 Solution

During partial entry, resistive magnetic braking gives $v=g\tau (1-e^{-t/\tau })$ with $\tau =mR/(B_{0}a{)}^2$. The lossless inductive limit gives $x=g(1-\cos \omega t)/{\omega }^2$, $v=g\sin \omega t/\omega$, with $\omega =B_{0}a/\sqrt{mL}$. Complete immersion changes the equations; the solution below distinguishes reversal from full entry.

Full derivation: solution.