In a uniform magnetic field of strength 0 . 15 T , a short bar magnet of magnetic moment m = 0 . 32 J T - 1 is placed. The potential energy of the magnet in its unstable equilibrium position is

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For the unstable equilibrium, the angle between the magnetic moment and magnetic field is 180^o. ( $∵$ In this position it will be in a direction perpendicular to magnetic field thus maximum torque will act on it.)

$θ = 18 0^{o}$

Potential energy of the magnet

U = - mB cos 180^o

= - 0.32 x 0.15 (- 1) = 4.8 x 10^-2J

Thus, for the unstable equilibrium the potential energy is 4.8 x 10^-2J.

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