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, What is the potential energy of the magnet in unstable position?

• A. $4.8\times 10^{- 2}J$
• B. $9.6\times 10^{- 2} \, J$
• C. $2.4\times 10^{- 2} \, J$
• 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 to magnetic field thus no torque will act on it.)
$\theta = 1 8 0^{\text{o}}$
The potential energy of the magnet
$U=-mBcos180^{^\circ }$
$=-0.32 \times 0.15(-1)=4.8 \times 10^{-2} \mathrm{~J}$
Thus, for the unstable equilibrium, the potential energy is $4.8\times 10^{- 2}J$ .