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Magnetic moment of bar magnet is $\mathrm{M}$. The work done to turn the magnet by $90^{\circ}$ of magnet in direction of magnetic field $\mathrm{B}$ will be
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The correct answer is:
$\mathrm{MB}$
Work done, $\mathrm{W}=\mathrm{MB}(1-\cos \theta)$
$\theta=90^{\circ}$
$\mathrm{W}=\mathrm{MB}$
$\theta=90^{\circ}$
$\mathrm{W}=\mathrm{MB}$
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