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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 √k W. Find k.

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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 $\sqrt{k} W$. Find $k$.

Magnetism and Matter

Solution:

$\tau=M B \sin 60^{\circ}$ ....(i)
$W=M B\left(1-\cos 60^{\circ}\right)$....(ii)
From (i) and (ii), we get
$\frac{\tau}{W}=\frac{\sqrt{3} / 2}{1 / 2}$
$ \Rightarrow \tau=W \sqrt{3}$