PH317 MAGNETISM
WINTER Quarter 2009-2010
Solution to Homework Assignment 8
Assigned: Monday 2/01/10
Due: Monday 2/08/10
A circular wire loop of radius b lies in the z=0 plane and centered
about the origin. The wire carries a constant current I. A second circular
wire loop of radius a lies in the z=c plane and is also centered on the
z-axis. In other words, the two planes are parallel. The second loop
contains no current. Use the Neumann formula, Eq. 7.23, page 311 to calculate
the system's mutual inductance. Do this for values of c=0 m, and
c=3.00 m. Also
a=1.00 m
b=2.00 m
For the integration command in Maple use Int (not int). Also, surround the
integration command with evalf.
SOLUTION
Let the lower loop be labeled 1 and the upper loop be labeled 2. The Neumann
formula is
M21
=
mo
4p
ó (ç) õ
ó (ç) õ
dl1· dl2
r
The equation for r in cartesian coordinates is
r
=
Ö
(x2-x1)2+(y2-y1)2+ c2
Because of the circular symmetry, it would be wise to use polar
coordinates. The transformation equations are
x1 = bcosf1
y1 = bsinf1
x2 = acosf2
y2 = asinf2
The distance equation becomes
r
=
Ö
(acosf2-bcosf1)2+(asinf2 -bsinf1)2+c2
The dot product in the numerator can be written
dl1·dl2
=
(dl1)(dl2)cos(f2-f1)
=
(b df1)(a df2)cos( f2-f1)
=
abcos(f2-f1) df1df2
The Neumann formula becomes
M21
=
moab
4p
ó õ
2p
0
ó õ
2p
0
cos(f2-f1)df1df2
Ö
(acosf2-bcosf1)2+( asinf2-bsinf1)2+c2
Let Maple do the rest. The commands I used are
K:=muo*a*b/2*Pi
M21=evalf(K*Int(Int(subs(c=0,f),phi1=0..2*Pi),phi2=0..2*Pi));
M21=evalf(K*Int(Int(subs(c=3,f),phi1=0..2*Pi),phi2=0..2*Pi));
The results are:
c=0
M21=1.10×10-6 H
c=3
M21=1.57×10-7 H
The electric and magnetic fields in a light beam in a vacuum oscillate
with simple harmonic motion. Assume a beam is traveling in the z-direction
and the equations for these fields can be written as homogeneous plane waves
of the form
E
=
Eosin(kz-wt)
^
x
B
=
Bosin(kz-wt)
^
y
Show that under the following conditions,
k=w/c
Eo=cBo
c=1/
Ö
moeo
where, c is the speed of light in a vacuum, that the field expressions
given above satisfy Maxwell's equations given on page 326 in the box,
Eq. 7.39.
SOLUTION
(a) Show that the above fields satisfy
Ñ·E
=
0
The expression for the electric field can be written in functional form as
E = Ex(z,t)
^
x
Expand the divergence equation in cartesian coordinates.
Ñ·E
=
¶Ex
¶x
+
¶Ey
¶y
+
¶Ez
¶z
=
¶Ex
¶x
=
0
Maxwell's first equation is satisfied.
(b) Show that the above fields satisfy
Ñ·B
=
0
The expression for the magnetic field can be written in functional form as
B = By(z,t)
^
y
Ñ·B
=
¶By
¶y
=
0
Maxwell's second equation is satisfied.
(c) Show that the above fields satisfy
Ñ×E
=
-
¶B
¶t
The above equation expands to
¶Ex
¶z
^
y
=
-
¶By
¶t
^
y
Eokcos(kz-wt)
^
y
=
-[-Bowcos(kz-wt)]
^
y
(Boc)kcos(kz-wt)
^
y
=
Bo(kc)cos(kz-wt)
^
y
Both side are the same. Therefore, Maxwell's third equation is satisfied.
(d) Show that the above fields satisfy
Ñ×B
=
moJ+moeo
¶E
¶t
Since there are no currents present, J=0. Treat each of the remaining
terms, separately.
