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
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → - 0.17 J
A current loop is a magnetic dipole of moment \( m = NIA \). In a field its potential energy is
\( \displaystyle U = -\vec{m}\cdot\vec{B} = -mB\cos\theta \)
The energy is least when \( \theta = 0 \), the moment along the field, so \( U_{\min} = -mB \).
\( \displaystyle A = \pi r^{2} = \pi (0.03)^{2} = 2.83\times10^{-3}\,\mathrm{m^{2}} \)
\( \displaystyle m = NIA = 50\times2\times2.83\times10^{-3} = 0.283\,\mathrm{A\,m^{2}} \)
\( \displaystyle U_{\min} = -mB = -0.283\times0.6 = -0.17\,\mathrm{J} \)