How is E.M.F induced?

1. Dec 30, 2013

Entanglement

I know when the magnetic flux falling on a wire changes, emf is induced and a potential difference is created in the wire, I don't understand how can a change in the flux give electrons potential to create a current what exactly happens when flux changes should I just accept it as a fundamental law of nature or does it have an actual understood explanation?

Last edited: Dec 30, 2013
2. Dec 30, 2013

Philip Wood

When a wire moves, 'cutting' lines of magnetic flux, an emf is induced in it. This is because the free electrons, in the wire are moving, because the wire itself is moving, and experience motor effect (magnetic Lorentz) forces which urge them along the wire.

The other type of 'electromagnetic induction' is when the flux linking a stationary loop (real or imaginary) changes: an emf is induced in the loop. As far as I know, this has to be accepted as a fundamental law of nature; it is described by one of the four Maxwell's equations.

3. Dec 30, 2013

tiny-tim

Hi ElmorshedyDr!
Faraday's Law (one of Maxwell's equations) is usually written ∂B/∂t + x E = 0

So if ∂B/∂t is non-zero, so is x E

which means that E is not irrotational, ie there will be loops of continually increasing E, and a wire that follows such a loop will sustain a current.

btw, we can rewrite Faraday's Law as ∂ $\wedge$ (E;B) = 0 … essentially, the "four-dimensional curl" of the electromagnetic field is zero.

This is equivalent to saying that the electromagnetic field can be written as the "four-dimensional gradient" of a vector potential:

(E;B) = ∂ $\wedge$ (A,φ) ie B = x A and E = φ - dA/dt

4. Dec 30, 2013

Entanglement

Thanks a lot for being helpful but This is my first year to study electromagnetic induction (I have been like only 1 month) so I don't know anything about maxwells 4 equations I just want to know how does the change in flux make electrons move

5. Dec 30, 2013

tiny-tim

a change in flux always means a non-zero x E

it doesn't cause it, it isn't caused by it, they simply go together

(and a non-zero x E is bound to make electrons move!)

6. Dec 30, 2013

Staff: Mentor

A changing magnetic field is associated with an electric field, according to Maxwell's equations. (Many people and textbooks say "causes" instead of "is associated with", but some people object violently to that terminology, with some justification, so I prefer not to stir them up. ) The electric field is what makes the electrons move in an induced current (e.g. in an electric generator).

7. Dec 31, 2013

vanhees71

For moving bodies, it's saver to think in terms of forces acting on the electrons, which is (using Heavside-Lorentz units) given by
$$\vec{F}_{\text{mag}}=-e \frac{\vec{v}}{c} \times \vec{B}.$$
Here $-e$ is the charge of an electron.

In the stationary state, i.e., after transient motions of the electrons (caused by the forces due to both the magnetic and the induced electric field!) are gone, you have a charge separation at the ends of the moving rod compensating the magnetic force by the corresponding electric field, i.e.,
$$-e \vec{E}- e \frac{\vec{v}}{c} \times \vec{B}=0.$$
The induced electric field thus is thus
$$\vec{E}=-\frac{\vec{v}{c}}{c} \times \vec{B}.$$
You can also get this result, using the (correct!) integral form of Faraday's Law, which is equivalent to the local law. The already given local law,
$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0$$
can be integrated to
$$\frac{\mathrm{d}}{\mathrm{d} t} \int_{A} \mathrm{d}^2 \vec{F} \cdot \vec{B}=\int_{\partial A} \mathrm{d} \vec{x} \cdot \left (\vec{E} + \frac{\vec{v}}{c} \times \vec{B} \right).$$

The Wikipedia has an excellent article on the Law of Induction:

https://en.wikipedia.org/wiki/Electromagnetic_induction

Last edited: Dec 31, 2013
8. Dec 31, 2013

BruceW

yep. and if the OP'er is not familiar with the $\nabla$, then a change in flux always means a non-zero electric field. well, this depends on whether we're talking about a moving surface, but anyway, that's the general idea.

9. Dec 31, 2013

Andrew Mason

Maxwell's equations can tell you when an induced electric potential will occur but do not really help one understand why, physically, it occurs. It was not until Einstein that this was understood.

This text explains, using special relativity, why the induced potential occurs.

AM

10. Dec 31, 2013

Philip Wood

AM Thank you for this text. Very nice reference if you don't have Purcell (or Resnick's Intro to Special Relativity). And I agree that Special Relativity sheds huge amounts of light on electromagnetism, but I don't find in the text any specific treatment of induced potential. Have I missed it?

11. Dec 31, 2013

Andrew Mason

No. You are right. Faraday's law is not specifically addressed. Section 2.2, explains how moving charges in an overall neutral conductor provide a net electric field as seen by a charge that is in perpendicular motion relative to the conductor. That is somewhat related, I think, to induction. Anyhow, it conveys the idea that magnetic fields are really just relativistic effects of the electric fields of charges moving within otherwise neutral conductors. From that one can see how a moving charge cutting through a magnetic field experiences an electric potential.

I did not want to refer the OP to a second year text on EM. If you know of any better papers that explain this at a lower level let me know.

AM

12. Dec 31, 2013

WannabeNewton

Technically Purcell is a first year text on electromagnetism, at least in the US where it's used in first year honors physics classes. I think it's one of those texts that should be referenced anyways so that people can be made aware of one of the most elegant textbooks on electromagnetism :)

It's worth getting anyways because the chapter on electromagnetic induction has a beautiful and well-diagrammed derivation of induction using special relativity.

13. Jan 1, 2014

vanhees71

It is VERY misleading to talk about an electric potential here, because the induced electric field has no potential since Faraday's Law tells you that its curl is not vanishing
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.$$
The general electromagnetic field needs both a scalar and a vector potential (or in relativistically covariant language a four-vector potential),
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A}-\vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}$$
or in covariant form
$$F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}.$$

14. Jan 1, 2014

Andrew Mason

That is an important point. The induced field is non-conservative so the concept of potential, or potential difference between two points, does not apply. The induced emf between two points depends on the path between them.

AM