Download Chapter 30 Maxwell`s Equations and Electromagnetic Waves

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Chapter 30
Maxwell’s Equations and Electromagnetic Waves
Conceptual Problems
1
•
(a)
(b)
The displacement current has different units than the conduction current.
Displacement current only exists if the electric field in the region is
changing with time.
In an oscillating LC circuit, no displacement current exists between the
capacitor plates when the capacitor is momentarily fully charged.
In an oscillating LC circuit, no displacement current exists between the
capacitor plates when the capacitor is momentarily uncharged.
(c)
(d)
True or false:
Comment [MSOffice1]: oscillating
LC circuit okay?
Comment [EPM2]: DAVID: Original
part (c) has been deleted.
(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 .
True or false:
3
•
(a)
Maxwell’s equations apply only to electric and magnetic fields that are
constant over time.
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.
(b)
(c)
(d)
(a) False. Maxwell’s equations apply to both time-independent and timedependent fields.
Deleted: 195
197
198
Chapter 30
(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.
If a red light beam, a green light beam, and a violet light beam, all
9
•
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.
Estimation and Approximation
One of the first successful satellites launched by the United States in
13 ••
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 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:
Fr = Pr A
where A is the cross-sectional area of
the balloon.
Maxwell’s Equations and Electromagnetic Waves 199
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:
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
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
=
m⎞
Fg
⎛
⎛
3 kg ⎞ ⎛
8 m⎞
2 ⎜1.00 × 10 3 ⎟ ⎜ 9.81 2 ⎟ (1 mm)⎜ 2.998 × 10
⎟
s⎠
m ⎠⎝
s ⎠
⎝
⎝
Maxwell’s Displacement Current
15 •
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.
200
Chapter 30
Q
(a) Express the electric field strength
between the plates of the parallelplate capacitor:
E=
Differentiate this expression with
respect to time to obtain an
expression for the rate of change of
the electric field strength:
1 dQ
dE d ⎡ Q ⎤
I
=
= ⎢
⎥=
dt dt ⎣∈0 A ⎦ ∈0 A dt ∈0 A
∈0 A
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 ×10 V/m ⋅ s
14
(b) Express the displacement current
Id :
I d =∈0
dφe
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
There is a current of 10 A in a resistor that is connected in series with a
19 ••
parallel plate capacitor. The plates of the capacitor have an area of 0.50 m2, and
no dielectric exists between the plates. (a) What is the displacement current
between the plates? (b) What is the rate of change of the electric field strength
r r
between the plates? (c) Find the value of the line integral ∫ B ⋅ d l , where the
C
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.
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
Maxwell’s Equations and Electromagnetic Waves 201
dφe
d
dE
=∈ 0 [EA] =∈0 A
dt
dt
dt
(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
⎝
= 2.3 × 1012
r
⎞
⎟⎟ 0.50 m 2
⎠
(
V
m⋅s
r
(c) Apply Ampere’s law to a circular
path of radius r between the plates
and parallel to their surfaces to
obtain:
∫ B ⋅ dl = μ I
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
B
∫C ⋅ d l = A I d
(
0 enclosed
C
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
Electric Dipole Radiation
27 ••• 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
202
Chapter 30
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
From the definition of intensity we
have:
(1)
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
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
4
3
Maxwell’s Equations and Electromagnetic Waves 203
Energy and Momentum in an Electromagnetic Wave
The amplitude of an electromagnetic wave’s electric field is 400 V/m.
31 •
Find the wave’s (a) rms electric field strength, (b) rms magnetic field strength,
(c) intensity and (d) radiation pressure (Pr).
Picture the Problem The rms values of the electric and magnetic fields are found
from their amplitudes by dividing by the square root of two. The rms values of the
electric and magnetic field strengths are related according to Brms = Erms/c. We can
find the intensity of the radiation using I = ErmsBrms/μ0 and the radiation pressure
using Pr = I/c.
(a) Relate Erms to E0:
Erms =
E0
2
=
400 V/m
= 282.8 V/m
2
= 283 V/m
(b) Find Brms from Erms:
Brms =
E rms
282.8 V/m
=
c
2.998 × 108 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 × 10 8 m/s
(a) For a given distance from a radiating electric dipole, at what angle
35 •
(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?
204
Chapter 30
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.
