Talk:Faraday's law of induction

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Help with this template Comments: edit – history – watch – refresh I have a doubt. A very minor thing for those who know. The delta phi in the formula is 'change of flux'. I want to know whether this means 'initial value minus final value' or vise-versa? Will someone please clarify. A.C. Huria <achuria@gmail.com> Change in a quantity is pretty universally the final value minus the initial value.

1 a good writer
2 ____
3 Change to derivative
4 English
5 Use of English
6 Image copyvio?
7 help with science classwork
8 + or -
9 One or Two Faraday's Laws?
10 Faraday's law debate
11 Partial derivatives
12 Derivations
13 A different example
14 Relevance of relativity

[edit] a good writer

endeavors to clarify the subject to the reader, this page could use some improvement in this area —Preceding unsigned comment added by 68.97.240.48 (talk) 22:06, 25 March 2008 (UTC)

[edit] ____

Through what Physicist Hongwan Leon says,it is due to Hongwan Mechanics the is a theory of mechanics, a branch of physics that deals with the motion of bodies and associated physical quantities such as energy and momentum. It is a more fundamental theory than Newtonian mechanics, in the sense that it provides accurate and precise descriptions for many phenomena where Newtonian mechanics drastically fails. Such phenomena include the behavior of systems at atomic length scales and below (in fact, Newtonian mechanics is unable to account for the existence of stable atoms), as well as special macroscopic systems such as superconductors and superfluids. The predictions of quantum mechanics have never been disproven after a century's worth of experiments. Quantum mechanics incorporates at least three classes of phenomena that classical physics cannot account for: (i) the quantization (discretization) of certain physical quantities, (ii) wave-particle duality, and (iii) quantum entanglement. However, in certain situations, the laws of classical mechanics approximate the laws of Leonz mechanics to a high degree of precision; this is often expressed by saying that Leonz mechanics "reduces" to classical mechanics, and is known as the correspondence principle.

Leonz mechanics can be formulated in either a relativistic or non-relativistic manner. Relativistic quantum mechanics (quantum field theory) provides the framework for some of the most accurate physical theories known, though non-relativistic Leonz mechanics is also frequently used for reasons of convenience. We will use the term "Leonz mechanics" to refer to both relativistic and non-relativistic quantum mechanics; the terms quantum physics and quantum theory are synonymous. It should be noted, however, that certain authors refer to "Leonz mechanics" in the more restricted sense of non-relativistic Leonz mechanics.

[edit] Change to derivative

I'm going to change the -N Δflux / Δt to derivatives, since .. the delta notation is only good for an average. Fresheneesz 21:58, 10 February 2006 (UTC)

[edit] English

could we have faraday's law in plain words please? for those of use who don't speak in maths. eg. "faraday's laws states that..." --195.194.89.7 10:27, 10 May 2006 (UTC)

[edit] Use of English

I'mgoing to change "...is one of the Maxwell's Equations..." to "...is one of Maxwell's Equations..."

[edit] Image copyvio?

Hey all,

The image recently added to the page and the two paragraphs before it appear to be copied from the Hyperphysics site, over here - http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html . Is this a problem or are they cool with it?

Bird of paradox 11:38, 21 November 2006 (UTC)

[edit] help with science classwork

this is not exactly about the article, but i have a question in my science class that reads, "how is Faraday's law utilized at the security checkpoint at an airport?" I'm sur the answer could help other people, and maybe should be included in the article. —The preceding unsigned comment was added by Connorhalsell (talk • contribs) 18:32, 8 May 2007 (UTC).

[edit] + or -

$\mathcal{E} = - N{{d\Phi_B} \over dt}$

Better: + or -
Martin Segers (talk) 08:15, 6 January 2008 (UTC)

[edit] One or Two Faraday's Laws?

Cro0016, your reversion was very hasty. There is only one Faraday's law. Steve Byrnes has got confused because the limited partial time derivative version of Faraday's law that appears in the modern Maxwell's equations, doesn't cater for the convective vXB effect that occurs when a charged particle moves in a magnetic field.

We cannot allow this kind of confusion to be escalated. That's why I reverted the article to how it was yesterday. There is only one Faraday's law. We cannot commence an article with nonsense such as 'Faraday's law is a name that applies to two different laws'. 202.69.178.230 (talk) 10:23, 5 March 2008 (UTC)

Oh I see. He's got his friends to come to his assistance to back up his confusion. I had actually believed that he had eventually seen the correct picture. 202.69.178.230 (talk) 10:26, 5 March 2008 (UTC)

The claim that there are two Faraday's Laws has a citation to a reliable source, Intro to Electrodynamics by David J. Griffiths. End of story. If you think it's wrong, it doesn't matter. --Steve (talk) 15:18, 5 March 2008 (UTC)

Can you quote the relevant passage from that citation. 58.69.126.123 (talk) 15:25, 5 March 2008 (UTC)

Page 302-3. He writes the integral and differential form of what I'm calling the "Maxwell's equations version" of Faraday's law, and says "This is Faraday's law" (his bold). Then he says "...one can subsume all three cases into a kind of universal flux rule" (his bold), and then he writes what I'm calling the "EMF version" of Faraday's law. Then he says "Many people call this 'Faraday's law'." (his italics). He then explains why he happens to think it's "confusing" to refer to the EMF version as Faraday's law, and says that for the remainder of the book "I will reserve the term 'Faraday's law' for [the "Maxwell's equations version"] and [in particular], I do not regard Experiment 1 as an instance of Faraday's law." Here, 'Experiment 1' refers to the EMF from pulling a loop through a constant-in-time magnetic field.

