The work done in turning a magnet of magnetic moment $$M$$ by an angle of $${90^ \circ }$$ from the meridian, is n times the corresponding work done to turn it through an angle of $${60^ \circ }.$$ The value of $$n$$ is given by

Solution :
Work done in rotating the dipole from $${\theta _1}$$ to $${\theta _2}$$ is
\[W = - MB\left( {\cos {\theta _2} - \cos {\theta _1}} \right)\,\,\,\left( {\begin{array}{*{20}{r}} {M = {\rm{Magnetic\,moment}}}\\ {B = {\rm{Magnetic\,induction}}} \end{array}} \right)\]
Case I
$${W_1} = - MB\left( {\cos {{90}^ \circ } - \cos {0^ \circ }} \right) = MB\,......\left( {\text{i}} \right)$$
Case II
$$\eqalign{ & {W_2} = - MB\left( {\cos {{60}^ \circ } - \cos {0^ \circ }} \right) \cr & = - MB\left( {\frac{1}{2} - 1} \right) = \frac{1}{2}MB\,......\left( {{\text{ii}}} \right) \cr} $$
From Eqs. (i) and (ii), $${W_2} = \frac{1}{2}{W_1}$$
As $${W_1} = n{W_2}$$
$$\therefore n = 2$$