The self‑inductance $L$ of a long solenoid of $N$ turns, length $\ell$, cross‑sectional area $A$, and vacuum permeability ${\mu }_0$ is given by:

Step-by-step solution

1. Write the magnetic field inside a solenoid: $B={\mu }_{0}nI$, where $n=N/\ell$ is the turn density.
2. Flux through one turn is $\Phi =BA={\mu }_{0}nIA$.
3. Total flux linkage for $N$ turns is $N\Phi =N({\mu }_{0}nIA)={\mu }_0{N}^{2}AI/\ell$.
4. By definition $N\Phi =LI$, so $L={\mu }_0{N}^{2}A/\ell$.
5. Compare with the options; option (a) matches the derived expression.

Concept tested: self‑inductance of solenoid

Published 6 September 2026 · How this solution was prepared and checked