Derive an expression for the torque acting on a current-carrying loop suspended in a uniform magnetic field.

(N/A) Consider a rectangular coil $ABCD$ of length $b$ and width $a$ carrying current $I$, suspended in a uniform magnetic field $\overrightarrow{B}$ such that its axis is perpendicular to the field.
The area of the coil is $A = a b$.
The magnetic force on a current-carrying conductor is given by $\vec{F} = I(\vec{L} \times \vec{B})$.
For the arms $AD$ and $BC$, the current is parallel or anti-parallel to the magnetic field $\vec{B}$. Thus, the force on these arms is $F = I L B \sin(0^{\circ}) = 0$.
For the arms $AB$ and $CD$, the current is perpendicular to the magnetic field $\vec{B}$.
The force on arm $AB$ is $\vec{F_1} = I b B \sin(90^{\circ}) = I b B$ (directed into the plane).
The force on arm $CD$ is $\vec{F_2} = I b B \sin(90^{\circ}) = I b B$ (directed out of the plane).
These two equal and opposite forces $\vec{F_1}$ and $\vec{F_2}$ form a couple that exerts a torque $\tau$ on the loop.
The torque is given by $\tau = \text{Force} \times \text{perpendicular distance}$.
$\tau = (I b B) \times (a) = I (ab) B = I A B$.
If the coil has $N$ turns, the total torque is $\tau = N I A B$.