Ñ×B
=
-
¶By
¶z
^
x
=
-Bokcos(kz-wt)
^
x
¶E
¶t
=
-Eowcos(kz-wt)
^
x
Inserting into Maxwell's fourth equation,
-Bokcos(kz-wt)
^
x
=
moeo
é ë
-Eowcos(kz-wt)
^
x
ù û
Bokcos(kz-wt)
=
moeoEowcos (kz-wt)
The question is: does Bok=moeoEow? Start with
the left-hand-side and use the conditional equations.
Bok
=
(Eo/c)(w/c)
=
Eow
c2
=
Eowmoeo
Therefore, Maxwell's fourth equation is satisfied.
Text: 7.33
SOLUTION
(a) The induced electric field in the cavity is, from problem 7.16
E(s,t)
=
moIow
2p
sin(wt) ln
æ è
a
s
ö ø
^
z
The displacement current density is
Jd
=
eo
¶E
¶t
=
moeoIow2
2p
cos(wt) ln
æ è
a
s
ö ø
^
z
=
moeoIw2
2p
ln
æ è
a
s
ö ø
^
z
º
k2I
2p
ln
æ è
a
s
ö ø
^
z
(b) The differential area is a concentric annular ring of area
da = 2ps ds
^
z
The displacement current is then
Id
=
ó õ
Jd·da
=
ó õ
a
0
é ë
moeoIw2
2p
ln
æ è
a
s
ö ø
^
z
ù û
·
é ë
2ps ds
^
z
ù û
=
moeow2I
ó õ
a
0
s ln(a/s)ds
The diameter of the central wire is to be neglected. That is why the lower
limit was taken to be zero. The answer to the integral can be found by
integration by parts or simply have Maple do it. The answer is a2/4.
The current density becomes
Id
=
1
4
moeow2Ia2
(c) Look at the ratio of currents using the result of part (b).
Id
I
=
1
4
moeow2a2
Using the reduction formula, moeo=1/c2,
Id
I
=
æ è
wa
2c
ö ø
2
Using the information supplied in the problem,
Id/I=1/100
a=10-3 m
1
100
=
æ è
wa
2c
ö ø
2
1
10
=
wa
2c
w
=
2c
10a
=
2(2.998×108 m/s)
10(1.00×10-3 m)
=
6×1010 rad/s
f
=
w
2p
»
1010 Hz
This frequency is in the microwave region.
text 7.42
SOLUTION
(a) Use Faraday's law in differential form.
Ñ×E
=
-
¶B
¶t
Since, E=0 inside a perfect conductor, the curl is zero.
Therefore,
-
¶B
¶t
=
0
B
=
f(r)
The magnetic field inside is independent of time but could still depend on
space.
(b) Use Faraday's law in integral form.
ó (ç) õ
E·dl
=
-
dF
dt
Again, since E=0 inside the perfect conductor, the left-hand-side is
zero.
dF
dt
=
0
F
=
const
(c) Use Ampere's modified law in differential form.
Ñ×B
=
moJ + moeo
¶E
¶t
Since B=0 inside a superconductor (Meissner effect), the curl of
the field is zero. Also, since the electric field is zero, its time
derivative is also zero. This implies that there is no conduction current
inside the volume of the conductor. Therefore, any existing current must be on
the surface. Because of the Meissner effect, there is a current.
(d) Use the results of example 5.11, page 236. The field inside the
spinning sphere is
Bin
=
2
3
mowsR
The surface current density is
K
=
sv
=
s w×R
=
swR sinq
^
f
We need to replace the quantity swR in the surface current
density expression. Let
Bin
=
-Bo (diamagneticbehavior)
Bo
=
-
2
3
mowsR
wsR
=
-
3Bo
2mo
Inserting this result into the surface current equation
K
=
-
3Bo
2mo
sinq
^
f
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version 3.85. On 10 Feb 2010, 17:22.