PHYS 3123
Georgia Tech, Spring 2017
Homework 9 Solutions
Instructor: David Ballantyne
(Thomas) Forrest Kieffer
1. (11.10) An insulating circular ring (radius b) lies in the xy plane, centered at the origin. It carries a linear charge density
λ = λ0 sin φ, where λ0 is a constant and φ is the usual azimuthal angle. The ring is now set spinning at a constant angular
velocity ω about the z axis. Calculate the power radiated.
Solution: We need to calculate that dipole moment of the spinning ring. If the ring rotates counterclockwise, the dipole
moment as a function of time will be p(t) = p0 [cos (ωt)ŷ − sin (ωt)x̂], where p0 is the dipole moment at time t = 0.
We calculate p0 to be
Z
p0 =
0
0
0
2
Z2π
λ(φ )r dl = λ0 b
sin φ0 [cos φ0 x̂ + sin φ0 ŷ]dφ0 = πλ0 b2 ŷ
0
Noting that p̈ = −ω 2 p and using Eq. 11.60 we find
Prad =
µ0
πµ0 λ20 b4 ω 4
[p̈(t0 )]2 =
6πc
6c
2. (11.12) An electron is released from rest and falls under the influence of gravity. In the first centimeter, what fraction of the
potential energy lost is radiated away?
Solution: The electron is released from rest, so the distance its fallen as a function of time is given by y(t) = (1/2)gt2 .
The dipole moment of the electron when it has fallen a distance y(t) is p(t) = −ey(t) = −ey(t)ŷ = −(1/2)get2 ŷ.
Therefore p̈ = −geŷ and Eq. 11.60 then reads
Prad =
µ0
(ge)2
6πc
p
2d/g, so the energy radiated in falling a distance d is
s
µ0 (ge)2 2d
= Prad t =
6πc
g
The time it takes to fall a distance d is given by t =
Urad
The potential energy lost is Upot = me gd and so the fraction of potential energy radiated away is
s
r
Urad
µ0 g 2 e2 2d 1
µ0 e2
2g
=
=
Upot
6πc
g me gd
6πme c d
Plugging in d = 0.01[m] and the standard SI values for e, µ0 , me , c, and g we find
Urad
= 2.76 × 10−22
Upot
3. (11.23) A radio tower rises to height h above flat horizontal ground. At the top is the magnetic dipole antenna, of radius b,
with its axis vertical. FM station KRUD broadcasts from this antenna at (angular) frequency ω, with a total radiated power
P (that’s averaged, of course, over a full cycle). Neighbors have complained about problems they attribute to excessive
radiation from the tower - interference with their stereosystems, mechanical garage doors opening and closing mysteriously,
and a variety of suspicious medical problems. But the city engineer who measured the radiation level at the base of the
tower found it to be well below the accepted standard. You have been hired by the Neighborhood Association to assess the
engineer’s report.
(a) In terms of the variables given (not all of which may be relevant), find the formula for the intensity of the radiation at
ground level, a distance R from the base of the tower. You may assume that b c/ω h. [Note: We are interested
PHYS 3123
Georgia Tech, Spring 2017
Homework 9 Solutions
Instructor: David Ballantyne
(Thomas) Forrest Kieffer
only in the magnitude of the radiation, not in the direction - when measurements are taken, the detector will be aimed
directly at the attenna.]
(b) How far from the base of the tower should the engineer have made the measurement? What is the formula for the
intensity at this location?
(c) KRUD’s actual power output is 35 kilowatts, its frequency is 90 MHz, the antenna’s radius is 6 cm, and the height of
the tower is 200 m. The city’s radioemmision limit is 200 microwatts/cm2 . Is KRUD in compliance?
Solution:
(a) The antenna is describing as a magnetic dipole of radius b oscillating at angular frequency ω. Therefore, using the
results of §11.1.3, the Poynting vector is
µ0 m20 ω 4 sin2 θ
hSi =
r̂
32π 2 c3
r2
In terms of the total power radiated,
P =
µ0 m20 ω 4
12πc3
the Poytning vector is
hSi =
3P sin2 θ
r̂
8π r2
Since we are
√ at a distance R from the base of a tower of height h, simple trigonometry tells us that sin θ = R/r
where r = R2 + h2 . Therefore the intensity as a function of R is
I(R) = |hSi| =
3P
R2
8π (R2 + h2 )2
(b) As a function of R, the intensity reaches is maximum at R = h (verifying this is simple exercise of setting
dI/dR = 0 and solving). At the maximum,
I(R = h) =
3P
32πh2
(c) Plugging the given values in we find
I(R = h) =
3(35 × 103 )
= 0.026 W/m2 = 2.6 µW/cm2
32π(200)2
We see that KRUD is well within compliance.
Page 2
PHYS 3123
Georgia Tech, Spring 2017
Homework 9 Solutions
Instructor: David Ballantyne
(Thomas) Forrest Kieffer
4. (11.24) As a model for electric quadrupole radiation, consider two oppositely oriented oscillating electric dipoles, separated
by a distance d, as shown in Fig. 11.19. Use the results of Sect. 11.1.2 for the potentials of each dipole, but note that they
are not located at the origin. Keeping only the terms of first order in d:
(a) Find the scalar and vector potentials.
(b) Find the electric and magnetic fields.
