔE
= ε0 A
Δt
Δt
→
I
ΔE
(3.8 A)
V
= D =
= 1.7 × 1015
12
2
2
2
−
Δt ε 0 A (8.85 × 10 C /N ⋅ m )(0.0160 m)
m ⋅s
The current in the wires is the rate at which charge is accumulating on the plates and also is the
displacement current in the capacitor. Because the location in question is outside the capacitor, use the
expression for the magnetic field of a long wire, Eq. 20–6.
B=
μ0 I ⎛ μ 0
=
2π R ⎜⎝ 4π
⎞ 2 I (10−7 T ⋅ m/A) 2 (32.0 × 10−3 A)
= 6.40 × 10−8 T
⎟ R =
(0.100 m)
⎠
After the capacitor is fully charged, all currents will be zero, so the magnetic field will be zero.
5.
Use Eq. 22–2.
E0
E
0.72 × 10−4 V/m
= c → B0 = 0 =
= 2.4 × 10−13 T
8
B0
c
3.00 × 10 m/s
6.
Use Eq. 22–2.
E0
= c → E0 = B0 c = (10.5 × 10−9 T)(3.00 × 108 m/s) = 3.15 V/m
B0
7.
The frequency of the two fields must be the same: 90.0 kHz.
The rms strength of the electric field is found from Eq. 22–2.
Erms = cBrms = (3.00 × 108 m/s)(7.75 × 10−9 T) = 2.33 V/m
The electric field is perpendicular to both the direction of travel and the magnetic field, so the electric
field oscillates along the horizontal north–south line.
8.
Use the relationship d = υ t to find the time.
d = υt
9.
→
t=
d
υ
=
(1.50 × 1011 m)
(3.00 × 108 m/s)
= 5.00 × 102 s = 8.33 min
We assume that the only significant time is the transit time from the Earth to the Moon, at the speed of
light. We use the Earth–Moon distance but subtract the Earth and Moon radii since the astronaut and
the receiver are on the surfaces of the bodies.
d = υt
→
t=
d
υ
d Earth- − rEarth − rMoon
=
Moon
c
=
(384 × 106 m − 6.38 × 106 m − 1.74 × 106 m)
(3.00 × 108 m/s)
= 1.25 s
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Electromagnetic Waves
10.
22-5
The frequency of the wave is given by Eq. 22–4.
f =
c
λ
=
(3.00 × 108 m/s)
(720 × 10−9 m)
= 4.2 × 1014 Hz
This frequency is just inside the red end of the visible region, so it is red visible light.
11.
The wavelength of the wave is calculated from Eq. 22–4.
λ=
c (3.00 × 108 m/s)
=
= 4.20 × 10−7 m
f (7.14 × 1014 Hz)
This wavelength is violet visible light.
12.
The radio frequency is found from Eq. 22–4.
f =
13.
c
λ
=
(3.0 × 108 m/s)
= 6.1× 106 Hz
(49 m)
Use Eq. 22–4 to find the frequency of the microwave.
c=λf
14.
(a)
→
15.
16.
λ
=
(3.00 × 108 m/s)
(1.50 × 10−2 m)
= 2.00 × 1010 Hz
λ=
c (3.00 × 108 m/s)
=
= 3.00 × 105 m
f
(1.00 × 103 Hz)
Again use Eq. 22–4, with the speed of sound in place of the speed of light.
υ =λf
(c)
c
Use Eq. 22–4 to find the wavelength.
c=λf
(b)
f =
→
→
λ=
υ
f
=
(341 m/s)
(1.00 × 103 Hz)
= 0.341 m
No, you cannot hear a 1000-Hz EM wave. It takes a pressure wave to excite the auditory system.
However, if you applied the 1000-Hz EM wave to a speaker, you could hear the 1000-Hz
pressure wave produced by the speaker.
Use Eq. 22–4 to find the desired wavelength and frequency.
c (3.000 × 108 m/s)
=
= 1.319 × 10−2 m
f
(22.75 × 109 Hz)
(a)
c=λf
→
λ=
(b)
c=λf
→
f =
(a)
The radio waves travel at the speed of light, so Δd = υΔt. The distance is found from the radii of
the orbits. For times when Mars is nearest the Earth, the radii should be subtracted.
