A single-turn coil in the shape of a perfect square of side $a$ is suspended between the pole pieces

Question

A single-turn coil in the shape of a perfect square of side $a$ is suspended between the pole pieces of a permanent magnet such that the magnetic field $\vec{B}$ is strictly in the plane of the coil. If a current $I$ passing through the square induces a torque $	au$, the side length $a$ of the square coil is

Professor PC Thomas Classes · Answer

(b) : The torque on a current-carrying coil in a magnetic field is $	au = NIAB \sin	heta$. Here, the number of turns $N = 1$, the area of the square coil is $A = a^2$, and the angle between the area vector (perpendicular to the plane) and the magnetic field is $	heta = 90^\circ$ (since the field is in the plane of the coil). Thus, $	au = 1 	imes I 	imes a^2 	imes B 	imes \sin(90^\circ) = Ia^2B$. Rearranging to solve for the side length $a$, we get $a^2 = \frac{	au}{IB}$, which implies $a = \sqrt{\frac{	au}{IB}}$.