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55. The magnetic dipole moment is
~µ
=
µ
(0
.
60
ˆ
i
−
0
.
80
ˆ
j), where
µ
=
NiA
=
Niπr
2
=1(0
.
20A)
π
(0
.
080m)
2
=
4
.
02
×
10
−
4
A
·
m
2
.He
re
i
is the current in the loop,
N
is the number of turns,
A
is the area of the loop,
and
r
is its radius.
(a) The torque is
~τ
=
~µ
×
~
B
=
µ
(0
.
60
ˆ
i
−
0
.
80
ˆ
j)
×
(0
.
25
ˆ
i+0
.
30
ˆ
k)
=
µ
h
(0
.
60)(0
.
30)(
ˆ
i
×
ˆ
k)
−
(0
.
80)(0
.
25)(
ˆ
j
×
ˆ
i)
−
(0
.
80)(0
.
30)(
ˆ
j
×
ˆ
k)
i
=
µ
[
−
0
.
18
ˆ
j+0
.
20
ˆ
k
−
0
.
24
ˆ
i]
.
Here
ˆ
i
×
ˆ
k=
−
ˆ
j,
ˆ
j
×
ˆ
i=
−
ˆ
k, and
ˆ
j
×
ˆ
k=
ˆ
i are used. We also use
ˆ
i
×
ˆ
i = 0. Now, we substitute the
value for
µ
to obtain
~τ
=
³
−
0
.
97
×
10
−
4
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Current

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