Question

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

• A. $4.8\times 10^{- 2}J$
• B. $9.6\times 10^{- 2}J$
• C. $2.8\times 10^{- 2}J$
• D. $1.2\times 10^{- 2}J$

Answer 1

Answer: (A)

For the unstable equilibrium, the angle between the magnetic moment and magnetic field is 180^o. ( $\because $ In this position it will be in a direction perpendicular to magnetic field thus maximum torque will act on it.)
$\theta = 1 8 0^{\text{o}}$
Potential energy of the magnet
U = - mB cos 180^o
= - 0.32 x 0.15 (- 1) = 4.8 x 10^-2 J
Thus, for the unstable equilibrium the potential energy is 4.8 x 10^-2 J.