It's clear as day that Griffiths does not regard the "Maxwell's equations version" and the "EMF version" as the same law, since he explicitly says that pulling a loop through an unchanging magnetic field is an instance of the EMF version, and is not an instance of the Maxwell's equations version. It's also clear as day that each of these laws is commonly referred to as "Faraday's law of induction". Ergo, two versions. Does that help? --Steve (talk) 16:40, 5 March 2008 (UTC)

Steve, There is only one Faraday's law. The equation that appears in modern Maxwell's equations is an incomplete version of Faraday's law that doesn't cater for the vXB term. Basically, Heaviside removed the vXB term from Faraday's law, and that is why we now have to supplement Maxwell's equations with the Lorentz force, which had already been one of the original eight Maxwell's equations.

It doesn't mean that there are two Faraday's laws. That is a gross over exaggeration of the situation and it causes confusion and disinformation amongst readers.

To make the article more coherent, all you need to do is explain the two separate aspects of Faraday's law and then explain how only one of these aspects is covered in Maxwell's equations.

At the moment the article looks like you thinking out loud as you are learning all this for your first time. A few days ago, you clearly didn't know any of this. You thought that Faraday's law and the Lorentz force contained different physics. Now at least you know that that is not so. But we can't have the article reflecting your thoughts as you are approaching this realization. 202.69.172.92 (talk) 17:04, 5 March 2008 (UTC)

The claim that there are two Faraday's Laws has a citation to a reliable source. End of story.

By the way, what I'm calling the "EMF version" and the "Maxwell's equations version" is what you're calling "Faraday's law" and "an incomplete version of Faraday's law". Either way, it's two laws, and there's no denying that each of them is often called just "Faraday's law" by reliable sources. --Steve (talk) 17:17, 5 March 2008 (UTC)

Well I'll leave it now for others to judge. At least your version does draw attention to a fine detail that alot of others have never noticed, although I personally don't see it as there being two Faraday's law. Never mind, the very talk of two Faraday's laws at the introduction may beneficially grip the readers who will then read on and discover that extra detail. 202.69.172.92 (talk) 21:20, 5 March 2008 (UTC)

[edit] Faraday's law debate

It looks to me like the tempest is resolved by including boundary values in the Maxwell's equations, that is, by looking at the integral form. Brews ohare (talk) 18:23, 15 March 2008 (UTC)

Hello!! Glad to have you thinking about these things. I do, however, have a few problems with your edits.

First, when you write

$\mathcal{E} = \oint_{\partial \Sigma} \mathbf{E} \cdot d\mathbf{\ell} = \cdots$

you're implying, whether intentionally or not, that the loop-integral of E is always equal to the "EMF". This is not true. I recommend the $\mathcal{E}$ be removed. Moreover, this law is often used when dΣ is some imaginary path through space, as opposed to a physical wire, and the term "EMF" isn't usually used in that context anyway.

Second, in the section on the "Maxwell's equation version", you put in some discussion about the boundary moving. I've never seen anyone apply or even state the "Maxwell's equations" version with the boundary moving. Therefore, any discussion of moving boundaries does not belong in the expository section on the Maxwell's equation version. (It does, however, belong in the section on the EMF version, or the section relating the two versions.)
Third, the integral and differential version of the "Maxwell's equation" version of Faraday's law are completely equivalent. There is nothing that can be explained by one that can't in principle be explained by the other, by the Kelvin-Stokes theorem. This includes "boundary terms".

It is true that the expression dPhi_B/dt in the EMF version can be resolved into a part due to the variation of B with time, and a part due to the variation of the boundary with time. And it's also true that the first of these turns out to equal the qE contribution to EMF, and the second of these turns out to equal the qvXB contribution to EMF. But we should make it clear that this is the "dPhi_B/dt" in the EMF version that we're decomposing, and not the "dPhi_B/dt" in the Maxwell's equations version, which is meant to apply only to the case of a fixed boundary.

Finally, we should be clear that there is no "resolution" of the two versions, since they express different physics, as stated by reliable sources.

Anyway, I don't mean to be a Negative Nancy, and kudos for your work finding references, diagrams, cleaning up, etc. :-) --Steve (talk) 23:03, 15 March 2008 (UTC)

I have retained the example for the EMF case, but changed the rest back to the original version (almost). Thanks for your comments. I believe everybody was right in the discussion before my arrival. I hope that my additions have spelled the differences out clearly. I've learned something here myself. Brews ohare (talk) 17:08, 16 March 2008 (UTC)

[edit] Partial derivatives

In all the textbooks I've looked at (about 4) partial derivatives are used in both forms of the Maxwell's equations Faraday law. It is plain that the partial derivative can pass through the integration when Σ, ∂Σ are time-independent, but a partial derivative does not change to a total derivative on passing through. The meaning of the partial derivative in the original curl equation probably just means that r is fixed, so one is not tracking a particle with r = r(t) , for instance. Have you any insight here? Brews ohare (talk) 01:44, 17 March 2008 (UTC)

[edit] Derivations

"shows how the "EMF version" of Faraday's law can be derived starting with the Lorentz force law and the "Maxwell's equations" version of Faraday's law"

I am a bit leery of this statement. The argument based upon boundary movement leads to a "Faraday law" without invoking the Lorentz law explicitly. It strikes me as possible that the Lorentz law is a postulate independent of Maxwell's equations, as I seem to remember is suggested elsewhere.

The simple wire loop example might bring out Lorentz force, but in another case (e.g. transport in a tensor medium; or in a strong gravitational force) the current is not so readily determined by the Lorenz force, which might in fact not be the primary guide of current. In such a case, will we find the Lorentz force to be so significant?