Δt =
c
λ
=
(3.00 × 108 m/s)
(0.12 × 10−9 m)
= 2.5 × 1018 Hz
Δd (230 × 109 m − 149.6 × 109 m)
=
= 268 s ≈ 300 s ≈ 4 min
c
(3.000 × 108 m/s)
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22-6
Chapter 22
(b)
For times when Mars is farthest from Earth, the radii should be subtracted.
Δt =
17.
Δd (230 × 109 m + 149.6 × 109 m)
=
= 1265 s ≈ 1300 s ≈ 21 min
c
(3.000 × 108 m/s)
We convert the units from light-years to meters.
d = (4.2 ly)(3.00 × 108 m/s)(3.16 × 107 s/yr) = 4.0 × 1016 m
18.
The distance is the rate (speed of light) times the time (1 year), with the appropriate units.
d = (3.00 × 108 m/s) (3.156 × 107 s/yr) = 9.47 × 1015 m
19.
The length of the pulse is Δd = cΔt , so the number of wavelengths in this length is found by dividing
this length by the wavelength.
N=
(cΔt )
λ
=
(3.00 × 108 m/s)(34 × 10−12 s)
(1062 × 10−9 m)
= 9600 wavelengths
The time for the length of the pulse to be one wavelength is the wavelength divided by the speed of light.
Δt ′ =
20.
(1062 × 10−9 m)
8
(3.00 × 10 m/s)
= 3.54 × 10−15 s
1
Δθ 8 (2π rad) (π rad)c (π rad)(3.00 × 108 m/s)
=
=
=
= 3400 rad/s (3.2 × 104 rev/ min)
3
Δt
(2Δx /c)
8Δx
8(35 × 10 m)
The mirror must rotate at least 1/8 of a revolution in the time it takes the light beam to travel twice the
length of the room.
t=
22.
c
=
The eight-sided mirror would have to rotate 1/8 of a revolution for the succeeding mirror to be in
position to reflect the light in the proper direction. During this time the light must travel to the opposite
mirror and back.
ω=
21.
λ
(2)(12 m)
8
(3.0 × 10 m/s)
= 8.0 × 10−8 s; ω =
( 18 revolution)
(8.0 × 10
−8
s)
= 1.6 × 106 revolutions/s
The average energy transferred across unit area per unit time is the average intensity and is given by
Eq. 22–8.
I = 12 ε 0 cE02 = 12 (8.85 × 10−12 C2 /N ⋅ m 2 )(3.00 × 108 m/s)(0.0225 V/m) 2 = 6.72 × 10−7 W/m 2
23.
The energy per unit area per unit time is the same as the magnitude of the average intensity, as in Eq.
22–8. Use the given area and energy with the intensity to solve for the time.
I=
Δt =
1
2
cB02
μ0
=
μ0 ΔU
2
cBrms
A
1
2
=
c( 2 Brms ) 2
μ0
=
2
cBrms
μ0
=
ΔU
AΔt
(4π × 10−7 T ⋅ m/A)(365 J)
(3.00 × 108 m/s)(22.5 × 10−9 T)2 (1.00 × 10−4 m 2 )
= 3.02 × 107 s ≈ 350 days
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Electromagnetic Waves
24.
22-7
The energy per unit area per unit time is the same as the magnitude of the average intensity, as
in Eq. 22–8. Use the given area and time with the intensity to solve for the energy.
2
I = 12 ε 0 cE02 = 12 ε 0 c( 2 Erms )2 = ε 0 cErms
=
ΔU
2
A
= cε 0 Erms
Δt
ΔU
AΔt
→
= (3.00 × 108 m/s)(8.85 × 10−12 C2 /N ⋅ m 2 )(0.0308 V/m) 2 (1.00 × 10−4 m 2 )(3600 s/h)
= 9.07 × 10−7 J/h
25.
The intensity is the power per unit area. The area is the surface area of a sphere, since the wave is
spreading spherically. Use Eq. 22–8 to find the rms value of the electric field.
