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 I d = ∈0 dφe dt , Id is zero if
dφe dt = 0 .
(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 dφe dt = 0 . At the moment when the capacitor is
momentarily uncharged, dE/dt ≠ 0 and so dφe dt ≠ 0 .
2
•
Using SI units, show that I d = ∈0 dφe dt has units of current.
Determine the Concept We need to show that ∈0 dφe dt has units of amperes.
We can accomplish this by substituting the SI units of ∈0 and dφe dt and
simplifying the resulting expression.
C
V 2
⋅m
C
C2 V C V s
C2 m
⋅
=
⋅ = ⋅ = ⋅V = = A
s
s
N s N s V
N⋅m
C
[SSM]
True or false:
3
•
(a)
Maxwell’s equations apply only to electric and magnetic fields that are
constant over time.
2829
2830 Chapter 30
(b)
(c)
(d)
The electromagnetic wave equation can be derived from Maxwell’s
equations.
Electromagnetic waves are transverse waves.
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 timedependent 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
∫
S
Bn dA = μ0 qm, inside
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
Maxwell’s Equations and Electromagnetic Waves 2831
(b) (1) microwaves (2) green light (3) ultraviolet light
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.
r
r r
8
•
Show that the expression E × B μ0 for the Poynting vector S
(Equation 30-21) has units of watts per square meter (the SI units for
electromagnetic wave intensity).
2832 Chapter 30
r r
Determine the Concept We can that E × B μ0 has units of W/m2 by substituting
r r
the SI units of E , B and μ0 and simplifying the resulting expression.
N
N
C
N⋅m
J
⋅T
⋅
N
A
N
m
W
C
= C = ⋅ =
⋅ s2 = s2 = s2 =
T⋅m m C m
C m
m
m
m2
A
A
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 ( p = U c ). 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
r
EB
I= S = 0 0.
av
2 μ0
The intensity of an electromagnetic
wave is given by:
r
I= S
Because E0 = cB0:
r
S
r
This result tells us that S
av
av
av
=
E0 B0
2 μ0
E02
=
2cμ0
∝ E02 independently of the wavelength of the
electromagnetic radiation. Thus
(d )
is correct.
Maxwell’s Equations and Electromagnetic Waves 2833
11 •
Two sinusoidal plane electromagnetic waves are identical except that
wave A has a peak electric field that is three times the peak electric field of wave
B. How do their intensities compare? (a) I A = 13 I B (b) I A = 91 I B (c) I A = 3I B
(d) I A = 9I B (e) You cannot determine how their intensities compare from the data
given.
Determine the Concept The intensity of an electromagnetic wave is given by
r
EB
I= S = 0 0.
av
2 μ0
Express the intensities of the two
waves:
Dividing the first of these equations
by the second and simplifying yields:
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:
IA =
E0,A B0,A
2μ 0
and I B =
E 0 , B B0 , B
2μ 0
E 0 , A B0 , A
E 0 , A B0 , A
2μ 0
IA
=
=
E 0 , B B0 , B
IB
E 0 , B B0 , B
2μ 0
I A (3E0, B )(3B0, B )
=
= 9 ⇒ I A = 9I B
IB
E0, B B0, B
(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 2nd 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.
2834 Chapter 30
(a) Apply Newton’s 2nd law to the
atom to obtain:
Fr = ma
(1)
where Fr is the radiation force exerted
by the laser beam.
The radiation pressure Pr and
intensity of the beam I are related
according to:
Solve for Fr to obtain:
Substitute for Fr in equation (1) to
obtain:
Pr =
Fr I
=
A c
Fr =
IA Iλ2
=
c
c
Iλ2
Iλ2
= ma ⇒ a =
c
mc
Substitute numerical values and evaluate a:
a=
(10 W/m )(780 nm)
⎛
⎞
g
1 mol
⎜⎜ 85
⎟ (2.998 × 10
×
mol 6.022 × 10 particles ⎟
2
2
23
⎝
8
⎠
m/s
)
= 1.44 × 10 5 m/s 2
= 1.4 × 10 5 m/s 2
(b) Using the definition of
acceleration, express the stopping
time Δt of the atom:
Because vfinal ≈ 0:
Using the rms speed as the initial
speed of an atom, relate vinitial to the
Δt =
vfinal − vinitial
a
Δt ≈
− v initial
a
vinitial = v rms =
3kT
m
temperature of the gas:
Substitute in the expression for the
stopping time to obtain:
Δt = −
1 3kT
a m
Substitute numerical values and evaluate Δt:
Δt = −
1
− 1.44 × 105 m/s 2
(
)
3 1.38 × 10 −23 J/K (300 K )
= 2.1 ms
⎞
⎛
g
1 mol
⎟⎟
⎜⎜ 85
×
23
⎝ mol 6.022 × 10 particles ⎠
Maxwell’s Equations and Electromagnetic Waves 2835
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,
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 the 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:
Because the radiation from the Sun is
reflected, the radiation pressure is
twice what it would be if it were
absorbed:
Substituting for Pr and A yields:
The gravitational force acting on the
balloon when it is in a near-Earth
orbit is approximately its weight at
the surface of Earth:
Because the surface area of the
balloon is 4π r 2 = π d 2 :
Express the ratio of the radiationpressure force to the gravitational
force and simplify to obtain:
Fr = Pr A
where A is the cross-sectional area of
the balloon.
Pr =
2I
c
Fr =
2 I 14 πd 2 πd 2 I
=
c
2c
(
)
Fg = wballoon = mballoon g = ρ MylarVMylar g
= ρ Mylar Asurface, ballon t g
where t is the thickness of the Mylar
skin of the balloon.
Fg = πρ Mylar d 2 t g
πd 2 I
Fr
I
2c
=
=
2
Fg πρ Mylar d t g 2 ρ Mylart gc
2836 Chapter 30
Assuming the thickness of the Mylar skin of the balloon to be 1 mm, substitute
numerical values and evaluate Fr/Fg:
kW
1.35 2
Fr
m
=
≈ 2 × 10 −7
kg
m
m
Fg
⎛
⎞⎛
⎞
⎛
⎞
2 ⎜1.00 × 10 3 3 ⎟ ⎜ 9.81 2 ⎟ (1 mm )⎜ 2.998 × 10 8 ⎟
s⎠
m ⎠⎝
s ⎠
⎝
⎝
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 PS A 2π r 2 c 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 PS A 2π r 2 c
(
)
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:
The intensity of the solar radiation on
P
the sail is given by I = s 2 .
4πr
Substituting for I yields:
2 IA
c
where A is the area of the sail.
Fr = Pr A =
Fr =
2 Ps A
Ps A
=
2
4π r c
2π r 2 c
Maxwell’s Equations and Electromagnetic Waves 2837
(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:
PS A
PS A
Fr
2π r 2 c
=
=
Fg GmM S 2π cGmM S
r2
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:
⎛π
2⎞
W )⎜ (15 m ) ⎟
Fr
⎝4
⎠
=
2
Fg
m ⎞⎛
N⋅m ⎞
⎛
⎟ (500 kg ) (1.99 × 10 30 kg )
2π ⎜ 2.998 × 10 8 ⎟⎜⎜ 6.673 × 10 −11
2 ⎟
s
kg ⎠
⎝
⎠⎝
(3.8 × 10
26
= 5.4 × 10 − 4
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 parallelplate capacitor:
E=
Q
∈0 A
2838 Chapter 30
1 dQ
dE d ⎡ Q ⎤
I
= ⎢
=
⎥=
dt dt ⎣∈0 A ⎦ ∈0 A dt ∈0 A
Differentiate this expression with
respect to time to obtain an
expression for the rate of change of
the electric field strength:
Substitute numerical values and evaluate dE/dt:
5.0 A
dE
=
= 3.40 ×1014 V/m ⋅ s
2
2
2
−12
dt
8.854 ×10 C / N ⋅ m π (0.023 m )
(
)
= 3.4 ×1014 V/m ⋅ s
(b) Express the displacement current
Id:
I d =∈ 0
d φe
dt
Substitute for the electric flux to
obtain:
I d =∈ 0
d
[EA] =∈0 A dE
dt
dt
Substitute numerical values and evaluate Id:
(
)
(
)
I d = 8.854 × 10 −12 C 2 / N ⋅ m 2 π (0.023 m ) 3.40 × 1014 V/m ⋅ s = 5.0 A
2
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:
I d =∈ 0
dφ e
dt
Substitute for the electric flux to
obtain:
I d =∈ 0
d
[EA] =∈0 A dE
dt
dt
Because E = (0.050 N/C)sin 2000t :
d
[(0.050 N/C)sin 2000t ]
dt
= 2000 s -1 ∈ 0 A(0.050 N/C ) cos 2000t
I d =∈ 0 A
(
)
Maxwell’s Equations and Electromagnetic Waves 2839
Id will have its maximum value
when cos 2000t = 1. Hence:
I d, max = (2000 s -1 )∈ 0 A(0.050 N/C )
Substitute numerical values and evaluate Id,max:
⎛
C2
I d, max = (2000 s −1 )⎜⎜ 8.854 × 10 −12
N ⋅ m2
⎝
⎞
N⎞
⎛
⎟⎟ (1.00 m 2 )⎜ 0.050 ⎟ = 0.89 nA
C⎠
⎝
⎠
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:
Assuming that the displacement
current is uniformly distributed:
r r
B
∫ ⋅ d l = 2πrB = μ0 I enclosed = μ0 I
C
Id
r2
I
I
=
Id
=
⇒
π r 2 π R2
R2
where R is the radius of the circular
plates.
