Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 4 / 7A / Solution

With consistent flux and loop orientations, Faraday's law is $V=-d\Phi /dt$. The magnetic force on a current-carrying wire is $d\mathbf{F}=Id\mathbf{\ell }\times \mathbf{B}$.

Let $z(t)$ be the top edge height and $v=-\dot{z}$ the downward speed. While $0<z<l$, only the top horizontal edge lies in the field, and the linked magnetic flux is $Bwz$. Therefore the induced electromotive force has magnitude $Bwv$, and Ohm's law gives current magnitude $I=Bwv/R$. Its force opposes the downward motion and has upward magnitude $IBw=B^2{w}^{2}v/R$. Forces on the vertical sides cancel horizontally. HenceFor a release from rest, $v=v_{\mathrm{terminal}}(1-e^{-B^2{w}^{2}t/(mR)})$. The terminal speed is the formal limit while partial overlap is maintained; after the entire loop leaves the field this magnetic braking term vanishes.