Question

A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.16 T experiences a torque of magnitude 0.032 J.

(a) Estimate the magnetic moment of the magnet.

(b) If the bar were free to rotate, which orientations would correspond to its (i) stable, and (ii) unstable equilibrium? What is its potential energy in the field for cases (i) and (ii)?

Answer 1

(a) The torque on a magnetic dipole of magnetic moment M in an external field B is

τ = MB sin θ

where θ is the angle between M and B

Here τ = 0.032 J, B = 0.16 T, θ = 30°

∴ M = \(\frac{0.032}{0.16 \times \frac{1}{2}}\)  = 0.40 JT^-1.

The direction of M is from the south pole of the magnet to its north pole.

(b) Potential energy of a magnetic dipole in an external field B is given by U = - MB

(i) The bar is in stable equilibrium when its magnetic moment m is parallel to B (θ = 0); its potential energy then is minimum:

U = - MB = 0.40 x 0.16 = - 0 064 J

(ii) The bar is in unstable equilibrium when M is antiparallel to B (θ = 180°); its potential energy then is maximum.