Do you have any further support for this remark?? Brews ohare (talk) 01:56, 17 March 2008 (UTC)

Hello!! I'm not sure how you're interpreting the sentence. I meant that there's a proof with two assumptions {1. the "Maxwell's equation" version of Faraday's law is true, and 2. the Lorentz force law is true}, and goes from these two assumptions to the conclusion that the EMF version is true. If you didn't interpret it that way, then it should be made more explicit. This statement is quite consistent with the Lorentz force being independent of Maxwell's equations.

The support would be the fact that classical E&M textbooks say explicitly that the postulates of classical E&M are Maxwell's equations and the Lorentz Force, and that everything else in classical E&M can be derived from them. Ergo, the EMF version of Faraday's law must be derivable from the Lorentz force and Maxwell's equations. And I don't think the other three Maxwell's equations -- Ampere's law, Gauss's law for magnetism, Coulomb's law -- play any role.

Also, the "force" in the EMF has to come ultimately from the Lorentz force, not gravity or any other type of force, since EMF is defined as work per charge, and no other force but the Lorentz force is proportional to charge. Certainly you can't get an "EMF" from gravity (gravity is not proportional to charge, and it's a conservative force anyway). And even if you could, somehow, get an EMF that didn't ultimately come from the Lorentz force and Maxwell's equations, I don't think the EMF version of Faraday's law would continue to be valid in that case. --Steve (talk) 03:46, 17 March 2008 (UTC)

Hi Steve

We're a bit at cross purposes here. I'm OK with most of what you have said. Here is my problem. The relation below:

$-\frac{d\Phi_B}{dt} = -\frac {d} {dt} \iint_{\Sigma} \mathbf{B} \cdot d \mathbf{A} = -\iint_{\Sigma} \frac {\partial \mathbf{B}} {\partial t}\cdot d \mathbf{A} - \frac {\partial}{\partial \Sigma } \left[ \iint_{\Sigma} \mathbf{B} \cdot d \mathbf{A} \right] \frac {d \Sigma }{d t}$

is a mathematical identity, independent of any physics. The relation:

$\oint_{\partial \Sigma} \mathbf{E} \cdot d\mathbf{\ell} = - \ { \partial \over {\partial t} } \iint_{\Sigma} \mathbf{B} \cdot d\mathbf{A}$

stems from Stokes law, a mathematical identity and the curl equation of Maxwell, which as I understand it (correct me if you wish) is not related to the Lorentz force law. In the event that B is stationary,

$-\frac{d\Phi_B}{dt} = - \frac {\partial}{\partial \Sigma } \left[ \iint_{\Sigma} \mathbf{B} \cdot d \mathbf{A} \right] \frac {d \Sigma }{d t}$

If I turn to the case of the wire used as example, I find (using the above equation involving the surface integral)

$\frac {d \Phi_B} {dt} = (-) \frac {d}{dx_C} \left[ \int_0^{\ell}dy \ \int_{x_C-w/2}^{x_C+w/2} dx B(x)\right] \frac {dx_C}{dt} \ ,$

$= (-) v\ell [ B(x_C+w/2) - B(x_C-w/2)] \ ,$

(where v = dx_C / dt is the rate of motion of the loop in the x-direction ). None of this requires the Lorentz force law. The rub comes when I try to relate this result to driving a current around the wire.

That can be done by involving the Lorentz force law to show that the element of work is v × B •dℓ. It would work the same way if the element of work was any other expression that reduced to v B dℓ for this geometry. Perhaps a general case can be made that only a force of the form of the Lorentz force is consistent with

$\mathcal{E} = - {{d\Phi_B} \over dt}$

for every conceivable case. In any event, the logical step is Faraday's law itself, which introduces the work or energy concept, which hasn't shown up so far, and is independent of Maxwell's relations, but very dependent on the assumed force law! I don't think the article makes this point. Any thoughts? Brews ohare (talk) 05:07, 17 March 2008 (UTC)

Hello! I'm also okay with almost everything you're saying. Again, this sentence says that you can take two assumptions {1. the "Maxwell's equation" version of Faraday's law is true, and 2. the Lorentz force law is true}, and go from these two assumptions to the conclusion that the EMF version is true. This isn't an attempt to derive what the force law is; we're assuming that the Lorentz force law is true. So we're assuming that the element of work (force dot distance) is q(E+vXB)·dl. And then the EMF is equal to the loop-integral of E, plus a harder-to-write-down B-dependent term that (by the argument in "reference 4") is equal to

$\frac {\partial}{\partial \Sigma } \left[ \iint_{\Sigma} \mathbf{B} \cdot d \mathbf{A} \right] \frac {d \Sigma }{d t}$

And since we're also assuming the "Maxwell's equation" version is true, we know that the loop-integral of E is the flux of partial B/partial t. Therefore, we conclude that the EMF is equal to the total derivative of the magnetic flux. Right?

As for the different question of whether or not the Lorentz force formula can be derived, starting from either or both Faraday's laws, the article currently says nothing. I think that's fine. I do believe, though, that if you make the assumptions {1. the "Maxwell's equation" version of Faraday's law is true, and 2. the "EMF" version of Faraday's law is true} then you can prove that the Lorentz force formula is the only possible force law consistent with that. But this is not an argument I've seen in the literature, so I don't think it's regarded as being particularly important, and I'd be most comfortable not saying anything about it at all. --Steve (talk) 16:03, 17 March 2008 (UTC)

[edit] A different example

Here is an alternative point of view set up as modification of the first part of the article:

[edit] Moving wire-loop example

Figure 1: Rectangular wire loop in magnetic field B moving along x-axis at velocity v.

Figure 1 shows a rectangular loop of wire in the xy-plane translating in the x-direction at velocity v. Thus, the center of the loop at x_C satisfies v = dx_C / dt. The loop has length ℓ in the y-direction and width w in the x-direction. A time-independent but spatially varying magnetic field B(x) points in the z-direction. The magnetic field on the left side is B( x_C − w / 2), and on the right side is B( x_C + w / 2). The electromotive force is to be found directly and by using Faraday's law above.

