= Solution
<Faraday's law> is $\mathcal E=\oint_C(E+v\times B)\cdot dl=-d\Phi_B/dt$. Here $I=-L(B\dot x+x\dot B)/R$ (sign relative to the chosen circuit orientation), and the bar force is $ILB$. For $B=B_0/(1+t)$ this gives the displayed equation with $\alpha=B_0^2L^2/(mR)$. At $\alpha=1$, the independent solutions are $u$ and $ue^{1/u}$, $u=1+t$; the data give $x=x_0ue^{1/u-1}$. Thus $\dot x\to x_0/e$ and total mechanical work is $mx_0^2/(2e^2)$. The current circulates to reinforce the decaying $+z$ flux, and its force drives increasing area, both opposing the flux decrease as <Lenz's law> requires.
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