Solution
= Solution
During partial entry, resistive <magnetic braking> gives $v=g\tau(1-e^{-t/\tau})$ with $\tau=mR/(B_0a)^2$. The lossless inductive limit gives $x=g(1-\cos\omega t)/\omega^2$, $v=g\sin\omega t/\omega$, with $\omega=B_0a/\sqrt{mL}$. Complete immersion changes the equations; the solution below distinguishes reversal from full entry.
Full derivation: <past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2010/paper-2a/section-b/b7/solution>.