(c) Find the Poynting vector and the power radiated. Sketch the intensity profile as a function of θ.
Solution:
(a) We label quantities associated with the “top” dipole in Fig. 11.19 with a + and quantities associated with the
“bottom” dipole in Fig. 11.19 with a −. Note that each dipole is located a distance d/2 from the origin. Using
Eq. 11.14 for the scalar potential, we find for each dipole
p0 ω
cos θ±
V± = ∓
sin [ω(t − r± /c)]
4π0 c
r±
where r± is the distance between the center of the top/bottom dipole and the observation location r and θ± is the
angle between the top/bottom dipole and the y-axis. In the approximation d r, r± is given by
p
p
d
cos θ
r± = r2 + (d/2)2 ∓ 2r(d/2) cos θ ' r 1 − ∓(d/r) cos θ ' r 1 ∓
2r
and
1
1
=
r±
r
1±
d
cos θ
2r
In the same approximation, cos θ± is given by
r cos θ ∓ (d/2)
d 1
d
d
d
cos θ± =
' r cos θ ∓
1±
cos θ = cos θ ±
cos2 θ ∓
r±
2r r
2r
2r
2r
d
d
= cos θ ∓ (1 − cos2 θ) = cos θ ∓
sin2 θ
2r
2r
Lastly we need to approximate sin [ω(t − r± /c)]. Using t0 = t − r/c and sin(α ± β) = sin α cos β ± cos α sin β,
we find
r
d
ωd
sin [ω(t − r± /c)] = sin ω t −
1 − ∓ cos θ
= sin ωt0 ±
cos θ
c
2r
2c
ωd
ωd
ωd
= sin (ωt0 ) cos
cos θ ± cos (ωt0 ) sin
cos θ ' sin (ωt0 ) ±
cos θ cos (ωt0 )
2c
2c
2c
where in the last step we used the approximation d c/ω. Plugging all this into V± we find
p0 ω
d
d
ωd
V± = ∓
1±
cos θ
cos θ ∓
sin2 θ sin (ωt0 ) ±
cos θ cos (ωt0 )
4π0 cr
2r
2r
2c
p0 ω
ωd
d
=∓
cos θ sin (ωt0 ) ±
cos2 θ cos (ωt0 ) ± (cos2 θ − sin2 θ) sin (ωt0 )
4π0 cr
2c
2r
Since the Maxwell equations are linear PDEs, the total potential is given by Vtot = V+ + V− . Using the formula
for V± above and the radiation zone approximation r c/ω we find
Vtot = −
p0 ω 2 d
cos2 θ cos [ω(t − r/c)]
4π0 c2 r
Page 3
PHYS 3123
Georgia Tech, Spring 2017
Instructor: David Ballantyne
(Thomas) Forrest Kieffer
Homework 9 Solutions
Likewise, using Eq. 11.17 the vector potential for each dipole is
µ0 p0 ω
sin [ω(t − r± /c)]ẑ
4πr±
µ0 p0 ω
ωd
d
∓
sin (ωt0 ) ±
cos θ cos (ωt0 ) ±
cos θ sin (ωt0 ) ẑ
4πr
2c
2r
A± = ∓
Therefore the total vector potential in the radiation zone is
Atot = A+ + A− = −
µ0 p0 ω 2 d
cos θ cos [ω(t − r/c)]ẑ
4πcr
(b) We let α = −µ0 p0 ω 2 d/4π to make the calculations below less messy. We have
∇V = ∂t V r̂ + r−1 ∂θ V θ̂
ω
1
cos θ sin θ
2
cos [ω(t − r/c)]θ̂
= α cos θ − 2 cos [ω(t − r/c)] +
sin [ω(t − r/c)] r̂ − 2α
r
rc
r2
which in the radiation zone reduces to
∇V = α
ω cos2 θ
sin [ω(t − r/c)]r̂
c r
Likewise,
∂t A = −
αω cos θ
αω cos θ
sin [ω(t − r/c)]ẑ = −
sin [ω(t − r/c)](cos θr̂ − sin θθ̂)
c r
c r
Thus, E is
E = −∇V − ∂t A = −
αω
sin θ cos θ sin [ω(t − r/c)]θ̂
cr
To find B we need to compute ∇ × A in the radiation zone:
B=∇×A=
1
αω
[∂r (rAθ ) − ∂θ Ar ] φ̂ = − 2 sin θ cos θ sin [ω(t − r/c)]φ̂
r
c r
(c) Noting that B = c−1 (r̂ × E) and E · r̂ = 0, we find
E2
1
1
E × (r̂ × E) =
r̂
(E × B) =
µ0
µ0 c
µ0 c
o2
1 n αω
=
sin θ cos θ sin [ω(t − r/c)] r̂
µ0 c rc
S=
Since hsin2 i = 1/2, the intensity is given by
I = |hSi| =
2
1 αω
sin θ cos θ
2µ0 c rc
We omit sketching the intensity profile as a function of θ.
The power radiated is
Z
P =
1 αω 2
hSi · da =
2µ0 c cR
r=R
Z2πZπ
0
sin3 θ cos2 θR2 dθdφ =
0
Expressing this without α, we find
P =
µ0 p20 d2 ω 6
60πc3
Page 4
1 αω 2 8π
2µ0 c c
15