I Unveiling the Proof of Induced EMF Formula: E = ∫ (v x B) · dl

AI Thread Summary
The formula E = ∫(v × B) · dl is derived from the Lorentz force and represents the energy per unit charge in electromagnetic fields. It is applicable in general scenarios, particularly when considering the relationship between electric fields and magnetic fields as described by Faraday's Law. The integral form of Faraday's Law can be expressed in terms of magnetic flux and is valid for time-dependent surfaces. The discussion emphasizes the importance of using the correct notation, as E represents potential difference or electromotive force (emf). Understanding this connection is crucial for applying the formula accurately in electrodynamics.
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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.
 
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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)
$$\text {EMF} = V_{AB} = \int_A^B {\vec F\over q} \cdot d\vec l $$##\ ##
 
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BvU said:
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?
 
phantomvommand said:
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).

##\ ##
 
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
 
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