# 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 =
BANIsin

q 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 = 0.5x

px0.5

^{2}x5x4xsin(

q)

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

sin(

q) = 3.93/[0.5x3.142x0.5

^{2}x5x4] = 3.93/7.853 =
0.5

Therefore F = 30

^{o}.