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 = 12
( )=
45°
1
2
(
)
37 ••
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 cos2 [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 ×10 9 s −1
λ
= 10.0 m −1 ⇒ λ = 0.628 m
Maxwell’s Equations and Electromagnetic Waves 205
Solving for f yields:
f =
r
(c) Express the magnitude of S in
terms of E:
3.00 ×109 s −1
= 477 MHz
2π
r E2
r
⇒ E = μ0c S
S =
μ0c
Substitute numerical values and evaluate E:
E=
(4π ×10
−7
)(
)(
)
N/A 2 2.998 × 108 m/s 100 W/m 2 = 194.1 V/m
Because
r
S ( x, t ) = (100 W/m 2 )cos 2 [kx − ω t ] iˆ
r 1 r r
E×B:
and S =
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.
194.1 V/m
= 647.4 nT
2.998 × 108 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
A pulsed laser fires a 1000-MW pulse that has a 200-ns duration at a
39 ••
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.)
206
Chapter 30
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:
θ
L
L cos θ
m
h
Kf − Ki + U f − U i = 0
or, because Ui = Kf = 0 and K i =
−
U f = mgh = mgL(1 − cos θ )
Substitute for Uf:
−
pi2
+ mgL(1 − cos θ ) = 0
2m
⎛
θ = cos −1 ⎜⎜1 −
⎝
Use conservation of momentum to
relate the momentum of the
electromagnetic pulse to the initial
momentum pi of the pendulum bob:
Substituting for pi gives:
pi2
,
2m
pi2
+ Uf = 0
2m
Uf is given by:
Solve for θ to obtain:
Ug = 0
pi2 ⎞
⎟
2m 2 gL ⎟⎠
U PΔt
=
= pi
c
c
where Δt is the duration of the pulse.
pem wave =
2
⎡
P 2 (Δt ) ⎤
θ = cos ⎢1 − 2 2 ⎥
⎣ 2m c gL ⎦
−1
Maxwell’s Equations and Electromagnetic Waves 207
Substitute numerical values and evaluate θ :
⎡
θ = cos −1 ⎢1 −
⎢⎣
2(10.0 mg )
2
(1000 MW )2 (200 ns )2
⎤
⎥ = 6.10°
2.998 × 108 m/s 9.81 m/s 2 (0.0400 m )⎥⎦
(
)(
2
)
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.)
(a) Estimate the force on Earth due to the pressure of the radiation on
41 ••
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 =
Iπ R 2
c
⇒ Fr, Earth = Pr, Earth A
A
where A is the cross-sectional area of
Earth.
Deleted: ¶
208
Chapter 30
Substitute numerical values and
evaluate Fr:
π (1.37 kW/m 2 )(6.37 × 10 6 m )
2
Fr, Earth =
2.998 × 108 m/s
= 5.825 × 10 8 N
= 5.83 × 10 8 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
3.529 × 10 22 N
Fg, Earth
or
Fr, Earth = (1.65 × 10 −14 )Fg, Earth
Pr, Mars =
Fr, Mars
Pr, Mars =
I Mars
c
Fr, Mars =
2
I Marsπ RMars
c
⇒ Fr, Mars = Pr, Mars A
A
where A is the cross-sectional area of
Mars.
I Mars ⎛ rearth
=⎜
I earth ⎜⎝ rMars
Fr, Mars
2
⎞
⎛r
⎟⎟ ⇒ I Mars = I earth ⎜⎜ earth
⎠
⎝ rMars
2
I π RMars
= earth
c
⎛ rearth
⎜⎜
⎝ rMars
⎞
⎟⎟
⎠
2
⎞
⎟⎟
⎠
2
Maxwell’s Equations and Electromagnetic Waves 209
Substitute numerical values and evaluate Fr, Mars:
π (1.37 kW/m 2 )(3.40 × 10 3 km ) ⎛ 1.50 × 1011 m ⎞
2
Fr, Mars =
2.998 × 10 8 m/s
The gravitational force exerted on
Mars by the Sun is given by:
2
⎜⎜
⎟⎟ = 7.18 × 10 7 N
11
⎝ 2.28 × 10 m ⎠
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 m)
2
11
= 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
45 ••
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.
210
Chapter 30
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
and
=
=
v
∂ u2
∂ x2 ∂ u 2
∂t2
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
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
and
=
=
v
∂ x2 ∂ u 2
∂ t2
∂ u2
Divide the first of these equations by
the second to obtain:
∂2f
∂ x2
1
∂2 f
1 ∂2 f
= 2⇒
= 2
2
2
∂x
v ∂ t2
∂ f
v
∂t2
General Problems
47 ••
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?
Maxwell’s Equations and Electromagnetic Waves 211
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.
The emf induced in the antenna is
given by Faraday’s law:
ε = − dφm
(
)
d
d r
B ⋅ Anˆ = − (BA)
dt
dt
dt
d
dB
= −A
= −πR 2 (B0 sin ωt )
dt
dt
2
= −πR ωB0 cos ωt = −ε peak cos ωt
where
=−
ε peak = πR 2ωB0 and R is the
radius of the loop antenna..
