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  4. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is

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Q. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^{\circ}$ is $W$. Now the torque required to keep the magnet in this new position is

NEETNEET 2016Magnetism and Matter

Solution:

$W = MB \left(\cos 0^{\circ} - \cos60^{\circ}\right) $
$ W = MB \left(1- \frac{1}{2}\right) = \frac{MB}{2}$
Required torque for this position
$ \tau = MB \sin\theta $
$ = MB \sin60^{\circ}$
$ = \frac{\sqrt{3}}{2} MB = \sqrt{3} W$