PH317 MAGNETISM
WINTER Quarter 2009-2010
Solution to Homework Assignment 5
Assigned: Wednesday 1/13/10
Due: Tuesday 1/19/10
Text: 6.1
SOLUTION
Setup the rectangular axes with the x-axis pointing out of the page. Take
the y-axis along the r-direction (directed from circle to square). Take
the z-axis straight up. The vectors involved are:
mC
=
pa2
^
z
mS
=
Ib2
^
y
^
r
=
^
y
Also,
r=y
The magnetic field at the square dipole due to the circle dipole is
(Eq. 5.87, page 246)
BC
=
mo
4p
1
y3
é ë
3
æ è
mC ·
^
y
ö ø
^
y
-mC
ù û
=
mo
4p
1
y3
é ë
3
æ è
pa2
^
z
·
^
y
ö ø
^
y
-mC
ù û
=
-
momC
4py3
=
-
moIa2
4y3
^
z
The torque on the square loop is
NS
=
mS×BC
=
æ è
Ib2
^
y
ö ø
×
æ è
-
moIa2
4y3
^
z
ö ø
=
-
mo(Iab)2
4y3
^
x
Looking in along the x-axis, the loop would rotate clockwise with the final
magnetic moment being
mS
=
-Ib2
^
z
Text: 6.3
SOLUTION
(a) We don't know anything about what makes up m1. Since a
current appears in Eq. 6.2 on page 257, take dipole m2 to be a
current loop of radius R whose axis coincides with the x-axis. Both
dipoles lie on, and point in, the x-direction. Then, symbolically,
m2 = m2
^
x
= pR2I
^
x
Use Fig. 6.4 on page 257 as a reference.
F=2pIRB1cosq
where q is the angle between B1 and the y-axis. Connect
dipole m1 to a point on the current loop (in the x-y plane).
Designate this vector displacement by r along with the unit vector
[^(r)]. The angle between the x-axis and r is f.
B1cosq
=
B1·
^
y
=
mo
4p
1
r3
é ë
3
æ è
m1·
^
r
ö ø
æ è
^
r
·
^
y
ö ø
-m1·
^
y
ù û
=
mo
4p
1
r3
é ë
3m1
æ è
^
x
·
^
r
ö ø
æ è
^
r
·
^
y
ö ø
-m1
^
x
·
^
y
ù û
The dot products are
^
x
·
^
r
=
cosf
^
r
·
^
y
=
sinf
^
x
·
^
y
=
0
The expression for B1cosq reduces to
B1cosq
=
3mom1
4pr3
sinfcosf
The trig functions become
sinf
=
R
r
cosf
=
æ Ö
1-
R2
r2
sinfcosf
=
R
Ö
r2-R2
r2
The component becomes
B1cosq
=
3mom1R
Ö
r2-R2
4pr5
The force becomes
F
=
2pmo IR
é ê
ë
3mom1R
Ö
r2-R2
4pr5
ù ú
û
=
3mo(pR2I)m1
Ö
r2-R2
2pr5
For r >> R, the force equation reduces to
F
=
3mo(pR2I)m1
2pr4
=
3mom1m2
2pr4
(b) The second method.
F
=
Ñ(m2·B)
=
(m2·Ñ)B
The first part can be written
m2·Ñ
=
æ è
m2
^
x
ö ø
·
æ è
¶
¶x
^
x
+
¶
¶y
^
y
+
¶
¶z
^
z
ö ø
=
m2
¶
¶x
The force expression becomes
F
=
æ è
m2
¶
¶x
ö ø
ì í
î
mo
4p
1
x3
é ë
3m1 (
^
x
·
^
x
)
^
x
-m1
^
x
ù û
ü ý
þ
=
mom1m2
2p
¶
¶x
æ è
1
x3
ö ø
^
x
=
-
3mom1m2
2px4
^
x
=
-
3mom1m2
2pr4
^
x
Text: 6.12
SOLUTION
(a) Given
M = ks
^
z
Jb
=
Ñ×M
=
-
¶M
¶s
^
f
=
-k
^
f
Kb
=
M×
^
n
=
(ks
^
z
)×
^
s
=
ks
^
f
ê ê
s=R
=
kR
^
f
Since there are no free currents H=0. The results indicate a
superposition of solenoids with two currents going in opposing directions.
Therefore, The magnetic field outside the cylinder is zero.
Bout = 0
Inside, use Ampere's law to find Bin.
ó (ç) õ
Bin·dl
=
moIencl
Choose a rectangular path for the amperian loop with two of the sides parallel
to the side of the cylinder (one side inside the cylinder, the other side
outside the cylinder). Let this side have a length l.
Binl
=
moIencl
Iencl
=
ó õ
Jb·da + Kbl
=
mo
é ë
ó õ
R
s
æ è
-k
^
f
ö ø
·
æ è
l ds¢
^
f
ö ø
+ kRl
ù û
=
mo[-kl(R-s)+kRl]
=
mokls
Bin
=
moks
Bin
=
moks
^
z
(b) The equation to use is
B = moH+moM
Since the magnetic intensity is zero,
Bin
=
moM
=
moks
^
z
Given Wout derived in class for the spinning sphere, calculate
Hout and Bout.
SOLUTION
Wout
=
MR3cosq
3r2
Hout
=
-ÑWout
=
-
é ë
¶Wout
¶r
^
r
+
1
r
¶Wout
¶q
^
q
ù û
¶Wout
¶r
=
-
2MR3cosq
3r3
¶Wout
¶q
=
-
MR3sinq
3r2
The outside magnetic intensity is
Hout
=
-
é ë
-
2MR3cosq
3r3
^
r
+
1
r
æ è
-
MR3sinq
3r2
ö ø
^
q
ù û
=
MR3
3r3
æ è
2cosq
^
r
+sinq
^
q
ö ø
The outside magnetic field is
Bout
=
moH
=
moMR3
3r3
æ è
2cosq
^
r
+sinq
^
q
ö ø
Turn in the magnetic dipole calculation from the measurements taken in
lab on Friday.
SOLUTION
Substitute measured values of md and zo into the equation below
for the magnetic dipole moment.
m
=
zo2
æ Ö
2pmdg
3mo
Answers will vary depending on your respective measurements of md and
zo.
File translated from
TEX
by
TTH,
version 3.85. On 20 Jan 2010, 10:13.