A wire of length ' $L$ ' carries a current ' $I$ '. If the wire is turned into a square coil of single turn,…

Maximum torque on a square coil

The magnetic dipole moment determines the torque experienced by a current-carrying loop in a magnetic field.

Using a wire of length $L$, the side length of the square coil is $a = \frac{L}{4}$, giving an area $A = a^2 = \frac{L^2}{16}$.

For a single-turn coil, the magnetic dipole moment is $M = IA = \frac{IL^2}{16}$.

The torque magnitude is $\tau = MB\sin\theta$, which reaches its maximum $\tau_{\text{max}} = MB$ when $\sin\theta = 1$.

Therefore, $\tau_{\text{max}} = \frac{IBL^2}{16}$.

This result corresponds to option D.