Prove that the formula for motional emf
$\displaystyle\epsilon=\oint(\vec{v}\times\vec{B})\cdot d\vec{l}$
reduces to Faraday's law
$\displaystyle\epsilon=-\frac{d\Phi_B}{dt}$
if $\vec{B}$ does not vary with time.
My attempt:
$\displaystyle\Phi_B=\iint\vec{B}\cdot d\vec{A}$
We need to prove that $\displaystyle\oint(\vec{v}\times\vec{B})\cdot d\vec{l}=-\frac{d}{dt}\iint\vec{B}\cdot d\vec{A}$
We start with the left-hand side.
$\displaystyle\oint(\vec{v}\times\vec{B})\cdot d\vec{l}=\iint\left(\vec{\nabla}\times(\vec{v}\times\vec{B})\right)\cdot d\vec{A}$ (Using Stokes' theorem)
But I don't know how to find $\displaystyle\vec{\nabla}\times(\vec{v}\times\vec{B})$. Can someone nudge me in the right direction?
The mathematical background for your physical problem is the integral formula $$\frac{d}{dt}\int\limits_{\Sigma(t)} d\vec{A} \cdot \vec{F}(t,\vec{x})=\int\limits_{\Sigma(t)}d\vec{A}\cdot \frac{\partial \vec{F}(t,\vec{x})}{\partial t}+ \!\int\limits_{\Sigma(t)} d\vec{A} \cdot \vec{v}(t, \vec{x})\, \vec{\nabla}\cdot \vec{F}(t, \vec{x}) \\- \! \! \!\int\limits_{\partial \Sigma(t)}d\vec{x} \cdot \left(\vec{v}(t, \vec{x}) \times \vec{F}(t, \vec{x})\right), $$ where $\Sigma(t)$ is a time (parameter) dependent two-dimensional orientable manifold with boundary $\partial \Sigma(t)$. $\vec{v}(t, \vec{x})$ denotes the velocity of the manifold $\Sigma(t)$ measured at time $t$ at the point $\vec{x}$ in the (moving) domain of integration $\Sigma(t)$ and $\partial \Sigma(t)$, respectively. The time (parameter) dependent vector field $\vec{F}(t,\vec{x})$ together with $\Sigma(t)$ are assumed to be subject to certain "nice" properties, usually fulfilled in physical applications.
Let us motivate this formula by giving a rough (non rigorous) sketch of its proof. The occurence of the first term on the right-hand side of the formula (the one with $\partial \vec{F}(t, \vec{x})/\partial t$) is obvious and we restrict ourselves to the special case of a time-independent vector field $\vec{F}(\vec{x})$. We consider the difference $$\int\limits_{\Sigma(t+h)} \!\! \!d \vec{A}\cdot \vec{F}(\vec{x})- \! \!\int\limits_{\Sigma(t)} d \vec{A} \cdot \vec{F}(\vec{x}) = \! \! \!\int\limits_{\partial V(t,h)}\! \! d \vec{A} \cdot \vec{F}(\vec{x}) - \! \! \! \! \int\limits_{\Delta(t,h)} \! \! d \vec{A} \cdot \vec{F}(\vec{x}) \\ = \! \! \! \! \int\limits_{V(t,h)}\! \! d^3x \, \, \vec{\nabla}\cdot \vec{F}(t, \vec{x})-\! \! \!\int\limits_{\Delta(t,h)} d \vec{A} \cdot \vec{F}(\vec{x}), $$ where the three-dimensional volume $V(t,h)$ is bounded by the closed surface $\partial V(t,h)$ consisting of $\Sigma(t+h)$, $\Sigma(t)$ and the surface $\Delta(t,h)$ generated by the movement of $\partial \Sigma(t)$ to $\partial\Sigma(t+h)$. In the last step, the integral theorem of Gauss was applied to the first integral. Keeping only terms linear in $h$, we find $$\int\limits_{V(t,h)} \! \! d^3 x \, \, \vec{\nabla}\cdot \vec{F}(\vec{x}) = \int\limits_{\Sigma(t)} d \vec{A} \cdot \vec{v}(t, \vec{x}) h \, \vec{\nabla} \cdot \vec{F}(\vec{x})+ \mathcal{O}(h^2) $$ and $$\int\limits_{\Delta(t,h)} \! \! d \vec{A} \cdot \vec{F}(\vec{x}) = \int\limits_{\partial \Sigma(t)} \left(d \vec{x} \times \vec{v}(t, \vec{x}) h \right) \cdot \vec{F}(\vec{x}) + \mathcal{O}(h^2) = h \int\limits_{\partial \Sigma(t)} d \vec{x} \cdot \left( \vec{v}(t, \vec{x}) \times \vec{F}(\vec{x}) \right)+ \mathcal{O}(h^2), $$ reproducing the second and third term of the claimed formula in the limit $$ \lim_{h \to 0} \, \frac{\int\limits_{\Sigma(t+h)}\! \! d \vec{A} \cdot \vec{F}(\vec{x}) - \int\limits_{\Sigma(t)} d \vec{A} \cdot \vec{F}(\vec{x})}{h}.$$
Turning now to the physical application, we set $\vec{F}(t, \vec{x})= \vec{B}(t, \vec{x})$, where the magnetic field $\vec{B}(t, \vec{x})$ satisfies the two Maxwell equations $\vec{\nabla} \cdot\vec{B}(t,\vec{x}) =0$ and $\vec{\nabla} \times \vec{E}(t, \vec{x}) = - \partial \vec{B}(t, \vec{x})/ \partial t$, resulting in the integral version $$ \frac{d}{dt} \int \limits_{\Sigma(t)} \! d \vec{A} \cdot \vec{B}(t, \vec{x}) = -\int\limits_{\Sigma(t)} \! d \vec{A}\cdot \left( \vec{\nabla} \times \vec{E}(t, \vec{x}) \right) - \int\limits_{\partial \Sigma(t)} \! d \vec{x} \cdot \left(\vec{v}(t, \vec{x}) \times \vec{B}(t, \vec{x}) \right) \\= - \int\limits_{\partial \Sigma(t)} \! d \vec{x} \left( \vec{E}(t, \vec{x})+ \vec{v}(t, \vec{x}) \times \vec{B}(t, \vec{x}) \right), $$ where Stokes' theorem was used in the last step. In the special case of a time-independent magnetic field $\vec{B}(\vec{x})$, the formula simplifies to $$ \frac{d}{dt} \int\limits_{\Sigma(t)} \! d \vec{A} \cdot \vec{B}(\vec{x})= - \! \! \int\limits_{\partial \Sigma(t)}\! d \vec{x} \cdot \left(\vec{v}(t, \vec{x}) \times \vec{B}(\vec{x}) \right).$$