= Solution
In Cartesian coordinates the <magnetic field> is $(bx,by,-2bz)$. Its <divergence> is $b+b-2b=0$, and all components of its <curl> vanish. Thus it satisfies the static vacuum magnetic equations $\nabla\cdot\mathbf B=0$ and $\nabla\times\mathbf B=0$, with no <electric current> in the region.
Fix clockwise orientation as viewed from above, looking down the positive $z$-axis. Its positive area normal is $-\mathbf e_z$. The loop's external <magnetic flux> is therefore
$$
\boxed{\Phi_B=2\pi ba^2z=Kz,\qquad K=2\pi ba^2.}
$$
With the opposite normal its flux has the opposite sign. On the circle $\mathbf B=ba\mathbf e_r-2bz\mathbf e_z$ and clockwise line element is $d\boldsymbol\ell=-a\mathbf e_\varphi d\varphi$. The <Lorentz force> element is
$$
I\,d\boldsymbol\ell\times\mathbf B=Iba^2\mathbf e_z\,d\varphi+2Ibaz\mathbf e_r\,d\varphi.
$$
The radial terms cancel around the circle, so
$$
\boxed{\mathbf F_B=KI\mathbf e_z.}
$$
This fixes the current/flux/force signs together.
Neglect <self-inductance> initially, so <Faraday's law> and <Ohm's law> give $RI=-K\dot z$. Including gravity then gives
$$
\boxed{m\ddot z=-mg-\frac{K^2}{R}\dot z.}
$$
For $R>0$ and $K\ne0$, the terminal solution is $z=z_0-vt$ with
$$
\boxed{v=\frac{mgR}{K^2}=\frac{mgR}{4\pi^2b^2a^4}.}
$$
The general <velocity> relaxes exponentially to $-v$ on time scale $mR/K^2$. In the terminal state $I=Kv/R$, and the Joule-heating rate is
$$
I^2R=\frac{K^2v^2}{R}=mgv=-\frac{d}{dt}(mgz).
$$
Thus all the lost <gravitational potential energy> becomes resistive heat when kinetic <energy> is constant.
Within the zero-inductance approximation $v\to0$ as $R\to0$; finite-speed motion would require an unbounded current. This limit is singular, so it is important to describe the physical ideal-conductor limit separately. With finite <self-inductance> $\mathcal L$, the circuit equation is $\mathcal L\dot I+RI=-K\dot z$. At $R=0$, total linked flux $\mathcal LI+Kz$ is conserved and there is no <Joule heating>. If $I=0$ initially at $z_0$, then
$$
I=-\frac K{\mathcal L}(z-z_0),\qquad
m\ddot z=-mg-\frac{K^2}{\mathcal L}(z-z_0).
$$
This <lossless limit of magnetic loop braking> has undamped oscillations about $z_0-mg\mathcal L/K^2$, with magnetic <energy> storing and returning mechanical <energy>. It has no generic dissipative terminal descent. Thus the vanishing terminal speed is the formal resistive-model limit, not a justification for dropping inductance in a perfectly conducting moving loop.
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