= Solution
For a moving circuit, <Faraday's law> states that the <electromotive force>
$$
\mathcal E=\oint_C(\mathbf E+\mathbf v\times\mathbf B)\cdot d\boldsymbol\ell
$$
equals $-d\Phi_B/dt$, where $\Phi_B=\int_S\mathbf B\cdot d\mathbf S$ is the <magnetic flux> through any spanning surface. Choose the upward normal. The horizontal square has
$$
\Phi_B=(-2bz)(2a)^2=-8ba^2z,
$$
so <Ohm's law> gives the induced current
$$
\boxed{I=\frac{\mathcal E}{R}=\frac{8ba^2}{R}\dot z}
$$
with positive current taken counterclockwise from above.
On the edge $x=a$, $d\boldsymbol\ell=dy\,\mathbf e_y$. The <magnetic force on a current-carrying wire> is $d\mathbf F=I,d\boldsymbol\ell\times\mathbf B$, hence
$$
\mathbf F_{x=a}
=Ib\int_{-a}^a\mathbf e_y\times(a\mathbf e_x+y\mathbf e_y-2z\mathbf e_z)\,dy
=\boxed{-4abIz\,\mathbf e_x-2a^2bI\,\mathbf e_z}.
$$
The ratio of the horizontal to vertical magnitudes is $2|z|/a$, so the directed force makes angle $\tan^{-1}(2z/a)$ with the vertical, with the sign fixing the horizontal side.
The horizontal forces on opposite edges cancel, while all four vertical contributions add:
$$
\boxed{\mathbf F_{\rm em}=-8ba^2I\,\mathbf e_z
=-\frac{(8ba^2)^2}{R}\dot z\,\mathbf e_z}.
$$
This is an electromagnetic drag force opposing the motion, as required by <Lenz's law>. Including gravity gives
$$
m\ddot z=-mg-\frac{(8ba^2)^2}{R}\dot z.
$$
At the <terminal speed of a falling conducting loop>[terminal velocity], $\ddot z=0$, and the constant downward speed has magnitude
$$
\boxed{v_{\rm terminal}=\frac{Rmg}{(8ba^2)^2}}.
$$
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