Chap 30 SM

Prévia do material em texto

839 
Chapter 30 
Maxwell’s Equations and Electromagnetic Waves 
 
Conceptual Problems 
 
1 • [SSM] True or false: 
 
(a) The displacement current has different units than the conduction current. 
(b) Displacement current only exists if the electric field in the region is 
changing with time. 
(c) In an oscillating LC circuit, no displacement current exists between the 
capacitor plates when the capacitor is momentarily fully charged. 
(d) In an oscillating LC circuit, no displacement current exists between the 
capacitor plates when the capacitor is momentarily uncharged. 
 
(a) False. Like those of conduction current, the units of displacement current are 
C/s. 
 
(b) True. Because displacement current is given by dtdI e0d φ∈= , Id is zero if 
0e =dtdφ . 
 
(c) True. When the capacitor is fully charged, the electric flux is momentarily a 
maximum (its rate of change is zero) and, consequently, the displacement current 
between the plates of the capacitor is zero. 
 
(d) False. Id is zero if 0e =dtdφ . At the moment when the capacitor is 
momentarily uncharged, dE/dt ≠ 0 and so 0e ≠dtdφ . 
 
2 • Using SI units, show that dtdI e0d φ∈= has units of current. 
 
Determine the Concept We need to show that dtd e0 φ∈ has units of amperes. 
We can accomplish this by substituting the SI units of 0∈ and dtd eφ and 
simplifying the resulting expression. 
 
A
s
C
s
m
C
N
mN
C
2
2
==
⋅
⋅⋅ 
 
3 • [SSM] True or false: 
 
(a) Maxwell’s equations apply only to electric and magnetic fields that are 
constant over time. 
(b) The electromagnetic wave equation can be derived from Maxwell’s 
equations. 
 Chapter 30 
 
 
840 
(c) Electromagnetic waves are transverse waves. 
(d) The electric and magnetic fields of an electromagnetic wave in free space 
are in phase. 
 
(a) False. Maxwell’s equations apply to both time-independent and time-
dependent fields. 
 
(b) True. One can use Faraday’s law and the modified version of Ampere’s law to 
derive the wave equation. 
 
(c) True. Both the electric and magnetic fields of an electromagnetic wave 
oscillate at right angles to the direction of propagation of the wave. 
 
(d) True. 
 
4 • Theorists have speculated about the existence of magnetic monopoles, 
and several experimental searches for such monopoles have occurred. Suppose 
magnetic monopoles were found and that the magnetic field at a distance r from a 
monopole of strength qm is given by B = (μ0/4π)qm/r2. Modify the Gauss’s law for 
magnetism equation to be consistent with such a discovery. 
 
Determine the Concept Gauss’s law for magnetism would become 
inside m,0S n
qdAB μ=∫ where qm, inside is the total magnetic charge inside the 
Gaussian surface. Note that Gauss’s law for electricity follows from the existence 
of electric monopoles (charges), and the electric field due to a point charge 
follows from the inverse-square nature of Coulomb’s law. 
 
5 • (a) For each of the following pairs of electromagnetic waves, which 
has the higher frequency: (1) visible light or X rays, (2) green light or red light, 
(3) infrared waves or red light. (b) For each of the following pairs of 
electromagnetic waves, which has the longer wavelength: (1) visible light or 
microwaves, (2) green light or ultraviolet light, (3) gamma rays or ultraviolet 
light. 
 
Determine the Concept Refer to Table 30-1 to rank order the frequencies and 
wavelengths of the given electromagnetic radiation. 
 
(a) (1) X rays (2) green light (3) red light 
 
(b) (1) microwaves (2) green light (3) ultraviolet light 
 
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
841
6 • The detection of radio waves can be accomplished with either an 
electric dipole antenna or a loop antenna. True or false: 
 
(a) The electric dipole antenna works according to Faraday’s law. 
(b) If a linearly polarized radio wave is approaching you head on such that its 
electric field oscillates vertically, to best detect this wave the normal to a 
loop antenna’s plane should be oriented so that it points either right or left. 
(c) If a linearly polarized radio wave is approaching you such that its electric 
field oscillates in a horizontal plane, to best detect this wave using an dipole 
antenna the antenna should be oriented vertically. 
 
(a) False. A dipole antenna is oriented parallel to the electric field of an incoming 
wave so that the wave can induce an alternating current in the antenna. 
 
(b) True. A loop antenna is oriented perpendicular to the magnetic field of an 
incoming wave so that the changing magnetic flux through the loop can induce a 
current in the loop. Orienting the loop antenna’s plane so that it points either right 
or left satisfies this condition. 
 
(c) False. The dipole antenna needs to be oriented parallel to the electric field of 
an incoming wave so that the wave can induce an alternating current in the 
antenna. 
 
7 • A transmitter emits electromagnetic waves using an electric dipole 
antenna oriented vertically. (a) A receiver to detect these waves also uses an 
electric dipole antenna that is one mile from the transmitting antenna and at the 
same altitude. How should the receiver’s electric dipole antenna be oriented for 
optimum signal reception? (b) A receiver to detect these waves uses a loop 
antenna that is one mile from the transmitting antenna and at the same altitude. 
How should the loop antenna be oriented for optimum signal reception? 
 
Determine the Concept 
(a) The electric dipole antenna should be oriented vertically. 
 
(b) The loop antenna and the electric dipole transmitting antenna should be in the 
same vertical plane. 
 
8 • Show that the expression 0μBE
rr × for the Poynting vector Sr 
(Equation 30-21) has units of watts per square meter (the SI units for 
electromagnetic wave intensity). 
 
Determine the Concept We can that 0μBE
rr × has units of W/m2 by substituting 
the SI units of E
r
, B
r
 and 0μ and simplifying the resulting expression. 
 Chapter 30 
 
 
842 
2222 m
W
m
s
J
m
s
mN
m
s
C
C
mN
m
A
C
N
A
m
C
N
A
mT
T
C
N
==
⋅
=⋅⋅=⋅==⋅
⋅
 
 
9 • [SSM] If a red light beam, a green light beam, and a violet light 
beam, all traveling in empty space, have the same intensity, which light beam 
carries more momentum? (a) the red light beam, (b) the green light beam, (c) the 
violet light beam, (d) They all have the same momentum. (e) You cannot 
determine which beam carries the most momentum from the data given. 
 
Determine the Concept The momentum of an electromagnetic wave is directly 
proportional to its energy ( cUp = ). Because the intensity of a wave is its energy 
per unit area and per unit time (the average value of its Poynting vector), waves 
with equal intensity have equal energy and equal momentum. ( )d is correct. 
 
10 • If a red light plane wave, a green light plane wave, and a violet light 
plane wave, all traveling in empty space, have the same intensity, which wave has 
the largest peak electric field? (a) the red light wave, (b) the green light wave, 
(c) the violet light wave, (d) They all have the same peak electric field. (e) You 
cannot determine the largest peak electric field from the data given. 
 
Determine the Concept The intensity of an electromagnetic wave is given by 
0
00
av 2μ
BEI == Sr . 
 
The intensity of an electromagnetic 
wave is given by: 
0
00
av 2μ
BEI == Sr 
 
Because E0 = cB0: 
0
2
0
av 2 μc
E=Sr 
 
This result tells us that 20av E∝S
r
independently of the wavelength of the 
electromagnetic radiation. Thus ( )d is correct. 
 
11 • Two sinusoidal plane electromagnetic waves are identical except that 
wave A has a peak electric field that is three times thepeak electric field of wave 
B. How do their intensities compare? (a) IA = 13 IB (b) IA = 19 IB (c) IA = 3IB 
(d) IA = 9IB (e) You cannot determine how their intensities compare from the data 
given. 
 
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
843
Determine the Concept The intensity of an electromagnetic wave is given by 
0
00
av 2μ
BEI == Sr . 
 
Express the intensities of the two 
waves: 
0
A,0A,0
A 2μ
BE
I = and
0
B,0B,0
B 2μ
BE
I = 
 
Dividing the first of these equations 
by the second and simplifying yields: 
 
B,0B,0
A,0A,0
0
B,0B,0
0
A,0A,0
B
A
2
2
BE
BE
BE
BE
I
I ==
μ
μ
 
 
Because wave A has a peak electric 
field that is three times that of wave 
B, the peak magnetic field of A is 
also three times that of wave B. 
Hence: 
( )( )
9
33
B,0B,0
B,0B,0
B
A ==
BE
BE
I
I ⇒ BA 9II = 
( )d is correct. 
 
Estimation and Approximation 
 
12 •• In laser cooling and trapping, the forces associated with radiation 
pressure are used to slow down atoms from thermal speeds of hundreds of meters 
per second at room temperature to speeds of a few meters per second or slower. 
An isolated atom will absorb only radiation of specific frequencies. If the 
frequency of the laser-beam radiation is tuned so that the target atoms will absorb 
the radiation, then the radiation is absorbed during a process called resonant 
absorption. The cross-sectional area of the atom for resonant absorption is 
approximately equal to λ2, where λ is the wavelength of the laser light. 
(a) Estimate the acceleration of a rubidium atom (molar mass 85 g/mol) in a laser 
beam whose wavelength is 780 nm and intensity is 10 W/m2. (b) About how long 
would it take such a light beam to slow a rubidium atom in a gas at room 
temperature (300 K) to near-zero speed? 
 
Picture the Problem We can use Newton’s second law to express the 
acceleration of an atom in terms of the net force acting on the atom and the 
relationship between radiation pressure and the intensity of the beam to find the 
net force. Once we know the acceleration of an atom, we can use the definition of 
acceleration to find the stopping time for a rubidium atom at room temperature. 
 
(a) Apply Newton’s second law to 
the atom to obtain: 
 
maF =r (1) 
where Fr is the radiation force exerted 
by the laser beam. 
 
