A current of 2 A flows through a plane circular coil of radius 3 cm and having 50 turns. The coil has been placed in a uniform magnetic field of 0.6 T. The minimum potential energy of the coil will be

- 0.17 J

The correct answer is Option (3) → - 0.17 J

Magnetic dipole moment:

$M = N I A = 50 \times 2 \times \pi (0.03)^2 = 0.2827 \,\text{A·m}^2$

Minimum potential energy (when $\theta = 0^\circ$):

$U_{min} = - M B = - (0.2827)(0.6) = -0.1696 \,\text{J}$

Answer: $-0.17 \,\text{J}$ (approx)