I=
P
(1800 W)
=
= 5.730 W/m 2 ≈ 5.7 W/m 2
A [4π (5.0 m) 2 ]
2
I = 12 ε 0 cE02 = 12 ε 0 c( 2 Erms ) 2 = ε 0 cErms
I
Erms =
26.
(a)
cε 0
=
→
5.730 W/m 2
(3.00 × 108 m/s)(8.85 × 10−12 C2 /N ⋅ m 2 )
= 46 V/m
We find E0 using Eq. 22–2 with υ = c.
E0 = cB0 = (3.00 × 108 m/s) (2.2 × 10−7 T) = 66 V/m
(b)
The average power per unit area is the intensity, given by Eq. 22–8.
I=
27.
E0 B0 (66 V/m)(2.2 × 10−7 T)
=
= 5.8 W/m 2
(2 μ0 ) [2(4π × 10−7 N ⋅ s 2 /C2 )]
Use Eq. 22–6a (the instantaneous value of energy density) with Eq. 22–7.
u = ε0E2
28.
→
I
I
1350 W/m 2
u = 12 ε 0 E02 = ; U = uV = V =
(1.00 m3 ) = 4.50 × 10−6 J
8
c
c
3.00 × 10 m/s
The power output per unit area is the intensity. Use Eq. 22–8 with rms values.
I=
Erms =
P 1
2
= ε 0 cE02 = 12 ε 0 c( 2 Erms ) 2 = ε 0 cErms
A 2
→
0.0158 W
P
=
3
2
−
Acε 0
π (1.20 × 10 m) (3.00 × 108 m/s)(8.85 × 10−12 C2 /N ⋅ m 2 )
= 1146.9 V/m ≈ 1150 V/m
Brms =
29.
Erms
1146.9 V/m
=
= 3.82 × 10−6 T
8
c
3.00 × 10 m/s
The Sun is assumed to radiate uniformly, so use the intensity at the Earth to calculate the total power
output of the sun. Use a sphere centered at the Sun with a radius equal to the Earth’s orbit radius. The
1350 W/m 2 is the average intensity.
I=
P
A
→
P = IA = 4π R 2 I = 4π (1.496 × 1011 m) 2 (1350 W/m 2 ) = 3.80 × 1026 W
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22-8
30.
Chapter 22
(a)
The energy emitted in each pulse is the laser’s power output times the duration of the pulse.
P=
(b)
ΔW
Δt
→
ΔW = PΔt = (1.5 × 1011 W)(1.0 × 10−9 s) = 150 J
We find the rms electric field from the intensity, which is the power per unit area. Use Eq. 22–8a
with rms values. The area to use is the cross-sectional area of the cylindrical beam.
I=
P 1
2
= ε 0 cE02 = 12 ε 0 c( 2 Erms ) 2 = ε 0 cErms
A 2
→
P
(1.5 × 1011 W)
=
Acε 0
π (2.2 ×10−3 m)2 (3.00 × 108 m/s)(8.85 × 10−12 C2 /N ⋅ m 2 )
Erms =
= 1.9 × 109 V/m
31.
We assume the light energy is all absorbed, so use Eq. 22–10a. The intensity is the power radiated over
a spherical surface area.
25 W
I 4π (9.5 × 10−2 m) 2
= 7.348 × 10−7 N/m 2 ≈ 7.3 × 10−7 N/m 2
Pressure = P = =
c
(3.00 × 108 m/s)
The force on the fingertip is the pressure times the area of the fingertip. We approximate the area of a
fingertip to be 1.0 cm 2 .
F = PA = (7.348 × 10−7 N/m 2 ) (1.0 × 10−4 m 2 ) = 7.3 × 10−11 N
32.
The intensity from a point source is inversely proportional to the distance from the source.
2
rSun-
I Earth
(7.78 × 1011 m) 2
Jupiter
= 2
=
= 27.0
I Jupiter
(1.496 × 1011 m) 2
rSunEarth
So it would take an area of 27 m 2 at Jupiter to collect the same radiation as a 1.0-m 2 solar panel at
the Earth.
33.
We convert the horsepower rating to watts and then use the intensity to relate the wattage to the needed
area.
I=
34.