 Chapter 30 
 
 
844 
The radiation pressure Pr and 
intensity of the beam I are related 
according to: 
 
c
I
A
FP == rr 
Solve for Fr to obtain: 
 c
I
c
IAF
2
r
λ== 
 
Substitute for Fr in equation (1) to 
obtain: mac
I =
2λ ⇒
mc
Ia
2λ= 
 
Substitute numerical values and evaluate a: 
 ( )( )
( )
25
25
8
23
22
m/s104.1
m/s1044.1
m/s10998.2
particles106.022
mol1
mol
g85
nm780W/m10
×=
×=
×⎟⎟⎠
⎞
⎜⎜⎝
⎛
××
=a
 
 
(b) Using the definition of 
acceleration, express the stopping 
time Δt of the atom: 
 
a
vvt initialfinal −=Δ 
Because finalv ≈ 0: 
 a
v
t initial
−≈Δ 
 
Using the rms speed as the initial 
speed of an atom, relate initialv to the 
temperature of the gas: 
 
m
kTvv 3rmsinitial == 
Substitute in the expression for the 
stopping time to obtain: m
kT
a
t 31−=Δ 
 
Substitute numerical values and evaluate Δt: 
 ( )( ) ms1.2
particles106.022
mol 1
mol
g85
K300J/K1038.13
m/s1044.1
1
Δ
23
23
25 =
⎟⎟⎠
⎞
⎜⎜⎝
⎛
××
×
×−−=
−
t 
 
13 •• [SSM] One of the first successful satellites launched by the United 
States in the 1950s was essentially a large spherical (aluminized) Mylar balloon 
from which radio signals were reflected. After several orbits around Earth, 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
845
scientists noticed that the orbit itself was changing with time. They eventually 
determined that radiation pressure from the sunlight was causing the orbit of this 
object to change—a phenomenon not taken into account in planning the mission. 
Estimate the ratio of the radiation-pressure force by the sunlight on the satellite to 
the gravitational force by Earth’s gravity on the satellite. 
 
Picture the Problem We can use the definition of pressure to express the 
radiation force on the balloon. We’ll assume that the gravitational force on the 
balloon is approximately its weight at the surface of Earth, that the density of 
Mylar is approximately that of water and that the area receiving the radiation from 
the sunlight is the cross-sectional area of the balloon. 
 
The radiation force acting on the 
balloon is given by: 
 
APF rr = 
where A is the cross-sectional area of 
the balloon. 
 
Because the radiation from the Sun is 
reflected, the radiation pressure is 
twice what it would be if it were 
absorbed: 
 
c
IP 2r = 
Substituting for Pr and A yields: 
 
( )
c
Id
c
dIF
2
2 2241
r
ππ == 
 
The gravitational force acting on the 
balloon when it is in a near-Earth 
orbit is approximately its weight at 
the surface of Earth: 
 
gtA
gVgmwF
ballon surface,Mylar
MylarMylarballoonballoong
ρ
ρ
=
===
 
where t is the thickness of the Mylar 
skin of the balloon. 
 
Because the surface area of the 
balloon is 224 dr ππ = : 
 
gtdF 2Mylarg πρ= 
Express the ratio of the radiation-
pressure force to the gravitational 
force and simplify to obtain: gct
I
gtd
c
Id
F
F
Mylar
2
Mylar
2
g
r
2
2
ρπρ
π
== 
 
 Chapter 30 
 
 
846 
 
Assuming the thickness of the Mylar skin of the balloon to be 1 mm, substitute 
numerical values and evaluate Fr/Fg: 
 
( )
7
8
23
3
2
g
r 102
s
m10998.2mm 1
s
m81.9
m
kg1000.12
m
kW35.1
−×≈
⎟⎠
⎞⎜⎝
⎛ ×⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛ ×
=
F
F 
 
14 •• Some science fiction writers have described solar sails that could 
propel interstellar spaceships. Imagine a giant sail on a spacecraft subjected to 
radiation pressure from our Sun. (a) Explain why this arrangement works better if 
the sail is highly reflective rather than highly absorptive. (b) If the sail is assumed 
highly reflective, show that the force exerted by the sunlight on the spacecraft is 
given by ( )crAP 2S 2π where PS is the power output of the Sun (3.8 × 1026 W), A 
is the surface area of the sail, m is the total mass of the spacecraft, r is the distance 
from the Sun, and c is the speed of light. (Assume the area of the sail is much 
larger than the area of the spacecraft so that all the force is due to radiation 
pressure on the sail only.) (c) Using a reasonable value for A, compare the force 
on the spacecraft due to the radiation pressure and the force on the spacecraft due 
to the gravitational pull of the Sun. Does the result imply that such a system will 
work? Explain your answer. 
 
Picture the Problem (b) We can use the definition of radiation pressure to show 
that the force exerted by the sunlight on the spacecraft is given by ( )crAP 2S 2π 
where PS is the power output of the Sun (3.8 × 1026 W), A is the surface area of 
the sail, m is the total mass of the spacecraft, r is the distance from the Sun, and c 
is the speed of light. 
 
(a) If the sail is highly reflective rather than highly absorptive, the radiation force 
is doubled. 
 
(b) Because the sail is highly 
reflective: 
 
c
IAAPF 2rr == 
where A is the area of the sail. 
 
The intensity of the solar radiation on 
the sail is given by 2
s
4 r
PI π= . 
Substituting for I yields: 
cr
AP
cr
APF 2
s
2
s
r 24
2
ππ == 
 
 Maxwell’s Equations and ElectromagneticWaves 
 
 
847
(c) Express the ratio of the force on 
the spacecraft due to the radiation 
pressure and the force on the 
spacecraft due the gravitational force 
of the Sun on the spacecraft: 
 
S
S
2
S
2
S
g
r
2
2
cGmM
AP
r
GmM
cr
AP
F
F
π
π == 
 
Assuming a 15-m diameter circular sail and a 500-kg spacecraft (values found 
using the internet), substitute numerical values and evaluate the ratio of the 
accelerations: 
 
( ) ( )
( )( )
4
30
2
2
118
226
g
104.5
kg 1099.1kg 500
kg
mN 10673.6
s
m 10998.22
m 15
4
 W108.3
−
−
×=
×⎟⎟⎠
⎞
⎜⎜⎝
⎛ ⋅×⎟⎠
⎞⎜⎝
⎛ ×
⎟⎠
⎞⎜⎝
⎛×
=
π
π
F
Fr
 
for A = 177 m2 and m = 500 kg. 
 
This scheme is not likely to work effectively. For any reasonable spacecraft mass, 
the surface mass density of the sail would to be extremely small (experimental 
sails have area densities of approximately 3 g/m2) and the sail would have to be 
huge. Additionally, unless struts are built into the sail, it would collapse during 
use. 
 
Maxwell’s Displacement Current 
 
15 • [SSM] A parallel-plate capacitor has circular plates and no dielectric 
between the plates. Each plate has a radius equal to 2.3 cm and the plates are 
separated by 1.1 mm. Charge is flowing onto the upper plate (and off of the lower 
plate) at a rate of 5.0 A. (a) Find the rate of change of the electric field strength in 
the region between the plates. (b) Compute the displacement current in the region 
between the plates and show that it equals 5.0 A. 
 
Picture the Problem We can differentiate the expression for the electric field 
between the plates of a parallel-plate capacitor to find the rate of change of the 
electric field strength and the definitions of the conduction current and electric 
flux to compute Id. 
 
(a) Express the electric field strength 
between the plates of the parallel-
plate capacitor: 
 
A
QE
0∈= 
 
 Chapter 30 
 
 
848 
Differentiate this expression with 
respect to time to obtain an 
expression for the rate of change of 
the electric field strength: 
 
A
I
dt
dQ
AA
Q
dt
d
dt
dE
000
1
∈∈∈ ==⎥⎦
⎤⎢⎣
⎡= 
Substitute numerical values and evaluate dE/dt: 
 
( ) ( )
sV/m104.3
sV/m1040.3
m023.0mN/C10854.8
A0.5
14
14
22212
⋅×=
⋅×=⋅×= − πdt
dE
 
 
(b) Express the displacement current 
Id: dt
dI e0d
φ∈= 
 
Substitute for the electric flux to 
obtain: 
 
[ ]
dt
dEAEA
dt
dI 00d ∈∈ == 
 
Substitute numerical values and evaluate Id: 
 ( ) ( ) ( ) A0.5sV/m1040.3m023.0mN/C10854.8 1422212d =⋅×⋅×= − πI 
 
16 • In a region of space, the electric field varies with time as 
E = (0.050 N/C) sin (ωt), where ω = 2000 rad/s. Find the peak displacement 
current through a surface that is perpendicular to the electric field and has an area 
equal to 1.00 m2. 
 
Picture the Problem We can express the displacement current in terms of the 
electric flux and differentiate the resulting expression to obtain Id in terms of 
dE/dt. 
 
The displacement current Id is 
given by: dt
dI e0d
φ∈= 
 
Substitute for the electric flux to 
obtain: 
[ ]
dt
dEAEA
dt
dI 00d ∈∈ == 
 
Because ( ) tE 2000sinN/C050.0= : 
 
( )[ ]
( ) ( ) tA
t
dt
dAI
2000cosN/C050.0s2000
2000sinN/C050.0
0
1-
0d
∈
∈
=
=
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
849
Id will have its maximum value 
when cos 2000t = 1. Hence: 
 
( ) ( )N/C050.0s2000 0-1maxd, AI ∈= 
Substitute numerical values and evaluate Id,max: 
 
( ) ( ) nA89.0
C
N050.0m00.1
mN
C10854.8s2000 22
2
121
max d, =⎟⎠
⎞⎜⎝
⎛⎟⎟⎠
⎞
⎜⎜⎝
⎛
⋅×=
−−I 
 
17 • For Problem 15, show that the magnetic field strength between the 
plates a distance r from the axis through the centers of both plates is given by 
B = (1.9 × 10–3 T/m)r. 
 
Picture the Problem We can use Ampere’s law to a circular path of radius r 
between the plates and parallel to their surfaces to obtain an expression relating B 
to the current enclosed by the amperian loop. Assuming that the displacement 
current is uniformly distributed between the plates, we can relate the displacement 
current enclosed by the circular loop to the conduction current I. 
 
Apply Ampere’s law to a circular 
path of radius r between the plates 
and parallel to their surfaces to 
obtain: 
 
IIrBd 0enclosed0C 2 μμπ ===⋅∫ lrrB 
Assuming that the displacement 
current is uniformly distributed: 
 
2
d
2 R
I
r
I
ππ = ⇒ d2
2
I
R
rI = 
where R is the radius of the circular 
plates. 
 
Substituting for I yields: 
d2
2
02 I
R
rrB μπ = ⇒ d202 IR
rB π
μ= 
 
Substitute numerical values and 
evaluate B: ( ) ( )( )( )
r
rrB
⎟⎠
⎞⎜⎝
⎛ ×=
×=
−
−
m
T109.1
m023.02
A0.5A/N104
3
2
27
π
π
 
 
18 •• The capacitors referred to in this problem have only empty space 
between the plates. (a) Show that a parallel-plate capacitor has a displacement 
current in the region between its plates that is given by Id = C dV/dt, where C is 
the capacitance and V is the potential difference between the plates. (b) A 5.00-nF 
 Chapter 30 
 
 
850 
parallel-plate capacitor is connected to an ideal ac generator so the potential 
difference between the plates is given by V = V0 cos ωt, where V0 = 3.00 V and 
ω = 500π rad/s. Find the displacement current in the region between the plates as 
a function of time. 
 
