Question

A short bar magnet of magnetic moment m = 0.32 JT^-1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its
(a) stable, and
(b) unstable equilibrium? What is the potential energy of the magnet in each case?

Answer 1

(a) When is parallel to g, the magnet is in stable equilibrium
U = -MB 

= -0.32 x 0.15 

= -4.8 x 10^-2 J

(b) When \(\vec M\) is antiparallel to g, the magnet is in unstable equilibrium.
U = MB 

= 0.32 x 0.15
= 4.8 x 10^-12 J.