PROBLEM 6.1
KNOWN:
A current loop having
N = 20
A = 1 in
2
= 0.000645 m
2
I = 0.02 A
B
= 0.4 Wb/m
2
FIND:
Torque on the current loop, T
µ
.
ASSUMPTIONS:
The magnetic field is oriented at an angle of 90
o
to the current flow direction.
SOLUTION:
From (6.3)
T
NIAB
µ
α
=
sin
The maximum torque occurs when sin
α
= 1, which yields
(
)
(
)(
)
max
2
2
20
0.02 A
0.000645 m
0.4 Wb/m
T
NIAB
µ
=
=
max
4
1.03
10
N-m
T
µ
−
=
×

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