P
A
→
A=
P
=
I
100 hp ×
746 W
1 hp
200 W/m
2
= 373 m 2 ≈ 400 m 2
Use Eq. 22–4. Note that the higher frequencies have the shorter wavelengths.
(a)
For FM radio we have the following.
λ=
c (3.00 × 108 m/s)
c (3.00 × 108 m/s)
=
=
2.78
m
to
=
=
= 3.41 m
λ
f
f
(1.08 × 108 Hz)
(8.8 × 107 Hz)
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electromagnetic Waves
(b)
For AM radio we have the following.
c (3.00 × 108 m/s)
c (3.00 × 108 m/s)
λ
=
=
180
m
to
=
=
= 561 m
f
f
(1.7 × 106 Hz)
(5.35 ×105 Hz)
λ=
35.
Use Eq. 22–4.
c (3.00 × 108 m/s)
=
= 0.16 m
f
(1.9 × 109 Hz)
λ=
36.
22-9
The frequencies are 980 kHz on the AM dial and 98.1 MHz on the FM dial. From Eq. 22–4, c = f λ ,
we see that the lower frequency will have the longer wavelength: the AM station.
When we form the ratio of wavelengths, we get
λ2 f1 (98.1× 106 Hz)
=
=
= 100
λ1 f 2 (980 × 103 Hz)
The factor is good to 2 significant figures.
37.
Each wavelength is found by dividing the speed of light by the frequency, as in Eq. 22–4.
λ2 =
Channel 2:
Channel 51: λ51 =
c (3.00 × 108 m/s)
=
= 5.56 m
f 2 (54.0 × 106 Hz)
(3.00 × 108 m/s)
c
=
= 0.434 m
f51 (692 × 106 Hz)
38.
The resonant frequency of an LC circuit is given by Eq. 21–19. We assume that the inductance is
constant and form the ratio of the two frequencies.
1
2
2
2π LC1
⎛ f1 ⎞
⎛ 550 kHz ⎞
f1
C2
→ C2 = ⎜
=
=
⎟ C1 = ⎜
⎟ (2500 pF) = 290 pF
1
f2
C1
⎝ 1610 kHz ⎠
⎝ f2 ⎠
2π LC 2
39.
The resonant frequency of an LC circuit is given by Eq. 21–19. Solve for the capacitance.
f =
40.
1
2π LC
→
C=
1
2
2
4π f F
=
1
4π (98.3 × 10 Hz) 2 (1.8 × 10−6 H)
2
6
= 1.5 × 10−12 F
The resonant frequency of an LC circuit is given by Eq. 21–19. Solve for the inductance.
f =
1
2π LC
L1 =
L2 =
1
4π 2 f12C
1
4π
2
f 22 C
→
=
=
L=
1
2
4π f 2C
1
4π 2 (88 × 106 Hz) 2 (810 × 10−12 F)
1
2
6
2
4π (108 × 10 Hz) (810 × 10
−12
F)
= 4.0 × 10−9 H
= 2.7 × 10−9 H
The range of inductances is 2.7 × 10−9 H ≤ L ≤ 4.0 × 10−9 H .
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22-10
41.
Chapter 22
(a)
The minimum value of C corresponds with the higher frequency, according to Eq. 21–19.
f =
L=
(b)
2
2
4π f C
→
=
1
6
1
2π LC
→
C=
1
2
2
4π f F
=
1
4π (14.0 ×10 Hz) 2 (1.31×10−6 H)
2
P 1
2
= ε 0 cE02 = 12 ε 0 c( 2 Erms ) 2 = ε 0 cErms
A 2
Erms =
6
= 9.9 × 10−11 F
→
P
1.3 × 104 W
=
= 1.665 V/m ≈ 1.7 V/m
Acε 0
π (750 m) 2 (3.00 × 108 m/s) (8.85 × 10−12 C2 /N ⋅ m 2 )
The electric field is found from the desired voltage and the length of the antenna. Then use that electric
field to calculate the intensity.
Erms =
I=
Vrms 1.00 × 10−3 V
=
= 6.25 × 10−4 V/m
d
1.60 m
V2
P 1
2
= 2 ε 0 cE02 = 12 ε 0 c( 2 Erms ) 2 = ε 0 cErms
= ε 0 c rms
A
d2
= (3.00 × 108 m/s)(8.85 × 10−12 C2 /N ⋅ m 2 )
44.
