Question

A bar magnet of magnetic moment 6J/T is aligned at 60 degrees with a uniform external magnetic field of 0.44T . Calculate (a) the work done in turning the magnet to align its magnetic moment (i)Normal to the magnetic field (ii) opposite to the magnetic field and (b) the torque on the magnet in the final orientation in case (ii)

Answer 1

Formula of work done is given by
W = - MB[cosθf - cosθi]
Here M is the magnetic moment ,
B is the magnetic field
θf is final angle subtended by bar magnet with magnetic field
θi is the initial angle subtended by bar magnet with magnetic field.

(a) Normal to the magnetic field
θf = 90° , θi = 60° , M = 6J/T and B = 0.44T
so, work done = - 6 × 0.44 [ cos90° - cos60°]
= - 2.64 × [ 0 - 1/2 ] = 1.32 J

(b) opposite to the magnetic field
θf = 180° , θi = 60°
∴ work done = - 6 × 0.44 [ cos180° - cos60° ]
= -2.64 × (-1 - 1/2) = 2.64 × 3/2 = 1.32 × 3 = 3.96 J