Substituting for I yields:
Substitute numerical values and
evaluate B:
2π rB =
B(r ) =
μ0 r 2
R
2
Id ⇒ B =
(4π × 10
−7
μ0r
Id
2π R 2
)
N / A 2 (5.0 A )
2π (0.023 m )
2
r
T⎞
⎛
= ⎜1.9 × 10 −3 ⎟ r
m⎠
⎝
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
2840 Chapter 30
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-coreparallel-plate capacitor to show that Id = C dV/dt.
(a) Use its definition to express the
displacement current Id:
I d =∈0
d φe
dt
Substitute for the electric flux to
obtain:
I d =∈ 0
d
[EA] =∈0 A dE
dt
dt
Because E = V/d:
I d =∈0 A
The capacitance of an air-coreparallel-plate capacitor whose plates
have area A and that are separated by
a distance d is given by:
C=
Substituting yields:
d
dt
⎡V ⎤ ∈ 0 A dV
⎢⎣ d ⎥⎦ = d dt
∈0 A
d
Id = C
dV
dt
(b) Substitute in the expression derived in (a) to obtain:
I d = (5.00 nF)
(
)
d
[(3.00 V ) cos 500π t ] = −(5.00 nF)(3.00 V ) 500π s −1 sin 500π t
dt
= − (23.6 μA )sin 500π t
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
r r
strength between the plates? (c) Find the value of the line integral ∫ B ⋅ d l , where
C
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 2841
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
r r
Ampere’s law to evaluate B ⋅ d l around the given path.
(a) From conservation of charge we
know that:
I d = I = 10 A
(b) Express the displacement current
Id:
I d =∈ 0
Substituting for dE/dt yields:
I
dE
= d
dt ∈ 0 A
Substitute numerical values and
evaluate dE/dt:
10 A
dE
=
dt ⎛
C2
⎜⎜ 8.85 × 10 −12
N ⋅ m2
⎝
V
= 2.3 × 1012
m⋅s
dφ e
d
dE
=∈ 0 [EA] =∈ 0 A
dt
dt
dt
⎞
⎟⎟ 0.50 m 2
⎠
(
r r
B
∫ ⋅ d l = μ0 I enclosed
(c) Apply Ampere’s law to a circular
path of radius r between the plates
and parallel to their surfaces to
obtain:
C
Assuming that the displacement
current is uniformly distributed and
letting A represent the area of the
circular plates yields:
I enclosed I d
π r2
I
=
Id
⇒
=
enclosed
A
A
π r2
Substitute for I enclosed to obtain:
r r μ 0π r 2
∫CB ⋅ d l = A I d
Substitute numerical values and evaluate
(
)
r r
B
∫ ⋅ dl :
C
)
r r 4π ×10−7 N / A 2 π (0.10 m )2 (10 A )
B
= 0.79 μT ⋅ m
∫C ⋅ d l =
0.50 m 2
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
2842 Chapter 30
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
μ Ia
1
.
that the magnitude of the magnetic field at point P is given by B = 0
2
2π R R + 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
32
Qπr 2
r + a 2 dr . (c) Use the result from Part (b) to
strip is given by E x dA =
(
)
∈0
show that the total electric flux φe through a circular surface S of radius R. is
given by φe =
Q⎛
1−
∈o ⎜⎝
and show that I + Id = I
a
a2 + R
a
⎞
. (d) Find the displacement current Id through S,
2 ⎟
⎠
(e) Finally, show that the generalized form of
a2 + R 2
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:
Substitute for sinθ1 and sinθ2 to
obtain:
(b) Express the electric flux through
the circular strip of radius r and
width dr in the yz plane:
The electric field due to the dipole is:
μ0 I
(sin θ1 + sin θ 2 )
4π R
where
a
sin θ1 = sin θ 2 =
2
R + a2
BP =
μ0 I
2a
4π R R 2 + a 2
1
μ 0 Ia
=
2π R R 2 + a 2
BP =
dφe = E x dA = E x (2π rdr )
Ex =
2kQ
2kQa
cos θ1 =
32
2
2
r +a
r + a2
2
(
)
Maxwell’s Equations and Electromagnetic Waves 2843
Substitute for Ex to obtain:
dφe = Ex dA =
=
2
+ a2
2Qa
(
4π ∈0 r 2 + a 2
=
(c) Multiply both sides of the
expression for dφe by ∈0:
(r
2kQa
32
(r
Qa
2
+ a2
)
32
32
32
Qa
∈0 (r 2 + a 2 )
∈0 dφe =
)
)
(2π rdr )
(2π rdr )
rdr
rdr
Integrate r from 0 to R to obtain:
R
∈0 φe = Qa ∫
0
(r
rdr
2
+a
)
2 32
⎛
⎛
1⎞
a
−1
= Qa⎜⎜
+ ⎟⎟ = Q⎜⎜1 −
2
2
a⎠
R2 + a2
⎝ R +a
⎝
(d) The displacement current is
defined to be:
⎞
⎟
⎟
⎠
dφe d ⎡ ⎛
a
= ⎢Q⎜⎜1 −
2
dt dt ⎢⎣ ⎝
R + a2
⎛
⎞ dQ
a
⎟
= ⎜⎜1 −
2
2 ⎟ dt
R +a ⎠
⎝
⎞
⎛
a
⎟
= − I ⎜⎜1 −
2
2 ⎟
R +a ⎠
⎝
I d =∈0
The total current is the sum of I and
Id:
⎛
a
I + I d = I − I ⎜⎜1 −
R2 + a2
⎝
a
= I
2
R + a2
(e) Apply Equation 30-4 (the
generalized form of Ampere’s law)
to obtain:
∫
Solving for B yields:
B=
C
⎞
⎟
⎟
⎠
r r
B ⋅ d l = 2πRB = μ 0 (I + I d )
μ0
(I + I d )
2π R
⎞⎤
⎟⎥
⎟
⎠⎥⎦
2844 Chapter 30
Substitute for I + Id from (d) to
obtain:
B=
=
μ0 ⎛
a
⎜I
⎜
2π R ⎝
R2 + a2
⎞
⎟
⎟
⎠
1
μ0 Ia
2
2π R R + a 2
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:
λ yellow-green = 580 nm
The corresponding frequency is:
f yellow-green =
c
λ yellow-green
2.998 × 10 8 m/s
=
580 nm
= 5.17 × 1014 Hz
22 •
(a) What is the frequency of microwave radiation that has a 3.00-cmlong 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:
Substitute numerical values and
evaluate f:
f =
f =
c
λ
2.998 × 108 m/s
= 1.00 × 1010 Hz
−2
3.00 × 10 m
= 10.0 GHz
Maxwell’s Equations and Electromagnetic Waves 2845
(b) The ratio of the shortest
wavelength green light to the shortest
wavelength red light is:
λshortest green 520 nm
≈
= 0.84
λshortest red
620 nm
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:
f =
c
λ
=
2.998 × 10 8 m/s
0.100 × 10 −9 m
= 3.00 × 1018 Hz
(b) The frequency of light with a
wavelength of 550 nm is:
f =
2.998 ×108 m/s
= 5.45 ×1014 Hz
550 nm
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 :
I (θ , r ) =
C
sin 2 θ
2
r
where C is a constant.
(1)
2846 Chapter 30
Express I(90°,10 m):
I (90°,10 m ) = I1 =
=
C
100 m 2
Solving for C yields:
C = (100 m 2 ) I1
Substitute in equation (1) to obtain:
(100 m )I
I (θ , r ) =
2
r
(a) Evaluate equation (2) for
r = 30 m and θ = 90°:
sin 2 θ
2
(30 m )
1
9
sin 2 90°
1
sin 2 45°
I1
(100 m )I
I (45°,10 m ) =
2
(10 m )
1
2
2
I1
(100 m ) I
I (30°,20 m ) =
2
(20 m )
=
(2)
1
2
=
(c) Evaluate equation (2) for
r = 20 m and θ = 30°:
1
2
(100 m ) I
I (90°,30 m ) =
=
(b) Evaluate equation (2) for
r = 10 m and θ = 45°:
C
sin 2 90°
2
(10 m )
1
16
2
1
sin 2 30°
I1
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 :
I (θ , r ) =
C
sin 2 θ
2
r
where C is a constant.
(1)
Maxwell’s Equations and Electromagnetic Waves 2847
Express I(90°,10 m):
I (90°,10 m ) = I1 =
=
C
100 m 2
Solving for C yields:
C = (100 m 2 ) I1
Substitute in equation (1) to obtain:
(100 m )I
I (θ , r ) =
2
Solve for θ to obtain:
(b) For θ = 45° and I(θ,r) = I1:
(100 m )I
=
2
I1
(5.0 m )
1
2
sin 2 θ
(2)
sin 2 θ ⇒ sin 2 θ =
1
4
θ = sin −1 ( 12 ) = 30°
(100 m ) I
=
2
I1
r
(
r=
1
2
1
2
or
r 2 = 12 100 m 2
Solve for r to obtain:
1
2
r
(a) For r = 5 m and I(θ,r) = I1:
C
sin 2 90°
2
(10 m )
sin 2 45°
)
(100 m ) =
2
7.1m
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 :
I (θ , r ) =
C
sin 2 θ
2
r
where C is a constant.