= 1.31× 10−6 H ≈ 1.3 × 10−6 H
4π (15.0 × 10 Hz) 2 (86 × 10−12 F)
2
The rms electric field strength of the beam can be found from the average intensity, which is the power
per unit area. Use Eq. 22–8.
I=
43.
2π LC
1
The maximum value of C corresponds with the lower frequency.
f =
42.
1
(1.00 × 10−3 V) 2
(1.60 m)
2
= 1.04 × 10−9 W/m 2
We ignore the time for the sound to travel to the microphone. Find the difference between the time for
sound to travel to the balcony and for a radio wave to travel 1200 km.
⎛d
Δt = tradio − tsound = ⎜ radio
⎝ c
⎞ ⎛ dsound ⎞ ⎛ 1.2 × 106 m ⎞ ⎛ 50.0 m ⎞
⎟−⎜
⎟=⎜
⎟ = −0.142 s,
⎟−⎜υ
⎠ ⎝ sound ⎠ ⎜⎝ 3.00 × 108 m/s ⎟⎠ ⎝ 343 m/s ⎠
so the person at the radio hears the voice 0.142 s sooner.
45.
The ratio of distance to time must be the speed of light.
9
Δx
Δx
3m
−8 ⎛ 10 ns ⎞
= c → Δt =
=
=
×
= 10 ns
1
10
s
⎜
⎜ 1 s ⎟⎟
Δt
c
3.0 × 108 m/s
⎝
⎠
46.
The length is found from the speed of light and the duration of the burst.
Δx = cΔt = (3.00 × 108 m/s)(10−8 s) = 3 m
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Electromagnetic Waves
47.
22-11
The time consists of the time for the radio signal to travel to Earth and the time for the sound to travel
from the loudspeaker. We use 343 m/s for the speed of sound.
⎞ ⎛ 3.84 × 108 m ⎞ ⎛ 28 m ⎞
⎛d
⎞ ⎛d
t = tradio + tsound = ⎜ radio ⎟ + ⎜ sound ⎟ = ⎜
⎟+⎜
⎟ = 1.36 s
⎝ c ⎠ ⎝ υsound ⎠ ⎜⎝ 3.00 × 108 m/s ⎟⎠ ⎝ 343 m/s ⎠
Note that about 5% of the time is for the sound wave.
The microphone was inside the helmet. The Moon has essentially no atmosphere, so there would not
be sound waves in free space on the Moon. The helmet is filled with breathable air, so the microphone
has to be inside the helmet so that the sound waves will travel in the air around the astronaut’s head,
inside the helmet.
48.
The time travel delay is the distance divided by the speed of radio waves (which is the speed of light).
t=
49.
d
3 × 106 m
=
= 0.01 s
c 3.00 × 108 m/s
After the change occurred, we would find out when the change in radiation reached the Earth, traveling
at the speed of light.
Δt =
50.
(a)
d
(1.50 × 1014 m)
=
= 5.00 × 102 s = 8.33 min
c (3.00 × 108 m/s)
The instantaneous energy density is given by Eq. 22–6a. We convert it to average values and use
the rms value of the associated electric field.
u = ε0E2
u
Erms =
(b)
→
ε0
=
2
u = 12 ε 0 E 2 = 12 ε 0 ( 2 Erms ) 2 = ε 0 Erms
4 × 10−14 J/m3
→
= 0.0672 V/m ≈ 0.07 V/m
8.85 ×10−12 C2 /N ⋅ m 2
The distance to receive a comparable value can be found using the average intensity.
I=
P
2
4π r
1
r=
Erms
2
= 12 ε 0 cE02 = 12 ε 0 c( 2 Erms ) 2 = ε 0 cErms
→
P
1
7500 W
=
4πε 0 c 0.0672 V/m 4π (8.85 × 10−12 C2 /N ⋅ m 2 )(3.00 × 108 m/s)
= 7055 m ≈ 7 km
51.
