Self-Inductance & Energy question from MHT CET 2025 (21 Apr, Shift 1)

Answer: (a)

1. $L = \dfrac{\mu_0 N^2 \pi r^2}{l}$. Wire length $w = N(2\pi r)$, so $Nr = \dfrac{w}{2\pi}$.
2. $L = \dfrac{\mu_0 \pi}{l}\left(\dfrac{w}{2\pi}\right)^2 = \dfrac{\mu_0 w^2}{4\pi l}$
3. $w = \left[\dfrac{4\pi l L}{\mu_0}\right]^{1/2}$
4. So the answer is (a).