Question

a bar magnet of magnetic moment 6 j/t is aligned at 60 with a uniform external magnetic field of 0·44 t. 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

Answer:

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