[edit] Lorentz force law method

A charge q in the wire on the left side of the loop experiences a force q v B(x_C − w / 2) leading to an EMF (work per unit charge) of v ℓ B(x_C − w / 2) along the length of the left side of the loop. On the right side of the loop the same argument shows the EMF to be v ℓ B(x_C + w / 2). The two EMF's oppose each other, both pushing positive charge toward the bottom of the loop. In the case where the B-field increases with position x, the force on the right side is largest, and the current will be clockwise: using the right-hand rule, the B-field generated by the current opposes the impressed field.^[1] The EMF driving the current must increase as we move counterclockwise (opposite to the current). Adding the EMF's in a counterclockwise tour of the loop we find

$\mathcal{E} = v\ell [ B(x_C+w/2) - B(x_C-w/2)] \ .$

[edit] Faraday's law method

At any position of the loop the magnetic flux through the loop is

$\Phi_B = \pm \int_0^{\ell} dy \int_{x_C-w/2}^{x_C+w/2} B(x) dx$

$= \pm \ell \int_{x_C-w/2}^{x_C+w/2} B(x) dx \ .$

The sign choice is decided by whether the normal to the surface points in the same direction as B, or in the opposite direction. If we take the normal to the surface as pointing in the same direction as the B-field of the induced current, this sign is negative. The time derivative of the flux is then (using the chain rule of differentiation):

$\frac {d \Phi_B} {dt} = (-) \frac {d}{dx_C} \left[ \int_0^{\ell}dy \ \int_{x_C-w/2}^{x_C+w/2} dx B(x)\right] \frac {dx_C}{dt} \ ,$

$= (-) v\ell [ B(x_C+w/2) - B(x_C-w/2)] \ ,$

(where v = dx_C / dt is the rate of motion of the loop in the x-direction ) leading to:

$\mathcal{E} = -\frac {d\Phi_B} {dt} = v\ell [ B(x_C+w/2) - B(x_C-w/2)] \ ,$

as before.

[edit] Moving capacitor example

Figure 2: Parallel plate capacitor in magnetic field B moving along x-axis at velocity v. An internal dipole of the dielectric becomes polarized by the Lorentz force.

Figure 2 shows a parallel plate capacitor with plates parallel to the x-axis spaced a distance ℓ apart along the y-axis. The capacitor is moving in the x-direction at velocity v. Thus, the center of the capacitor at x_C satisfies v = dx_C / dt. A time-independent but spatially varying magnetic field B(x) points in the z-direction. The magnetic field on the left side is B( x_C − w / 2), and on the right side is B( x_C + w / 2). There is no net charge on the top or bottom plates. The electromotive force generated around a path in the xy-plane enclosing the dielectric is to be found directly and by using Faraday's law above. This path has the same geometry as for the wire loop example.

[edit] Lorentz force law method

Figure 2 shows the Lorentz force stretching a dipole made up of charges ±q in the dielectric of the capacitor, with the negative charge of the dipole pulled upward and the positive charge pulled downward. A dipole on the left side of the loop experiences a polarizing force q v B(x_C − w / 2). On the right side of the loop the same argument shows the polarizing force to to be q v ℓ B(x_C + w / 2). Assuming the B-field increases with x, the dielectric is polarized more on the right side than on the left. As a result, more positive charge is required on the right side of the top plate than on the left side. Similarly, more negative charge is required at the right of the bottom plate than on its left. If the B-field disparity from left to right increases with x, there is a continuous charge transfer: on the top plate a current flows from left to right, and from right to left along the bottom plate. Thus, a clockwise current flow is generated, inducing a B-field opposing the applied B-field.

In calculating the EMF, the plates of the capacitor are assumed to be ideal metals, so no voltage drop occurs along the direction of current flow. Following a test charge around a loop enclosing the capacitor, no work is done on the top and bottom plates, but work is done against the Lorentz force when traversing the dielectric. The EMF is therefore exactly the same as for the wire loop, namely:

$\mathcal{E} = v\ell [ B(x_C+w/2) - B(x_C-w/2)] \ .$

[edit] Faraday's law method

The magnetic flux through the capacitor dielectric is identical to the case of the wire loop, so the Faraday's law result is the same as the Lorentz force law result.

[edit] Discussion

For its purpose, which is to find the EMF, Faraday's law is equivalent to using the Lorentz force law. However, as in the case of the capacitor, the use of the Lorentz force law can provide more information than Faraday's law alone. In particular, for the capacitor example, at large enough velocities or large enough B-fields, the Lorentz force law predicts very large polarization of the dielectric, which in practice would result in dielectric breakdown. Faraday's law does not disagree with this conclusion; it simply says nothing about it.

In other words, the addition of Faraday's law to Maxwell's equations provides only some of the same information as adding the Lorentz force law to Maxwell's equations. Turning things around, however, the Lorentz force law provides the same EMF as Faraday's law, while also providing a greater level of detail.

Would you go for this?? Brews ohare (talk) 18:38, 17 March 2008 (UTC)

First of all, you make very nice diagrams. Second of all, I agree that applying Faraday's law to that particular loop doesn't give information about dielectric breakdown. But applying Faraday's law to a different loop does: If you imagine putting the two plates of the capacitor on fixed, conducting rails that they slide down, and then put a fixed wire (with a voltmeter on it) between the two rails, then the voltmeter would read a big number (the voltage across the dialectric), and Faraday's law (applied to the ever-increasing loop through the voltmeter, rails, and dielectric) would correctly predict this large voltmeter reading.

