= Solution
At <terminal velocity>, the upward magnetic <force> balances the <weight>. Substitute the <electric current> from the preceding solution:
$$
Mg=IaB=\frac{B^2a^2}{R}v_s,\qquad M=\rho_m\pi a d_w^2,\qquad R=\frac{16\rho_ca}{\pi d_w^2}.
$$
The geometric factors cancel, giving
$$
\boxed{v_s=\frac{MgR}{B^2a^2}=\frac{16\rho_m\rho_cg}{B^2}=2.39\times10^{-2}\ {\rm m\,s^{-1}}.}
$$
Here $g=9.81\ {\rm m\,s^{-2}}$, $B=1\ {\rm T}$, $\rho_m=8960\ {\rm kg\,m^{-3}}$ and $\rho_c=1.7\times10^{-8}\ {\rm \Omega\,m}$. As numerical checks, the specified $a=0.08\ {\rm m}$ and $d_w=0.001\ {\rm m}$ give $M=2.252\times10^{-3}\ {\rm kg}$, $R=6.926\times10^{-3}\ {\rm \Omega}$ and $I_s=0.276\ {\rm A}$.
The gravitational <power> lost, $Mgv_s$, equals the <Joule heating> $I_s^2R$, consistent with <energy conservation>. This <terminal velocity> applies only while the top arm remains in the <magnetic field>; after it exits, the magnetic braking ceases, apart from any brief <inductive transient>.
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