$A$ 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 $(ii)?$

(N/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,$\theta_{1} = 0^{\circ}$.
Final angle between the axis and the magnetic field,$\theta_{2} = 90^{\circ}$.
The work required to make the magnetic moment normal to the direction of the magnetic field is given by:
$W = -MB(\cos \theta_{2} - \cos \theta_{1})$
$W = -1.5 \times 0.22 \times (\cos 90^{\circ} - \cos 0^{\circ})$
$W = -0.33 \times (0 - 1) = 0.33 \, J$.
$(ii)$ Initial angle,$\theta_{1} = 0^{\circ}$.
Final angle,$\theta_{2} = 180^{\circ}$.
The work required to make the magnetic moment opposite to the direction of the magnetic field is:
$W = -1.5 \times 0.22 \times (\cos 180^{\circ} - \cos 0^{\circ})$
$W = -0.33 \times (-1 - 1) = 0.66 \, J$.
$(b)$ Torque is given by $\tau = MB \sin \theta$.
For case $(i)$,$\theta = 90^{\circ}$:
$\tau = 1.5 \times 0.22 \times \sin 90^{\circ} = 0.33 \, N \cdot m$.
For case $(ii)$,$\theta = 180^{\circ}$:
$\tau = 1.5 \times 0.22 \times \sin 180^{\circ} = 0 \, N \cdot m$.