Picture the Problem We can use the definitions of the displacement current and 
electric flux, together with the expression for the capacitance of an air-core-
parallel-plate capacitor to show that Id = C dV/dt. 
 
(a) Use its definition to express the 
displacement current Id: dt
dI e0d
φ∈= 
 
Substitute for the electric flux to 
obtain: 
[ ]
dt
dEAEA
dt
dI 00d ∈∈ == 
 
Because E = V/d: 
dt
dV
d
A
d
V
dt
dAI 00d
∈∈ =⎥⎦
⎤⎢⎣
⎡= 
 
The capacitance of an air-core-
parallel-plate capacitor whose plates 
have area A and that are separated by 
a distance d is given by: 
 
d
AC 0∈= 
Substituting yields: 
dt
dVCI =d 
 
(b) Substitute in the expression derived in (a) to obtain: 
 
( ) ( )[ ] ( )( )( )
( ) t
tt
dt
dI
πμ
πππ
500sinA6.23
500sins500V00.3nF00.5500cosV00.3nF00.5 1d
−=
−== −
 
 
19 •• [SSM] There is a current of 10 A in a resistor that is connected in 
series with a parallel plate capacitor. The plates of the capacitor have an area of 
0.50 m2, and no dielectric exists between the plates. (a) What is the displacement 
current between the plates? (b) What is the rate of change of the electric field 
strength between the plates? (c) Find the value of the line integral ∫ ⋅C dB lrr , where 
the integration path C is a 10-cm-radius circle that lies in a plane that is parallel 
with the plates and is completely within the region between them. 
 
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
851
Picture the Problem We can use the conservation of charge to find Id, the 
definitions of the displacement current and electric flux to find dE/dt, and 
Ampere’s law to evaluate l
rr
d⋅B around the given path. 
 
(a) From conservation of charge we 
know that: 
 
A10d == II 
(b) Express the displacement current 
Id: 
[ ]
dt
dEAEA
dt
d
dt
dI 00e0d ∈∈φ∈ === 
 
Substituting for dE/dt yields: 
 A
I
dt
dE
0
d
∈= 
 
Substitute numerical values and 
evaluate dE/dt: 
 ( )
sm
V103.2
m50.0
mN
C1085.8
A10
12
2
2
2
12
⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⋅×
=
−dt
dE
 
 
(c) Apply Ampere’s law to a circular 
path of radius r between the plates 
and parallelto their surfaces to 
obtain: 
 
enclosed0C
Id μ=⋅∫ lrrB 
Assuming that the displacement 
current is uniformly distributed and 
letting A represent the area of the 
circular plates yields: 
 
A
I
r
I d
2
enclosed =π ⇒ d
2
enclosed IA
rI π= 
 
Substitute for enclosedI to obtain: 
d
2
0
C
I
A
rd πμ=⋅∫ lrrB 
 
Substitute numerical values and evaluate ∫ ⋅C lrr dB : 
 ( ) ( ) ( ) mT79.0
m50.0
A10m10.0A/N104
2
227
C
⋅=×=⋅
−∫ μππlrr dB 
 
20 ••• Demonstrate the validity of the generalized form of Ampère’s law 
(Equation 30-4) by showing that it gives the same result as the Biot–Savart law 
(Equation 27-3) in a specified situation. Figure 30-13 shows two momentarily 
 Chapter 30 
 
 
852 
equal but opposite point charges (+Q and –Q) on the x axis at x = –a and x = +a, 
respectively. At the same instant there is a current I in the wire connecting them, 
as shown. Point P is on the y axis at y = R. (a) Use the Biot–Savart law to show 
that the magnitude of the magnetic field at point P is given byB = μ0Ia
2πR
1
R 2 + a2 . 
(b) Now consider a circular strip of radius r and width dr in the x = 0 plane that 
has its center at the origin. Show that the flux of the electric field through this 
strip is given by ( ) drarrQdAEx 2322
0
+= ∈
π . (c) Use the result from Part (b) to 
show that the total electric flux φe through a circular surface S of radius R. is 
given by 
 
φe = Q∈o 1−
a
a2 + R 2
⎛
⎝⎜
⎞
⎠⎟ . (d) Find the displacement current Id through S, 
and show that 
 
I + Id = I a
a2 + R 2 (e) Finally, show that the generalized form of 
Ampere’s law (Equation 30-4) gives the same result for the magnitude of the 
magnetic field as found in Part (a). 
 
Picture the Problem We can follow the step-by-step instructions in the problem 
statement to show that Equation 30-4 gives the same result for B as that given in 
Part (a). 
 
(a) Express the magnetic field 
strength at P using the expression for 
B due to a straight wire segment: 
 
( )210 sinsin4 θθπ
μ +=
R
IBP 
where 
2221
sinsin
aR
a
+== θθ 
 
Substitute for sinθ1 and sinθ2 to 
obtain: 
22
0
22
0
1
2
2
4
aRR
Ia
aR
a
R
IBP
+=
+=
π
μ
π
μ
 
 
(b) Express the electric flux through 
the circular strip of radius r and 
width dr in the yz plane: 
 
( )rdrEdAEd xx πφ 2e == 
The electric field due to the dipole is: ( ) 2322122 2cos2 ar kQaar kQEx +=+= θ 
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
853
Substitute for Ex to obtain: ( ) ( )
( ) ( )
( ) rdrar Qa
rdr
ar
Qa
rdr
ar
kQadAEd x
2322
0
2322
0
2322e
2
4
2
22
+=
+=
+==
∈
π∈π
πφ
 
 
(c) Multiply both sides of the 
expression for dφe by ∈0: 
 
( ) rdrar Qad 2322e0 +=φ∈ 
Integrate r from 0 to R to obtain: 
 
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−=⎟
⎟
⎠
⎞
⎜⎜⎝
⎛ ++
−=+= ∫ 22220 2322e0 1
11
aR
aQ
aaR
Qa
ar
rdrQa
R
φ∈ 
 
(d) The displacement current is 
defined to be: 
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−==
22
22
22
e
0d
1
1
1
aR
aI
dt
dQ
aR
a
aR
aQ
dt
d
dt
dI φ∈
 
 
The total current is the sum of I and 
Id: 
22
22d
1
aR
aI
aR
aIIII
+=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+−−=+
 
 
(e) Apply Equation 30-4 (the 
generalized form of Ampere’s law) 
to obtain: 
 
( )∫ +==⋅C IIRBd d02 μπlrrB 
Solving for B yields: ( )d02 IIRB += π
μ 
 
 Chapter 30 
 
 
854 
Substitute for I + Id from (d) to 
obtain: 
22
0
22
0
1
2
2
aRR
Ia
aR
aI
R
B
+=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+=
π
μ
π
μ
 
 
Maxwell’s Equations and the Electromagnetic Spectrum 
 
21 • The color of the dominant light from the Sun is in the yellow-green 
region of the visible spectrum. Estimate the wavelength and frequency of the 
dominant light emitted by our Sun. HINT: See Table 30-1. 
 
Picture the Problem We can find both the wavelength and frequency of the 
dominant light emitted by our Sun in Table 30-1. 
 
Because the radiation from the Sun is 
yellow-green dominant, the dominant 
wavelength is approximately: 
 
nm 580green-yellow =λ 
The corresponding frequency is: 
Hz 1017.5
nm 580
m/s 10998.2
14
8
green-yellow
green-yellow
×=
×=
= λ
cf
 
 
22 • (a) What is the frequency of microwave radiation that has a 3.00-cm-
long wavelength? (b) Using Table 30-1, estimate the ratio of the shortest 
wavelength of green light to the shortest wavelength of red light. 
 
Picture the Problem We can use c = fλ to find the frequency corresponding to 
the given wavelength. 
 
(a) The frequency of an 
electromagnetic wave is the ratio of 
the speed of light in a vacuum to the 
wavelength of the wave: 
 
λ
cf = 
 
Substitute numerical values and 
evaluate f: 
GHz99.9
Hz10993.9
m1000.3
m/s10998.2 10
2
8
=
×=×
×= −f
 
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
855
(b) The ratio of the shortest 
wavelength green light to the shortest 
wavelength red light is: 
84.0
nm 620
nm 520
redshortest 
greenshortest =≈λ
λ
 
 
23 • (a) What is the frequency of an X ray that has a 0.100-nm-long 
wavelength? (b) The human eye is sensitive to light that has a wavelength equal to 
550 nm. What is the color and frequency of this light? Comment on how this 
answer compares to your answer for Problem 21. 
 
Picture the Problem We can use c = fλ to find the frequency corresponding to 
the given wavelengths and consult Table 30-1 to determine the color of light with 
a wavelength of 550 nm. 
 
(a) The frequency of an X ray with a 
wavelength of 0.100 nm is: 
 Hz1000.3
m10100.0
m/s10998.2
18
9
8
×=
×
×== −λ
cf
 
 
(b) The frequency of light with a 
wavelength of 550 nm is: Hz1045.5nm550
m/s10998.2 148 ×=×=f 
 
Consulting Table 30-1, we see that the color of light that has a wavelength of 
550 nm is yellow-green. This result is consistent with those of Problem 21 and is 
close to the wavelength of the peak output of the Sun. Because we see naturally 
by reflected sunlight, this result is not surprising. 
 
Electric Dipole Radiation 
 
24 •• Suppose a radiating electric dipole lies along the z axis. Let I1 be the 
intensity of the radiation at a distance of 10 m and at angle of 90º. Find the 
intensity (in terms of I1) at (a) a distance of 30 m and an angle of 90º, (b) a 
distance of 10 m and an angle of 45º, and (c) a distance of 20 m and an angle of 
30º. 
 
Picture the Problem We can use the intensity I1 at a distance r = 10 m and at an 
angle θ = 90° to find the constant in the expression for the intensity of radiation 
from an electric dipole and then use the resulting equation to find the intensity at 
the given distances and angles. 
 
Express the intensity of radiation as a 
function of r and θ : 
 
( ) θθ 22 sin, r
CrI = (1) 
where C is a constant. 
 
