Derive an expression for the torque acting on a current-carrying loop which subtends an angle $\theta$ with a uniform magnetic field.

(N/A) As shown in the figure,the plane $ABCD$ is not along the magnetic field but makes an angle with it.
We take the angle between the field and the normal to the coil to be angle $\theta$.
The forces on the arms $BC$ and $DA$ are equal,opposite,and act along the axis of the coil,which connects the centers of mass of $BC$ and $DA$. Being collinear along the axis,they cancel each other,resulting in no net force or torque.
The forces on arms $AB$ and $CD$ are $\overrightarrow{F}_{1}$ and $\overrightarrow{F}_{2}$ respectively.
They are also equal and opposite with magnitude $F_{1} = F_{2} = I b B$.
The figure illustrates a view of the arrangement from the $AD$ side,showing these two forces constituting a couple. The magnitude of the torque on the loop is,
$\tau = \tau_{1} + \tau_{2}$
$\tau = F_{1} \left( \frac{a}{2} \sin \theta \right) + F_{2} \left( \frac{a}{2} \sin \theta \right)$
$[\because \tau = (\text{magnitude of force}) \times (\text{perpendicular distance from reference points})]$
$\tau = (I b B) \left( \frac{a}{2} \sin \theta \right) + (I b B) \left( \frac{a}{2} \sin \theta \right)$
$\tau = I (a b) B \sin \theta$
$\tau = I A B \sin \theta$