The work done for rotating a magnet with magnetic dipole moment m, by 90° from its magnetic meridian is n times the work done to rotate it by 60°, find value of n.

Explanation

Suppose the work done for relating a magnet with magnetic dipole moment m, by 90° from its magnetic meridian is `W_1`, then
`W_1 = mB (1- cos theta_(1) )= mB (1- cos 90^(@) )= mB` and the work done to rotate it by 60° is W
`therefore W_(2) = mB ((1)/(2)) = (mB)/(2)`
Now according to condition given `W_(1) = nW_2`
`mB= n ((mB)/( 2))`
`therefore n=2`