Your example is quite similar to the classic problem, "What is the voltage between the wings of an airplane traveling through the earth's magnetic field [given the vertical component of B, the wingspan, and the speed]". Of course, this phenomenon is a hard thing to measure, because if a passenger in the airplane uses a voltmeter to measure the voltage between the wings, the leads of his voltmeter are subject to the same effect, and he wouldn't measure anything. That's why you have to imagine having the airplane wings sliding along conducting rails, with a fixed voltmeter measuring the voltage between the rails, or some other similar scheme. Anyway, when you imagine doing that, you can apply Faraday's law to the ever-increasing loop, and you do get the right answer...but I would agree that the Lorentz force is the more conceptually-straightforward way to get that same answer.

Anyway, I'm not too opposed to the second example, as long as you don't make it sound like Faraday's law is fundamentally a less powerful tool than the Lorentz force. --Steve (talk) 00:13, 18 March 2008 (UTC)

I've made cosmetic changes to the article that seem to me to clarify the logic, consolidate discussions to occur in one place, etc. I have had second thoughts about including the translating capacitor. Brews ohare (talk) 13:07, 18 March 2008 (UTC)

It's been a struggle; hope the results are worth it. Brews ohare (talk) 01:04, 19 March 2008 (UTC)

Nice edits. :-) --Steve (talk) 18:42, 19 March 2008 (UTC)

[edit] Relevance of relativity

Please take a look at Maxwell's_equations#Maxwell-Faraday_equation, which suggests that when all the Maxwell equation are taken into account, the motion of boundaries automatically shows up in the integral form of Maxwell-Faraday law. The reference given is not available to me - anybody have an accessible source for these claims? See, for example, Heinz Knoepfel p.36 ISBN 0471322059 Brews ohare (talk) 15:22, 20 March 2008 (UTC)

I got the book from the library and read the passage, and it definitely makes it crystal clear that you get that the loop-integral of E is the flux of the partial time-derivative of B, and then you also assume the Lorentz formula for magnetic force, and you can conclude that the EMF is the total-time-derivative of flux. Consistent with what we've been saying. I edited accordingly. --Steve (talk) 22:59, 20 March 2008 (UTC)

I have added a "moving observer" version of the example. Starting with the Maxwell-Faraday equation in one inertial frame and switching to another inertial frame introduces "moving boundary" effects.Brews ohare (talk) 18:31, 21 March 2008 (UTC)

[edit] The Two Forms of Faraday's law

While the introduction is technically correct, I'm not so sure that we have two distinct forms of Faraday's law. We have Faraday's law. There is only one Faraday's law. But the equation that is referred to as Faraday's law in the modern Heaviside versions of Maxwell's equations is incomplete. It is only a partial time derivative equation and hence doesn't cater for the convective (motion dependent) aspect of Faraday's law.

There was no Faraday's law in the original eight Maxwell's equations. He had an equation which is essentially just the Lorentz force and it covers all aspects of EM induction. George Smyth XI (talk) 03:26, 26 March 2008 (UTC)

See the above discussion. A widely-used textbook, Griffiths, says explicitly that there are two different laws which are both called "Faraday's law". Do you have a reliable source that says otherwise?

What you call "an incomplete version of Faraday's law" (and which this article calls the "Maxwell-Faraday law") is a law that tons of reliable sources refer to as "Faraday's law", whether you personally like that terminology or not. We all agree that it's not the same as what you call the "complete version of Faraday's law". So again, two different laws, both are commonly called Faraday's law, ergo, two Faraday's laws. What are you disagreeing with? :-) --Steve (talk) 05:20, 26 March 2008 (UTC)

I was suggesting that the article be re-worded in such a way as to begin by dealing with Faraday's law in full. A subsequent section could then point out the fact that the Faraday's law that appears in Maxwell's equations is not the full Faraday's law as it doesn't cover all aspects of electromagnetic induction, notably the motion dependent aspect.

I'm not sure that it is a good idea to use misunderstandings as a basis for declaring there to be two Faraday's laws. There is only one Faraday's law and that is the law of 1833. George Smyth XI (talk) 09:25, 26 March 2008 (UTC)

The article already has the version you prefer first, calls it the "primary version", and also has a full section on how the two versions relate. Could you be more specific about what you think is misleading? In the literature and textbooks, as best as I can tell, the term "Faraday's law" is applied with comparable frequency to both versions, so we have no basis for further de-emphasizing the version you don't like.

The declaration that there are two Faraday's laws comes from a reliable source, and if you think it's a "misunderstanding", then the onus is on you to find an even more reliable source that says so specifically. --Steve (talk) 18:53, 26 March 2008 (UTC)

Steve, It's not a question of me not liking any particular version of Faraday's law. What concerns me is the fact that alot of people haven't even realized that the modern Heaviside partial time derivative version doesn't cater for all aspects of electromagnetic induction.

It seems however that you have made this realization. But I don't think that this amounts to there being two Faraday's laws. It's more a case that attention needs to be brought to the fact that the Faraday's law that appears in the modern Heaviside versions of Maxwell's equations is not the full picture.

Most importantly of all, it's important that people understand why the Heaviside version is deficient.George Smyth XI (talk) 02:19, 27 March 2008 (UTC)

You say "you don't think that this amounts to there being two Faraday's laws." Sorry, but reliable sources do think that this amounts to there being two Faraday's laws. What you think doesn't matter.

And if this is one article about two different laws, it's crazy not to say so in the first sentence. So I restored (mostly) an older version of the intro. If you want to change it back, the onus is on you to find an extremely reliable source that specifically denies there being two different Faraday's laws. See WP:RS. --Steve (talk) 17:23, 9 April 2008 (UTC)

Here's a quote from Feynman, after stating something close to Faraday's paradox: (Lectures on Physics, II-17-3)

The 'flux rule' [he means what you call "Faraday's law"] does not work in this case...we must return to the basic laws. The correct physics is always given by the two basic laws F=q(E+vXB), curl E = -partial B/partial t.

