A circular loop carrying a current' I' is made of a wire of length 'L'. A uniform magnetic field 'B' exists parallel to the plane of the loop. The torque experienced by the circular loop will be

The length of the wire is $L$. For a circular loop, the circumference is equal to $L$.

$2\pi R = L \Rightarrow R = \dfrac{L}{2\pi}$

The area of the circular loop is given by:

$A = \pi R^2 = \pi \left(\dfrac{L}{2\pi}\right)^2 = \dfrac{L^2}{4\pi}$

The magnetic dipole moment of the current-carrying loop is:

$M = IA = \dfrac{IL^2}{4\pi}$

The magnetic field $B$ is parallel to the plane of the loop, which means the angle $\theta$ between the area vector (magnetic moment $\vec{M}$) and the magnetic field $\vec{B}$ is $90^{\circ}$.

The torque experienced by the loop is:

$\tau = MB \sin\theta = \left(\dfrac{IL^2}{4\pi}\right) B \sin(90^{\circ})$

$\tau = \dfrac{IL^2B}{4\pi}$

Answer: $\dfrac{IL^2B}{4\pi}$