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).

Magnetic moment M = 6 J/T
Magnetic field B = 0.44 T
(a) :
(i) : Work done in rotating the magnet normal to the magnetic field
W = MB(cosθ₁ - cosθ₂)
where θ₁ = 60° and θ₂ = 90°
=> W = 6 × 0.44(cos60° - cos90°) = 6 × 0.44(0.5 - 0) = 1.32 J
Work done in rotating the magnet opposite to the magnetic field
W = MB(cosθ₁ - cosθ₂)
where θ₁ = 60° and θ₂ = 180°
=> W = 6 × 0.44(cos60° - cos180°) = 6 × 0.44(0.5 + 1) = 3.96 J
(b) : Torque in case (ii)
τ = MB sinθ₂
=> τ = 6 × 0.44 × sin180° = 0