A square shaped plane coil of area `100 cm^(2)` turns a carries a steady current of 5 A . It is placed in a uniform magnetic field of 0.2 T acting perpendicular to the plane of the coil. Calculate the torque on the coil when its when makes an angle of `60^@` with the direction of the field. In which orinetation will the coil be in stable equilibrium?

Explanation

Here area of coil `A = 100 cm^2 = 10^(-2) m^2, ` number of turns N = 200, current I = 5 A, magnetic field B = 0.2 T. As plane of coil makes an angle of `60^@` with the direction of the field hence angle between `vecA and vecB` is `theta = 90^@ - 60^@ = 30^@`
`:.` Torque `tau = NA IB sin theta = 200 xx 10^(-2) xx 5 xx 0.2 xx sin 30^@ = 1 N-m`
The coil will be in stable equilibrium when plane of the coil is perpendicular to the direction of magnetic field B so that `vecA and vecB` are parallel and torque `tau = 0`.