 Chapter 30 
 
 
856 
Express I(90°,10 m): 
 
( ) ( )
2
2
21
m100
90sin
m10
m10,90
C
CII
=
°==°
 
 
Solving for C yields: 
 
( ) 12m100 IC = 
 
Substitute in equation (1) to obtain: 
 ( ) ( ) θθ 22 1
2
sinm100,
r
IrI = (2) 
 
(a) Evaluate equation (2) for 
r = 30 m and θ = 90°: 
 
( ) ( )( )
19
1
2
2
1
2
90sin
m30
m100m30,90
I
II
=
°=°
 
 
(b) Evaluate equation (2) for 
r = 10 m and θ = 45°: 
 
( ) ( )( )
12
1
2
2
1
2
45sin
m10
m100m10,45
I
II
=
°=°
 
 
(c) Evaluate equation (2) for 
r = 20 m and θ = 30°: 
 
( ) ( )()
116
1
2
2
1
2
30sin
m20
m100m20,30
I
II
=
°=°
 
 
25 •• (a) For the situation described in Problem 24, at what angle is the 
intensity at a distance of 5.0 m equal to I1? (b) At what distance is the intensity 
equal to I1 when θ = 45º? 
 
Picture the Problem We can use the intensity I1 at a distance r = 10 m and at an 
angle θ = 90° to find the constant in the expression for the intensity of radiation 
from an electric dipole and then use the resulting equation to find the angle for a 
given intensity and distance and the distance corresponding to a given intensity 
and angle. 
 
Express the intensity of radiation as a 
function of r and θ : 
 
( ) θθ 22 sin, r
CrI = (1) 
where C is a constant. 
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
857
Express I(90°,10 m): 
 
( ) ( )
2
2
21
m100
90sin
m10
m10,90
C
CII
=
°==°
 
 
Solving for C yields: 
 
( ) 12m100 IC = 
 
Substitute in equation (1) to obtain: 
 ( ) ( ) θθ 22 1
2
sinm100,
r
IrI = (2) 
 
(a) For r = 5 m and I(θ,r) = I1: ( )
( ) θ22 1
2
1 sinm0.5
m100 II = ⇒ 412sin =θ 
 
Solve for θ to obtain: 
 
( ) °== − 30sin 211θ 
 
(b) For θ = 45° and I(θ,r) = I1: ( ) °= 45sinm100 22 121 r II 
or ( )2212 m100=r 
 
Solve for r to obtain: ( ) m1.7m100 221 ==r 
 
26 •• You and your engineering crew are in charge of setting up a wireless 
telephone network for a village in a mountainous region. The transmitting antenna 
of one station is an electric dipole antenna located atop a mountain 2.00 km above 
sea level. There is a nearby mountain that is 4.00 km from the antenna and is also 
2.00 km above sea level. At that location, one member of the crew measures the 
intensity of the signal to be 4.00 × 10–12 W/m2. What should be the intensity of the 
signal at the village that is located at sea level and 1.50 km from the transmitter? 
 
Picture the Problem We can use the intensity I at a distance r = 4.00 km and at 
an angle θ = 90° to find the constant in the expression for the intensity of 
radiation from an electric dipole and then use the resulting equation to find the 
intensity at sea level and 1.50 km from the transmitter. 
 
Express the intensity of radiation as a 
function of r and θ : 
 
( ) θθ 22 sin, r
CrI = (1) 
where C is a constant. 
 
 Chapter 30 
 
 
858 
Use the given data to obtain: 
 ( )
( )2
2
2
212
km00.4
90sin
km00.4
W/m104
C
C
=
°=× −
 
 
Solving for C yields: 
 
( ) ( )
W1040.6
W/m1000.4km00.4
5
2122
−
−
×=
×=C
 
 
Substitute in equation (1) to obtain: 
 ( ) θθ 22
5
sinW1040.6,
r
rI
−×= (2) 
 
For a point at sea level and 1.50 km 
from the transmitter: 
 
°=⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 1.53
km1.50
km00.2tan 1θ
 
Evaluate I(53.1°,1.50 km): 
 
( ) ( ) 222
5
pW/m2.181.53sin
km50.1
W1040.6km5.1,1.53 =°×=°
−
I 
 
27 ••• [SSM] A radio station that uses a vertical electric dipole antenna 
broadcasts at a frequency of 1.20 MHz and has a total power output of 500 kW. 
Calculate the intensity of the signal at a horizontal distance of 120 km from the 
station. 
 
Picture the Problem The intensity of radiation from an electric dipole is given by 
C(sin2θ)/r2, where C is a constant whose units are those of power, r is the distance 
from the dipole to the point of interest, and θ is the angle between the antenna 
and the position vector .rr We can integrate the intensity to express the total power 
radiated by the antenna and use this result to evaluate C. Knowing C we can find 
the intensity at a horizontal distance of 120 km. 
 
Express the intensity of the signal as 
a function of r and θ : ( ) 2
2sin,
r
CrI θθ = 
 
At a horizontal distance of 120 km 
from the station: 
 
( ) ( )
( )2
2
2
km120
km120
90sin90,km120
C
CI
=
°=°
 (1) 
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
859
 
From the definition of intensity we 
have: 
 
IdAdP = 
and 
( )dArIP ∫∫= θ,tot 
where, in polar coordinates, 
φθθ ddrdA sin2= 
 
Substitute for dA to obtain: 
 ( ) φθθθ
π π
ddrrIP sin, 2
2
0 0
tot ∫ ∫= 
 
Substitute for I(r,θ): φθθ
ππ
ddCP ∫ ∫= 2
0 0
3
tot sin 
 
From integral tables we find that: ( )]
3
42sincossin 0
2
3
1
0
3 =+−=∫ ππ θθθθd 
 
Substitute and integrate with respect 
to φ to obtain: [ ] CCdCP 3
8
3
4
3
4 2
0
2
0
tot
πφφ π
π
=== ∫ 
 
Solving for C yields: 
tot8
3 PC π= 
 
Substitute for Ptot and evaluate C to 
obtain: 
 
( ) kW68.59kW500
8
3 == πC 
Substituting for C in equation (1) 
and evaluating I(120 km, 90°): ( ) ( )
2
2
W/m14.4
km120
kW68.5990 ,km120
μ=
=°I
 
 
28 ••• Regulations require that licensed radio stations have limits on their 
broadcast power so as to avoid interference with signals from distant stations. 
You are in charge of checking compliance with the law. At a distance of 30.0 km 
from a radio station that broadcasts from a single vertical electric dipole antenna 
at a frequency of 800 kHz, the intensity of the electromagnetic wave is 
2.00 × 10–13 W/m2. What is the total power radiated by the station? 
 
Picture the Problem The intensity of radiation from an electric dipole is given by 
C(sin2θ)/r2, where C is a constant whose units are those of power, r is the distance 
from the dipole to the point of interest, and θ is the angle between the electric 
dipole moment and the position vector .rr We can integrate the intensity to express 
 Chapter 30 
 
 
860 
the total power radiated by the antenna and use this result to evaluate C. Knowing 
C we can find the total power radiated by the station. 
 
From the definition of intensity we 
have: 
 
IdAdP = 
and 
( )dArIP ∫∫= θ,tot 
where, in polar coordinates, 
φθθ ddrdA sin2= 
 
Substitute for dA to obtain: 
 ( ) φθθθ
π π
ddrrIP sin, 2
2
0 0
tot ∫ ∫= (1) 
 
Express the intensity of the signal as 
a function of r and θ : ( ) 2
2sin,
r
CrI θθ = (2) 
 
Substitute for I(r,θ) in equation (1) to 
obtain: φθθ
ππ
ddCP ∫ ∫=
2
0 0
3
tot sin 
 
From integral tables we find that: ( )]
3
42sincossin 0
2
3
1
0
3 =+−=∫ π
π
θθθθd 
 
Substitute and integrate with respect 
to φ to obtain: [ ] CCdCP 3
8
3
4
3
4 2
0
2
0
tot
πφφ π
π
=== ∫ 
 
From equation (2) we have: 
 
( )
θ
θ
2
2
sin
, rrIC = 
 
Substitute for C in the expression for 
Ptot to obtain: 
 
( )
θ
θπ
2
2
tot sin
,
3
8 rrIP = 
or, because θ = 90°, 
( ) 2tot 3
8 rrIP π= 
 
Substitute numerical values and 
evaluate Ptot: 
( )( )
mW51.1
km0.30W/m1000.2
3
8 2213
tot
=
×= −πP
 
 
29 ••• A small private plane approaching an airport is flying at an altitude of 
2.50 km above sea level. As a flight controller at the airport, you know your 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
861
system uses a vertical electric dipole antenna to transmit 100 W at 24.0 MHz. 
What is the intensity of the signal at the plane’s receiving antenna when the plane 
is 4.00 km from the airport? Assume the airport is at sea level. 
 
Picture the Problem The intensity of radiation from the airport’s vertical dipole 
antenna is given by C(sin2θ)/r2, where C is a constant whose units are those of 
power, r is the distance from the dipole to the point of interest, and θ is the angle 
between the electric dipole moment and the position vector .rr We can integrate the 
intensityto express the total power radiated by the antenna and use this result to 
evaluate C. Knowing C we can find the intensity of the signal at the plane’s 
elevation and distance from the airport. 
 
Express the intensity of the signal as 
a function of r and θ : ( ) 2
2sin,
r
CrI θθ = (1) 
 
From the definition of intensity we 
have: 
 
IdAdP = 
and 
( )dArIP ∫∫= θ,tot 
where, in polar coordinates, 
φθθ ddrdA sin2= 
 
Substitute for dA to obtain: 
 ( ) φθθθ
π π
ddrrIP sin, 2
2
0 0
tot ∫ ∫= 
 
Substituting for I(r,θ) yields: φθθ
ππ
ddCP ∫ ∫=
2
0 0
3
tot sin 
 
From integral tables we find that: ( )]
3
42sincossin 0
2
3
1
0
3 =+−=∫ π
π
θθθθd 
 
Substitute and integrate with respect 
to φ to obtain: [ ] CCdCP 3
8
3
4
3
4 2
0
2
0
tot
πφφ π
π
=== ∫ 
 
Solving for C yields: 
tot8
3 PC π= 
 
Substitute for C in equation (1) to 
obtain: ( ) 2
2
tot sin
8
3,
r
PrI θπθ = 
 
 Chapter 30 
 
 
862 
At the elevation of the plane: 
 °=⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 0.58
m2500
m4000tan 1θ 
and 
( ) ( ) m4717m4000m2500 22 =+=r 
 
Substitute numerical values and 
evaluate I(4717 m, 58°): ( ) ( ) ( )
2
2
2
nW/m 386
m4717
0.58sin
8
W10030.58 ,m4717
=
°=° πI 
 
Energy and Momentum in an Electromagnetic Wave 
 
30 • An electromagnetic wave has an intensity of 100 W/m2. Find its 
(a) rms electric field strength, and (b) rms magnetic field strength. 
 
