Terminal speed of a falling conducting loop

If a loop of resistance $R$ has magnetic flux ${\Phi }_B=-kz$, then its induced current is $I=k\dot{z}/R$. When its magnetic force is $-kI$ in the vertical direction, its equation under gravity isso its downward terminal speed is $Rmg/k^2$.

For an external loop flux $Kz$, positive current orientation and upward Lorentz force $KI$, including self-inductance gives $\mathcal{L}\dot{I}+RI=-K\dot{z}$ and $m\ddot{z}=-mg+KI$. At zero resistance total linked flux is conserved. If initially $I=0$ at $z=z_0$, then $I=-K(z-z_0)/\mathcal{L}$ and the mechanical motion is an undamped oscillator about $z_0-mg\mathcal{L}/K^2$. The finite-resistance terminal speed tends to zero as $R\to 0$, but neglecting self-inductance is a singular approximation in that limit. A lossless loop does not dispose of energy through Joule heating.