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

[Meow12]

Prove that the formula for motional emf

[imath]\displaystyle\epsilon=\oint(\vec{v}\times\vec{B})\cdot d\vec{l}[/imath]

reduces to Faraday's law

[imath]\displaystyle\epsilon=-\frac{d\Phi_B}{dt}[/imath]

if [imath]\vec{B}[/imath] does not vary with time.

My attempt:

[imath]\displaystyle\Phi_B=\iint\vec{B}\cdot d\vec{A}[/imath]

We need to prove that [imath]\displaystyle\oint(\vec{v}\times\vec{B})\cdot d\vec{l}=-\frac{d}{dt}\iint\vec{B}\cdot d\vec{A}[/imath]

We start with the left-hand side.

[imath]\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}[/imath] (Using Stokes' theorem)

But I don't know how to find [imath]\displaystyle\vec{\nabla}\times(\vec{v}\times\vec{B})[/imath]. Can someone nudge me in the right direction?

 

[fresh_42]

Looks as if the Jacobi identity could help:
[math] \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} [/math]or the Graßmann identity
[math] \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} [/math]
Note that [imath] B [/imath] does not depend on time.