Question

A short bar magnet of magnetic moment m = 0.32 J T^-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

Moment of the bar magnet, M = 0.32 J T^-1
External magnetic field, B = 0.15 T
( a ) The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle $\theta$, between the bar magnet and the magnetic field is 0°. 
Potential energy of the system = -MBcos $\theta$
= -0.32 × 0.15cos 0°
= -4.8 × 10-2 J

---

( b) The bar magnet is oriented 180° to the magnetic field. Hence, it is in unstable
equilibrium. $\theta$ = 180°
Potential energy = − MB cos $\theta$
= -0.32 × 0.15cos180°
= 4.8 × 10^-2 J