MHTCET 22 Apr 2025 Shift 2 Paper

Concept:For a planar current loop in a uniform magnetic field, torque is proportional to the area of the loop.Explanation:The magnetic torque on a loop is given by $\tau = mB\sin\theta$, where $m = IA$ is the magnetic dipole moment.Since current $I$, magnetic field $B$, and angle $\theta$ are the same for both loops, the torque depends only on the enclosed area $A$.Let the total length of the wire be $L$.For the square loop, perimeter $L = 4a$, so side $a = \frac{L}{4}$ and area $A_{\mathrm{sq}} = \left(\frac{L}{4}\right)^2 = \frac{L^2}{16}$.For the circular loop, circumference $L = 2\pi r$, so radius $r = \frac{L}{2\pi}$ and area $A_{\mathrm{cir}} = \pi r^2 = \frac{L^2}{4\pi}$.Now compare the areas: $\frac{A_{\mathrm{cir}}}{A_{\mathrm{sq}}} = \frac{L^2 / 4\pi}{L^2 / 16} = \frac{4}{\pi} \approx 1.27$.Thus, $A_{\mathrm{cir}} > A_{\mathrm{sq}}$, which implies $\tau_{\mathrm{cir}} > \tau_{\mathrm{sq}}$.Answer:Circular loop experiences greater torque. Hence, the correct option is A.