A bar magnet of magnetic moment 1.5 J T⁻¹ lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)?

Ans (a).

Magnetic moment, M = 1.5 J T^-1 ,

Magnetic field strength, B = 0.22 T

(i) Initial angle between the axis and the magnetic field, 0₁ = 0°

Final angle between the axis and the magnetic field, 0₂ = 90°

The work required to make the magnetic moment normal to the direction of magnetic field is given as:

W = -MB(cos0₂ – cos 0₁)

= -1.5 x 0.22(cos 90° – cos 0°)

= -0.33(0-1)

= 0.33 J

(ii) Initial angle between the axis and the magnetic field, 0₁= 0°

Final angle between the axis and the magnetic field, 0₂ = 180°

The work required to make the magnetic moment opposite to the direction of magnetic field is given as:

W = -MB(cos0₂ – cos 0₁)

= -1.5 x 0.22(cos 180 – cos 0°)

= —0.33(—1 -1)

= 0.66 J

Ans (b).

For case (i):      0 = 0₂= 90⁰

Therefore ,Torque,       τ =  MBsin 6

= 1.5×0.22 sin 90°

= 0.33 J

For case (ii): 0 = 0₂= 180°

Therefore Torque,         τ = MB sin 0

= MBsin 180° = 0 J