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For motion to the right in $B_0{\mathbf{e}}_z$, $B_0>0$, the induced current is clockwise. Its own magnetic field points into the page inside the circuit, opposing the increase in external flux, as required by Lenz's law. As that current changes, its changing flux also produces an electromotive force opposing the change in current: the circuit has inductance.

Consequently the algebraic current law used in parts (i)-(iv) must be replaced by a dynamical circuit equation that includes the derivative of the self-flux. The induced current cannot in general adjust instantaneously to the bar velocity. Energy is temporarily stored in the self-field and can later return to the motion or be dissipated; the simple exponential braking law can change. Self-field forces and changes of inductance with circuit geometry may also matter. The magnetic self-field changes the transient dynamics but does not create energy: mechanical energy, magnetic-field energy and Joule heat must be included in the conservation balance. A quantitative answer would require the circuit's inductance and its dependence on $s$.

For a moving circuit, Faraday's law states that the electromotive forceequals $-d{\Phi }_B/dt$, 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 hasso Ohm's law gives the induced currentwith positive current taken counterclockwise from above.

On the edge $x=a$, $d\mathbf{\ell }=dy{\mathbf{e}}_y$. The magnetic force on a current-carrying wire is $d\mathbf{F}=I,d\mathbf{\ell }\times \mathbf{B}$, henceThe 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:This is an electromagnetic drag force opposing the motion, as required by Lenz's law. Including gravity givesAt the terminal velocity, $\ddot{z}=0$, and the constant downward speed has magnitude

At time $t$, let the centre of the loop have coordinate $Y=d+vt$. With normal $+{\mathbf{e}}_x$, the magnetic flux isFaraday's law and Ohm's law therefore give a current of magnitudeThe original flux points into the plane and decreases in magnitude as the loop recedes. By Lenz's law, the induced current reinforces the into-plane field: it is clockwise when viewed from the $+x$ side.

Faraday's law isChoose the positive circulation to be anticlockwise as viewed from $+z$. Since ${\Phi }_B=\pi B{r}^2$,

If $\dot{r}>0$, the current is clockwise and its field points in the $-z$ direction, reducing the field inside the loop. Since $d\mathbf{F}=Id\mathbf{l}\times \mathbf{B}$, the force due to the imposed field is radially inward. If $\dot{r}<0$, the current is anticlockwise, its field points in the $+z$ direction and increases the field inside, while the force is radially outward. In both cases the magnetic force opposes the radial motion, in accordance with Lenz's law.

Faraday's law is $\mathcal{E}={\oint }_C(E+v\times B)\cdot dl=-d{\Phi }_B/dt$. Here $I=-L(B\dot{x}+x\dot{B})/R$ (sign relative to the chosen circuit orientation), and the bar force is $ILB$. For $B=B_0/(1+t)$ this gives the displayed equation with $\alpha =B_0^2{L}^2/(mR)$. At $\alpha =1$, the independent solutions are $u$ and $u{e}^{1/u}$, $u=1+t$; the data give $x=x_{0}u{e}^{1/u-1}$. Thus $\dot{x}\to x_0/e$ and total mechanical work is $m{x}_0^2/(2e^2)$. The current circulates to reinforce the decaying $+z$ flux, and its force drives increasing area, both opposing the flux decrease as Lenz's law requires.