ε rms equals ε peak divided by the
square root of 2:
The intensity of the signal is given
by:
ε rms =
Substituting for B0 and ω in
equation (1) and simplifying yields:
2
=
πR 2ωB0
2
(1)
E 0 B0
2μ 0
or, because E 0 = cB0 ,
I=
I=
Solving for B0 yields:
ε peak
cB0 B0 B02 c
=
2μ 0
2μ 0
2μ 0 I
c
B0 =
ε rms =
πR 2 (2πf )
2μ 0 I
c
2
= 2π 2 R 2 f
μ0 I
c
Substitute numerical values and evaluate εrms:
−7
2
2
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
212
Chapter 30
51 ••
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 by
r
E = E cos(k x − ω t + δ )ˆj . For the resultant of these two waves, find (a) the
2
2 ,0
2
2
instantaneous Poynting vector and (b) the time-averaged Poynting vector.
(c) Repeat Parts (ra) 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
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 − ω 2t + δ )) ˆj
S ( x, t ) =
μ0
×
=
1
μ0c
1
(E1,0 cos(k1 x − ω1t ) + E2,0 cos(k 2 x − ω 2t + δ ))kˆ
c
(E
1
= μ0c
1, 0
[E
(
2
cos(k1 x − ω1t ) + E 2,0 cos(k 2 x − ω 2 t + δ )) ˆj × kˆ
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 =
1
2μ 0 c
[E
2
1, 0
]
+ E22, 0 iˆ
)
Maxwell’s Equations and Electromagnetic Waves 213
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 − ω 2t + δ )) ˆj
S ( x, t ) =
μ0
×
=
1
μ0c
1
(E1,0 cos(k1 x − ω1t ) − E2,0 cos(k 2 x + ω2t + δ ))kˆ
c
[E
2
1, 0
]
cos 2 (k1 x − ω1t ) − E 22,0 cos 2 (k 2 x + ω 2 t + δ ) iˆ
The time average of the square of the
cosine terms is ½:
r
S av =
[
]
1
E12,0 − E22,0 iˆ
2μ 0 c
55 ••• 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
r
over its cross-section. (a) Use Ohm’s law to relate the electric field E in the
r
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 Poynting vector
r
r
r r
S = E × B μ 0 at r = a (the edge of the conductor). In what direction is S ?
(
)
(d) Find the flux ∫ S n dA through the surface of the cylinder, and use this flux to
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.
214
Chapter 30
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
0 enclosed
C
= μ0 I
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:
r
μ0 I ˆ
B=
θ where θ̂ is a unit vector
2π a
Comment [EPM3]: David: Note
change of unit vectors from k-hat to
theta-hat.
perpendicular to iˆ and tangent to the
surface of the conducting cylinder.
Deleted:
(c) The Poynting vector is given
by:
r
r
Substitute for E and B and simplify
to obtain:
θˆ
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
Deleted: Substitute for
simplify to obtain:
where k̂ has been substituted for θ̂
because the calculation is being done at
the point where the axes are drawn in
the diagram.
Deleted:
¶
r
r
E and B and
r 1 ⎛ Iρ ⎞ ⎛ μ
⎜
⎟ iˆ × ⎜
S=
μ 0 ⎜⎝ π a 2 ⎟⎠ ⎜⎝ 2π
I 2ρ
= − 2 3 ˆj
2π a
Maxwell’s Equations and Electromagnetic Waves 215
Letting r̂ be a unit vector directed
radially outward from the axis of the
cylindrical conductor yields.
(d) The flux through the surface of
the conductor into the conductor is:
Substitute for Sn, the inward
r
component of S , and simplify to
obtain:
Because R =
ρL
A
=
ρL
:
πa 2
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.
∫ S dA =S (2π aL )
n
∫ S n dA =
∫ S dA =
n
I 2ρ
I 2 ρL
(
)
aL
2
=
π
π a2
2π 2 a 3
I 2R
Remarks: The equality of the two flow rates is a statement of the
conservation of energy.
59 ••• 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 2 + a 2
(
)
Deleted: ¶
216
Chapter 30
The elemental force dF on the
elemental ring of area 2π rdr is given
by:
dF =
P rdr
c r 2 + a2
=
Pardr
(
(
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 = ∞:
From integral tables:
∞
F=
∞
∫ (r
Pa
rdr
c ∫0 r 2 + a 2
(
rdr
2
0
+ a2
)
32
)
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