Proof of the induced EMF formula E = ∫ (v × B) · dl

[phantomvommand]

TL;DR

Why is $E = \int (\vec v \times \vec B) \cdot d \vec l$?

Why is $E = \int (\vec v \times \vec B) \cdot d \vec l$? This seems to be a general formula, and I would like to know its proof.

Thanks for all the help.

 

[BvU]

Are you familiar with the Lorentz force ?
Then the energy per unit charge (in SI units: Joule/Coulomb) follows from $\int {\vec F\over q} \cdot d\vec l$ .

PS it's a bit confusing to use the symbol $E$ for this; it's actually a potential difference (or emf)
$$\text {EMF} = V_{AB} = \int_A^B {\vec F\over q} \cdot d\vec l $$$\$

 

[phantomvommand]

Are you familiar with the Lorentz force ?
Then the energy per unit charge (in SI units: Joule/Coulomb) follows from $\int \vec F \cdot d\vec l$ .

$\$

Yes, I am. Thanks for the help, I never realized this connection. I suppose this formula can be used in general? Would I be more accurate to use a loop integral instead of an integral?

 

[BvU]

Yes, I am. Thanks for the help, I never realized this connection. I suppose this formula can be used in general? Would I be more accurate to use a loop integral instead of an integral?

The expression is correct, so it should be universally applicable (but for a loop I expect to get 0 : $\ V_{AA}\equiv 0$ ) .

Check out a few of the sections in the link (Force on a current-carrying wire, EMF).

$\$

 

[vanhees71]

As anything in electrodynamics the formula can be derived from Maxwell's equations in differential form, which are always valid. Here one integrates Faraday's Law (in SI units)
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}$$
over a surface $A$ with boundary curve $\partial A$ with the usual orientation of the path given by the right-hand rule.
$$\int_A \mathrm{d}^2 \vec{f} \cdot \vec{\nabla} \times \vec{E}=\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}=-\int_A \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}.$$
Now the usual integral form of Faraday's Law is written in terms of the magnetic flux
$$\Phi_{B}=\int_{A} \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
If now the surface $A$ and its boundary $\partial A$ are time-dependent you have to use Reynold's transport theorem for surface integrals to take the time derivative. Together with $\vec{\nabla} \cdot \vec{B}$ this leads to the ONLY generally correct form of Faraday's Law in integral form:
$$\mathcal{E}=\int_{\partial A} \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B})=-\dot{\Phi}_B=-\frac{\mathrm{d}}{\mathrm{d} t} \Phi_B.$$
For a derivation of the corresponding Reynold's transport theorem, see

https://en.wikipedia.org/wiki/Faraday's_law_of_induction#Proof