Page 183 - Electromagnetics
P. 183
and by integration
I 2 ∞ y 2 2 µ 0
F 1 = µ 0 2 ˆ x dz 2 2 2 dy = I ˆ x dz.
2π −∞ [y + d /4] 2πd
The resulting force per unit length agrees with (3.175)when I 1 = I 2 = I.
Torque in a magnetostatic field. The torque exerted on a current-carrying conduc-
tor immersed in a magnetic field plays an important role in many engineering applica-
tions. If a rigid body is exposed to a force field of volume density dF(r), the torque on
that body about a certain origin is given by
T = r × dF dV (3.177)
V
where integration is performed over the body and r extends from the origin of torque.
If the force arises from the interaction of a current with a magnetostatic field, then
dF = J × B and
T = r × (J × B) dV. (3.178)
V
For a line current we can replace J dV with Idl to obtain
T = I r × (dl × B).
If B is uniform then by (B.7)we have
T = [J(r · B) − B(r · J)] dV.
V
The second term can be written as
3
B(r · J) dV = B x i J i dV = 0
V i=1 V
where (x 1 , x 2 , x 3 ) = (x, y, z), and where we have employed (3.27). Thus
3 3 3 3
T = J(r · B) dV = ˆ x j J j x i B i dV = B i ˆ x j J j x i dV.
V j=1 V i=1 i=1 j=1 V
We can replace the integral using (3.144)to get
3
3
1
1
T = ˆ x j B i [x i J j − x j J i ] dV =− B × (r × J) dV.
2 V j=1 i=1 2 V
Since B is uniform we have, by (3.145),
T = m × B (3.179)
where m is the dipole moment. For a planar loop we can use (3.147)to obtain
T = IAˆ n × B.
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