The work done in turning a magnet of magnetic moment ' M ' by an angle of 90^ from the meridian is ' n ' times the corresponding work done to turn it through an angle of 60^ where the value of ' n ' is ( 90^ =0, 60^ =0.5 )

Step-by-step solution

The work required to rotate a magnetic dipole moment M through angle in a uniform magnetic field B is expressed as W = MB( ₁ - ₂) where ₁ and ₂ denote the initial and final angular positions relative to the field direction. Given the magnet's initial orientation is in the meridian direction (taken to align with the magnetic field), we set ₁ = 0^ . The work to rotate through an angle of 90^ is then W_ 90^ = MB( 0^ - 90^ ) = MB(1 - 0) = MB . Similarly, the work to rotate through 60^ is W_ 60^ = MB( 0^ - 60^ ) = MB(1