(1)
2848 Chapter 30
Use the given data to obtain:
4 × 10 −12 W/m 2 =
Solving for C yields:
C = (4.00 km ) (4.00 × 10 −12 W/m 2 )
C
sin 2 90°
2
(4.00 km )
C
=
(4.00 km )2
2
= 6.40 × 10 −5 W
Substitute in equation (1) to obtain:
For a point at sea level and 1.50 km
from the transmitter:
6.40 ×10−5 W 2
I (θ , r ) =
sin θ
r2
(2)
⎛ 2.00 km ⎞
⎟⎟ = 53.1°
⎝ 1.50 km ⎠
θ = tan −1 ⎜⎜
Evaluate I(53.1°,1.50 km):
I (53.1°,1.5 km ) =
6.40 × 10 −5 W 2
sin 53.1° = 18.2 pW/m 2
2
(1.50 km)
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
r
and the position vector r . 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.
sin 2 θ
r2
Express the intensity of the signal as
a function of r and θ :
I (r , θ ) = C
At a horizontal distance of 120 km
from the station:
I (120 km,90°) = C
sin 2 90°
(120 km )2
C
=
(120 km )2
(1)
Maxwell’s Equations and Electromagnetic Waves 2849
From the definition of intensity we
have:
dP = IdA
and
Ptot = ∫∫ I (r , θ ) dA
where, in polar coordinates,
dA = r 2 sin θ dθ dφ
Substitute for dA to obtain:
2π π
Ptot =
∫ ∫ I (r ,θ ) r
2
sin θ dθ dφ
0 0
Substitute for I(r,θ):
2π π
Ptot = C ∫ ∫ sin 3 θ dθ dφ
0 0
From integral tables we find that:
π
∫ sin
3
θ dθ = − 13 cosθ (sin 2 θ + 2 )]0 =
π
0
4
3
2π
Substitute and integrate with respect
to φ to obtain:
4
4
8π
2π
Ptot = C ∫ dφ = C [φ ]0 =
C
3 0
3
3
Solving for C yields:
C=
3
Ptot
8π
Substitute for Ptot and evaluate C to
obtain:
C=
3
(500 kW ) = 59.68 kW
8π
Substituting for C in equation (1)
and evaluating I(120 km, 90°):
I (120 km, 90°) =
59.68 kW
(120 km )2
= 4.14 μW/m 2
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
r
dipole moment and the position vector r . We can integrate the intensity to express
2850 Chapter 30
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:
dP = IdA
and
Ptot = ∫∫ I (r , θ ) dA
where, in polar coordinates,
dA = r 2 sin θ dθ dφ
Substitute for dA to obtain:
2π π
Ptot =
∫ ∫ I (r ,θ ) r
2
sin θ dθ dφ (1)
0 0
sin 2 θ
r2
Express the intensity of the signal as
a function of r and θ :
I (r ,θ ) = C
Substitute for I(r,θ) in equation (1) to
obtain:
Ptot = C ∫ ∫ sin 3 θ dθ dφ
From integral tables we find that:
(2)
2π π
0 0
π
∫ sin
3
θ dθ = − 13 cosθ (sin 2 θ + 2)]0 =
π
0
Substitute and integrate with respect
to φ to obtain:
From equation (2) we have:
4
3
2π
Ptot =
C=
4
4
8π
2π
C ∫ dφ = C [φ ] 0 =
C
3 0
3
3
I (r ,θ )r 2
sin 2 θ
Substitute for C in the expression for
Ptot to obtain:
8π I (r ,θ )r 2
3 sin 2 θ
or, because θ = 90°,
8π
Ptot =
I (r )r 2
3
Substitute numerical values and
evaluate Ptot:
Ptot =
Ptot =
8π
2
2.00 × 10−13 W/m2 (30.0 km )
3
(
)
= 1.51 mW
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 2851
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
r
between the electric dipole moment and the position vector r . 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 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 :
sin 2 θ
I (r ,θ ) = C 2
r
From the definition of intensity we
have:
dP = IdA
and
Ptot = ∫∫ I (r , θ ) dA
(1)
where, in polar coordinates,
dA = r 2 sin θ dθ dφ
Substitute for dA to obtain:
2π π
Ptot =
∫ ∫ I (r ,θ ) r
2
sin θ dθ dφ
0 0
Substituting for I(r,θ) yields:
2π π
Ptot = C ∫ ∫ sin 3 θ dθ dφ
0 0
From integral tables we find that:
π
(
3
2
∫ sin θ dθ = − 13 cos θ sin θ + 2
)]
π
0
0
2π
Substitute and integrate with respect
to φ to obtain:
4
4
8π
2π
Ptot = C ∫ dφ = C [φ ] 0 =
C
3 0
3
3
Solving for C yields:
C=
Substitute for C in equation (1) to
obtain:
3Ptot sin 2 θ
I (r ,θ ) =
8π r 2
3
Ptot
8π
=
4
3
2852 Chapter 30
At the elevation of the plane:
⎛ 4000 m ⎞
⎟⎟ = 58.0°
⎝ 2500 m ⎠
θ = tan −1 ⎜⎜
and
r=
Substitute numerical values and
evaluate I(4717 m, 58°):
(2500 m )2 + (4000 m)2
= 4717 m
3(100 W ) sin 2 58.0°
I (4717 m, 58.0°) =
8π
(4717 m )2
= 386 nW/m 2
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.
B
(a) Relate the intensity of the
electromagnetic wave to Erms and
Brms:
B
I=
B
Erms Brms
μ0
or, because Brms = Erms/c,
E E c E2
I = rms rms = rms
μ0
μ0c
B
Erms = μ 0 cI
Solving for Erms yields:
Substitute numerical values and evaluate Erms:
E rms =
(4π ×10
−7
N/A 2 )(2.998 × 108 m/s)(100 W/m 2 ) = 194 V/m
(b) Express Brms in terms of Erms:
Brms =
Erms
c
Substitute numerical values and
evaluate Brms:
Brms =
194 V/m
= 647 nT
2.998 × 10 8 m/s
B
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 2853
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.
B
B
(a) Relate Erms to E0:
Erms =
E0
2
=
400 V/m
= 282.8 V/m
2
= 283 V/m
(b) Find Brms from Erms:
B
Brms =
E rms
282.8 V/m
=
c
2.998 × 10 8 m/s
= 0.9434 μT = 943 nT
(c) The intensity of an
electromagnetic wave is given by:
I=
Substitute numerical values and
evaluate I:
I=
Erms Brms
μ0
(282.8 V/m )(0.9434 μT )
4π × 10 −7 N/A 2
= 212.3 W/m 2 = 212 W/m 2
(d) Express the radiation pressure in
terms of the intensity of the wave:
Pr =
I
c
Substitute numerical values and
evaluate Pr:
Pr =
212.3 W/m 2
= 708 nPa
2.998 × 108 m/s
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 .
B
B
B
(a) Express Brms in terms of Erms:
B
Brms =
Erms
c
2854 Chapter 30
Substitute numerical values and
evaluate Brms:
Brms =
B
400 V/m
= 1.334 μT
2.998 ×10 8 m/s
= 1.33 μT
(b) The average energy density uav is
given by:
uav =
Substitute numerical values and
evaluate uav:
u av =
Erms Brms
μ0c
(400 V/m )(1.334 μT )
(4π ×10
−7
)(
N/A 2 2.998 × 10 8 m/s
)
= 1.417 μJ/m 3 = 1.42 μJ/m 3
(c) Express the intensity as the
product of the average energy
density and the speed of light in a
vacuum:
I = uavc
Substitute numerical values and
evaluate I:
I = 1.417 μJ/m 3 2.998 × 108 m/s
(
)(
)
= 425 W/m 2
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:
Fr = Pr A
Relate the radiation pressure to the
intensity of the wave:
Pr =
I
c
Substitute for Pr to obtain:
Fr =
IA
c
Maxwell’s Equations and Electromagnetic Waves 2855
Substitute numerical values and
evaluate Fr:
(200 W/m )(0.20 m)(0.30 m)
=
2
Fr
2.998 × 108 m/s
= 40 nN
(b) If the card reflects all of the
radiation incident on it, the force
exerted on the card is doubled:
Fr = 80 nN
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:
Fr = 2 Pr A cosθ
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:
Pr =
I
c
Substitute for Pr to obtain:
Fr =
2 IA cos θ
c
Substitute numerical values and evaluate Fr:
2(200 W/m 2 )(0.20 m )(0.30 m )cos 30°
= 69 nN
Fr =
2.998 ×108 m/s
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:
I (θ ) = I 0 sin 2 θ
(1)
where I0 is the maximum intensity.
2856 Chapter 30
For I = 12 I 0 :
1
2
Solving for θ yields:
θ = sin −1
(b) For I = 0.01I 0 :
0.01I 0 = I 0 sin 2 θ ⇒ sin 2 θ = 0.01
Solving for θ yields:
θ = sin −1 0.01 = 5.7°
I 0 = I 0 sin 2 θ ⇒ sin 2 θ =
( )=
45°
1
2
(
1
2
)
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 the pulse.
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 u =∈0 E 2 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:
L = cΔt
Substitute numerical values and
evaluate L:
L = 2.998 × 10 8 m/s (10.0 ns ) = 2.998 m
(b) The energy density within the
pulse is the energy of the beam per
unit volume:
u=
U
U
=
V π r2L
Substitute numerical values and
evaluate u:
u=
20.0 J
π (2.00 mm)2 (2.998 m )
(
)
= 3.00 m
= 530.9 kJ/m 3 = 531kJ/m 3
(c) E is related to u according to:
2
u =∈ 0 E rms
⇒ E rms =
u
∈0
Maxwell’s Equations and Electromagnetic Waves 2857
Substitute numerical values and
evaluate Erms:
E rms =
530.9 kJ/m 3
8.854 ×10 −12 C 2 / N ⋅ m 2
= 244.9 MV/m = 245 MV/m
Use Brms = Erms/c to find Brms:
B
B
Brms =
244.9 MV/m
= 0.817 T
2.998 × 10 8 m/s
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
r
S ( x, t ) = 100 W/m 2 cos 2 [kx − ωt ] iˆ , 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
r
function. We can find E from S = E 2 μ 0 c and B from B = E/c. Finally, we can
r
use the definition of the Poynting vector and the given expression for S to find
r
r
E and B .
(a) Because the argument of the cosine function is of the form kx − ωt , the wave
propagates in the +x direction.
