A bar magnet of magnetic moment 1.5 JT^–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)?

Given,
Magnetic moment of bar magnet, M = 1.5 JT^–1
Uniform magnetic field, B = 0.22 T 
(a) Work done on magnet 

(i) Amount of work required to turn the magnet normal to the field direction is, 

W₁ = – MB [cos 90° – cos 0°]
      = + MB
      = MB = 1.5 x 0.22
      = 0.33 N-m 

(ii) Amount of work required to turn the magnet in a direction opposite to the field direction is, 

W₂ = – MB [ cos 180° - cos 0°]
      = 2MB 
      = – MB [cos θ₂ – cos θ₁]
      = 2 MB = 2 x 0.33
      = 0.66 J 

(b) Torque acting on the magnet. 

(i) τ = MB sin 90°
       = MB
       = 0.33 J. 

It works in the direction that tends to align the magnetic moment vector along B. 

(ii) Torque acting on the magnet so as to align it's magnetic moment in a direction opposite to the field direction.
τ = MB sin θ
Here, θ = 180°
τ  = 1.5 x 0.22 x sin 180°
   = 1.5 x 0.22 x 0
   = 0.