 Solution

Orient the moving boundary $C(t)$ consistently with the normal on a spanning surface $S(t)$. The relevant Maxwell equations are $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$.

First hold the magnetic field fixed at time $t$ and compare the fluxes through the old and new surfaces. The closed surface consists of the new surface, the reversed old surface, and the joining ribbon. A parametrization of the ribbon by boundary arclength and time gives its outward oriented area vector as $d\mathbf{r}\times \mathbf{v}\delta t$. Thus its flux isBy the divergence theorem, the total flux through the closed surface is zero. Adding the field's explicit change in time consequently gives the moving-surface flux derivativeNow apply Stokes theorem and $\mathbf{B}\cdot (\mathbf{v}\times d\mathbf{r})=-(\mathbf{v}\times \mathbf{B})\cdot d\mathbf{r}$:Consequently the electromotive force, including the motional electromotive force, satisfiesThis derives the moving-circuit form of Faraday's law, including both changing fields and changing circuit geometry.