Question

A long piece of wire with a mass of 0.170 kg and total length of 4.00 m is used to make a square coil with a side of 0.100 m. The coil is hinged along a horizontal side, carries a 3.10 A current, and is placed in a vertical magnetic field with a magnitude of 0.0100 T. (a) Determine the angle that the plane of the coil makes with the vertical when the coil is in equilibrium. =______ degrees with the vertical

(b) Find the torque acting on the coil due to the magnetic force at equilibrium.
=_______ N*m

Answer 1

The magnetic torque is given by: N = r x F
where F = c IL x B
(I'm using c for # of coils, I for current, L for length of a side, B for the magnetic field)
So N = r x (cIL x B)

r has the same magnitude as L. The L vector is always perpendicular to B, so.

N(magnetic) = cIL^2B cos theta
where r is the distance from the hinge. The cosine comes about because you have max torque when the coil is vertical and the magnetic force is in the direction of rotation. You have no torque when the coil is horizontal and the magnetic force is toward the axis.

The gravitational torque is given by:

(weight/4 * 0 (hinge side) +
weight/4 * L (opposite side) +
2 * weight/4 * L/2 (adjacent sides)) * sin (theta)
= weight * L / 2 * sin (theta)

I just added up the weight of each side times the distance from the CM to the axis. The sin comes about because there's no torque when the loop hangs straight down (gravity pulls directly away from axis) and maximum torque when it's horizontal and gravity pulls it down.

Set the torques equal:

cIL^2B cos theta = weight * L / 2 * sin (theta)

theta = arctan (2 cILB / mg)

They give you the current (I), the length of a side (L), the magnetic field (B), the mass. You know g. c (number of coils) is the square's perimeter divided by the entire length of wire. Plugnchug.

Then plug that back in to get the magnetic and gravitational torques, which should be equal.