A magnet of magnetic dipole moment M is released in a uniform magnetic field of induction B from the position shown in the figure. Find its maximum kinetic energy during motion.

\frac{M B}{2}

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M B

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\frac{3 M B}{2}

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2 M B

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Solution

The correct option is C $\frac{3 M B}{2}$
K.E. is maximum when angle between $- \to M a n d \to B i s 0^{\circ}$
Now, $K_{i} + U_{i} = K_{f} + U_{f}$
$0 - M G c o s 120^{\circ} = K_{m a x} - M B c o s 0^{\circ} K_{m a x} = \frac{3}{2} M B$

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