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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 d φe dt , I d is zero if
d φe 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. I d is zero if d φe dt = 0. At the moment when the capacitor is
momentarily uncharged, dE / dt ≠ 0 and so d φe dt ≠ 0.
2 • Using SI units, show that I d= ∈ 0 d φe dt has units of current.
Determine the Concept We need to show that ∈ 0 d φe dt has units of amperes.
We can accomplish this by substituting the SI units of ∈ 0 and d φ e dt and
simplifying the resulting expression.
A
s
C
s
m C
N
N m
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.
840 Chapter 30
( 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 q m is given by B = ( μ 0 /4 π) q m/ r^2. 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 B n^ dA =^ μ^0 q m,inside where^ q m, 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
842 Chapter 30
(^2 22) m 2
W
m
s
J
m
s
N m
m
s
C
C
N m m
A
C
N
A
m
C
N
A
T m
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 ( p = Uc ). 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
0 0
av 2 μ
E B
I = S =
r .
The intensity of an electromagnetic wave is given by: (^0)
0 0
av 2 μ
E B
I = S =
r
Because E 0 = cB 0 :
0
2 0
av 2 c μ
E
S =
r
This result tells us that 02 av
S ∝ E
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 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 = 3 I B
( d ) I A = 9 I B ( 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
0 0
av 2 μ
E B
I = S =
r .
Express the intensities of the two waves: 0
0 ,A 0 ,A
A 2 μ
E B
I = and 0
0 ,B 0 ,B
B 2 μ
E B
I =
Dividing the first of these equations by the second and simplifying yields:
0 ,B 0 ,B
0 ,A 0 ,A
0
0 ,B 0 ,B
0
0 ,A 0 ,A
B
A
E B
E B
E B
E B
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:
0 ,B 0 ,B
0 ,B 0 ,B B
A = =
E B
E B
I
I
⇒ I A = 9 I 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 λ^2 , 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/m^2. ( 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:
F r (^) = ma (1) where F r is the radiation force exerted by the laser beam.
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:
F r (^) = P r A 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
I
P
r =
Substituting for P r and A yields: ( )
c
d I c
I d F 2
r
π π = =
The gravitational force acting on the balloon when it is in a near-Earth orbit is approximately its weight at the surface of Earth:
A t g
F w m g V g
Mylar surface, ballon
g balloon balloon Mylar Mylar
where t is the thickness of the Mylar skin of the balloon.
Because the surface area of the
balloon is 4 π r^2 = π d^2 :
F g = πρMylar d^2 t g
Express the ratio of the radiation- pressure force to the gravitational force and simplify to obtain: (^) tgc
I
d tg
c
d I
F
F
Mylar
2 Mylar
2
g
r 2
πρ ρ
π = =
846 Chapter 30
Assuming the thickness of the Mylar skin of the balloon to be 1 mm, substitute numerical values and evaluate F r/ F g :
7 8 3 2
3
2
g
r (^210)
s
m 1 mm 2. 998 10 s
m
- 81 m
kg
- 00 10
m
kW
- 35 ≈ × − ⎟ ⎠
⎛ ×
⎛ ×
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 P S A ( 2 π r^2 c )where P S 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 P S A ( 2 π r^2 c )
where P S 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
IA
F PA
r = r =
where A is the area of the sail.
The intensity of the solar radiation on
the sail is given by s 2 4 r
P
I
Substituting for I yields:
r c
PA
rc
PA
F (^) r s 2 s 2 4 2
848 Chapter 30
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 A A
Q
dt
d dt
dE 0 0 0
∈ ∈ ∈
Substitute numerical values and evaluate dE / dt :
4 10 V/m s
40 10 V/m s
854 10 C /N m 0. 023 m
5. 0 A
14
14 12 2 2 2
= × ⋅
= × ⋅
× ⋅
dt π
dE
( b ) Express the displacement current I d : dt
d I (^) d 0 e
Substitute for the electric flux to
obtain:
[ ]
dt
dE EA A dt
d
I d=∈ 0 = ∈ 0
Substitute numerical values and evaluate I d :
I d = ( 8. 854 × 10 −^12 C^2 /N⋅m^2 )π ( 0. 023 m) (^23. 40 × 1014 V/m⋅s) = 5. 0 A
16 • In a region of space, the electric field varies with time as
E = (0.050 N/C) sin ( ω t ), 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 m^2.
Picture the Problem We can express the displacement current in terms of the electric flux and differentiate the resulting expression to obtain I d in terms of
dE / dt.
The displacement current I d is given by: dt
d I (^) d 0 e
Substitute for the electric flux to
obtain:
[ ]
dt
dE EA A dt
d
I d=∈ 0 = ∈ 0
Because E = ( 0. 050 N/C) sin 2000 t :
[( ) ]
( ) A ( ) t
t dt
d I A
2000 s 0. 050 N/Ccos 2000
- 050 N/Csin 2000
0
d 0
Maxwell’s Equations and Electromagnetic Waves 849
I d will have its maximum value
when cos 2000 t = 1. Hence:
I d,max = ( 2000 s-1) ∈ 0 A ( 0. 050 N/C)
Substitute numerical values and evaluate I d,max :
( ) ( ) 0. 89 nA
C
N
- 00 m 0. 050 N m
C
2000 s 8. 854 10 2 2
2 1 12 d, max ⎟= ⎠
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:
∫C ⋅ d^ l=^2 π rB^ =μ^0 I enclosed=^ μ^0^ I
r r B
Assuming that the displacement
current is uniformly distributed: 2
d (^2) R
I
r
I
= ⇒ (^2) d
2 I R
r I =
where R is the radius of the circular plates.
