Outside: Sin Cos 2 12 Inside: 6: R R A Q
Outside: Sin Cos 2 12 Inside: 6: R R A Q
Outside: Sin Cos 2 12 Inside: 6: R R A Q
2
Due Friday, September 21, 2001
1. Consider the problem of a thin spherical shell of charge Q and radius a rotating with
constant angular frequency (Jackson 5.13). Take the z-axis to be the rotation axis. The
solution for the static magnetic field is:
Outside:
B out
0Q a 2
=
2 cos r + sin
3
12 r
Inside:
B in =
0Q
z
6a
Imagine now that is not constant but is varying slowly such that B is still given by the
above expressions.
a) Use Faradays law to find the induced electric field inside the shell.
Q r
40 r 2
Ein = 0
E out =
dl E =
or
da B
t S
B
(a sin )2
t
a sin B
Q sin
&
E=
= 0
2 t
12
2a sin E =
b) Evaluate the rate that the induced field does work on the charge using
dUmech
= d 3r E J
dt
V
This term represents the rate of change of mechanical energy. The
induced electric field is going to cause a reduction (increase) in
energy if the acceleration is positive (negative).
Q sin (r a )
J=
4 a
dU mech
Q sin & Q sin (r a )
= d 3r E J = d 3r 0
dt
12
4 a
V
V
dU mech
Q sin Q sin (r a )
&
= 2 drr 2 d cos 0
dt
12
4 a
1
1
& Q
& 4
0Q
0 aQ 2
2
= 2a
d
cos
sin
12 4a 1
24 3
1
2
d cos sin d cos 1 cos = 2
2 4
=
3 3
&
dU mech
0 aQ 2
=
dt
18
This equation says that if we give the sphere a positive
acceleration, it will radiate electromagnetic energy and thus lose
mechanical energy.
c) Evaluate the rate of change of stored electromagnetic energy using
2
d
B2
3 E
= d r
+
dt
dt V
2
0 2 0
This term represents the rate of change of stored electromagnetic
energy. If the acceleration is positive (negative), there is more
(less) stored energy.
dU field
dU field
dt
2
2
d
E
B
3
= d r
+
dt V
2
0
0
dU field
dt
&
d 1
d 1 0Q 4a 3 0 aQ 2
3
2
=
d
r
B
=
=
dt 20 V
dt 20 6a 3
27
2
d) Evaluate rate that energy enters the spherical volume by integrating the Poynting
vector over the surface of the shell and demonstrate explicitly that
d
U field + U mech = da (E H )
dt
S
(E B ) r = (E B ) r = E B
2
& sin 2
0Q sin & 0Q sin 0 Q
=
=
12
144 2 a
12 a
2
& sin 2
1
2a 2 1
0 Q 2
da (E B ) =
d cos
0 S
0 1
144 2a
2
& 4
&
0aQ 2
0aQ 2
=
=
72 3
54
This is the energy flowing into the spherical volume, which causes
a change in mechanical and/or field energy inside the volume.
Note that the volume we are talking about contains the shell itself
and its interior.
Finally, check that
& 0 aQ 2
&
&
0 aQ 2
0 aQ 2
+
=
18
27
54
2. (Jackson 7.2)
A plane wave is incident upon a layered interface as shown in the figure. The indices of
refraction of the three nonpermeable media are n1 , n2 , n3 . The thickness of the
intermediate layer is d. Each of the other media is semi-infinite.
(a) Calculate the transmission and reflection coefficients (ratios of transmitted and
reflected Poyntings flux to the incident flux), and sketch their behavior as a function of
frequency for n1 = 1, n2 = 2, n3 = 3; n1 = 3, n2 = 2, n3 = 1; and n1 = 2, n2 = 4, n3 = 1.
S ~ E2 ~ n E2
where E is the amplitude of the electric field
(Jackson 7.13 with = 1).
