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 ?

Explanation

Here m =0.32 J `T^(-1) and B = 0.15 T`
(a) Stable equilibrium orientation means the magnet is set parallel to the external magnetic field and in this position , the potential energy of the magnet is
`U_1 = -vecm * vecB = - mcos 0^@ =- mB =-0.32 xx 0.15 = -4.8 xx 10^(-2) J. `
(b) Unstable equilibrium orientation means the magnet is set anti-parallel to the external magnetic field i.e. `theta = pi` . In unstable equilibrium stable, the potential energy of the magnet will be
`U_2 = -vecm * vecB = - m B cos pi = -0.32xx0.15 xx (-1) = + 4.8xx10^(-2) J`.