Picture the Problem We can use Pr = I/c to find the radiation pressure. The 
intensity of the electromagnetic wave is related to the rms values of its electric 
and magnetic field strengths according to I = ErmsBrms/μ0, where Brms = Erms/c. 
 
(a) Relate the intensity of the 
electromagnetic wave to Erms and 
Brms: 
 
0
rmsrms
μ
BEI = 
or, because Brms = Erms/c, 
c
EcEEI
0
2
rms
0
rmsrms
μμ == 
 
Solving for Erms yields: cIE 0rms μ= 
 
Substitute numerical values and evaluate Erms: 
 ( )( )( ) V/m194W/m100m/s10998.2N/A104 2827rms =××= −πE 
 
(b) Express Brms in terms of Erms: 
c
EB rmsrms = 
 
Substitute numerical values and 
evaluate Brms: 
nT647
m/s10998.2
V/m194
8rms =×=B 
 
31 • [SSM] The amplitude of an electromagnetic wave’s electric field is 
400 V/m. Find the wave’s (a) rms electric field strength, (b) rms magnetic field 
strength, (c) intensity and (d) radiation pressure (Pr). 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
863
Picture the Problem The rms values of the electric and magnetic fields are found 
from their amplitudes by dividing by the square root of two. The rms values of the 
electric and magnetic field strengths are related according to Brms = Erms/c. We can 
find the intensity of the radiation using I = ErmsBrms/μ0 and the radiation pressure 
using Pr = I/c. 
 
(a) Relate Erms to E0: 
V/m283
V/m 8.282
2
V/m400
2
0
rms
=
=== EE
 
 
(b) Find Brms from Erms: 
nT 943T9434.0
m/s10998.2
V/m8.282
8
rms
rms
==
×==
μ
c
EB
 
 
(c) The intensity of an 
electromagnetic wave is given by: 
 
0
rmsrms
μ
BEI = 
Substitute numerical values and 
evaluate I: 
( )( )
22
27
W/m212W/m3.212
N/A104
T9434.0V/m8.282
==
×= −π
μI
 
 
(d) Express the radiation pressure in 
terms of the intensity of the wave: 
 
c
IP =r 
Substitute numerical values and 
evaluate Pr: 
nPa 708
m/s10998.2
W/m3.212
8
2
r =×=P 
 
32 • The rms value of an electromagnetic wave’s electric field strength is 
400 V/m. Find the wave’s (a) rms magnetic field strength, (b) average energy 
density, and (c) intensity. 
 
Picture the Problem Given Erms, we can find Brms using Brms = Erms/c. The 
average energy density of the wave is given by uav = ErmsBrms/μ0c and the intensity 
of the wave by I = uavc . 
 
(a) Express Brms in terms of Erms: 
c
EB rmsrms = 
 
 Chapter 30 
 
 
864 
Substitute numerical values and 
evaluate Brms: 
T33.1
T334.1
m/s10998.2
V/m400
8rms
μ
μ
=
=×=B 
 
(b) The average energy density uav is 
given by: 
 
c
BEu
0
rmsrms
av μ= 
Substitute numerical values and 
evaluate uav: 
 
( )( )( )( )
33
827av
J/m42.1J/m417.1
m/s10998.2N/A104
T334.1V/m400
μμ
π
μ
==
××= −u 
 
(c) Express the intensity as the 
product of the average energy 
density and the speed of light in a 
vacuum: 
 
cuI av= 
Substitute numerical values and 
evaluate I: 
( )( )
2
83
W/m425
m/s10998.2J/m417.1
=
×= μI
 
 
33 •• (a) An electromagnetic wave that has an intensity equal to 200 W/m2 
is normal to a black 20 cm by 30 cm rectangular card absorbs 100 percent of the 
wave. Find the force exerted on the card by the radiation. (b) Find the force 
exerted by the same wave if the card reflects 100 percent of the wave. 
 
Picture the Problem We can find the force exerted on the card using the 
definition of pressure and the relationship between radiation pressure and the 
intensity of the electromagnetic wave. Note that, when the card reflects all the 
radiation incident on it, conservation of momentum requires that the force is 
doubled. 
 
(a) Using the definition of pressure, 
express the force exerted on the card 
by the radiation: 
 
APF rr = 
Relate the radiation pressure to the 
intensity of the wave: 
 
c
IP =r 
Substitute for Pr to obtain: 
 c
IAF =r 
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
865
Substitute numerical values and 
evaluate Fr: 
 
( )( )( )
nN40
m/s10998.2
m30.0m20.0W/m200
8
2
r
=
×=F 
 
(b) If the card reflects all of the 
radiation incident on it, the force 
exerted on the card is doubled: 
nN80r =F 
 
34 •• Find the force exerted by the electromagnetic wave on the card in Part 
(b) of Problem 33 if both the incident and reflected rays are at angles of 30º to the 
normal. 
 
Picture the Problem Only the normal component of the radiation pressure exerts 
a force on the card. 
 
Using the definition of pressure, 
express the force exerted on the card 
by the radiation: 
 
θcos2 rr APF = 
where the factor of 2 is a consequence 
of the fact that the card reflects the 
radiation incident on it. 
 
Relate the radiation pressure to the 
intensity of the wave: 
 
c
IP =r 
Substitute for Pr to obtain: 
 c
IAF θcos2r = 
 
Substitute numerical values and evaluate Fr: 
 ( )( )( ) nN69
m/s10998.2
30cosm30.0m20.0W/m2002
8
2
r =×
°=F 
 
35 • [SSM] (a) For a given distance from a radiating electric dipole, at 
what angle (expressed as θ and measured from the dipole axis) is the intensity 
equal to 50 percent of the maximum intensity? (b) At what angle θ is the intensity 
equal to 1 percent of the maximum intensity? 
 
Picture the Problem At a fixed distance from the electric dipole, the intensity of 
radiation is a function θ alone. 
 
(a) The intensity of the radiation 
from the dipole is proportional to 
sin2θ: 
( ) θθ 20 sinII = (1) 
where I0 is the maximum intensity. 
 Chapter 30 
 
 
866 
For 021 II = : θ20021 sinII = ⇒ 212sin =θ 
 
Solving for θ yields: ( ) °== − 45sin 211θ 
 
(b) For 001.0 II = : θ200 sin01.0 II = ⇒ 01.0sin 2 =θ 
 
Solving for θ yields: ( ) °== − 7.501.0sin 1θ 
 
36 •• A laser pulse has an energy of 20.0 J and a beam radius of 2.00 mm. 
The pulse duration is 10.0 ns and the energy density is uniformly distributed 
within the pulse. (a) What is the spatial length of the pulse? (b) What is the 
energy density within the pulse? (c) Find the rms values of the electric and 
magnetic fields in thepulse. 
 
Picture the Problem The spatial length L of the pulse is the product of its speed c 
and duration Δt. We can find the energy density within the pulse using its 
definition (u = U/V). The electric amplitude of the pulse is related to the energy 
density in the beam according to 20 Eu ∈= and we can find B from E using 
B = E/c. 
 
(a) The spatial length L of the pulse 
is the product of its speed c and 
duration Δt: 
 
tcL Δ= 
Substitute numerical values and 
evaluate L: 
 
( )( )
m00.3
m 998.2ns0.10m/s10998.2 8
=
=×=L
 
 
(b) The energy density within the 
pulse is the energy of the beam per 
unit volume: 
 
Lr
U
V
Uu 2π== 
Substitute numerical values and 
evaluate u: 
 
( ) ( )
33
2
kJ/m531kJ/m9.530
m998.2mm00.2
J0.20
==
= πu 
 
(c) E is related to u according to: 
 
2
rms0 Eu ∈= ⇒
0
rms ∈
uE = 
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
867
Substitute numerical values and 
evaluate Erms: 
 
MV/m245MV/m9.244
mN/C10854.8
kJ/m9.530
2212
3
rms
==
⋅×= −E 
 
Use Brms = Erms/c to find Brms: 
 T817.0m/s10998.2
MV/m9.244
8rms =×=B 
 
37 •• [SSM] An electromagnetic plane wave has an electric field that is 
parallel to the y axis, and has a Poynting vector that is given by 
( ) ( ) [ ] iS ˆcos W/m100, 22 tkxtx ω−=r , where x is in meters, k = 10.0 rad/m, 
ω = 3.00 × 109 rad/s, and t is in seconds. (a) What is the direction of propagation 
of the wave? (b) Find the wavelength and frequency of the wave. (c) Find the 
electric and magnetic fields of the wave as functions of x and t. 
 
Picture the Problem We can determine the direction of propagation of the wave, 
its wavelength, and its frequency by examining the argument of the cosine 
function. We can find E from cE 0
2 μ=Sr and B from B = E/c. Finally, we can 
use the definition of the Poynting vector and the given expression for S
r
to find 
E
r
and B
r
. 
 
(a) Because the argument of the cosine function is of the form tkx ω− , the wave 
propagates in the +x direction. 
 
(b) Examining the argument of the 
cosine function, we note that the 
wave number k of the wave is: 
 
1m0.102 −== λ
πk ⇒ m628.0=λ 
Examining the argument of the 
cosine function, we note that the 
angular frequency ω of the wave 
is: 
 
19 s1000.32 −×== fπω 
Solving for f yields: 
 MHz4772
s1000.3 19 =×=
−
πf 
 
(c) Express the magnitude of S
r
in 
terms of E: 
 
c
E
0
2
μ=S
r ⇒ SrcE 0μ= 
 Chapter 30 
 
 
868 
 
Substitute numerical values and evaluate E: 
 ( )( )( ) V/m1.194W/m100m/s10998.2N/A104 2827 =××= −πE 
 
Because 
( ) ( ) [ ] iS ˆcosW/m100, 22 tkxtx ω−=r 
and BES
rrr ×=
0
1
μ : 
 
( ) ( ) [ ] jE ˆcosV/m194, tkxtx ω−=r 
where k = 10.0 rad/m and 
ω = 3.00 × 109 rad/s. 
 
