The self-inductance of a long air-cored solenoid of length $\ell$, cross-sectional area $A$ and total number of turns $N$ is:
- $\mu_0 N A \ell$
- $\mu_0 N^2 A \ell$
- $\mu_0 N^2 A / \ell$
- $\mu_0 N A / \ell$
Show answer and solution
Correct answer: C. $\mu_0 N^2 A / \ell$
For a long solenoid the field inside is $B = \mu_0 n I = \mu_0 (N/\ell) I$.
Flux through each turn is $BA$ and total linkage is $N(BA)$, giving $L = N\Phi/I = \mu_0 N^2 A / \ell$.
Equivalently $L = \mu_0 n^2 A \ell$ if $n = N/\ell$.