The light has the same intensity in all directions, so use a spherical geometry centered on the source to
find the value of the intensity, Then use Eq. 22–8 to find the magnitude of the electric field and Eq.
22–2 with υ = c to find the magnitude of the magnetic field.
I=
P
P
=
= 1 cε 0 E02
A 4π r 2 2
E0 =
B0 =
P
2
2π r cε 0
=
→
(18 W)
2π (2.50 m) (3.00 × 10 m/s)(8.85 × 10−12 C 2 /N ⋅ m 2 )
2
8
= 13.14 V/m ≈ 13 V/m
E0
(13.14 V/m)
=
= 4.4 × 10−8 T
c
(3.00 × 108 m/s)
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22-12
52.
Chapter 22
The radiation from the Sun has the same intensity in all directions, so the rate at which energy passes
through a sphere centered at the Sun is P = I (4π R 2 ). This rate must be the same at any distance from
the Sun. Use this fact to calculate the magnitude of the intensity at Mars, and then use the intensity
vector to calculate the rms magnitude of the electric field at Mars.
2
2
I Mars (4π RMars
) = I Earth (4π REarth
) →
⎛ R2
= I Earth ⎜ Earth
⎜ R2
⎝ Mars
I Mars
I Earth ⎛ REarth ⎞
1350 W/m 2
⎛ 1 ⎞
⎜
⎟=
⎜
⎟ = 469 V/m
−
8
12
2
2
cε 0 ⎝ RMars ⎠
(3.00 × 10 m/s)(8.85 × 10
C /N ⋅ m ) ⎝ 1.52 ⎠
Erms =
Mars
53.
(a)
2
⎞ 1
⎛
⎞
2
2
= 12 ε 0 c ⎜ 2 Erms ⎟ = ε 0 cErms
⎟ = 2 ε 0 cE0
⎜
⎟
⎟
Mars
Mars
Mars
⎝
⎠
⎠
Since intensity is energy per unit time per unit area, the energy is found by multiplying the
intensity by the area of the antenna and the elapsed time.
2
⎛ 0.33 m ⎞
−10
J
ΔU = IAΔt = (1.0 × 10−13 W/m 2 )π ⎜
⎟ (4.0 h)(3600 s/h) = 1.2 × 10
⎝ 2 ⎠
(b)
The electric field amplitude can be found from the intensity, using Eq. 22–8. The magnitude of
the magnetic field is then found from Eq. 22–2 with υ = c.
I = 12 ε 0 cE02
→
2I
2(1.0 × 10−13 W/m 2 )
=
= 8.679 × 10−6 V/m
ε 0c
(8.85 × 10−12 C2 /N ⋅ m 2 )(3.00 × 108 m/s)
E0 =
≈ 8.7 × 10−6 V/m
B0 =
E0 8.679 × 10−6 V/m
=
= 2.9 × 10−14 T
c
3.00 × 108 m/s
54.
According to Fig. 22–8, satellite TV waves have a wavelength of approximately 0.03 m. Example 22–2
says that the wavelength is 4 times the antenna’s length. So the antenna might be 0.075 m in length.
55.
Use the relationship between average intensity (the magnitude of the Poynting vector) and electric
field strength, as given by Eq. 22–8. Also use the fact that intensity is power per unit area. We assume
that the power is spherically symmetric about the source.
I = 12 ε 0 cE0 2 =
r=
P
2πε 0 cE0
2
P
P
=
A 4π r 2
=
→
25, 000 W
2π (8.85 × 10
−12
2
C /N ⋅ m 2 ) (3.00 × 108 m/s)(0.020 V/m) 2
= 61, 200 m
≈ 61 km
Thus, to receive the transmission one should be within 61 km of the station.
56.
(a)
The radiation pressure when there is total reflection is given in Eq. 22–10b.
P=
2 I avg
c
=
2(1350 W/m 2 )
8
3.00 × 10 m/s
= 9.00 × 10−6 N/m 2
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Electromagnetic Waves
(b)
The force on the sail is the pressure times the area; then use Newton’s second law to calculate
the acceleration.
