A short bar magnet of magnetic moment $m=0.45 J/T$ is placed in a uniform magnetic field of 0.22T. If the bar is free to rotate in the plane of the field, what is the potential energy of the magnet in stable and unstable equilibrium respectively?

$9.8×10^{-2} J$ (Unstable); $-9.8×10^{-2} J$, (Stable)

The correct answer is Option (4) → $9.8×10^{-2} J$ (Unstable); $-9.8×10^{-2} J$, (Stable)

The potential energy (U) of a magnetic dipole in a uniform magnetic field -

$U=-mB\cos θ$

where,

Magnetic moment (m) = 0.45 J/T

Magnetic field (B) = 0.22 T

θ = Angle between magnetic moment vector and magnetic field

for stable equilibrium, $θ=0°$

$U_{stable}=-mB\cos 0$

$=-(0.45×0.22)$

$≃-9.8×10^{-2}J$

for unstable equilibrium, $θ=180°$

$U_{unstable}=mB(\cos 180)=mB$

$≃9.8×10^{-2}J$