Current Loop as a Magnetic Dipole question from MHT CET 2026 (11 Apr, Shift 1)

Answer: (a)

1. $\tau_{max} = NiAB$. With $N$ turns: $N(2\pi r) = L \Rightarrow r = \frac{L}{2\pi N}$.
2. $NA = N\pi\frac{L^2}{4\pi^2N^2} = \frac{L^2}{4\pi N}$, so $\tau_{max} = \frac{iBL^2}{4\pi N} \propto \frac{1}{N}$.
3. Torque is largest for the smallest $N$.
4. So the answer is (a) $1$.