2π
(b) Examining the argument of the
cosine function, we note that the
wave number k of the wave is:
k=
Examining the argument of the
cosine function, we note that the
angular frequency ω of the wave
is:
ω = 2πf = 3.00 ×109 s −1
Solving for f yields:
3.00 ×10 9 s −1
f =
= 477 MHz
2π
r
(c) Express the magnitude of S in
terms of E:
λ
= 10.0 m −1 ⇒ λ = 0.628 m
r E2
r
S =
⇒ E = μ0c S
μ0c
2858 Chapter 30
Substitute numerical values and evaluate E:
(4π ×10
E=
−7
)(
Because
r
S ( x, t ) = 100 W/m 2 cos 2 [kx − ω t ] iˆ
r 1 r r
and S =
E×B:
(
)(
)
N/A 2 2.998 ×108 m/s 100 W/m2 = 194.1V/m
)
μ0
r
E ( x, t ) = (194 V/m ) cos[kx − ωt ] ˆj
where k = 10.0 rad/m and
ω = 3.00 × 109 rad/s.
194.1V/m
= 647.4 nT
2.998 × 10 8 m/s
Use B = E/c to evaluate B:
B=
r 1 r r
Because S =
E × B , the direction
r
B ( x, t ) =
r
of B must be such that the cross
r
r
product of E with B is in the
positive x direction:
where k = 10.0 rad/m and
ω = 3.00 × 109 rad/s.
μ0
(647 nT )cos[kx − ωt ] kˆ
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
r
plates. In (c), we’ll demonstrate that the flux of S 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:
I d =∈0
dφe
d
dE
=∈0 ( AE ) =∈0 A
dt
dt
dt
Maxwell’s Equations and Electromagnetic Waves 2859
The electric field strength between
the plates of the capacitor is given
by:
Substituting for E yields:
E=
Q
∈0 A
where Q is the instantaneous charge on
the capacitor plates.
I d =∈0 A
d Q
dQ
=
= I
dt ∈0 A dt
r
r
(b) Because E is perpendicular to the plates of the capacitor and B is tangent to
circles that are concentric and whose center is through the middle of the capacitor
r
plates, S points radially inward toward the center of the capacitor.
(c) The Poynting vector is:
r 1 r r
S=
E×B
μ0
r
Letting the direction of E be the +x
direction:
Apply Ampere’s law to a closed
circular path of radius R ≤ b to
obtain:
(1)
r
E = Eiˆ
where E is the electric field strength
between the plates of the capacitor.
B(2π R ) = μ0 I d
dφe
d
= μ0 ∈0 EA
dt
dt
dE
= μ 0 ∈0 π R 2
dt
Substituting for Id and simplifying
yields:
B(2πR ) = μ0 ∈0
Solve for B to obtain:
B=
μ0 ∈0
2
R
dE
dt
and
r
μ ∈ dE ˆ
B=− 0 0 R
j
2
dt
where ĵ is a unit vector that is tangent
to the concentric circles.
2860 Chapter 30
r
r
Substitute for B and E in equation
(1) and simplify to obtain:
r
E
x
(
R̂
r
B
y
r
S
R
The rate at which energy is stored
in the capacitor is:
Because E =
r ⎛ 1 ⎞ ⎛ μ ∈ dE ⎞
S = ⎜⎜ E ⎟⎟iˆ × ⎜ − 0 0 R
⎟ ˆj
dt
μ
2
⎠
⎝ 0 ⎠ ⎝
∈ E dE ˆ ˆ
= 0
R i ×−j
2 dt
∈ E dE ˆ
= − 0
RR
2 dt
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, R̂ is a unit vector
pointing radially outward from the line
joining the centers of the plates, and b
is the radius of the plates.
Q
:
∈0 A
(
dU d
d
= (uV ) = V
∈0 E 2
dt dt
dt
dE
= Ad ∈0 E
dt
Substituting for E and simplifying
yields:
Because
)
⎛ Q ⎞d ⎛ Q ⎞
dU
⎟⎟
⎟⎟ ⎜⎜
= Ad ∈0 ⎜⎜
dt
⎝∈0 A ⎠ dt ⎝∈0 A ⎠
=
Consider a cylindrical surface of
length d and radius b. Because
r
S points inward, the energy flowing
into the solenoid per unit time is:
)
Qd dQ Qd
=
I
∈0 A dt ∈0 A
∫ S dA = S (2π bd )
n
dE ⎞
⎛
= ⎜ 12 ∈0 Eb
⎟ (2π bd )
dt ⎠
⎝
dE
= π ∈0 Eb 2 d
dt
⎛ Q ⎞ 2 d ⎛ Q ⎞
⎜⎜
⎟⎟b d ⎜⎜
⎟
dt ⎝ ∈0 A ⎟⎠
⎝ ∈0 A ⎠
2
⎛ Q ⎞ b d dQ Qd
=π⎜ 2 ⎟
=
I
⎝ πb ⎠ ∈0 A dt ∈0 A
∫ S dA = π ∈
n
0
r
S
dA
=
dU
dt
S
,
we’ve
proved
that
the
flux
of
into the region
∫ n
between the capacitor is equal to the rate of change of the energy stored in the
capacitor.
Maxwell’s Equations and Electromagnetic Waves 2861
39 ••
[SSM] A pulsed laser fires a 1000-MW pulse that has a 200-ns
duration at a small object that has 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.
Apply conservation of energy to
obtain:
Uf is given by:
Substitute for Uf:
Solve for θ to obtain:
θ
L
L cos θ
m
h
Ug = 0
K f − Ki + U f − U i = 0
pi2
,
or, because Ui = Kf = 0 and K i =
2m
p2
− i + Uf = 0
2m
U f = mgh = mgL(1 − cosθ )
−
pi2
+ mgL(1 − cos θ ) = 0
2m
⎛
θ = cos −1 ⎜⎜1 −
⎝
pi2 ⎞
⎟
2m 2 gL ⎟⎠
2862 Chapter 30
Use conservation of momentum to
relate the momentum of the
electromagnetic pulse to the initial
momentum pi of the pendulum bob:
U P Δt
=
= pi
c
c
where Δt is the duration of the pulse.
pem wave =
2
⎡
P 2 (Δt ) ⎤
θ = cos −1 ⎢1 − 2 2 ⎥
⎣ 2m c gL ⎦
Substitute numerical values and evaluate θ :
Substitute for pi:
2
2
⎤
⎡
(
1000 MW ) (200 ns )
θ = cos ⎢1 −
⎥
2
2
8
2
⎣⎢ 2(10.0 mg ) 2.998 × 10 m/s 9.81 m/s (0.0400 m )⎦⎥
−1
(
)(
)
= 6.10 × 10−3 degrees
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.)
The mirrors used in a particular type of laser are 99.99% reflecting.
40 ••
(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:
P=
15 W
= 1.5 × 10 5 W
1.0 × 10 − 4
Maxwell’s Equations and Electromagnetic Waves 2863
(b) Use the definition of radiation
pressure to obtain:
Fr
A
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:
2I 2P
=
c
Ac
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:
Fr 2 P
2P
⇒ Fr =
=
A Ac
c
Substitute numerical values and
evaluate Fr:
2 1.5 × 105 W
Fr =
= 1.0 mN
2.998 × 108 m/s
Pr =
Pr =
(
)
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:
Express the radiation pressure in
terms of the intensity of the
radiation I from the Sun:
Substituting for Pr, Earth and A yields:
Pr, Earth =
Fr, Earth
Pr, Earth =
I
c
⇒ Fr, Earth = Pr, Earth A
A
where A is the cross-sectional area of
Earth.
Fr, Earth
Iπ R 2
=
c
2864 Chapter 30
Substitute numerical values and
evaluate Fr:
π (1.37 kW/m 2 )(6.37 × 10 6 m )
2
Fr, Earth =
2.998 × 10 8 m/s
= 5.825 × 108 N
= 5.83 × 108 N
The gravitational force exerted
on Earth by the Sun is given by:
Gmsun mearth
r2
where r is the radius of Earth’s orbit.
Fg, Earth =
Substitute numerical values and evaluate Fg, Earth:
Fg, Earth =
(6.673 ×10
−11
)(
)(
N ⋅ m 2 / kg 2 1.99 × 10 30 kg 5.98 × 10 24 kg
(1.50 ×10
Express the ratio of the force due to
radiation pressure Fr, Earth to the
gravitational force Fg, Earth:
(b) The radiation pressure exerted
on Mars is given by:
Express the radiation pressure
on Mars in terms of the intensity
of the radiation IMars from the
sun:
Substituting for Pr, Mars and A yields:
Express the ratio of the solar
constant at Earth to the solar
constant at Mars:
Substitute for I Mars to obtain:
11
m
)
2
Fr, Earth
=
) = 3.529 ×10
22
N
5.825 × 10 8 N
= 1.65 × 10 −14
22
3.529 × 10 N
Fg, Earth
or
Fr, Earth = (1.65 × 10 −14 )Fg, Earth
Pr, Mars =
Fr, Mars
Pr, Mars =
I Mars
c
⇒ Fr, Mars = Pr, Mars A
A
where A is the cross-sectional area of
Mars.
Fr, Mars
2
I Marsπ RMars
=
c
I Mars ⎛ rearth
=⎜
I earth ⎜⎝ rMars
Fr, Mars
2
⎞
⎛r
⎟⎟ ⇒ I Mars = I earth ⎜⎜ earth
⎠
⎝ rMars
2
I earth π RMars
=
c
⎛ rearth
⎜⎜
⎝ rMars
⎞
⎟⎟
⎠
2
⎞
⎟⎟
⎠
2
Maxwell’s Equations and Electromagnetic Waves 2865
Substitute numerical values and evaluate Fr, Mars:
π (1.37 kW/m 2 )(3.40 × 10 3 km ) ⎛ 1.50 × 1011 m ⎞
⎟⎟ = 7.18 × 10 7 N
⎜⎜
=
8
11
2.998 × 10 m/s
⎝ 2.28 × 10 m ⎠
2
2
Fr, Mars
The gravitational force exerted on
Mars by the Sun is given by:
Gmsun mMars Gmsun (0.11mEarth )
=
r2
r2
where r is the radius of Mars’ orbit.
