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.
phantomvommand
Messages
287
Reaction score
39
TL;DR Summary
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.
 
Physics news on Phys.org
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)
$$\text {EMF} = V_{AB} = \int_A^B {\vec F\over q} \cdot d\vec l $$##\ ##
 
  • Like
Likes phantomvommand
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
 
  • Like
Likes phantomvommand
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
Back
Top