Question

A short bar magnet of moment 0.5.A m² is suspended in a horizontal magnetic field of 1000 gauss such that it can rotate freely in a horizontal plane. Calculate (i) the magnitude of the torque and (ii) the magnetic potential energy when its angular displacement with respect to the direction of the field is 10°.

Answer 1

Data: M = 0.5 A.m², B = 1000 gauss = 0.1 T, θ = 10°

The magnitude of the torque is

τ = MB sin θ

= (0.5)(0.1) sin 10°

= 0.05 x 0.1736

= 8.68 × 10⁻³ N.m

The magnetic potential energy of the bar magnet is

U_θ = MB cos θ

= (0.5)(0.1) cos 10°

= 0.05 x 0.9848

= 0.04924 J