= Solution
For a fixed contour, <Faraday's law of induction> is
$$
\oint_C\mathbf E\cdot d\mathbf l=-\frac{d}{dt}\int_\Sigma\mathbf B\cdot d\mathbf S,\qquad\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}.
$$
The second equation is the differential Maxwell equation, related to the first by <Stokes theorem>. For a moving wire, the relevant <motional electromotive force> is $\mathcal E=\oint(\mathbf E+\mathbf v_{\rm wire}\times\mathbf B)\cdot d\mathbf l=-d\Phi_B/dt$. Thus a static magnetic field can induce current in a moving circuit even when $\nabla\times\mathbf E=0$.
For <translational magnetic damping of a square wire loop>, while the upper horizontal leg remains in the field, a downward speed $v$ reduces the flux at rate $Bav$. Hence $|\mathcal E|=Bav$, $|I|=Bav/R$, and the upward <magnetic braking> force is $B^2a^2v/R$. With negligible electrical self-inductance,
$$
m\dot v=mg-\frac{B^2a^2}{R}v,\qquad\boxed{v_t=\frac{mgR}{B^2a^2},\quad v(t)=v_t(1-e^{-t/\tau_t}),\quad\tau_t=\frac{mR}{B^2a^2}.}
$$
Let the wire cross-sectional area be $S$, mass <density> $\rho_m$, and resistivity $\rho_e$. Since $m=4aS\rho_m$ and $R=4a\rho_e/S$,
$$
\boxed{\tau_t=\frac{16\rho_m\rho_e}{B^2}=2.403072\,\mathrm{ms},\qquad t_{95}=\tau_t\log20=7.20\,\mathrm{ms}.}
$$
These use the printed copper values and $B=1\,\mathrm T$. The wire dimensions cancel. The terminal-speed approximation is restricted to the interval before the upper leg leaves the field, as assumed in the question.
For <rotational magnetic damping of a square wire loop>, consider rotation about the central $x$ axis in a uniform field, the flux is $Ba^2\cos\theta$. The induced current produces torque $-B^2a^4\dot\theta\sin^2\theta/R$. A thin square loop has <moment of inertia> $\mathcal I_x=ma^2/6$: the two legs parallel to $x$ contribute $ma^2/8$, and the other two $ma^2/24$. Averaging $\sin^2\theta$ over a rotation gives
$$
\mathcal I_x\dot\omega=-\frac{B^2a^4}{2R}\omega,\qquad\boxed{\tau_{\rm rot,\omega}=\frac{2\mathcal I_xR}{B^2a^4}=\frac{\tau_t}{3}=0.801024\,\mathrm{ms}.}
$$
If “decay time” means rotational <kinetic energy> rather than angular speed, $K\propto\omega^2$ gives $\tau_{\rm rot,K}=\tau_t/6=0.400512\,\mathrm{ms}$. This distinction, and the assumed separation between rotation period and slow decay, are essential. The <electrical and mechanical time scales in magnetic braking> must be separated: electrical self-inductance is neglected in both calculations.
\Image[/past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1-loop-damping.png]
{title=The falling-loop speed deficit decays three times more slowly than the rotating-loop angular speed}
{source=paper-1-loop-damping.py}
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