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

Solution :
$$\because $$ Work done in rotating the magnet
$$W = MB\left( {\cos {\theta _0} - \cos \theta } \right)$$
Where, $$M =$$  magnetic moment of the magnet
$$B =$$  magnetic field
$$W = MB\left( {\cos {0^ \circ } - \cos {{60}^ \circ }} \right)$$
$$ = MB\left( {1 - \frac{1}{2}} \right) = \frac{{MB}}{2}$$
$$\therefore MB = 2W\,.......\left( {\text{i}} \right)$$
Torque on a magnet in this position is given by,
$$\eqalign{ & \tau = M \times B \cr & = MB \cdot \sin \theta = 2W \cdot \sin {60^ \circ }\,\,\left[ {{\text{from}}\,{\text{Eq}}{\text{.}}\left( {\text{i}} \right)} \right] \cr & = 2W\frac{{\sqrt 3 }}{2} = W\sqrt 3 \cr} $$