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For a fixed contour, Faraday's law of induction isThe second equation is the differential Maxwell equation, related to the first by Stokes theorem. For a moving wire, the relevant motional electromotive force is $\mathcal{E}=\oint (\mathbf{E}+{\mathbf{v}}_{\mathrm{wire}}\times \mathbf{B})\cdot d\mathbf{l}=-d{\Phi }_B/dt$. 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 $B^2{a}^{2}v/R$. With negligible electrical self-inductance,Let the wire cross-sectional area be $S$, mass density ${\rho }_m$, and resistivity ${\rho }_e$. Since $m=4aS{\rho }_m$ and $R=4a{\rho }_e/S$,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 $B{a}^2\cos \theta$. The induced current produces torque $-B^2{a}^4\dot{\theta }{\sin }^2\theta /R$. A thin square loop has moment of inertia ${\mathcal{I}}_x=m{a}^2/6$: the two legs parallel to $x$ contribute $m{a}^2/8$, and the other two $m{a}^2/24$. Averaging ${\sin }^2\theta$ over a rotation givesIf “decay time” means rotational kinetic energy rather than angular speed, $K\propto {\omega }^2$ gives ${\tau }_{\mathrm{rot},\mathrm{K}}={\tau }_t/6=0.400512\mathrm{ms}$. 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.