= Solution
<Faraday's law of induction> gives the <electromotive force> $\mathcal E=-d\Phi_B/dt$. For a fixed circuit, $\mathcal E=\oint\mathbf E\cdot d\mathbf l$; for moving material wire, include the motional term $\mathbf u\times\mathbf B$ in that line integral. <Lenz's law> fixes the sign: the induced response opposes the change of linked flux, rather than necessarily opposing the field itself.
For the long solenoid, <Ampère's law> gives $B\simeq\mu_0nI$ inside and negligible exterior field away from the ends. With $N=nD$ turns and area $\pi a^2$, the linked flux is $N\Phi_B=\mu_0n^2D\pi a^2I$. Hence the <self-inductance> is
$$
\boxed{L=\frac{N\Phi_B}{I}=\mu_0n^2\pi a^2D.}
$$
This assumes $D\gg a$, closely spaced turns, negligible end fringing and an air core with relative <magnetic permeability> close to one. A finite core permeability or significant fringing changes the answer.
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