Question

A magnet of total magnetic moment $10^{-2} \hat{i} \; A-m^2$ is placed in a time varying magnetic field, $B \hat{i} (\cos t\omega t)$ where $B = 1$Tesla and $\omega = 0.125 \; rad/s$. The work done for reversing the direction of the magnetic moment at $t = 1$ second, is :

• A. 0.007 J
• B. 0.02 J
• C. 0.01 J
• D. 0.028 J

Answer 1

Answer: (B)

0.02 J

Answer 2

Work done, $W = ( \Delta \vec{\mu} ) . \vec{B}$
$= 2 \times 10^{-2} \times 1 \; \cos(0.125)$
= $0.02\, J$