MAGNETISM
Solution to Exam 3
DUE: Monday 2/15/10 - In Class
Friday 2/12/10
Name
DL119: period 4
JFW
CM:
(1)
(25)
(2)
(25)
(3)
(25)
(4)
(25)
TOTAL
(100)
Give brief answers to the following questions.
(a) [5 pts] Is there an electric field inside a perfect conductor which
is carrying a current? Explain your reasoning.
ANS: There is no electric field inside a perfect conductor.
E
=
J
s
=
J
¥
=
0
(b) [5 pts] If there is no electric field in a superconductor, how
would you go about increasing the surface currents when an applied magnetic
field is present?
ANS: Change the applied magnetic field with time.
(c) [5 pts] After electromagnetic radiation is emitted from a source,
how does it sustain itself when traveling through space?
ANS: An electric field that changes with time induces a magnetic field which
varies with time, also. The time varying magnetic field induces an electric
field which likewise varies with time. They keep generating each other until
something absorbs the energy.
(d) [5 pts] For solutions to the wave equations which obey Maxwell's
equations, what conditions must be met? Keep it qualitative. Don't write down
any equations!
ANS: Boundary conditions must be satified to give a unique solution.
(e) [5 pts] How did Maxwell solve Ampere's constant current constraint?
ANS: The displacement current density term was added to include magnetic
fields induced by time-varying electric fields.
Two concentric metal spherical shells, of radius a and b,
respectively, are separated by a weakly conducting material of conductivity
s. The material is linear, isotropic but not homogenious producing a
conductivity which varies with position.
s(r) = asa/r
Assume a positive charge Q on the inner plate with an equal negative charge
on the outer plate.
(a) [10 pts] If the plates are maintained at a constant potential
difference, V, what current flows from one to the other?
I
=
ó õ
J·da
=
ó õ
sE·da
=
s
ó õ
E·da
=
sFE
=
sQ
eo
(b) [15 pts] Calculate the resistance between the shells (Hint: Use the
equation dR=dV/I, where I depends on the conductivity which varies with
position).
Calculate the potential difference between the shells. Since resistance is
positive, we want Va-Vb.
dR
=
dVab
I
=
E·dl
I
=
(E
^
r
)·(dr
^
r
)
I
=
Edr
I
=
(Q/4peo r2)dr
( asa Q/reo)
R
=
1
4pasa
ó õ
b
a
dr
r
=
ln(b/a)
4pasa
Refer to the results of the spinning sphere, Example 5.11 on
page 236. This problem requires the determination of the total energy stored
in the system (that means the energies stored inside and outside the sphere
are considered). However, I wan't you to use Eqn. 7.31 on page 318 with
Jdt replaced with Kda. With this form, you only need to
integrate over the surface to get the answer. The reason for not using the
volume integral is because the current density inside the sphere is
zero. (25 pts)
W
=
1
2
ó õ
Ain|surf·K da
The three parts to the integrand are
A|surf
=
1
3
mosw R2sinq
^
f
K
=
sv
=
sw×R
=
swRsinq
^
f
da
=
R2sinq dqdf
Assembling the integrand
W
=
1
2
ó õ
p
0
ó õ
2p
0
æ è
1
3
moswR2sinq
^
f
ö ø
·
æ è
swRsinq
^
f
ö ø
R2sinq dqdf
=
mow2s2R5
6
ó õ
2p
0
df
ó õ
p
0
sin3qdq
=
mow2s2R5
6
(2p)(4/3)
=
4pmow2s2R5
9
The electric and magnetic fields for a long coaxial transmission line
(see Fig. 7.39, page 319) are given by
E(s,f,z,t)
=
Acos(kz-wt)
s
^
s
B(s,f,z,t)
=
Acos(kz-wt)
cs
^
f
where, A is a constant.
(a) [15 pts] Show, by direct substitution, that the above fields
satisfy Maxwell's equations in the form
Ñ·E
=
0
Ñ·B
=
0
Ñ×E
=
-
¶B
¶t
Ñ×B
=
1
c2
¶E
¶t
and also satify the boundary conditions
E|| = 0
B^ = 0
Rewrite the fields as
E(s,f,z,t)=E(s,z,t)
^
s
=Es
^
s
B(s,f,z,t)=B(s,z,t)
^
f
=Bf
^
f
Ñ·E
=
1
s
¶
¶s
( sEs)
=
1
s
¶
¶s
é ë
s
æ è
A
s
ö ø
cos(kz-wt)
ù û
=
0
Ñ·B
=
1
s
¶Bf
¶f
=
0
Ñ×E
=
-
¶B
¶t
¶Es
¶z
^
f
-
1
s
¶Es
¶f
^
z
=
-
Bf
¶t
^
f
-
kAsin(kz-wt)
s
^
f
-
1
s
(0)
^
z
=
-
é ë
Awsin(kz-wt)
cs
ù û
^
f
-
kAsin(kz-wt)
s
^
f
=
-
kAsin(kz-wt)
s
^
f
Ñ×B
=
1
c2
¶E
¶t
-
¶Bf
¶z
^
s
+
1
s
¶
¶s
(sBf)
^
z
=
1
c2
¶Es
¶t
^
z
Aksin(kz-wt)
cs
^
s
+
1
s
¶
¶s
ì í
î
s
é ë
Acos(kz-wt)
cs
ù û
ü ý
þ
^
z
=
1
c2
Awsin(kz-wt)
s
^
s
Aksin(kz-wt)
cs
^
s
+
1
s
(0)
^
z
=
w
c
Asin(kz-wt)
cs
^
s
Aksin(kz-wt)
cs
^
s
=
Aksin(kz-wt)
cs
^
s
As far as the boundary conditons go, only Es and Bf exist.
Therefore,
E|| = Ez = 0
B^ = Bs = 0
(b) [10 pts] Use Gauss's law to find the charge density l(z,t)
on the inner conductor and Ampere's law to find the current on the inner
conductor.
From Gauss's law
FE
=
ó (ç) õ
E·da
=
ó õ
E·da + FE(caps)
=
ó õ
é ë
Acos(kz-wt)
s
^
s
ù û
·
æ è
2psdz
^
s
ö ø
+0
=
2pAlcos(kz-wt)
=
Qencl
eo
=
ll
eo
l(z,t)
=
2pAeocos(kz-wt)
Since the electric field is in the radial direction, we don't have to worry
about the displacement current along the wire: there isn't any.
ó (ç) õ
B·dl
=
moIencl
Rewrite B in terms of E.
B
=
1
c
^
k
×E
=
1
c
^
z
×
æ è
Es
^
s
ö ø
=
Es
c
^
f
Substitute into Ampere's law.
ó (ç) õ
æ è
Es
c
^
f
ö ø
·
æ è
sdf
^
f
ö ø
=
2ps Es
c
=
2pAcos(kz-wt)
c
I
=
2pAcos(kz-wt)
moc
File translated from
TEX
by
TTH,
version 3.85. On 18 Feb 2010, 18:04.