PH317 MAGNETISM
WINTER Quarter 2009-2010
Solution to Homework Assignment 3
Assigned: Thursday 12/17/09
Due: Tuesday 1/05/10
Text: 5.23
SOLUTION
The equation to deal with is
A=k
^
f
=Af
^
f
Two equations are needed:
B=Ñ×A
moJ=Ñ×B
Since, As=Az=0, the curl equation in cylindrical coordinates
reduces to
Ñ×A
=
1
s
¶
¶s
( sAf)
^
z
Since, Bs=Bf=0, the curl equation in cylindrical coordinates
reduces to
Ñ×B
=
-
¶Bz
¶s
^
f
The first expression gives B.
B
=
æ è
k
s
ö ø
^
z
The second expression gives J.
J
=
æ è
k
mos2
ö ø
^
f
The boundary conditions for the vector potential and the magnetic field
were derived in class. The results are:
A^above
=
A^below
A||above
=
A||below
B^above
=
B^below
B||above - B||below
=
mo K
Babove-Bbelow
=
mo(K×
^
n
)
¶Aabove
¶n
-
¶Abelow
¶n
=
-moK
where,
¶A
¶n
=
(Ñ×A)×
^
n
ê ê
surface
Apply all of the above boundary conditions to the charged spinning sphere
example worked out in class. See your notes for the A and B
expressions or check Example 5.11 on page 236.
SOLUTIONS
The field equations are:
Ain
=
1
3
moMrsinq
^
f
r £ R
Aout
=
moMR3sinq
3r2
^
f
r ³ R
Bin
=
2
3
moM
æ è
cosq
^
r
- sinq
^
q
ö ø
r < R
Bout
=
moMR3
3r3
æ è
2cosq
^
r
+ sinq
^
q
ö ø
r > R
(a) A^above = A^below
SOLUTION
Each side is to be evaluated at the surface.
Aout·
^
r
=
Ain·
^
r
Af(out)
^
f
·
^
r
=
Af(in)
^
f
·
^
r
moMR3sinq
3R2
(0)
=
moMRsinq
3
(0)
0
=
0 consistency
(b) A||above = A||below
SOLUTION
Aout·
^
f
=
Ain·
^
f
moMR3sinq
3R2
^
f
·
^
f
=
moMRsinq
3R
^
f
·
^
f
1
=
1 consistency
(c) B^above = B^below
SOLUTION
Bout·
^
r
=
Bin·
^
r
moM
3
(2cosq
^
r
+ sinq
^
q
)·
^
r
=
2moM
3
(cosq
^
r
-sinq
^
q
)·
^
r
2moMcosq
3
=
2moMcosq
3
consistency
(d) B||above - B||below = mo K
SOLUTION
Bout·
^
q
- Bin·
^
q
=
moK
moM
3
(2cosq
^
r
+sinq
^
q
)·
^
q
-
2moM
3
(cosq
^
r
-sinq
^
q
)·
^
q
=
moMsinq
3
+
2moMsinq
3
=
moMsinq
=
moK
K
=
Msinq
(e) Babove-Bbelow = mo(K× [^(n)])
SOLUTION
Bout-Bin
=
mo(K
^
f
×
^
r
)
=
moK
^
q
moMsinq
3
^
q
+
2moMsinq
3
^
q
=
moK
^
q
Msinq
=
K consistency
(f) [(¶Aabove)/(¶n)]-[(¶ Abelow)/(¶n)] = -moK
SOLUTION
(Ñ×A)out×
^
n
ê ê
R
-(Ñ×A)in×
^
n
ê ê
R
=
-moK
Bout×
^
r
ê ê
R
-Bin×
^
r
ê ê
R
=
-moK
moMR3
3r3
(2cosq
^
r
+ sinq
^
q
)×
^
r
-
2moM
3
(cosq
^
r
- sinq
^
q
)×
^
r
=
-moK
-
moMsinq
3
^
f
-
2moMsinq
3
^
f
=
-moK
-moMsinq
^
f
=
-moK
K
=
Msinq
^
f
consistency
File translated from
TEX
by
TTH,
version 3.85. On 16 Jan 2010, 11:44.