Fg, Mars =
Substitute numerical values and evaluate Fg
Fg, Mars =
(6.673 ×10
−11
)(
)
(
N ⋅ m 2 / kg 2 1.99 × 10 30 kg (0.11) 5.98 × 10 24 kg
(2.28 ×10
11
m
)
)
2
= 1.68 × 10 21 N
Express the ratio of the force due to
radiation pressure Fr, Mars to the
gravitational force Fg, Mars:
Fr, Mars
=
Fg, Mars
or
Fr, Mars =
7.18 × 10 7 N
= 4.27 × 10 −14
21
1.68 × 10 N
(4.27 × 10 )F
−14
g, Mars
(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 R 2 / R 3 = R −1 . Because Mars is smaller than Earth, the ratio
is larger.
The Wave Equation for Electromagnetic Waves
Show by direct substitution that Equation 30-8a is satisfied by the
42 •
wave function Ey = E0 sin (kx − ω t ) = 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.
2866 Chapter 30
Differentiate E y = E0 sin (kx − ωt )
∂E y
∂
[E0 sin(kx − ωt )]
∂x
∂x
= kE0 cos(kx − ωt )
with respect to x:
=
Evaluate the second partial
derivative of Ey with respect to x:
∂2Ey
Differentiate E y = E0 sin (kx − ωt )
∂E y
∂x
2
=
Evaluate the second partial
derivative of Ey with respect to t:
∂2Ey
Divide equation (1) by equation (2)
to obtain:
∂2 Ey
•
(1)
∂
[E0 sin(kx − ωt )]
∂t
∂t
= −ωE0 cos(kx − ωt )
with respect to t:
43
∂
[kE0 cos(kx − ωt )]
∂x
= −k 2 E0 sin( kx − ωt )
=
∂t
2
∂
[− ωE0 cos(kx − ωt )] (2)
∂t
= −ω 2 E 0 sin( kx − ωt )
=
2
2
∂x 2 = − k E0 sin (kx − ωt ) = k
∂ 2 E y − ω 2 E0 sin (kx − ωt ) ω 2
∂t 2
or
2
∂2Ey k 2 ∂2 Ey
1 ∂ Ey
= 2
= 2
ω ∂t 2
c ∂t 2
∂x 2
provided c = ω/k.
Use the values of μ0 and ∈0 in SI units to compute 1 ∈0 μ0 and
show that it is equal to 3.00 × 108 m/s.
Picture the Problem Substitute numerical values and evaluate c:
c=
(4π ×10
1
−7
N/A
2
)(8.854 ×10
−12
C / N⋅m
2
2
)
= 3.00 ×108 m/s
r
••
(a) Use Maxwell’s equations to show for a plane wave, in which E
r
∂B y
∂E z
∂E z ∂B y
=
= μ0 ∈0
and
.
and B are independent of y and z, that
∂t
∂t
∂x
∂x
(b) Show that Ez and By also satisfy the wave equation.
44
B
Maxwell’s Equations and Electromagnetic Waves 2867
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.
∂E z
Δx
∂x
In Figure 30-5, replace Bz by Ez. For
Δx small:
E z ( x2 ) = E z ( x1 ) +
r
Evaluate the line integral of E
around the rectangular area ΔxΔz:
∫ E ⋅ dl ≈ −
Express the magnetic flux through
the same area:
∫ B dA = B ΔxΔz
Apply Faraday’s law to obtain:
r r
∂
∂
E
∫ ⋅ d l ≈ − ∂ t ∫S Bn dA = − ∂ t (By ΔxΔz )
∂B y
=−
ΔxΔz
∂t
∂B
∂E
− z ΔxΔz = − y ΔxΔz
∂x
∂t
or
∂E z ∂B y
=
∂x
∂t
B
Substitute in equation (1) to obtain:
In Figure 30-6, replace Ey by By and
r
evaluate the line integral of B
around the rectangular area ΔxΔz:
B
Evaluate these integrals to obtain:
r
S
n
∂E z
ΔxΔz
∂x
(1)
y
r r
B
∫ ⋅ d l = μ 0 ∈0
∫
S
En dA
provided there are no conduction
currents.
∂B y
∂x
(b) Using the first result obtained in
(a), find the second partial derivative
of Ez with respect to x:
r
= μ0 ∈0
∂ ⎛ ∂E z
⎜
∂x ⎝ ∂x
∂E z
∂t
⎞ ∂ ⎛ ∂B y
⎟ = ⎜⎜
⎠ ∂x ⎝ ∂ t
or
∂ ⎛ ∂B
∂ 2 Ez
= ⎜⎜ y
2
∂ t ⎝ ∂x
∂x
⎞
⎟⎟
⎠
⎞
⎟⎟
⎠
2868 Chapter 30
Use the second result obtained in (a)
to obtain:
∂ ⎛
∂E z ⎞
∂ 2 Ez
∂ 2 Ez
⎜
⎟
μ
ε
μ
∈
=
=
0 0
0
0
∂ t ⎜⎝
∂ t ⎟⎠
∂ t2
∂x 2
or, because μ0∈0 = 1/c2,
1 ∂ 2 Ez
∂ 2 Ez
.
=
c2 ∂ t 2
∂x 2
Using the second result obtained in
(a), find the second partial derivative
of By with respect to x:
B
∂ ⎛ ∂B y
⎜
∂x ⎜⎝ ∂x
or
∂ 2 By
∂x
Use the first result obtained in (a) to
obtain:
2
⎞
∂ ⎛ ∂E z ⎞
⎟⎟ = μ 0 ∈0
⎜
⎟
∂x ⎜⎝ ∂ t ⎟⎠
⎠
= μ 0 ∈0
∂ ⎛ ∂Ez ⎞
⎜
⎟
∂ t ⎝ ∂x ⎠
∂ 2 By
⎞
⎟
μ
∈
=
0
0
⎟
∂t2
∂x 2
⎠
or, because μ0∈0 = 1/c2,
2
∂ 2 By
1 ∂ By
.
=
c2 ∂ t 2
∂x 2
∂ 2 By
= μ 0 ∈0
∂ ⎛ ∂B y
⎜
∂ t ⎜⎝ ∂ t
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:
∂ f ∂u ∂ f ∂ f
=
=
∂ x ∂ x ∂u ∂u
and
∂ f ∂u ∂ f
∂f
=
= −v
∂t
∂t ∂u
∂u
Express the second derivatives of
f with respect to x and t to obtain:
2
∂2 f ∂2 f
∂2 f
2 ∂ f
=
=
v
and
∂ x2 ∂ u 2
∂t2
∂ u2
Divide the first of these equations by
the second to obtain:
∂2f
∂ x2
1
∂2 f
1 ∂2 f
=
⇒
=
∂ x2 v2 ∂ t 2
v2
∂2f
∂t2
Maxwell’s Equations and Electromagnetic Waves 2869
Let u = x + vt. Then:
∂ f ∂u ∂ f ∂ f
=
=
∂ x ∂ x ∂u ∂u
and
∂ f ∂u ∂ f
∂f
=
=v
∂t
∂t ∂u
∂u
Express the second derivatives of
f with respect to x and t to obtain:
2
∂2 f ∂2 f
∂2 f
2 ∂ f
=
=v
and
∂ x2 ∂ u 2
∂t2
∂ u2
Divide the first of these equations by
the second to obtain:
∂2f
∂ x2
1
∂2 f
1 ∂2 f
=
⇒
=
∂ x2 v2 ∂ t 2
v2
∂2f
∂t2
General Problems
46 •
An electromagnetic wave has a frequency of 100 MHz and is traveling
r
in a vacuum. The magnetic field is given by B (z, t ) = 1.00 × 10 −8 T cos(kz − ωt )iˆ .
(a) Find the wavelength and the direction of propagation of this wave. (b) Find
r
the electric field vector E (z, t ) . (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
r
results to obtain E (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:
Substitute numerical values and
evaluate λ:
λ=
c
f
λ=
2.998 × 108 m/s
= 3.00 m
100MHz
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.
2870 Chapter 30
(
r
(b) Express the amplitude of E :
)(
E = cB = 2.998 ×108 m/s 10 −8 T
= 3.00 V/m
)
ω = 2π f = 2π (100 MHz) = 6.28 × 108 s −1
Find the angular frequency and
wave number of the wave:
and
k=
2π
λ
=
2π
= 2.09 m −1
3.00 m
r
r
Because S is in the positive z direction, E must be in the negative y direction in
order to satisfy the Poynting vector expression:
[(
)]
r
E (z , t ) = − (3.00 V/m ) cos 2.09 m −1 z − 6.28 ×108 s −1 t ˆj
) (
(c) Use its definition to express and evaluate the Poynting vector:
(
)
r
1 r r − (3.00 V/m) 10−8 T
S (z , t ) =
E×B =
cos 2 2.09 m −1 z − 6.28 ×108 s −1 t ˆj × iˆ
−7
2
μ0
4π ×10 N/A
or
[(
[(
) ](
) (
)
)]
r
S (z , t ) = 23.9 mW/m 2 cos 2 2.09 m −1 z − 6.28 × 108 s −1 t kˆ
(
)
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:
) (
r
I=S=
1
2
(23.9 mW/m )
2
= 11.9 mW/m 2
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 ε rms = AωB0 2 , 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
ε rms as a function of the intensity.