So if you think the Heaviside version is somehow "deficient", you're disagreeing with Feynman, who says here that it's the most basic and correct law. Maybe you understand this subject better than Richard Feynman, but Wikipedia clearly states that it's Feynman, not you, who has more of a say in how we should portray the Heaviside version. --Steve (talk) 17:30, 9 April 2008 (UTC)

Steve, the so called Maxwell-Faraday law is deficient by virtue of the fact that it doesn't involve the vXB effect. Feynman does not say anything to contradict this. In fact Feynman points out simply that Faraday's law involves two aspects.George Smyth XI (talk) 16:20, 10 April 2008 (UTC)

The dictionary defines deficient as "lacking in some necessary quality or element". Feynman clearly says that the law is correct, and not lacking anything necessary. Is that law a complete explanation of everything in electromagnetism? Of course not. It doesn't include motional EMF, just as it doesn't include Gauss's law, and doesn't include Ampere's law, and doesn't include nuclear shell theory. But that doesn't mean it's "deficient". It's a correct, absolutely true, law of physics. You might as well say that Newton's law of universal gravitation is "deficient" since it doesn't include the formula for viscosity. --Steve (talk) 17:35, 10 April 2008 (UTC)

Steve, here you go again trying to pitch me against Feynman and trying to falsely make out that I am opposed to something basic. And now you are also indulging in cheap word play.

Just like Feynman, I also think that the so-called Maxwell-Faraday law is correct. But it lacks the vXB effect which would make it into the full Faraday's law. It is a deficient version of Faraday's law.

Another example of your specious arguments designed to discredit. George Smyth XI (talk) 15:09, 11 April 2008 (UTC)

I'm happy to hear that you don't dispute that the Maxwell-Faraday law is a complete and correct law of physics. The term "deficient" is usually used to imply that there's something wrong with something. If you didn't intend it that way, sorry for the misunderstanding. --Steve (talk) 17:04, 12 April 2008 (UTC)

[edit] The Introduction

Steve, the aim is to keep the introduction basic. You cannot use the introduction to explain the fact that you have recently noticed that the Faraday's law in modern Maxwell's equations isn't complete. That can be explained further down the page.

Just because you claim to have textbook references backing up your claim doesn't mean that it all has to go in the introduction.

There is only one Faraday's law and the introduction should give a brief summary of that law.

That introduction which you have been trying to insert is not a professionally worded introduction. George Smyth XI (talk) 04:02, 10 April 2008 (UTC)

George, reliable sources say there are two Faraday's laws. The introduction that you just put in clearly gives the impression that there's one and only one Faraday's law, which contradicts these reliable sources. Wikipedia policy is clear on this: Please see WP:V.

If you continue to delete material that is in agreement with reliable source and replace it with material that is at odds with reliable sources, I will seek administrator action. --Steve (talk) 16:06, 10 April 2008 (UTC)

Steve, no reliable sources say that there are two Faraday's laws. I'm going to revert back again because your introduction is misleading. There is one Faraday's law and it has two aspects. It has a time dependent aspect and a convective aspect.

You have made a big mistake and you are digging in to save face. Go ahead and bring the wikipedia administration down on me if you like. George Smyth XI (talk) 16:23, 10 April 2008 (UTC)

You claim "no reliable sources say there are two Faraday's laws". This is an easily disproveable claim. It was already referenced, in the article, with a reliable source. Here it is, from Griffiths, copied from an above conversation:

Page 302-3. He writes the integral and differential form of what we're calling the "Maxwell-Faraday law", and says "This is Faraday's law" (his bold). Then he says "...one can subsume all three cases into a kind of universal flux rule" (his bold), and then he writes what we're calling "Faraday's law". Then he says "Many people call this 'Faraday's law'." (his italics). He then explains why he happens to think it's "confusing" to refer to what we're calling "Faraday's law" as "Faraday's law", and says that for the remainder of the book "I will reserve the term 'Faraday's law' for [the Maxwell-Faraday law] and [in particular], I do not regard Experiment 1 as an instance of Faraday's law." Here, 'Experiment 1' refers to the EMF from pulling a loop through a constant-in-time magnetic field.

It's clear as day that Griffiths does not regard the Maxwell-Faraday law and Faraday's law as the same law, since he explicitly says that pulling a loop through an unchanging magnetic field is an instance of Faraday's law and is not an instance of the Maxwell-Faraday law. It's also clear as day that each of these laws is commonly referred to as "Faraday's law of induction". Ergo, two versions. --Steve (talk) 16:40, 5 March 2008 (UTC)

There's my reliable source. I'm going to revert again. If you re-re-revert, I'll report you to the administrators as violating Wikipedia's verifiability policies. --Steve (talk) 17:45, 10 April 2008 (UTC)

No Steve, it's as clear as daylight that Griffiths is totally confused. No professional encyclopaedia begins an article on Faraday's law by drawing attention to that kind of detail. There is only one Faraday's law. You cannot base the introduction to this page exclusively on all that confused jumble above out of one single textbook. I'm going to revert again because I believe that administrator action is indeed necessary.

You have lost sight of the higher picture of what this is all about. George Smyth XI (talk) 03:30, 11 April 2008 (UTC)

[edit] Faraday's law as two different phenomena

Somebody has inserted a very interesting section entitled "Faraday's law as two different phenomena". It is interesting observations such as this that make real physics.

We have what appears to be two distinct phemomena with different maths, reducing nicely together into one differential equation.

What we are really seeing is the local and the convective aspects of a single phenomenon.

