Use Stokes' theorem (physics): Prove formula for motional emf reduces to Faraday's law if...

[Meow12]

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?

 

[fresh_42]

Looks as if the Jacobi identity could help:
$$\begin{gathered} \vec{\nabla} \times \left( \vec{v} \times \vec{B} \right)-\vec{v} \times \left( \vec{\nabla} \times \vec{B} \right)+ \vec{B} \times \left( \vec{\nabla} \times \vec{v} \right)=\vec{0} \end{gathered}$$or the Graßmann identity
$$\begin{gathered} \vec{\nabla} \times \left( \vec{v} \times \vec{B} \right)=\left(\vec{\nabla}\cdot \vec{B}\right)\vec{v}-\left(\vec{\nabla}\cdot \vec{v}\right)\vec{B} \end{gathered}$$
Note that $B$ does not depend on time.