= Solution
Project <Newton's second law> down the slope. The normal reaction has no component along the rails, while the magnetic term opposes motion:
$$
\boxed{m\frac{dv}{dt}=mg\sin\phi-IBs\cos\phi
=mg\sin\phi-\frac{B^2s^2\cos^2\phi}{R}v.}
$$
This first-order linear <differential equation> has the release-from-rest solution
$$
v(t)=\frac{mgR\sin\phi}{B^2s^2\cos^2\phi}\left(1-e^{-t/\tau_m}\right),\qquad
\tau_m=\frac{mR}{B^2s^2\cos^2\phi}.
$$
The electrical <self-inductance> has been neglected; retaining it introduces a coupled <electric current> equation rather than this instantaneous resistive response.
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