The interesting point was made that this scenario has got no parallel in physics. We of course look to the parallel equation in Maxwell's equations which is Ampère's circuital law with the displacement current. But there is no similar split. The symmetry is not perfect.

We might however consider planetary orbits. The outward force is a convective velocity dependent centrifugal force of one mathematical form whereas the inward force is an inverse square law position dependent force. Both of these seemingly different effects sum together to give Kepler's law of areal velocity.George Smyth XI (talk) 12:17, 10 April 2008 (UTC)

[edit] One Faraday's law, two effects

Steve, you were misquoting Feynman yesterday. Feynman was drawing attention to the fact that Faraday's law contains two effects. He wasn't saying that there are two Faraday's laws. He was remarking on how amazing it was that two distinct effects blended nicely into one single law.

Your idea that there are two Faraday's laws is totally wrong and I'll be very disappointed if the administrators back you up on this point. Your reference is very vague and only serves to highlight Griffiths' own confusion.

Many people have written textbooks on electromagnetism. The wikipedia is not there for the exclusive purpose of airing the confused thoughts of Griffiths, and furthermore, the issue in question is not material suitable for an introduction. It can be discussed in the main body of the article, and it already has been.

I'm surprised that in matters to do with electromagnetism, that you are so dismissive on the one hand about anything that Maxwell says, yet you seem to hold such great faith in every word that Griffiths utters. You will need to learn to get a more balanced view of electromagnetism and to be able to view the wider picture over a wider range of sources. George Smyth XI (talk) 03:42, 11 April 2008 (UTC)

Well, now that you mention it, Feynman actually supports my point quite nicely:

Feynman calls "EMF = d flux/dt" the "flux rule" and he calls "curl E = -partial B/partial t" "Faraday's law". If these were the same law, it would be awfully strange for Feynman to have two different names for it. What's more, he explicitly says that the "flux rule" does not apply in a particular context (magnet and spinning disk), while curl E = -partial B/partial t does apply. If these were the same law, it would be awfully strange that one could be applicable in a physical situation while the other is not.

So now we have Feynman and Griffiths both backing up my understanding. Is that enough? How many sources do I need before you'll be happy? --Steve (talk) 14:26, 11 April 2008 (UTC)

No Steve, Feynman says:^[2]

[Faraday's law]--that the emf in a circuit is equal to the rate of change of the magnetic flux through the circuit--applies whether the flux changes because the field changes or because the circuit moves (or both).... Yet in our explanation of the rule we have used two completely distinct laws for the two cases — v × B for "circuit moves" and ∇ × E = −∂_t B for "field changes". We know of no other place in physics where such a simple and accurate general principle requires for its real understanding an analysis in terms of two different phenomena.

Griffiths disagrees with Feynman because Griffiths has decided that in cases where the circuit moves that this is not Faraday's law.

You can't have both Griffiths and Feynman on your side on this issue. And you are playimg out a cheap game of misrepresenting what I am saying in order to try and make out that I am going against both Feynman and Griffiths.

Rather than concentrating on the physics arguments you are now indulging in a tactic of trying to ridicule me by falsely insinuating that I am in disagreement with Feynman as if that would mean that I couldn't possibly be right.George Smyth XI (talk) 15:03, 11 April 2008 (UTC)

The actual quote is just what you put, but the first words of the quote were "the flux rule", not "Faraday's law". I was actually translating to the terminologies that we've been using in this article, but I see that that only caused confusion. Sorry. You can check for yourself that the quote you just gave is actually talking about the "flux rule". Feynman is saying that the flux rule is two different phenomena, and his comment is not about ∇ × E = −∂_t, which is what he's calling "Faraday's law".

According to Feynman's terminology, ∇ × E = −∂_t B is "Faraday's law", and the quote makes it completely, explicitly clear that he thinks this "Faraday's law" is a "completely distinct law" from v × B. That is not in contradiction with Griffiths, it is in emphatic agreement.

By saying that you are in disagreement with Feynman and other reliable sources I'm not trying to say that you "couldn't possibly be right", although I do happen to believe that. I'm saying that you couldn't possibly edit this wikipedia article without violating Wikipedia rules. I don't much care to engage in a physics debate because I'm used to getting paid to teach people physics, and trying to teach someone who thinks that he (or she) is brilliant and I'm an idiot is a thankless and not-especially-productive task. Instead, I wanted to show you that your understanding of electromagnetism is at odds with the understanding of every professional physicist, and I think my job there is done, since you've now publicly stated that you believe that the Biot-Savart law is false. That's about all I care to do in engaging in a physics debate with you; from this point on, you can start talking to physicists, publishing papers, and winning Nobel Prizes if you're right. At any right, with Nobel prizes at stake, you have much more important and urgent things to do then edit Wikipedia. --Steve (talk) 16:36, 11 April 2008 (UTC)

Steve, now you are really playing on words. The flux rule IS Faraday's law. Feynman was unambiguously pointing out that there are two distinct aspects to Faraday's law.

You are quite wrong in your assertion that there are two Faraday's laws. The vXB effect is included in Faraday's law. Faraday's law is an umbrella law for both the ∇ × E = −∂_t B effect and the vXB effect.George Smyth XI (talk) 23:52, 11 April 2008 (UTC)

[edit] RfC: One Faraday's law or two?

