= Solution
Taking curls gives
$$
B_1=(0,0,2b_1),\qquad B_2=(-b_2x,-b_2y,2b_2z).
$$
Both have zero divergence and curl, so they satisfy the static vacuum equations. Their upward fluxes through the loop are
$$
\Phi_1=2\pi b_1r^2,\qquad \Phi_2=2\pi b_2r^2z.
$$
For clockwise current $I$, the net force is zero in $B_1$, while in $B_2$
$$
F_z=-2\pi b_2Ir^2.
$$
Put $K=2\pi b_2r^2$. Motion in $B_1$ changes no flux, so no current is induced and $z=z_0-gt^2/2$. In $B_2$, using counterclockwise current as positive,
$$
I=-\frac{K\dot z}{R},\qquad F_z=-\frac{K^2}{R}\dot z.
$$
Thus
$$
m\ddot z=-mg-\frac{K^2}{R}\dot z.
$$
With $\Gamma=K^2/R$, the <velocity> tends to $-mg/\Gamma$ and
$$
z=z_0-\frac{mg}{\Gamma}\left[t-\frac m\Gamma(1-e^{-\Gamma t/m})\right].
$$
At terminal speed, $I^2R=mg|\dot z|$, so Joule loss equals gravitational-energy loss. For $R=0$, Ohm's quasistatic formula is singular: flux is conserved, persistent current stores magnetic energy, and <ideal> motion has no resistive terminal state (<inductance> must be retained).
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