Substituting for I yields: 2 d
2 2 0 I R
r rB
π = ⇒ 02 d
I
R
r B
Substitute numerical values and
evaluate B :
r
Br r
= ⎛^ ×
×
−
−
m
T
2 0. 023 m
4 10 N/A 5. 0 A
3
2
7 2
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 I d = C dV / dt , where C is the capacitance and V is the potential difference between the plates. ( b ) A 5.00-nF
Maxwell’s Equations and Electromagnetic Waves 851
Picture the Problem We can use the conservation of charge to find I d , the
definitions of the displacement current and electric flux to find dE / dt , and
Ampere’s law to evaluate l
r r B ⋅ d around the given path.
( a ) From conservation of charge we
know that:
I d = I = 10 A
( b ) Express the displacement current I d :
[ ]
dt
dE EA A dt
d dt
d
I d 0 e ∈ 0 ∈ 0
Substituting for dE / dt yields:
A
I
dt
dE 0
d
Substitute numerical values and evaluate dE / dt :
m s
V
- 50 m N m
C
10 A
12
2 2
2 12
= ×
×
dt −
dE
( c ) Apply Ampere’s law to a circular
path of radius r between the plates and parallel to their surfaces to
obtain:
∫C ⋅ d l=^ μ^0 I enclosed
r r B
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 (^) =
⇒ (^) d
2 enclosed I A
r I
Substitute for I (^) enclosed to obtain: d
2 0 C
I
A
r d
∫ ⋅ l=
r r B
Substitute numerical values and evaluate ∫ C ⋅ l
r r B d :
( ) (^ ) (^ )^
- 79 T m
- 50 m
4 10 N/A 0. 10 m 10 A 2
7 2 2 C
×
−
l
r r B d
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
852 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
that the magnitude of the magnetic field at point P is given by B =
μ 0 Ia
2 π R
R^2 + a^2
( 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 ( r a ) dr
Q r E (^) xdA 2 232 0
. ( 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
a a^2 + R^2
⎠⎟^
. ( d ) Find the displacement current I d through S ,
and show that I + I d = I
a a^2 + 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:
0 (sin 1 sin 2 )
R
I
BP
where
sin 1 sin (^2) R (^2) a 2 a
Substitute for sin θ 1 and sin θ 2 to
obtain:
2 2
0
2 2
0
R R a
Ia
R a
a R
I
BP
( b ) Express the electric flux through
the circular strip of radius r and width dr in the yz plane:
d φ e = ExdA = Ex ( 2 π rdr )
The electric field due to the dipole is:
2 2 1 2 232
cos
r a
kQa r a
kQ E (^) x
854 Chapter 30
Substitute for I + I d from ( d ) to
obtain:
2 2
0
2 2
0
R R a
Ia
R a
a I R
B
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:
λyellow- green= 580 nm
The corresponding frequency is:
- 17 10 Hz
580 nm
- 998 10 m/s
14
8
yellow-green
yellow-green
= ×
×
c f
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 = f λ 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:
c f =
Substitute numerical values and evaluate f :
99 GHz
993 10 Hz
00 10 m
998 10 m/s 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:
620 nm
520 nm shortestred
shortest green ≈ =
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 = f λ 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:
00 10 Hz
100 10 m
998 10 m/s
18
9
8
= ×
×
×
c f
( b ) The frequency of light with a
wavelength of 550 nm is:
45 10 Hz 550 nm
998 108 m/s 14 = ×
×
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 I 1 be the intensity of the radiation at a distance of 10 m and at angle of 90º. Find the intensity (in terms of I 1 ) 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 I 1 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 :
(θ , ) 2 sin^2 θ r
C
I r = (1)
where C is a constant.
Maxwell’s Equations and Electromagnetic Waves 857
Express I (90°,10 m):
2
2 (^12)
100 m
sin 90 10 m
90 , 10 m
C
C
I I
Solving for C yields: C =( 100 m^2 ) I 1
Substitute in equation (1) to obtain:
2 sin
100 m , r
I
I r = (2)
( a ) For r = 5 m and I ( θ, r ) = I 1 : ( )
2 1 sin
- 0 m
100 m I
I = ⇒ sin 2 θ= 41
Solve for θ to obtain: θ= sin −^1 ( 21 ) = 30 °
( b ) For θ = 45° and I ( θ, r ) = I 1 : ( )
= sin 45 ° 100 m 2 2
1
2 1 r
I
I
or
r^2 =^12 ( 100 m^2 )
Solve for r to obtain: ( 100 m^2 ) 7. 1 m
2 r =^1 =
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/m^2. 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 :
(θ , ) 2 sin^2 θ
r
C
I r = (1)
where C is a constant.
858 Chapter 30
Use the given data to obtain:
2
2 2
12 2
- 00 km
sin 90
- 00 km
4 10 W/m
C
C
× − = °
Solving for C yields: ( ) ( )
6. 40 10 W
- 00 km 4. 00 10 W/m 5
2 12 2 −
−
= ×
C = ×
Substitute in equation (1) to obtain:
5 sin
6. 40 10 W
r
I r
×^ −
For a point at sea level and 1.50 km
from the transmitter:
= −^53. 1
1.50km
- 00 km
θ tan 1
Evaluate I (53.1°,1.50 km):
2 2 2
5 sin 53. 1 18. 2 pW/m
- 50 km
6. 40 10 W
- 1 , 1. 5 km °=
×
− 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 (sin^2 θ)/ r^2 , 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 r.
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.
Express the intensity of the signal as
a function of r and θ :
sin^2 , r
Ir C
At a horizontal distance of 120 km from the station:
( )^2
2
2
120 km
120 km
sin 90 120 km, 90
C
I C
