= Solution
With consistent flux and loop orientations, <Faraday's law> is $V=-d\Phi/dt$. The <magnetic force on a current-carrying wire> is $d\mathbf F=I\,d\boldsymbol\ell\times\mathbf B$.
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 $IBw=B^2w^2v/R$. Forces on the vertical sides cancel horizontally. Hence
$$
\boxed{m\dot v=mg-\frac{B^2w^2}{R}v,\qquad
m\ddot z=-mg-\frac{B^2w^2}{R}\dot z,\qquad
v_{\rm terminal}=\frac{mgR}{B^2w^2}.}
$$
For a release from rest, $v=v_{\rm terminal}(1-e^{-B^2w^2t/(mR)})$. The terminal speed is the formal limit while partial overlap is maintained; after the entire loop leaves the field this magnetic braking term vanishes.
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