Two pure dipoles are positioned in the y-z plane as shown in the
figure. The distance between the dipoles is r. The angle between the
dipoles is 30o. The angle between r and the y-axis is 30o.
Determine the torque on dipole m2, due to the magnetic fiels
of dipole m1.
N2 = m2×B1
SOLUTION
N2
=
m2×B1
=
m2×
� �
�
mo
4pr3
� �
3
� �
m1·
^
r
� �
^
r
- m1
� �
� �
�
=
mo
4pr3
� �
3
� �
m1·
^
r
� �
� �
m2×
^
r
� �
-m2×m1
� �
=
mo
4pr3
� �
3(m1cos60o )
� �
-m2sin90o
^
x
� �
-
� �
-m1m2 sin30o
^
x
� �
� �
=
mo
^
x
4pr3
� �
-
3
2
m1m2+
1
2
m1m2
� �
=
-
mom1m2
4pr3
^
x
An infinitely long right-circular cylinder of radius R carries a
"frozen-in" magnetization in a plane perpendicular to the cylinder axis.
M = ks
^
f
where, k, is a positive constant and s is the perpendicular distance from
the axis. There is no free current anywhere.
(a) Calculate Jb, Kb, rb, sb.
SOLUTION
The bound current density is
Jb
=
�×M
=
1
s
�
�s
(sMf)
^
z
=
1
s
d
d s
[s(ks)]
^
z
=
2k
^
z
The bound surface current density is
Kb
=
M×
^
n
=
ks
^
f
×
^
s
� �
r=R
=
-kR
^
z
The bound volume charge density is
rb
=
-�·M
=
-
1
s
�Mf
�f
=
0
The bound surface charge density is
sb
=
^
n
·M
=
^
s
·
� �
ks
^
f
� �
=
ks
� �
^
s
·
^
f
� �
=
0
(b) Calculate the total bound current in the cylinder. Draw a figure
showing how the currents are directed.
SOLUTION
The total bound current in the cylinder is
Ib
=
� �
Jb·da + Kbl
=
� �
R
0
� �
2k
^
z
� �
·
� �
2psds
^
z
� �
-(2pR)(kR)
=
2pkR2-2pkR2
=
0
The bound current due to the volume current density flows up the inside of the
cylinder. The bound current due to the surface current density flows down
the surface.
(c) Calculate Bout and Bin by any convenient
method.
SOLUTION
Since there are no free currents the auxiliary field is zero. Also, there
is no magnetization outside the sample. Therefore, the magnetic field outside
the cylinder is
Bout = 0
Inside the cylinder,
Bin
=
moM
=
moks
^
f
A long right-circular cylindrical conductor of radius a and magnetic
permeability m has its axis coincident with the z-axis. The conductor
carries a uniformly distributed constant current I along the positive
z-axis.
(a) Using Ampere's law in the form
� �
H·dl
=
(Ifree)encl
derive the formula for the magnetic intensity (auxiliary field),
Hin, inside the cylinder. Use the right-hand-rule to determine
its direction. Use the standard cylindrical variables and unit vectors.
SOLUTION
� �
2p
0
� �
Hin
^
f
� �
·
� �
sdf
^
f
� �
=
If
� �
ps2
pa2
� �
Hin(2ps)
=
If
� �
s2
a2
� �
Hin
=
� �
If
2pa2
� �
s
Hin
=
� �
If
2pa2
� �
s
^
f
(b) Determine the formula for the internal magnetic field,
Bin.
SOLUTION
Bin
=
mHin
=
mo(1+cm)
� �
If
2pa2
� �
s
^
f
(c) Using the results of parts (a) and (b), determine the formula
for the magnetization, M.
SOLUTION
Bin
=
moHin + moM
M
=
1
mo
[Bin-moHin]
=
1
mo
[mo(1+cm)-mo ]Hin
=
cmHin
=
� �
cmIf
2pa2
� �
s
^
f
File translated from
TEX
by
TTH,
version 3.85. On 28 Jan 2010, 16:13.