Motional EMF question from MHT CET 2026 (11 Apr, Shift 1)

Answer: (b)

1. Falling with speed $v$, the loop has emf $e = Blv$ and current $I = \dfrac{Blv}{R}$.
2. Magnetic force on the side of length $l$: $F = BIl = \dfrac{B^2l^2v}{R}$, upward.
3. No acceleration: $\dfrac{B^2l^2v}{R} = mg$
4. So the answer is (b) $v = \dfrac{mgR}{B^2l^2}$.