Use B = E/c to evaluate B: 
 
nT 4.647
m/s10998.2
V/m1.194
8 =×=B 
 
Because BES
rrr ×=
0
1
μ , the direction 
of B
r
must be such that the cross 
product of E
r
 with B
r
is in the +x 
direction: 
( ) ( ) [ ] kB ˆcosnT 647, tkxtx ω−=r 
where k = 10.0 rad/m and 
ω = 3.00 × 109 rad/s. 
 
38 •• A parallel-plate capacitor is being charged. The capacitor consists of a 
pair of identical circular parallel plates that have radius b and a separation 
distance d. (a) Show that the displacement current in the capacitor gap has the 
same value as the conduction current in the capacitor leads. (b) What is the 
direction of the Poynting vector in the region between the capacitor plates? 
(c) Find an expression for the Poynting vector in this region and show that its flux 
into the region between the plates is equal to the rate of change of the energy 
stored in the capacitor. 
 
Picture the Problem We can use the expression for the electric field strength 
between the plates of the parallel-plate capacitor and the definition of the 
displacement current to show that the displacement current in the capacitor is 
equal to the conduction current in the capacitor leads. In (b) we can use the 
definition of the Poynting vector and the directions of the electric and magnetic 
fields to determine the direction of the Poynting vector between the capacitor 
plates. In (c), we’ll demonstrate that the flux of S
r
into the region between the 
plates is equal to the rate of change of the energy stored in the capacitor by 
evaluating these quantities separately and showing that they are equal. 
 
(a) The displacement current is 
proportional to the rate at which the 
flux is changing between the plates: 
 
( )
dt
dEAAE
dt
d
dt
dI 00e0d ∈∈φ∈ === 
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
869
The electric field strength between 
the plates of the capacitor is given 
by: 
A
QE
0∈= 
where Q is the instantaneous charge on 
the capacitor plates. 
 
Substituting for E yields: 
 
I
dt
dQ
A
Q
dt
dAI ===
0
0d ∈∈ 
 
(b) Because E
r
is perpendicular to the plates of the capacitor and B
r
is tangent to 
circles that are concentric and whose center is through the middle of the capacitor 
plates, S
r
points radially inward toward the center of the capacitor. 
 
(c) The Poynting vector is: 
 
BES
rrr ×=
0
1
μ (1) 
 
Letting the direction of E
r
 be the +x 
direction: 
 
iE ˆE=r 
where E is the electric field strength 
between the plates of the capacitor. 
 
Apply Ampere’s law to a closed 
circular path of radius R ≤ b to 
obtain: 
 
( ) d02 IRB μπ = 
Substituting for Id and simplifying 
yields: 
 
( )
dt
dER
EA
dt
d
dt
dRB
2
00
00
e
002
π∈μ
∈μφ∈μπ
=
==
 
 
Solve for B to obtain: 
 dt
dERB
2
00∈μ= 
and 
jB ˆ
2
00
dt
dER∈μ−=r 
where jˆ is a unit vector that is tangent 
to the concentric circles. 
 
 Chapter 30 
 
 
870 
Substitute for B
r
 and E
r
 in equation 
(1) and simplify to obtain: 
 
x
y
B
rEr
S
r
Rˆ
R
 
( )
R
ji
jiS
ˆ
2
ˆˆ
2
ˆ
2
ˆ1
0
0
00
0
R
dt
dEE
R
dt
dEE
dt
dERE
∈
∈
∈μ
μ
−=
−×=
⎟⎠
⎞⎜⎝
⎛−×⎟⎟⎠
⎞
⎜⎜⎝
⎛=r
 
where R ≤ b, E is the electric field 
strength between the plates, R is the 
radial distance from the line joining the 
centers of the plates, Rˆ is a unit vector 
pointing radially outward from the line 
joining the centers of the plates, and b 
is the radius of the plates. 
 
The rate at which energy is stored 
in the capacitor is: 
( ) ( )
dt
dEEAd
E
dt
dVuV
dt
d
dt
dU
0
2
0
∈
∈
=
==
 
 
Because 
A
QE
0∈= : 
 
I
A
Qd
dt
dQ
A
Qd
A
Q
dt
d
A
QAd
dt
dU
00
00
0
∈∈
∈∈∈
==
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
 
 
Consider a cylindrical surface of 
length d and radius b. Because 
S
r
points inward, the energy flowing 
into the solenoid per unit time is: 
( )
( )
dt
dEdEb
bd
dt
dEEb
bdSdAS
2
0
02
1
n
2
2
∈π
π∈
π
=
⎟⎠
⎞⎜⎝
⎛=
=∫
 
 
Substituting for E and simplifying 
yields: 
 
I
A
Qd
dt
dQ
A
db
b
Q
A
Q
dt
ddb
A
QdAS
00
2
2
0
2
0
0n
∈∈ππ
∈∈∈π
=⎟⎠
⎞⎜⎝
⎛=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛=∫
 
 
Because dtdUdAS =∫ n , we’ve proved that the flux of Sr into the region 
between the capacitor is equal to the rate of change of the energy stored in the 
capacitor. 
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
871
39 •• [SSM] A pulsed laser fires a 1000-MW pulse that has a 200-ns 
duration at a small object thathas a mass equal to 10.0 mg and is suspended by a 
fine fiber that is 4.00 cm long. If the radiation is completely absorbed by the 
object, what is the maximum angle of deflection of this pendulum? (Think of the 
system as a ballistic pendulum and assume the small object was hanging vertically 
before the radiation hit it.) 
 
Picture the Problem The diagram 
shows the displacement of the 
pendulum bob, through an angle θ, as a 
consequence of the complete absorption 
of the radiation incident on it. We can 
use conservation of energy (mechanical 
energy is conserved after the collision) 
to relate the maximum angle of 
deflection of the pendulum to the initial 
momentum of the pendulum bob. 
Because the displacement of the bob 
during the absorption of the pulse is 
negligible, we can use conservation of 
momentum (conserved during the 
collision) to equate the momentum of 
the electromagnetic pulse to the initial 
momentum of the bob. 
 
h
m
LL cos 
θ
θ
0g =U
 
 
Apply conservation of energy to 
obtain: 
 
0ifif =−+− UUKK 
or, because Ui = Kf = 0 and 
m
p
K
2
2
i
i = , 
0
2 f
2
i =+− U
m
p 
 
Uf is given by: 
 
( )θcos1f −== mgLmghU 
Substitute for Uf: 
 ( ) 0cos12
2
i =−+− θmgL
m
p 
 
Solve for θ to obtain: 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −= −
gLm
p
2
2
i1
2
1cosθ 
 
 Chapter 30 
 
 
872 
 
Use conservation of momentum to 
relate the momentum of the 
electromagnetic pulse to the initial 
momentum pi of the pendulum bob: 
 
i waveem pc
tP
c
Up =Δ== 
where Δt is the duration of the pulse. 
Substitute for pi: ( ) ⎥⎦
⎤⎢⎣
⎡ Δ−= −
gLcm
tP
22
22
1
2
1cosθ 
Substitute numerical values and evaluate θ : 
 
( ) ( )
( ) ( ) ( )( ) °=⎥⎥⎦
⎤
⎢⎢⎣
⎡
×−=
− 10.6
m0400.0m/s81.9m/s10998.2mg0.102
ns200MW10001cos
2282
22
1θ 
 
Remarks: The solution presented here is valid only if the displacement of the 
bob during the absorption of the pulse is negligible. (Otherwise, the 
horizontal component of the momentum of the pulse-bob system is not 
conserved during the collision.) We can show that the displacement during 
the pulse-bob collision is small by solving for the speed of the bob after 
absorbing the pulse. Applying conservation of momentum (mv = P(Δt)/c) and 
solving for v gives v = 6.67 × 10−7 m/s. This speed is so slow compared to c, 
we can conclude that the duration of the collision is extremely close to 200 ns 
(the time for the pulse to travel its own length). Traveling at 6.67 × 10−7 m/s 
for 200 ns, the bob would travel 1.33 × 10−13 m—a distance 1000 times 
smaller that the diameter of a hydrogen atom. (Because 6.67×10−7 m/s is the 
maximum speed of the bob during the collision, the bob would actually travel 
less than 1.33 × 10−13 m during the collision.) 
 
40 •• The mirrors used in a particular type of laser are 99.99% reflecting. 
(a) If the laser has an average output power of 15 W, what is the average power of 
the radiation incident on one of the mirrors? (b) What is the force due to radiation 
pressure on one of the mirrors? 
 
Picture the Problem We can use the definitions of pressure and the relationship 
between radiation pressure and the intensity of the radiation to find the force due 
to radiation pressure on one of the mirrors. 
 
(a) Because only about 0.01 percent 
of the energy inside the laser "leaks 
out", the average power of the 
radiation incident on one of the 
mirrors is: 
 
W105.1
101.0
W15 5
4 ×=×= −P 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
873
(b) Use the definition of radiation 
pressure to obtain: 
 
A
FP rr = 
where Fr is the force due to radiation 
pressure and A is the area of the mirror 
on which the radiation is incident. 
 
The radiation pressure is also related 
to the intensity of the radiation: Ac
P
c
IP 22r == 
where P is the power of the laser and 
the factor of 2 is due to the fact that the 
mirror is essentially totally reflecting. 
 
Equate the two expression for the 
radiation pressure and solve for Fr: 
 
Ac
P
A
F 2r = ⇒ 
c
PF 2r = 
Substitute numerical values and 
evaluate Fr: 
( ) mN0.1
m/s10998.2
W105.12
8
5
r =×
×=F 
 
41 •• [SSM] (a) Estimate the force on Earth due to the pressure of the 
radiation on Earth by the Sun, and compare this force to the gravitational force of 
the Sun on Earth. (At Earth’s orbit, the intensity of sunlight is 1.37 kW/m2.) 
(b). Repeat Part (a) for Mars which is at an average distance of 2.28 × 108 km 
from the Sun and has a radius of 3.40 × 103 km. (c) Which planet has the larger 
ratio of radiation pressure to gravitational attraction. 
 
Picture the Problem We can find the radiation pressure force from the definition 
of pressure and the relationship between the radiation pressure and the intensity of 
the radiation from the Sun. We can use Newton’s law of gravitation to find the 
gravitational force the Sun exerts on Earth and Mars. 
 
(a) The radiation pressure exerted on 
Earth is given by: 
 
A
F
P Earth r,Earth r, = ⇒ APF Earth r,Earth r, = 
where A is the cross-sectional area of 
Earth. 
 