B
Maxwell’s Equations and Electromagnetic Waves 2871
ε = − dφ m
(
)
d r
d
B ⋅ Anˆ = − (BA)
dt
dt
dt
dB
d
= −A
= −πR 2 (B0 sin ωt )
dt
dt
2
= −πR ωB0 cos ωt = −ε peak cos ωt
The emf induced in the antenna
is given by Faraday’s law:
where
=−
ε peak
= πR 2ωB0 and R is the
radius of the loop antenna..
ε rms equals ε peak divided by the
ε rms =
square root of 2:
The intensity of the signal is given
by:
ε peak
πR 2ωB0
=
2
2
(1)
E 0 B0
2μ 0
or, because E 0 = cB0 ,
I=
cB0 B0 B02 c
I=
=
2μ 0
2μ 0
Solving for B0 yields:
2μ 0 I
c
B0 =
B
Substituting for B0 and ω in
equation (1) and simplifying yields:
B
ε rms =
πR 2 (2πf )
2μ 0 I
c
2
= 2π 2 R 2 f
μ0 I
c
Substitute numerical values and evaluate εrms:
2
2
−7
ε rms = 2π 2 (0.100 m )2 (100 MHz ) (4π × 10 N/A )(84.0 μW/m ) =
2.998 × 10 m/s
2.6 mV
48 ••
The electric field strength from a radio station some distance from the
electric dipole transmitting antenna is given by
1.00 × 10−4 N/C cos 1.00 × 106 rad/s t , 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?
(
) (
)
2872 Chapter 30
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 ε = AωB0 , where A is the area of the loop, B0 is the
B
amplitude of the magnetic field, and ω is the angular frequency of the wave.
(a) Because E is independent of x:
V = El
where l is the length of the wire.
Substitute numerical values and
evaluate V:
V = 1.00 × 10− 4 N/C cos 106 t (0.500 m )
[(
)
= (50.0 μV ) cos 106 t
]
and Vpeak = 50.0 μV
(b) The maximum voltage induced in
a loop is given by:
Eliminate B0 in favor of E0 and
substitute for A to obtain:
ε = ωB0 A
where A is the area of the loop and B0 is
the amplitude of the magnetic field.
B
ε = ωE0π R
2
c
Substitute numerical values and evaluate ε:
ε = (1.00 ×10
6
)(
)
s −1 1.00 ×10− 4 N/C π (0.200 m )
= 41.9 nV
2.998 ×108 m/s
2
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 2873
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:
Express the conduction current Ic in
terms of the potential difference
between the plates and the resistance
of the wire:
Because Q = CV :
Substitute in equation (1):
The capacitance of a parallel-plate
capacitor with plate area A and plate
separation d is given by:
Substituting for C equation (2) gives:
I = Ic +
Ic =
dQ
dt
(1)
V V0
=
sin ωt
R R
dQ
dV
=C
= ωCV 0 cos ωt
dt
dt
I=
C=
V0
sin ωt + ωCV0 cos ωt
R
(2)
∈0 A ∈0 π a 2
d
=
d
⎛1
⎞
ω ∈0 π a 2
⎜
I = V0 ⎜ sin ωt +
cos ωt ⎟⎟
d
⎝R
⎠
2874 Chapter 30
(b) Apply the generalized form of
Ampere’s law to a circular path of
radius r centered within the plates of
the capacitor, where I'd is the
r r
B
∫ ⋅ d l = μ0 (I c + I'd )
C
displacement current through the flat
surface S bounded by the path and Ic
is the conduction current through the
same surface:
By symmetry the line integral is B
times the circumference of the circle
of radius r:
B(2π r ) = μ 0 (I c + I'd )
(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:
dφ e
d
=∈ 0 ( A'E )
dt
dt
dE
dE
=∈ 0 A'
=∈ 0 π r 2
dt
dt
provided (r ≤ a )
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:
E = σ ∈0 , where σ = Q A = Q π a 2
Substituting for E in the equation
for I'd gives:
I' d =∈ 0
(
so
E=
Q
∈0 π a 2
dE
d ⎛V ⎞
=∈0 π r 2 ⎜ ⎟
dt ⎝ d ⎠
dt
∈ π r 2 dV ∈0 π r 2 d
(V0 sin ωt )
= 0
=
d
dt
d
dt
∈0 π r 2
=ω
V0 cos ωt
d
I'd =∈0 π r 2
Solving for B in equation (3) and substituting for Ic and I'd yields:
B(r ) =
)
⎞
μ0 (I c + I'd ) μ0 ⎛ V0
∈ π r2
⎜⎜ sin ωt + ω 0
=
V0 cos ωt ⎟⎟
d
2π r
2π r ⎝ R
⎠
⎞
μ0V0 ⎛ 1
∈0 π r 2
⎜
=
sin ωt + ω
cos ωt ⎟⎟
⎜
d
2π r ⎝ R
⎠
Maxwell’s Equations and Electromagnetic Waves 2875
(c) Both the charge Q and the
conduction current Ic are in phase
with V. However, dQ/dt, which is
equal to the displacement current Id
through S for r ≥ a, lags V by 90°.
(Mathematically, cos ωt lags behind
sin ωt by 90°.) The voltage V leads
the current I = Ic + Id by phase angle
δ. The current relation is expressed
in terms of the current amplitudes:
The values of the conduction and
displacement current amplitudes
are obtained by comparison with
the answer to Part (a):
I = Ic + Id
or
I max sin (ωt + δ ) = I c, max sin ωt
+ I d, max cos ωt
I c, max =
V0
R
and
I d, max =
A phasor diagram for adding the
currents Ic and Id is shown to the
right. The conduction current Ic is in
phase with the voltage V across the
resistor and Id lags behind it by 90°:
ω ∈0 π a 2V0
d
I c, max
r
Ic
δ
I d,max
I max
r
Id
r
I
From the phasor diagram we have:
tan δ =
=
r
V
I d,max
I c,max
=
V0
ω ∈0 π a 2
d
V0 R
Rω ∈0 π a 2
d
so
⎛ Rω ∈0 π a 2 ⎞
⎟⎟
d
⎝
⎠
δ = tan −1 ⎜⎜
Remarks: The capacitor and the resistive wire are connected in parallel.
The potential difference across each of them is the applied voltage V0 sin ωt.
50 ••
A 20-kW beam of electromagnetic radiation is normal to a surface that
reflects 50 percent of the radiation. What is the force exerted by the radiation on
this surface?
2876 Chapter 30
Picture the Problem The total force on the surface is the sum of the force due to
the reflected radiation and the force due to the absorbed radiation. From the
conservation of momentum, the force due to the 10 kW that are reflected is twice
the force due to the 10 kW that are absorbed.
Express the total force on the
surface:
Ftot = Fr + Fa
Substitute for Fr and Fa to obtain:
Ftot =
2( 12 P ) 12 P 3P
+
=
c
c
2c
Substitute numerical values and
evaluate Ftot:
Ftot =
3(20 kW )
= 0.10 mN
2(2.998 ×108 m/s )
51 ••
[SSM] The electric fields of two
r harmonic electromagnetic waves of
angular frequency ω1 and ω2 are given by E1 = E1,0 cos(k1x − ω1t )ˆj and
r
by E2 = E2 ,0 cos(k2 x − ω2 t + δ )ˆj . For the resultant of these two waves, find (a) the
instantaneous Poynting vector and (b) the time-averaged Poynting vector.
(c) Repeat Parts (a)
r and (b) if the direction of propagation of the second wave is
reversed so that E2 = E2 ,0 cos(k2 x + ω 2t + δ )ˆj
Picture the Problem We can use the definition of the Poynting vector and the
r
r
relationship between B and E to find the instantaneous Poynting vectors for each
of the resultant wave motions and the fact that the time average of the cross
product term is zero for ω1 ≠ ω2, and ½ for the square of cosine function to find
the time-averaged Poynting vectors.
(a) Because both waves propagate in
the x direction:
Express B in terms of E1 and E2:
r
r r
E × B = μ 0 Siˆ ⇒ B = Bkˆ
B=
1
(E1 + E2 )
c
Substitute for E1 and E2 to obtain:
r
1
B ( x, t ) = [E1, 0 cos (k1 x − ω1t ) + E 2, 0 cos (k 2 x − ω 2 t + δ )]kˆ
c
Maxwell’s Equations and Electromagnetic Waves 2877
The instantaneous Poynting vector for the resultant wave motion is given by:
r
1
(E1,0 cos(k1 x − ω1t ) + E2,0 cos(k 2 x − ω2t + δ )) ˆj
S ( x, t ) =
μ0
×
=
1
(E1,0 cos(k1 x − ω1t ) + E2,0 cos(k 2 x − ω 2t + δ ))kˆ
c
(
(E
1
μ0c
2
cos(k1 x − ω1t ) + E 2, 0 cos(k 2 x − ω 2 t + δ )) ˆj × kˆ
1, 0
1
= μ0c
[E
2
1, 0
)
cos 2 (k1 x − ω1t ) + 2 E1,0 E 2,0 cos(k1 x − ω1t )
]
× cos(k 2 x − ω 2 t + δ ) + E 22,0 cos 2 (k 2 x − ω 2 t + δ ) iˆ
(b) The time average of the cross
product term is zero for ω1 ≠ ω2, and
the time average of the square of the
cosine terms is ½:
r
S av =
[E
2μ c
1
2
1, 0
]
+ E22,0 iˆ
0
r
(c) In this case B2 = − Bkˆ because the wave with k = k2 propagates in the
− iˆ direction. The magnetic field is then:
r
1
B ( x, t ) = [E1, 0 cos (k1 x − ω1t ) − E 2, 0 cos(k 2 x + ω 2 t + δ )]kˆ
c
The instantaneous Poynting vector for the resultant wave motion is given by:
r
1
(E1,0 cos(k1 x − ω1t ) + E2,0 cos(k 2 x − ω2t + δ )) ˆj
S ( x, t ) =
μ0
×
=
1
(E1,0 cos(k1 x − ω1t ) − E2,0 cos(k 2 x + ω2t + δ ))kˆ
c
[E
μc
1
2
1, 0
]
cos 2 (k1 x − ω1t ) − E 22,0 cos 2 (k 2 x + ω 2 t + δ ) iˆ
0
The time average of the square of the
cosine terms is ½:
r
S av =
[E
2μ c
1
2
1, 0
]
− E22, 0 iˆ
0
∂Bz
∂E
= μo ∈0 n (Equation 30-10) follows from
∂x
∂t
r r
∂E y
∫CB ⋅ d l = −μ 0 ∈0 ∫S ∂t dA (Equation 30-6d with I = 0) by integrating along a
52
••
Show that
2878 Chapter 30
suitable curve C and over a suitable surface S in a manner that parallels the
derivation of Equation 30-9.
