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A magnet of total magnetic moment $10^{- 2} \hat{i} {Am}^{2}$ is placed in a time varying magnetic field $B \hat{i} (cosωt),$ where B=1T and $ω = 0 .125 rad / s$ . The work done for reversing the direction of the magnetic moment at t=1s is

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$0.007 J$

$0.028 J$

$0.0198 J$

$0.01 J$

answer is C.

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Detailed Solution

Work done in reversing dipole i

$W = 2 M B$

Where M= magnetic dipole moment = $10^{- 2} A m^{2}$

And

B = external field = $B \cos w t = 1 \cos (0.125 \times 1)$

$= \cos (7^{0}) = 0.992$

$∴ W = 2 \times 10^{- 2} \times 0.992 = 0.0198 J$

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