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The self-inductance of a long air-cored solenoid of length $\ell$, cross-sectional area $A$ and total number of turns $N$ is:
A$\mu_0 N A \ell$
B$\mu_0 N^2 A \ell$
C$\mu_0 N^2 A / \ell$
D$\mu_0 N A / \ell$
Answer & 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 $n = N/\ell$.
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