Picture the Problem We’ll choose the
curve with sides Δx and Δz in the xy
plane shown in the diagram and apply
Equation 30-6d to show that
∂ Ey
∂ Bz
= − μ 0 ∈0
.
∂x
∂t
Because Δx is very small, we can
approximate the difference in Bz
at the points x1 and x2 by:
Bz ( x2 ) − Bz ( x1 ) = ΔB ≈
Then:
∫ B ⋅ dl ≈ μ
B
r
C
r
0
∈0
∂E y
∂t
∂Bz
Δx
∂x
ΔxΔz
The flux of the electric field through
this curve is approximately:
∫
Apply Faraday’s law to obtain:
∂E y
∂Bz
ΔxΔz = − μ 0 ∈0
ΔxΔz
∂x
∂t
or
∂E y
∂Bz
= − μ 0 ∈0
∂t
∂x
S
En dA = E y ΔxΔy
53 ••
For your backpacking excursions, you have purchased a radio capable
of detecting a signal as weak as 1.00 × 10–14 W/m2. This radio has a 2000-turn
coil antenna that has a radius of 1.00 cm wound on an iron core that increases the
magnetic field by a factor of 200. The broadcast frequency of the radio station is
1400 kHz. (a) What is the peak magnetic field strength of an electromagnetic
wave of this minimum intensity? (b) What is the peak emf that it is capable of
inducing in the antenna? (c) What would be the peak emf induced in a straight
2.00-m long metal wire oriented parallel to the direction of the electric field?
Picture the Problem We can use the relationship between the average value of
the Poynting vector (the intensity), E0, and B0 to find B0. The application of
Faraday’s law will allow us to find the emf induced in the antenna. The emf
induced in a 2.00-m wire oriented in the direction of the electric field can be
found using ε = El and the relationship between E and B.
B
B
Maxwell’s Equations and Electromagnetic Waves 2879
(a) The intensity of the signal is
related the amplitude of the magnetic
field in the wave:
S av = I =
E0 B0 cB02
2μ0 I
=
⇒ B0 =
2μ0
2μ 0
c
Substitute numerical values and evaluate B0:
B
B0 =
(
)(
)
2 4π × 10 −7 N/A 2 1.00 × 10 −14 W/m 2
= 9.16 × 10 −15 T
8
2.998 × 10 m/s
(b) Apply Faraday’s law to the
antenna coil to obtain:
ε (t ) = d (BA) = A d (NK m B0 sin ωt )
dt
dt
= NK m AB0ω cos ωt
= ε peak cos ωt
where
Substitute numerical values and evaluate
ε peak = NK m AB0ω
ε peak :
ε peak = 2000(200)π (0.0100 m )2 (9.16 × 10 −15 T )[2π (1400 kHz )] =
(c) The voltage induced in the wire is
the product of its length l and the
amplitude of electric field E0:
101 mV
ε = El
E = cB = cB0 sin ωt
Relate E to B:
ε = clB0 sin ωt = ε peak sin ωt
where ε peak = clB0
Substitute for E to obtain:
Substitute numerical values and evaluate ε peak :
ε peak = (2.998 × 108 m/s)(2.00 m )(9.16 × 10 −15 T ) =
5.49 μV
54 ••
The intensity of the sunlight striking Earth’s upper atmosphere is
1.37 kW/m2. (a) Find the rms values of the magnetic and electric fields of this
light. (b) Find the average power output of the Sun. (c) Find the intensity and the
radiation pressure at the surface of the Sun.
Picture the Problem We can use I = ErmsBrms/μ0 and Brms = Erms/c to express Erms
in terms of I. We can then use Brms = Erms/c to find Brms. The average power output
B
B
B
B
2880 Chapter 30
of the Sun is given by Pav = 4πR 2 I where R is the Earth-Sun distance. The
intensity and the radiation pressure at the surface of the sun can be found from the
definitions of these physical quantities.
(a) The intensity of the radiation
is given by:
Erms Brms
I=
μ0
=
2
Erms
⇒ Erms = cμ 0 I
cμ0
Substitute numerical values and evaluate Erms :
Erms =
(2.998 ×10
8
)(
)(
)
m/s 4π × 10 −7 N/A 2 1.37 kW/m 2 = 718.4 V/m
= 718 V/m
Use Brms = E rms c to evaluate Brms :
Brms =
718.4 V/m
= 2.40 μT
2.998 × 10 8 m/s
(b) Express the average power
output of the Sun in terms of the
solar constant:
Pav = 4π R 2 I
where R is the earth-sun distance.
Substitute numerical values and
evaluate Pav:
Pav = 4π (1.50 × 1011 m ) (1.37 kW/m 2 )
2
= 3.874 × 10 26 W
= 3.87 × 10 26 W
(c) Express the intensity at the
surface of the Sun in terms of the
sun’s average power output and
radius r:
Substitute numerical values (see
Appendix B for the radius of the
Sun) and evaluate I at the surface
of the Sun:
Pav
4π r 2
I=
I=
3.874 × 10 26 W
(
4π 6.96 × 108 m
)
2
= 6.363 × 10 7 W/m 2
= 6.36 × 10 7 W/m 2
Express the radiation pressure in
terms of the intensity:
Pr =
Substitute numerical values and
evaluate Pr:
6.363 ×10 7 W/m 2
Pr =
= 0.212 Pa
2.998 ×108 m/s
I
c
Maxwell’s Equations and Electromagnetic Waves 2881
55 ••• [SSM] A conductor in the shape of a long solid cylinder that has a
length L, a radius a, and a resistivity ρ carries a steady current I that is uniformly
distributed over its cross-section. (a) Use Ohm’s law to relate the electric field
r
r
E in the conductor to I, ρ, and a. (b) Find the magnetic field B just outside the
conductor. (c) Use the results from Part (a) and Part (b) to compute the Poyntingr
r
r r
vector S = E × B μ 0 at r = a (the edge of the conductor). In what direction is S ?
(
(d) Find the flux
)
∫ S dA through the surface of the cylinder, and use this flux to
n
show that the rate of energy flow into the conductor equals I2R, where R is the
resistance of the cylinder.
Picture the Problem A side view of the cylindrical conductor is shown in the
diagram. Let the current be to the right (in the +x direction) and choose a
coordinate system in which the +y direction is radially outward from the axis of
the conductor. Then the +z direction is tangent to cylindrical surfaces that are
concentric with the axis of the conductor (out of the plane of the diagram at the
location indicated in the diagram). We can use Ohm’s law to relate the electric
field strength E in the conductor to I, ρ, and a and Ampere’s law to find the
r
r
magnetic field strength B just outside the conductor. Knowing E and B we can
r
find S and, using its normal component, show that the rate of energy flow into the
conductor equals I2R, where R is the resistance.
y
r
B
r
E
x
I
Axis of the conductor
a
(a) Apply Ohm’s law to the
cylindrical conductor to obtain:
r
Because E is in the same direction
as I:
IρL
Iρ
=
L = EL
A
π a2
Iρ
.
where E =
π a2
V = IR =
r
E=
Iρ ˆ
i where iˆ is a unit vector
π a2
in the direction of the current.
(b) Applying Ampere’s law to a
circular path of radius a at the
surface of the cylindrical conductor
yields:
r
r
∫ B ⋅ d l = B(2π a ) = μ I
C
0 enclosed
= μ0 I
2882 Chapter 30
μ0 I
2π a
Solve for the magnetic field strength
B to obtain:
B=
Apply a right-hand rule to determine
r
the direction of B at the point of
interest shown in the diagram:
μ0 I ˆ
θ where θ̂ is a unit vector
2π a
perpendicular to iˆ and tangent to the
r
B=
surface of the conducting cylinder.
(c) The Poynting vector is given
by:
r
r
Substitute for E and B and simplify
to obtain:
r 1 r r
S=
E×B
μ0
r 1 ⎛ Iρ ⎞ ⎛ μ 0 I ⎞
⎟ iˆ × ⎜
⎜
⎟ kˆ
S=
μ 0 ⎜⎝ π a 2 ⎟⎠ ⎜⎝ 2π a ⎟⎠
I 2ρ
= − 2 3 ˆj
2π a
Letting r̂ be a unit vector directed
radially outward from the axis of the
cylindrical conductor yields.
r
I 2ρ
S = − 2 3 rˆ where r̂ is a unit
2π a
vector directed radially outward away
from the axis of the conducting
cylinder.
(d) The flux through the surface of
the conductor into the conductor is:
∫ S dA =S (2π aL)
Substitute for Sn, the inward
r
component of S , and simplify to
obtain:
∫ Sn dA =
Because R =
ρL
A
=
ρL
:
πa 2
n
∫ S dA =
n
I 2ρ
I 2 ρL
(
)
2
=
π
aL
π a2
2π 2 a 3
I 2R
Remarks: The equality of the two flow rates is a statement of the
conservation of energy.