Express the radiation pressure in 
terms of the intensity of the 
radiation I from the Sun: 
 
c
IP =Earth r, 
Substituting for Pr, Earth and A yields: 
 c
RIF
2
Earth r,
π= 
 
 Chapter 30 
 
 
874 
Substitute numerical values and 
evaluate Fr: 
( )( )
N1083.5
N10825.5
m/s10998.2
m1037.6kW/m37.1
8
8
8
262
Earth r,
×=
×=
×
×= πF
 
 
The gravitational force exerted 
on Earth by the Sun is given by: 
 
2
earthsun
Earth g, r
mGm
F = 
where r is the radius of Earth’s orbit. 
 
Substitute numerical values and evaluate Fg, Earth: 
 ( )( )( )
( ) N10529.3m1050.1
kg1098.5kg1099.1kg/mN10673.6 22
211
24302211
Earth g, ×=×
××⋅×=
−
F 
 
Express the ratio of the force due to 
radiation pressure Fr, Earth to the 
gravitational force Fg, Earth: 
 
14
22
8
Earth g,
Earth r, 1065.1
N10529.3
N10825.5 −×=×
×=
F
F
 
or ( ) Earth g,14Earth r, 1065.1 FF −×= 
 
(b) The radiation pressure exerted 
on Mars is given by: 
 
A
F
P Mars r,Mars r, = ⇒ APF Mars r,Mars r, = 
where A is the cross-sectional area of 
Mars. 
 
Express the radiation pressure 
on Mars in terms of the intensity 
of the radiation IMars from the 
sun: 
 
c
I
P MarsMars r, = 
Substituting for Pr, Mars and A yields: 
 c
RIF
2
MarsMars
Mars r,
π= 
 
Express the ratio of the solar 
constant at Earth to the solar 
constant at Mars: 
 
2
Mars
earth
earth
Mars ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
r
r
I
I ⇒
2
Mars
earth
earthMars ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
r
rII 
Substitute for MarsI to obtain: 
 
2
Mars
earth
2
Marsearth
Mars r, ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
r
r
c
RIF
π
 
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
875
 
Substitute numerical values and evaluate Fr, Mars: 
 ( )( ) N1018.7
m1028.2
m1050.1
m/s10998.2
km1040.3kW/m37.1 7
2
11
11
8
232
Mars r, ×=⎟⎟⎠
⎞
⎜⎜⎝
⎛
×
×
×
×= πF 
 
The gravitational force exerted on 
Mars by the Sun is given by: 
 
( )
2
Earthsun
2
Marssun
Mars g,
11.0
r
mGm
r
mGmF == 
where r is the radius of Mars’ orbit. 
 
Substitute numerical values and evaluate Fg 
 ( )( )( )( )
( )
N1068.1
m1028.2
kg1098.511.0kg1099.1kg/mN10673.6
21
211
24302211
Mars g,
×=
×
××⋅×=
−
F
 
 
Express the ratio of the force due to 
radiation pressure Fr, Mars to the 
gravitational force Fg, Mars: 
 
14
21
7
Mars g,
Mars r, 1027.4
N1068.1
N1018.7−×=×
×=
F
F
 
or ( ) Mars g,14Mars r, 1027.4 FF −×= 
 
(c) Because the ratio of the radiation pressure force to the gravitational force is 
1.65 × 10−14 for Earth and 4.27 × 10−14 for Mars, Mars has the larger ratio. The 
reason that the ratio is higher for Mars is that the dependence of the radiation 
pressure on the distance from the Sun is the same for both forces (r−2), whereas 
the dependence on the radii of the planets is different. Radiation pressure varies as 
R2, whereas the gravitational force varies as R3 (assuming that the two planets 
have the same density, an assumption that is nearly true). Consequently, the ratio 
of the forces goes as 132 / −= RRR . Because Mars is smaller than Earth, the ratio 
is larger. 
 
The Wave Equation for Electromagnetic Waves 
 
42 • Show by direct substitution that Equation 30-8a is satisfied by the 
wave function 
 
Ey = E0 sin kx −ωt( )= E0 sin k x − ct( ) where c = ω/k. 
 
Picture the Problem We can show that Equation 30-8a is satisfied by the wave 
function Ey by showing that the ratio of ∂2Ey/∂x2 to ∂2Ey/∂t2 is 1/c2 where c = ω/k. 
 
 Chapter 30 
 
 
876 
Differentiate ( )tkxEEy ω−= sin0 
with respect to x: 
 
[ ]
)cos(
)sin(
0
0
tkxkE
tkxE
xx
Ey
ω
ω
−=
−∂
∂=∂
∂
 
 
Evaluate the second partial 
derivative of Ey with respect to x: 
 
[ ]
)sin(
)cos(
0
2
02
2
tkxEk
tkxkE
xx
Ey
ω
ω
−−=
−∂
∂=∂
∂
 (1) 
 
Differentiate ( )tkxEEy ω−= sin0 
with respect to t: 
 
[ ]
)cos(
)sin(
0
0
tkxE
tkxE
tt
Ey
ωω
ω
−−=
−∂
∂=∂
∂
 
 
Evaluate the second partial 
derivative of Ey with respect to t: 
 
[ ]
)sin(
)cos(
0
2
02
2
tkxE
tkxE
tt
Ey
ωω
ωω
−−=
−−∂
∂=∂
∂
 (2) 
 
Divide equation (1) by equation (2) 
to obtain: 
 
( )
( ) 2
2
0
2
0
2
2
2
2
2
sin
sin
ωωω
ω k
tkxE
tkxEk
t
E
x
E
y
y
=−−
−−=
∂
∂
∂
∂
 
or 
2
2
22
2
2
2
2
2 1
t
E
ct
Ek
x
E yyy
∂
∂=∂
∂=∂
∂
ω 
provided c = ω/k. 
 
43 • Use the values of μ0 and 0∈ in SI units to compute 001 μ∈ and 
show that it is equal to 3.00 × 108 m/s. 
 
Picture the Problem Substitute numerical values and evaluate c: 
 
( )( ) m/s1000.3mN/C10854.8N/A104
1 8
221227
×=⋅××= −−πc 
 
44 •• (a) Use Maxwell’s equations to show for a plane wave, in which E
r
 
and B
r
 are independent of y and z, that 
t
B
x
E yz
∂
∂=∂
∂ and 
t
E
x
B zy
∂
∂=∂
∂
00∈μ . 
(b) Show that Ez and By also satisfy the wave equation. 
 
 
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
877
Picture the Problem We can use Figures 30-5 and 30-6 and a derivation similar 
to that in the text to obtain the given results. 
 
In Figure 30-5, replace Bz by Ez. For 
Δx small: 
 
( ) ( ) x
x
ExExE zzz Δ∂
∂+= 12 
Evaluate the line integral of E
r
 
around the rectangular area ΔxΔz: 
 
zx
x
Ed z ΔΔ∂
∂−≈⋅∫ lrrE (1) 
Express the magnetic flux through 
the same area: 
 
∫ ΔΔ=S n zxBdAB y 
Apply Faraday’s law to obtain: ( )
zx
t
B
zxB
t
dAB
t
d
y
y
ΔΔ∂
∂−=
ΔΔ∂
∂−=∂
∂−≈⋅ ∫∫ S nlrrE
 
Substitute in equation (1) to obtain: 
zx
t
B
zx
x
E yz ΔΔ∂
∂−=ΔΔ∂
∂− 
or 
t
B
x
E yz
∂
∂=∂
∂ 
 
In Figure 30-6, replace Ey by By and 
evaluate the line integral of B
r
 
around the rectangular area ΔxΔz: 
 
∫ ∫=⋅ S n00 dAEd ∈μlrrB 
provided there are no conduction 
currents. 
 
Evaluate these integrals to obtain: 
 t
E
x
B zy
∂
∂=∂
∂
00∈μ 
 
(b) Using the first result obtained in 
(a), find the second partial derivative 
of Ez with respect to x: 
⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂
∂
∂=⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂
t
B
xx
E
x
yz 
or 
⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂
∂
∂=∂
∂
x
B
tx
E yz
2
2
 
 
 Chapter 30 
 
 
878 
 
Use the second result obtained in (a) 
to obtain: 
 
2
2
00002
2
t
E
t
E
tx
E zzz
∂
∂=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂
∂
∂=∂
∂ ∈μεμ 
or, because μ0∈0 = 1/c2, 
2
2
22
2 1
t
E
cx
E zz
∂
∂=∂
∂ . 
 
Using the second result obtained in 
(a), find the second partial derivative 
of By with respect to x: 
⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂
∂
∂=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂
∂
∂
t
E
xx
B
x
zy
00∈μ 
or 
⎟⎠
⎞⎜⎝
⎛
∂
∂
∂
∂=∂
∂
x
E
tx
B zy
002
2
∈μ 
 
Use the first result obtained in (a) to 
obtain: 
 
2
2
00002
2
t
B
t
B
tx
B yyy
∂
∂=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∂
∂
∂
∂=∂
∂ ∈μ∈μ 
or, because μ0∈0 = 1/c2, 
2
2
22
2 1
t
B
cx
B yy
∂
∂=∂
∂
. 
 
45 •• [SSM] Show that any function of the form y(x, t) = f(x – vt) or 
y(x, t) = g(x + vt) satisfies the wave Equation 30-7 
 
Picture the Problem We can show that these functions satisfy the wave 
equations by differentiating them twice (using the chain rule) with respect to x 
and t and equating the expressions for the second partial of f with respect to u. 
 
