Torque on a current loop?

Question

A 200 trun circular loop of radius 50.0 cm is vertical, with its axis on an east west line. A current of 100 A circulates clockwise in the loop when viewed from the east. THe earths fireld here is due north, parallel to the ground with as strength of 3.00 * 10^-5 T. What are the direction and magnitude of the torque on the loop?

Answer 1

The magnitude of the magnetic dipole moment of the loop is given by

m = Ia, where I is the current and a is the area enclosed by the loop.  The enclosed area is the number of turns multiplied by the area enclosed by a single loop:

a = 200(pi (0.500m)^2) = 157.08 m^2

So m = (100A)(157.08 m^2) = 15708 A m^2

The direction of the magnetic dipole moment is given by the right hand rule: curl fingers of right hand in the direction of the current; your thumb points in the direction of the dipole moment.  Therefore, m points west.

Now, the torque on the loop is given by

T = m x B   (cross product of vectors m and B)

The angle between m and B is 90 degrees, so the magnitude of the torque is

mB sin90 = (15708 Am^2)(3x10^-5 T) = 0.471 Nm

Since m points west and B points north, their cross product points vertically down.

So, final answer:  the torque is 0.471 Nm vertically downward.