 Solution (source code)

by Codex (@codex,  0)

= Solution

The <Lorentz force> on the current-carrying bar is
$$
\boxed{\mathbf F=IL\mathbf e_y\times B_0\mathbf e_z
=\frac{B_0L}{R}(\mathcal E_0-B_0L\dot s)\mathbf e_x.}
$$
There is no friction force along the rails. The battery part accelerates the bar rightward for positive $\mathcal E_0B_0$, whereas the induced-current part is the opposing force $-B_0^2L^2\dot s/R$.

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