= 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 is
$$
\int_{S'}\mathbf B\cdot d\mathbf S=-\delta t\oint_C\mathbf B\cdot(\mathbf v\times d\mathbf r).
$$
By 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 derivative
$$
\frac{d\Phi}{dt}=\int_{S(t)}\partial_t\mathbf B\cdot d\mathbf S+\oint_C\mathbf B\cdot(\mathbf v\times d\mathbf r).
$$
Now apply <Stokes theorem> and $\mathbf B\cdot(\mathbf v\times d\mathbf r)=-(\mathbf v\times\mathbf B)\cdot d\mathbf r$:
$$
\frac{d\Phi}{dt}=-\oint_C\mathbf E\cdot d\mathbf r-\oint_C(\mathbf v\times\mathbf B)\cdot d\mathbf r.
$$
Consequently the <electromotive force>, including the <motional electromotive force>, satisfies
$$
\boxed{\frac{d\Phi}{dt}=-\varepsilon.}
$$
This derives the moving-circuit form of <Faraday's law>, including both changing fields and changing circuit geometry.
Back to article page
