Force on a coil

A circular coil consists of 5 loops, each of diameter 1.0 m.
The coil is placed in an external magnetic field of 0.5T (telsa).
When the coil carries a current of 4.0 A, a torque of magnitude 3.93 Nm , acts on it . Find the angle between the normal to the plane of the coil and the direction of the magnetic field.

Answer

Firstly unless there is an opposing torque your coil will start spinning.

The formula for torque on a coil in a magnetic field is:

T = BANIsinq where B is the magnetic field intensity, A is the area of the coil, N the number of coils, I the current in the coil and q the angle between the normal to the plane of the coil and the magnetic field.

So:

3.93 = 0.5xpx0.5²x5x4xsin(q)

(Note that the radius of the coils is 0.5 m)

sin(q) = 3.93/[0.5x3.142x0.5²x5x4] = 3.93/7.853 = 0.5

Therefore F = 30^o.