Describe the fields in the different regions as
Incident (medium 1),
Ei = E0 eik 1 r i t
Reflected (medium 1),
E r = E1e ik 1 r it
intermediate (medium 2)
E 2 = E 2+ eik 2 r it + E 2 e ik 2 r it
transmitted (medium 3)
Et = E 3eik 3 r it
The boundary conditions are given by Jackson 7.37 (last two
equations). The incident angle is zero and thus, so is the refraction
angle.
i = r = 0
n1
E0
n2
incident
k
k2
E2
E1
reflected
n3
E3
E2+
transmitted
k3
k2
n
z=0
z=d
(Ei + E r ) n = E2 n
or
E0 + E1 = E 2+ + E2
(1)
(k1 Ei + k 1 E r ) n = (k 2 E 2 ) n
or
(2)
E 2 n = Et n
or
E 2+ eik 2d + E 2 e ik 2 d = E3 eik 3d
(3)
(k 2 E 2 ) n = (k 3 E t ) n
or
(4)
E 2+
E 2
(
n2 + n3 )
=
E e ik d ik d
3
2 n2
(n n )
= 2 3 E3 eik3 d + ik2 d
2n2
or
E0 =
=
(n1 + n2 )
2n1
E2 + +
(n1 n2 )
2n1
E2
2n1
2n2
and solving for E3 we get
2n1
2n2
4e ik 3 d
E3 =
E0
n2 n3 ik 2 d n2 n3 ik 2 d
1 + 1 + e
+ 1 1 e
n
n
1
2
n1 n2
Now solve for E1, again using (1) and (2)
(n n )
(n + n )
E1 = 1 2 E2 + + 1 2 E2
2n1
2n1
2n1
2n2
2n1
2n2
or
n2 n3 n2 n3 2ik 2 d
1 1 + + 1 + 1 e
n1 n2 n1 n2
E1 =
E0
n2 n3 n2 n3 2ik 2 d
1 + 1 + + 1 1 e
n1 n2 n1 n2
The reflection coefficient is
R=
E1
E0
Numerically, for n1 = 1, n 2 = 2, n 3 = 3,
2
5 3 2ik 2d
+ e
2 2
R=
15 1 2ik 2 d
+ e
2 2
2ik 2 d 2
5 3e
15 + e 2ik 2 d
e 2ik2 d = 1
corresponding to
d=
and
2 c c
=
=
n2 2d
For n1 = 3, n 2 = 2, n 3 = 1,
1 5 2 ik2 d
+ e
2
6
R=
5 1 2 ik2 d
+ e
2 6
e 2ik2 d = 1
corresponding to
d=
and
2ik 2 d 2
3 + 5e
15 + e 2ik 2d
2 c c
=
=
n2 4d
For n1 = 2, n 2 = 4, n 3 = 1,
5 9 2 ik 2d
+ e
4 4
R=
15 3 2 ik2 d
+ e
4 4
e 2ik2 d = 1
corresponding to
d=
2
and
2 c c
=
n2 4d
T=
n3 E3
n1 E0
2 ik 2 d 2
5 + 9e
15 3e 2ik 2d
Tranmission, n1 = 1, n 2 = 2, n 3 = 3
Reflection, n1 = 3, n 2 = 2, n 3 = 1
Transmission, n1 = 3, n 2 = 2, n 3 = 1
Reflection, n1 = 2, n 2 = 4, n 3 = 1
Transmission, n1 = 2, n 2 = 4, n 3 = 1
(b) The medium n1 is part of an optical system (e.g., a lens); medium n3 is air (n3 = 1). It
is desired to put an optical coating (medium n2 ) on the surface so that there is no reflected
wave for a frequency 0 . What thickness d and index of refraction n2 are necessary?
Ei
n1
n2
n3
Bi
For n 3 = 1,
n2
1 n
1
1 1 + + 1 + 2 1 e 2ik 2 d
n1 n2 n1 n2
R=
n2
1 n
1
1 + 1 + + 1 2 1 e 2ik 2 d
n1 n2 n1 n2
The reflection is minimum when
e 2ik2 d = 1
or
2 k2 d =
or
d=
=
2 k2 4
n2
1 n
1
1 1 + 1 + 2 1 = 0
n1 n2 n1 n2
Solve for n2 to get
n2 = n1