MHTCET 5 May 2025 Shift 2 Paper

Concept:A current-carrying loop experiences torque in a magnetic field, given by $\tau = I A B \sin\theta$.Maximum torque occurs when the plane of the loop is parallel to the field, i.e. $\sin\theta = 1$.Explanation:The total length of the wire is $L$, and it forms a square loop of single turn.Therefore, the perimeter of the square equals $L$.Side of the square is $a = \frac{L}{4}$.Area of the square is $A = a^{2} = \left(\frac{L}{4}\right)^{2} = \frac{L^{2}}{16}$.The magnetic moment of the loop is $\mu = I A$.Torque on the loop is $\tau = \mu B \sin\theta = I A B \sin\theta$.For maximum torque, $\sin\theta = 1$, so $\tau_{\max} = I A B$.Substituting the area: $\tau_{\max} = I \cdot \frac{L^{2}}{16} \cdot B = \frac{I B L^{2}}{16}$.Answer:$\tau_{\max} = \frac{I B L^{2}}{16}$Hence, the correct option is D.