P=
(c)
22-13
F ma
=
A
A
a=
→
PA
P
9.00 × 10−6 N/m 2
=
=
= 9 × 10−3 m/s 2
−3
2
m m /A
1× 10 kg/m
This answer is not an accurate description of the full situation. There is also a gravitational
attraction due to the Sun’s mass, and it might be big enough to change the acceleration
calculated here.
The required force is the total mass (the payload plus the sail) and is provided by the radiation
pressure times the area of the sail.
P=
A=
F ma
=
A
A
A=
→
2
ma (mpayload + msail )a (mpayload + (0.001 kg/m ) A)a
=
=
P
P
P
mpayload a
2
[ P − (0.001 kg/m )a]
=
→
(100 kg)(1× 10−3 m/s 2 )
[(9.00 × 10
−6
N/m 2 ) − (0.001 kg/m 2 )(1× 10−3 m/s 2 )]
= 1.25 × 104 m 2 ≈ 1× 104 m 2
Again, the Sun’s gravitational attraction should also be figured into this before we claim to have
a full understanding of the situation.
57.
(a)
The radio waves have the same intensity in all directions. The power crossing a given area is the
intensity times the area. The intensity is the total power through the area of a sphere centered at
the source.
⎛ P
P = IA = ⎜ 0
⎝ Atotal
(b)
⎞
35, 000 W
(1.0 m 2 ) = 2.785 × 10−3 W ≈ 2.8 mW
⎟A=
3
2
4π (1.0 × 10 m)
⎠
We find the rms value of the electric field from the intensity, using Eq. 22–8.
I=
P
4π r 2
Erms =
2
= 12 ε 0 cE02 = 12 ε 0 c( 2 Erms ) 2 = ε 0 cErms
P
2
4π r cε 0
=
35, 000 W
4π (1.0 × 10 m) (3.00 × 108 m/s) (8.85 × 10−12 C2 /N ⋅ m 2 )
3
2
= 1.024 V/m ≈ 1.0 V/m
(c)
The voltage over the length of the antenna is the electric field times the length of the antenna.
Vrms = Erms d = (1.024 V/m) (1.0 m) = 1.0 V
(d)
We calculate the electric field at the new distance, and then calculate the voltage.
Erms =
P
2
4π r cε 0
=
35, 000 W
4π (5.0 × 10 m) (3.00 × 108 m/s) (8.85 × 10−12 C2 /N ⋅ m 2 )
4
2
= 2.048 × 10−2 V/m; Vrms = Erms d = (2.048 × 10−2 V/m) (1.0 m) = 2.0 × 10−2 V
58.
The light has the same intensity in all directions. Use a spherical geometry centered at the source with
the definition of the average intensity, Eq. 22–8. We also use Eq. 22–3.
I=
⎛ 1
P0
P
= 0 2 = 12 cε 0 E0 2 = 12 c ⎜ 2
⎜c μ
A 4π r
0
⎝
⎞
⎟ E0
⎟
⎠
→
⎛
⎞
P
⎟ E0 = 0 2
⎟
4π r
⎝ c μ0 ⎠
1
1c
⎜
2 ⎜ 2
→
E0 =
μ0 cP0
2π r 2
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22-14
59.
Chapter 22
The power output of the antenna would be the intensity at a given distance from the antenna times the
area of a sphere surrounding the antenna. The intensity is given by Eq. 22–8.
I=
P 1
= cε 0 E02
A 2
P = 4π r 2 I = 2π r 2 cε 0 E02 = 2π (0.65 m) 2 (3.00 × 108 m/s) (8.85 × 10−12 C2 /N ⋅ m 2 )(3 × 106 V/m) 2
≈ 6 × 1010 W
This is many orders of magnitude higher than the power output of commercial radio stations, which
are no higher than tens of kilowatts.
60.
The wavelength of the AM radio signal is given by Eq. 22–4, λ =
(a)
1λ
2
= 12 (300 m) = 150 m
(b)
1λ
4
= 14 (300 m) = 75 m
c (3.0 × 108 m/s)
=
= 300 m.
f
(106 Hz)
So there must be another way to pick up this signal—we don’t see antennas this long for AM radio.
61.
We find the average intensity at the 12-km distance from Eq. 22–8. Then the power can be found from
the intensity, assuming that the signal is spherically symmetric, using the surface area of a sphere of
radius 12 km.
