Q. 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 = l Tesla and $\omega = 0.125 \; rad/s$. The work done for reversing the direction of the magnetic moment at t = 1 second, is :

Solution:

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

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