Starting from Maxwell's equations, deduce that

$$\frac{d \Phi}{d t}=-\mathcal{E}$$

for a moving circuit $C$, where $\Phi$ is the flux of $\mathbf{B}$ through the circuit and where the electromotive force $\mathcal{E}$ is defined to be

$$\mathcal{E}=\oint_{\mathcal{C}}(\mathbf{E}+\mathbf{v} \times \mathbf{B}) \cdot \mathbf{d} \mathbf{r}$$

where $\mathbf{v}=\mathbf{v}(\mathbf{r})$ denotes the velocity of a point $\mathbf{r}$ on $C$.

[Hint: Consider the closed surface consisting of the surface $S(t)$ bounded by $C(t)$, the surface $S(t+\delta t)$ bounded by $C(t+\delta t)$ and the surface $S^{\prime}$ stretching from $C(t)$ to $C(t+\delta t)$. Show that the flux of $\mathbf{B}$ through $S^{\prime}$ is $-\delta t \oint_{C} \mathbf{B} \cdot(\mathbf{v} \times \mathbf{d r})$.]