 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 $B\simeq {\mu }_{0}nI$ inside and negligible exterior field away from the ends. With $N=nD$ turns and area $\pi a^2$, the linked flux is $N{\Phi }_B={\mu }_0{n}^{2}D\pi a^{2}I$. Hence the self-inductance isThis 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.