56 ••• A long solenoid that has n turns per unit length carries a current that
increases linearly with time. The solenoid has radius R, length L, and the current I
in the windings is given by I = at. (a) Find the induced electric field at a distance r
< R from the central axis
r of the solenoid. (b) Find the magnitude and direction of
the Poynting vector S at r = R (just inside the solenoid windings). (c) Calculate
Maxwell’s Equations and Electromagnetic Waves 2883
the flux
∫ S dA into the region inside the solenoid, and show that this flux equals
n
the rate of increase of the magnetic energy inside the solenoid.
Picture the Problem An end view of the solenoid is shown in the diagram. Let
the current be clockwise and choose a coordinate system in which the +x direction
is tangent to cylindrical surfaces concentric with the axis of the solenoid. Note
that the +z direction is out of the plane of the diagram. We can use Faraday’s law
to express the induced electric field at a distance r < R from the solenoid axis in
r
terms of the rate of change of magnetic flux and B = nμ0 at to express B in terms
of the current in the windings of the solenoid. We can use the results of (a) to find
r
the Poynting vector S at the cylindrical surface r = R just inside the solenoid
windings. In Part (c) we’ll use the definition of flux and the expression for the
r
magnetic energy in a given region to show that the flux of S into the solenoid
equals the rate of increase of the magnetic energy inside the solenoid.
y
r
E
×
r
B
x
r
I
Axis of the
solenoid
R
r
r
(a) Apply Faraday’s law to a circular
path of radius r < R to relate the
magnitude of the induced electric
field to the magnitude of the rate of
change of the magnetic flux:
∫ E ⋅ d l = E (2π r ) = −
Solving for E yields:
E=−
C
1 dφ m
2π r dt
dφ m
dt
(1)
2884 Chapter 30
Express the magnetic field strength
inside a long solenoid:
B = nμ 0 I = nμ0 at
The magnetic flux through a circle of
radius r is:
φm = BA = nμ 0 atπ r 2
Substitute for φm in equation (1) and
simplify to obtain:
r
The direction of E is such that it
produces an emf that opposes the
increase in the current, so if the
r
current is clockwise, then E is in the
opposite direction and:
E=−
nμ a r
d
nμ 0 atπ r 2 = − 0
2π r dt
2
1
[
]
r
E = − 12 nμ 0 a rθˆ
where θ̂ is a unit vector that is tangent
to the circles that are concentric with
the axis of the solenoid.
r
(b) Express S at r = R:
r 1 r r
S=
E×B
r
The magnitude of B is given by:
B = nμ 0 I = nμ 0 at
Applying a right-hand rule yields:
r
B = −(nμ 0 at ) kˆ
r
r
Substitute for E and B in equation
(2) and simplify to obtain:
r 1 ⎛ nμ 0 a r ⎞
S=
⎜−
⎟ iˆ × (− nμ 0 at ) kˆ
2 ⎠
μ0 ⎝
μ0
(2)
n 2 μ 0 a 2 Rt ˆ
=−
j
2
r r
Because E × B is a vector that
points toward the axis of the
r
solenoid, we can also write S as:
(c) Consider a cylindrical surface of
length L and radius R. Because
r
S points inward, the energy flowing
into the solenoid per unit time is:
r
S = − 12 n 2 μ0 a 2 Rtrˆ
where r̂ is a unit vector that points
radially outward—away from the axis
of the solenoid.
⎛ n 2 μ 0 a 2 Rt ⎞
∫ Sn dA =2π RLS = 2π RL⎜⎜⎝ 2 ⎟⎟⎠
= n 2π μ 0 R 2 La 2t
Maxwell’s Equations and Electromagnetic Waves 2885
Express the magnetic energy in the
solenoid:
U B = u mV =
(
B2
π R2L
2μ0
)
(
μ 0 nat )2
(π R 2 L )
=
2μ0
=
Evaluate dUB/dt:
B
n 2π μ 0 R 2 La 2t 2
2
dU B d ⎡ n 2π μ 0 R 2 La 2 t 2 ⎤
= ⎢
⎥
dt
dt ⎣
2
⎦
= n 2π μ 0 R 2 La 2 t = ∫ S n dA
Remarks: The equality of the two flow rates is a statement of the
conservation of energy.
57 ••• Small particles are be blown out of the solar system by the radiation
pressure of sunlight. Assume that each particle is spherical, has a radius r, has a
density of 1.00 g/cm3, and absorbs all the radiation in a cross-sectional area of
πr2. Assume the particles are located at some distance d from the Sun, which has
a total power output of 3.83 × 1026 W. (a) What is the critical value for the radius
r of the particle for which the radiation force of repulsion just balances the
gravitational force of attraction to the Sun? (b) Do particles that have radii larger
than the critical value get ejected from the solar system, or is it only particles that
have radii smaller than the critical value that get ejected? Explain your answer.
Picture the Problem We can use a condition for translational equilibrium to
obtain an expression relating the forces due to gravity and radiation pressure that
act on the particles. We can express the force due to radiation pressure in terms of
the radiation pressure and the effective cross sectional area of the particles and the
radiation pressure in terms of the intensity of the solar radiation. We can solve the
resulting equation for r.
(a) Apply the condition for
translational equilibrium to the
particle:
Fr − Fg = 0
or, since Fr = PrA and Fg = mg,
GM s m
(1)
Pr A −
=0
R2
The radiation pressure Pr depends on
the intensity of the radiation I:
Pr =
The intensity of the solar radiation at
a distance R is:
I=
I
c
P
4π R 2
2886 Chapter 30
Substitute for I to obtain:
Pr =
P
4π R 2 c
3
4
P
2
3 π r ρ GM s
r
−
=0
π
R2
4π R 2c
(
Substitute for Pr, A, and m in
equation (1):
Solve for r to obtain:
r=
)
3P
16π ρ c GM s
Substitute numerical values and evaluate r:
r=
(
16π 1.00 g/cm
3
(
)
3 3.83 × 10 26 W
2.998 × 108 m/s 6.673 × 10 −11 N ⋅ m 2 / kg 2 1.99 × 1030 kg
)(
)(
)(
)
= 574 nm
(b) Because both the gravitational and radiation pressure forces decrease as the
square of the distance from the Sun, it is then a comparison of grain mass to grain
area. Since mass is proportional to volume and thus varies with the cube of the
radius, the larger grains have more mass and thus experience a stronger
gravitational than radiation-pressure force. The critical radius is an upper limit
and so particles smaller than that radius will be blown out.
58 ••• When an electromagnetic wave at normal incidence on a perfectly
conducting surface is reflected, the electric field of the reflected wave at the
reflecting surface is equal and opposite to the electric field of the incident wave at
the reflecting surface. (a) Explain why this assertion is valid. (b) Show that the
superposition of incident and reflected waves results in a standing wave. (c) Are
the magnetic fields of the incident waves and reflected waves at the reflecting
surface equal and opposite as well? Explain your answer.
Picture the Problem
r
(a) At a perfectly conducting surface E = 0 . Therefore, the sum of the electric
r
r
fields of the incident and reflected wave must add to zero, and so E i = − E r .
(b) Let the incident and reflected
waves be described by:
Ei = E 0 y cos(ωt − kx )
and
E r = − E 0 y cos(ωt + kx )
Maxwell’s Equations and Electromagnetic Waves 2887
Use the trigonometric identity cos(α + β) = cosαcosβ − sinαsinβ to obtain:
Ei + E r = E 0 y cos(ωt − kx ) − E 0 y cos(ωt + kx ) = E 0 y [cos(ωt − kx ) − cos(ωt + kx )]
= E 0 y [cos ωt cos(− kx ) − sin ωt sin (− kx ) − cos ωt cos kx + sin ωt sin (kx )]
= E 0 y [cos ωt cos kx + sin ωt sin kx − cos ωt cos kx + sin ωt sin kx ]
= 2 E 0 y sin ωt sin kx , the equation of a standing wave.
r
r
r r
(c) Because E × B = μ 0 S and S is in the direction of propagation of the wave, we
r
see that for the incident wave Bi = Bz cos(ωt − kx) . Since both S and Ey are
reversed for the reflected wave Br = Bz cos(ωt + kx) . So the magnetic field vectors
are in the direction at the reflecting surface and add at that surface.
r
r
Hence B = 2 Br .
59 ••• [SSM] An intense point source of light radiates 1.00 MW
isotropically (uniformly in all directions). The source is located 1.00 m above an
infinite, perfectly reflecting plane. Determine the force that the radiation pressure
exerts on the plane.
Picture the Problem Let the point source be a distance a above the plane.
Consider a ring of radius r and thickness dr in the plane and centered at the point
directly below the light source. Express the force on this elemental ring and
integrate the resulting expression to obtain F.
The intensity anywhere along this
infinitesimal ring is given by:
P
4π r + a 2
The elemental force dF on the
elemental ring of area 2π rdr is given
by:
dF =
P rdr
c r 2 + a2
=
Pardr
(
2
)
(
(
c r 2 + a2
)
a
r + a2
2
)
32
where we have taken into account that
only the normal component of the
incident radiation contributes to the
force on the plane, and that the plane is
a perfectly reflecting plane.
Integrate dF from r = 0 to r = ∞:
∞
F=
Pa
rdr
∫
2
c 0 (r + a 2 )3 2
2888 Chapter 30
From integral tables:
∞
∫ (r
0
rdr
2
+ a2
)
32
∞
⎤
1
=
⎥ =
2
2
r + a ⎦0 a
−1
Substitute to obtain:
F=
Pa ⎛ 1 ⎞ P
⎜ ⎟=
c ⎝a⎠ c
Substitute numerical values and
evaluate F:
F=
1.00 MW
= 3.34 mN
2.998 × 10 8 m/s