spins with constant angular velocity w about an axis passing through
the center of the sphere (There is no material present). Take the axis of
rotation as the z-axis. The vector potentials both inside and outside the
sphere are:
Ain
=
mowr
2
rsinq
æ è
R2
3
-
r2
5
ö ø
^
f
Aout
=
mowrR5
15
sinq
r2
^
f
Maple will be used to do all of the calculations below. See the code in file
ph317t2.mws.
(a) [14 pts] From the vector potentials, determine the magnetic field
equations both inside and outside the sphere.
SOLUTION
Bin
=
Ñ×Ain
=
mowr
15
(5R2-3r2)cosq
^
r
-
mowr
15
(5R2 -6r2)sinq
^
q
Bout
=
Ñ×Aout
=
2mowrR5
15r3
cosq
^
r
+
mowrR5
15r3
sinq
^
q
=
mowrR5
15r3
æ è
2cosq
^
r
+ sinq
^
q
ö ø
(b) [8 pts] Why would an external observer consider the spinning sphere
to be a dipole?
SOLUTION
The equation for the magnetic field outside the sphere is just that of a
dipole. See Eq. 5.86 on page 246.
(c) [14 pts] Using the boundary condition
B||above-B||below
=
moK
determine the value of the surface current density, K.
SOLUTION
Dot [^(q)] into Bin and Bout to obtain the
parallel components. Evaluate each expression at r=R.
B||above
=
Bout·
^
q
ê ê
r=R
=
mowrR5
15r3
sinq
ê ê
r=R
=
mowrR2
15
sinq
B||below
=
Bin·
^
q
ê ê
r=R
=
-
mowr (5R2-6r2)
15
sinq
ê ê
r=R
=
mowrR2
15
sinq
Solving for the surface current density,
moK
=
mowrR2
15
sinq-
mowrR2
15
sinq
=
0
K
=
0
(d) [14 pts] Calculate the current density inside the sphere.
SOLUTION
J
=
1
mo
Ñ×Bin
=
wrrsinq
^
f
File translated from
TEX
by
TTH,
version 3.85. On 28 Jan 2010, 16:57.