Let u = x − vt. Then: 
u
f
u
f
x
u
x
f
∂
∂=∂
∂
∂
∂=∂
∂ 
and 
u
fv
u
f
t
u
t
f
∂
∂−=∂
∂
∂
∂=∂
∂ 
 
Express the second derivatives of 
f with respect to x and t to obtain: 2
2
2
2
u
f
x
f
∂
∂=∂
∂ and 2
2
2
2
2
u
fv
t
f
∂
∂=∂
∂ 
 
Divide the first of these equations by 
the second to obtain: 
 2
2
2
2
2
1
v
t
f
x
f
=
∂
∂
∂
∂
⇒ 2
2
22
2 1
t
f
vx
f
∂
∂=∂
∂ 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
879
Let u = x + vt. Then: 
u
f
u
f
x
u
x
f
∂
∂=∂
∂
∂
∂=∂
∂ 
and 
u
fv
u
f
t
u
t
f
∂
∂=∂
∂
∂
∂=∂
∂ 
 
Express the second derivatives of 
f with respect to x and t to obtain: 2
2
2
2
u
f
x
f
∂
∂=∂
∂ and 2
2
2
2
2
u
fv
t
f
∂
∂=∂
∂ 
 
Divide the first of these equations by 
the second to obtain: 
 2
2
2
2
2
1
v
t
f
x
f
=
∂
∂
∂
∂
⇒ 2
2
22
2 1
t
f
vx
f
∂
∂=∂
∂ 
 
General Problems 
 
46 • An electromagnetic wave has a frequency of 100 MHz and is traveling 
in a vacuum. The magnetic field is given by ( ) ( ) ( )iB ˆcosT 1000.1, 8 tkztz ω−×= −r . 
(a) Find the wavelength and the direction of propagation of this wave. (b) Find 
the electric field vector ( )tz,Er . (c) Determine the Poynting vector, and use it to 
find the intensity of this wave. 
 
Picture the Problem We can use c = fλ to find the wavelength. Examination of 
the argument of the cosine function will reveal the direction of propagation of the 
wave. We can find the magnitude, wave number, and angular frequency of the 
electric vector from the given information and the result of (a) and use these 
results to obtain E
r
(z, t). Finally, we can use its definition to find the Poynting 
vector. 
 
(a) Relate the wavelength of the 
wave to its frequency and the speed 
of light: 
 
f
c=λ 
Substitute numerical values and 
evaluate λ: m00.3MHz100
m/s10998.2 8 =×=λ 
 
From the sign of the argument of the cosine function and the spatial dependence 
on z, we can conclude that the wave propagates in the +z direction. 
 
 Chapter 30 
 
 
880 
(b) Express the amplitude of E
r
: ( )( )
V/m00.3
T10m/s10998.2 88
=
×== −cBE
 
 
Find the angular frequency and 
wave number of the wave: 
 
( ) 18 s1028.6MHz10022 −×=== ππω f
and 
1m09.2
m00.3
22 −=== πλ
πk 
 
Because S
r
is in the positive z direction, E
r
must bein the negative y direction in 
order to satisfy the Poynting vector expression: 
 
( ) ( ) ( ) ( )[ ] jE ˆs1028.6m09.2cosV/m00.3, 181 tztz −− ×−−=r 
 
(c) Use its definition to express and evaluate the Poynting vector: 
 
( ) ( )( ) ( ) ( )[ ]( )ijBES ˆˆs1028.6m09.2cos
N/A104
T10V/m00.31, 181227
8
0
××−×
−=×= −−−
−
tztz πμ
rrr
 
or 
( ) ( ) ( ) ( )[ ]kS ˆs1028.6m09.2cosmW/m9.23, 18122 tztz −− ×−=r 
 
The intensity of the wave is the 
average magnitude of the Poynting 
vector. The average value of the 
square of the cosine function is 1/2: 
( )
2
2
2
1
mW/m9.11
mW/m9.23
=
== SrI
 
 
47 •• [SSM] A circular loop of wire can be used to detect electromagnetic 
waves. Suppose the signal strength from a 100-MHz FM radio station 100 km 
distant is 4.0 μW/m2, and suppose the signal is vertically polarized. What is the 
maximum rms voltage induced in your antenna, assuming your antenna is a 
10.0-cm-radius loop? 
 
Picture the Problem We can use Faraday’s law to show that the maximum rms 
voltage induced in the loop is given by ,20rms BAωε = where A is the area of 
the loop, B0 is the amplitude of the magnetic field, and ω is the angular frequency 
of the wave. Relating the intensity of the radiation to B0 will allow us to express 
rmsε as a function of the intensity. 
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
881
The emf induced in the antenna is 
given by Faraday’s law: 
 
( ) ( )
( )
ttBR
tB
dt
dR
dt
dBA
BA
dt
dA
dt
d
dt
d
ωωωπ
ωπ
φ
ε
ε
coscos
sin
ˆ
peak0
2
0
2
m
−=−=
−=−=
−=⋅−=−= nBr
 
where 0
2
peak BR ωπε = and R is the 
radius of the loop antenna.. 
 
rmsε equals peakε divided by the 
square root of 2: 22
0
2
peak
rms
BR ωπεε == (1) 
 
The intensity of the signal is given 
by: 0
00
2μ
BEI = 
or, because 00 cBE = , 
0
2
0
0
00
22 μμ
cBBcBI == 
 
Solving for B0 yields: 
c
IB 00
2μ= 
 
Substituting for B0 and ω in 
equation (1) and simplifying yields: ( )
c
IfR
c
IfR
022
02
rms
2
2
2
2
μπ
μππε
=
=
 
 
Substitute numerical values and evaluate εrms: 
 
( ) ( ) ( )( ) mV .62
m/s10998.2
W/m0.4N/A104MHz100m100.02 8
227
22
rms =×
×=
− μππε 
 
48 •• The electric field strength from a radio station some distance from the 
electric dipole transmitting antenna is given by ( ) ( )trad/s 1000.1cosN/C 1000.1 64 ×× − , where t is in seconds. (a) What peak 
voltage is picked up on a 50.0-cm long wire oriented parallel with the electric 
field direction? (b) What is the maximum voltage that can be induced by this 
electromagnetic wave in a conducting loop of radius 20.0 cm? What orientation of 
the loop does this require? 
 
 Chapter 30 
 
 
882 
Picture the Problem The voltage induced in the piece of wire is the product of 
the electric field and the length of the wire. The maximum rms voltage induced in 
the loop is given by ,0BAωε = where A is the area of the loop, B0 is the 
amplitude of the magnetic field, and ω is the angular frequency of the wave. 
 
(a) Because E is independent of x: lEV = 
where l is the length of the wire. 
 
Substitute numerical values and 
evaluate V: 
( )[ ]( )
( ) t
tV
6
64
10cosV0.50
m500.010cosN/C1000.1
μ=
×= −
 
and V 0.50peak μ=V 
 
(b) The maximum voltage induced in 
a loop is given by: 
 
AB0ωε = 
where A is the area of the loop and B0 is 
the amplitude of the magnetic field. 
 
Eliminate B0 in favor of E0 and 
substitute for A to obtain: 
 
c
RE 20πωε = 
Substitute numerical values and evaluate ε: 
 ( )( ) ( ) nV9.41
m/s10998.2
m200.0N/C1000.1s1000.1
8
2416
=×
××=
−− πε 
 
The loop antenna should be oriented so the transmitting antenna lies in the plane 
of the loop. 
 
49 ••• A parallel-plate capacitor has circular plates of radius a that are 
separated by a distance d. In the gap between the two plates is a thin straight wire 
of resistance R that connects the centers of the two plates. A time-varying voltage 
given by V0 sin ωt is applied across the plates. (a) What is the current drawn by 
this capacitor? (b) What is the magnetic field as a function of the radial distance r 
from the centerline within the capacitor plates? (c) What is the phase angle 
between the current drawn by the capacitor and the applied voltage? 
 
Picture the Problem Some of the charge entering the capacitor passes through 
the resistive wire while the rest of it accumulates on the upper plate. The total 
current is the rate at which the charge passes through the resistive wire plus the 
rate at which it accumulates on the upper plate. The magnetic field between the 
capacitor plates is due to both the current in the resistive wire and the 
displacement current though a surface bounded by a circle a distance r from the 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
883
resistive wire. The phase difference between the current drawn by the capacitor 
and the applied voltage may be calculated using a phasor diagram. 
 
 
(a) The current drawn by the 
capacitor is the sum of the 
conduction current through the 
resistance wire and dQ/dt, where Q 
is the charge on the upper plate of 
the capacitor: 
 
dt
dQII += c (1) 
Express the conduction current Ic in 
terms of the potential difference 
between the plates and the resistance 
of the wire: 
 
t
R
V
R
VI ωsin0c == 
 
Because CVQ = : 
 
tCV
dt
dVC
dt
dQ ωω cos0== 
 
Substitute in equation (1): 
 
tCVt
R
V
I ωωω cossin 00 += (2) 
 
The capacitance of a parallel-plate 
capacitor with plate area A and plate 
separation d is given by: 
 
d
a
d
AC
2
00 π∈∈ == 
 
Substituting for C in equation (2) 
gives: ⎟⎟⎠
⎞
⎜⎜⎝
⎛ += t
d
at
R
VI ωπ∈ωω cossin1
2
0
0 
 
 Chapter 30 
 
 
884 
 
(b) Apply the generalized form of 
Ampere’s law to a circular path of 
radius r centered within the plates of 
the capacitor, where dI' is the 
displacement current through the flat 
surface S bounded by the path and Ic 
is the conduction current through the 
same surface: 
 
( )dc0C I'Id +=⋅∫ μlrrB 
 
 
 
By symmetry the line integral is B 
times the circumference of the circle 
of radius r: 
 
( ) ( )dc02 I'IrB += μπ (3) 
In the region between the capacitor 
plates there is a uniform electric field 
due to the surface charges +Q and –
Q. The associated displacement 
current through S is: 
 
( )
dt
dEr
dt
dEA'
A'E
dt
d
dt
dI'
2
00
0
e
0d
π∈∈
∈φ∈
==
==
 
provided ( )ar ≤ 
 
To evaluate the displacement current 
we first must evaluate E everywhere 
on S. Near the surface of a 
conductor, where σ is the surface 
charge density: 
 
0∈σ=E , where ( )2aQAQ πσ == 
so 
2
0 a
QE π∈= 
 
Substituting for E in the equation 
for dI' gives: 
( )
tV
d
r
tV
dt
d
d
r
dt
dV
d
r
d
V
dt
dr
dt
dErI'
ωπ∈ω
ωπ∈π∈
π∈π∈
cos
sin
0
2
0
0
2
0
2
0
2
0
2
0d
=
==
⎟⎠
⎞⎜⎝
⎛==
 
 
Solving for B in equation (3) and substituting for Ic and dI' yields: 
 
( ) ( )
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=+=
t
d
rt
Rr
V
tV
d
rt
R
V
rr
I'IrB
ωπ∈ωωπ
μ
ωπ∈ωωπ
μ
π
μ
cossin1
2
cossin
22
2
000
0
2
000dc0
 
 
 Maxwell’s Equations and Electromagnetic Waves 
 
 
885
 
(c) Both the charge Q and the 
conduction current Ic are in phase 
with V. However, dQ/dt, which is 
equal to the displacement