I = 12 ε 0 cE02 = 12 (8.85 × 10−12 C2 /N ⋅ m 2 ) (3 × 108 m/s) (0.12 V/m) 2 = 1.9116 × 10−5 W/m 2
P = I A = (1.9116 × 10−5 W/m 2 )4π (12 × 103 m) 2 = 35 kW
Solutions to Search and Learn Problems
1.
In each case, the required area is the power requirement of the device divided by 20% of the intensity
of the sunlight.
(a)
A=
P 50 × 10−3 W
=
= 2.5 × 10−4 m 2 ≈ 3 cm 2
I
200 W/m 2
(b)
A=
P
1500 W
=
= 7.5 m 2 ≈ 8 m 2 (to 1 significant figure)
I 200 W/m 2
(c)
A=
P 40 hp(746 W/hp)
=
= 149 m 2 ≈ 100 m 2
I
200 W/m 2
(d)
Calculator: A typical graphing calculator is about 18 cm × 8 cm, which is about 140 cm 2 , so
yes, the solar panel can be mounted directly on the calculator. Hair dryer: A house of floor area
1000 ft 2 would have a roof area on the order of 100 m 2 , so yes, a solar panel on the roof
should be able to power the hair dryer. Car: This would require a square panel of side length
about 12 m, which is much larger than the car. So no, this panel could not be mounted on a car
and used for real-time power.
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Electromagnetic Waves
2.
(a)
2I
, where the intensity is the power per unit
c
area delivered to the suit. The force is the pressure times the area.
From Eq. 22–10b, the radiation pressure is P =
Flaser = PA =
(b)
22-15
2
IA =
c
2 ⎛ 3.0 W ⎞
2(3.0 W)
(π r 2 ) =
= 2.0 × 10−8 N
8
c ⎜⎝ π r 2 ⎟⎠
3.00 × 10 m/s
Use Newton’s law of universal gravitation, Eq. 6–1, to estimate the gravitational force. We take
the 30-m distance as having 2 significant figures.
Fgrav = G
mshuttle mastronaut
r2
= (6.67 × 10−11 N ⋅ m 2 /kg 2 )
(1.03 ×105 kg) (120 kg)
(30 m) 2
= 9.160 × 10−7 N ≈ 9.2 × 10−7 N
3.
(c)
The gravity force is larger, by a factor of approximately 50.
(a)
The intensity from a point source is inversely proportional to the distance from the source.
2
rStarISun
Earth
= 2
IStar rSun-
→
rStar- = rSunEarth
Earth
ISun
1350 W/m 2
= (1.496 × 1011 m)
IStar
1× 10−23 W/m 2
Earth
⎛
⎞
1 ly
⎜⎜
⎟⎟
15
⎝ 9.46 × 10 m ⎠
= 1.84 ×108 ly ≈ 2 × 108 ly
(b)
Compare this distance to the galactic size.
rStar-
Earth
galactic size
=
1.84 × 108 ly
1× 105 ly
= 1840 ≈ 2000
The distance to the star is about 2000 times the size of our galaxy.
4.
(a)
Since the cylinder is absorbing the radiation, the pressure (P) is given by Eq. 22–10a, where the
intensity is the power ( Pavg ) divided by the cross-sectional area. The force is the pressure
multiplied by the cross-sectional area of the cylinder, or the power absorbed divided by the speed
of light.
P=
I Pavg
1.0 W
=
=
= 4244 W/m 2 ≈ 4 × 103 W/m 2
c
cA (3.00 × 108 m/s)π (5 × 10−7 m) 2
F = PA =
(b)
Pavg
cA
A=
Pavg
c
=
1.0 W
8
3.00 × 10 m/s
= 3.3 × 10−9 N
The acceleration of the cylindrical particle will be the force on it (due to radiation pressure)
divided by its mass. The mass of the particle is its volume times the density of water.
a=
F
F
3.3 × 10−9 N
=
=
= 8.49 × 106 m/s 2 ≈ 8 × 106 m/s 2
m ρH Oπ r 3 (1000 kg/m 2 )π (5 × 10−7 m)3
2
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