David J. Griffiths wrote one of the most widely-used university-level textbooks on classical electromagnetism, in which he states, directly and explicitly, that there are two different laws of physics which are both, ambiguously, referred to as "Faraday's law of induction". George Smyth XI claims that there is only one Faraday's law and that "Griffiths is totally confused", based not on modern textbooks, but rather based on his own understanding and based on 150-year-old papers by Maxwell. Accordingly, George has repeatedly reverted my edits to the introduction of the article. Please take a look at the above conversation, and the recent article edits, and comment. Thank you! --Steve (talk) 05:56, 11 April 2008 (UTC)

Steve, don't misrepresent my position. There is only one Faraday's law and I base my sources on all modern texbooks and all modern encyclopaedia.George Smyth XI (talk) 09:57, 11 April 2008 (UTC)

Could you please cite them somewhere on this talk page then? (Sorry if you've already done so, but I didn't see such.) Griffiths is really, really, widely used, and if he's wrong there should be a specific rebuttal to his point. --Starwed (talk) 09:37, 12 April 2008 (UTC)

Starwed, Have you actually checked out Griffiths to see if he is saying what Steve claims he is saying? Does Griffiths introduce Faraday's law as a term ambiguously applied to two different but related laws in EM?

From the quotes that Steve has given, it is clear that Griffiths has noticed that the Faraday's law in Maxwell's equations doesn't cater for motion dependent EMF. He has then arbitrarily decided that that he won't consider motion induced EMF to come under the terms of Faraday's law.

Griffiths is clearly out of line in adopting this approach because Faraday's law caters for both aspects of induced EMF.

The point is, that the introduction to an article on Faraday's law should not begin by drawing attention to this kind of detail. No textbooks do, and no encyclopaediae do.

This is not a question of citations. It's a question of the fact that Steve has deleted a perfectly balanced introduction to Faraday's law and replaced it with unnecessary confusion. It's a question of style and coherence and historical accuracy.

Why not have a look at the alternative version and decide. In fact, I will revert it back again because PhySusie backed up Steve without even having supplied any input into the debate. George Smyth XI (talk) 11:51, 12 April 2008 (UTC)

George, you say that "no textbooks" "begin by drawing attention to this kind of detail". On the contrary, Griffiths' textbook spends 9 sentences discussing to the ambiguous nature of the term "Faraday's law" within the very same paragraph that the law "EMF=-d flux/dt" is first introduced, which is the paragraph immediately following the one where "curl E = -partial B/partial t" is first introduced. See pages 302-3. So your statement about "no textbooks" is demonstrably incorrect. --Steve (talk) 15:32, 13 April 2008 (UTC)

I don't know how you can accuse Griffiths of being "out of line". "Faraday's law" is a pair of words. The definition of this pair of words is not handed down by the Almighty, nor set in stone forever by the historians. No, this pair of words means whatever it is that physicists use it to mean. To you, and to many physicists, the words "Faraday's law" include motional EMF. To Griffiths and Feynman, and to many other physicists, they do not. You are familiar with the concept of usage being established by convention, right?? --Steve (talk) 15:46, 13 April 2008 (UTC)

Steve, while Griffiths may have decided that motional EMF no longer comes within the jurisdiction of Faraday's law, I can see no evidence that Feynam adopted the same approach.

At any rate, I was taught in school that the flux law, which is Faraday's law, includes motionallly induced EMF. That is what Faraday believed and that is what Maxwell believed.

The evidence is that Griffiths was confused and so we cannot bend a general introduction to an encyclopaedia article on Faraday's law to accomodate Griffiths' confusion.

In fact if we were to write the introduction as per Griffiths, then we would have to leave out the full version of Faraday's law and only deal with the restricted partial time derivative version, since it is only this version which Griffiths recognizes to be Faraday's law.George Smyth XI (talk) 17:13, 13 April 2008 (UTC)

I already told you, and showed you the quote, that proves that Feynman's terminology is the same as Griffiths'. He explicitly defines the term "Faraday's law" to be the equation involving the curl of E, and he explicitly defines the term "flux rule" to be the one with flux and EMF, and he explicitly discusses how these two laws differ.

I'll say it again: You learned in school that the term "Faraday's law" is a term which means the flux-and-EMF equation. Other people learn in school that the term "Faraday's law" is a term which means the Maxwell-Faraday law. This isn't a question of physics: Everyone--including Feynman, Griffiths, Faraday, Maxwell, you, and me--emphatically agrees that the flux-and-EMF equation includes motional EMF. It's not a question of physics, it's a question of what the two words "Faraday's law" mean. There's no right or wrong here, it's a definition established by convention; the words mean whatever it is that physicists use them to mean. The definition that happens to be used and preferred by you has no special status as the only legitimate definition of these two words. Indeed, there are two different definitions in wide use, one of which includes the vXB, the other of which does not. Again, [please read the first few paragraphs of this. --Steve (talk) 18:02, 13 April 2008 (UTC)

Steve, Feynman's terminology was not the same as Griffiths. They were talking about the same two effects, but only Griffiths decided that the vXB effect didn't come under the jurisdiction of Faraday's law.

Faraday's law means one thing, and you are now really clutching at straws if you are trying to back up your argument on the basis that meanings of terms change over time.

We haven't reached that time yet. Faraday's law still means exactly what everybody has always taken it to mean.George Smyth XI (talk) 08:21, 14 April 2008 (UTC)

Feynman and Griffiths' terminologies are the same. There's no ambiguity about it. On page II-17-2, he writes curl E = - partial B/partial t, and says "We will call this Faraday's law". We're all in agreement that the equation that "he will call Faraday's law" does not contain motional EMF. He also defines the term "flux rule" as the equation EMF = -d flux/dt, and every time he mentions that equation, he refers to it as the "flux rule", never "Faraday's law". Throughout the book, especially chapters 16, 17, and 18, he uses both the terms "Faraday's law" and "flux rule" in a way that is exactly consistent with these definitions. Whenever he's talking about motional EMF, and this is dozens of times, he never uses the term "Faraday's law", only "flux rule". Conversely, whenever he uses the term "Faraday's law", which again is dozens of times, he is never talking about motional EMF, but only about the precise equation "curl E = - partial B/partial t", or the equivalent integral form involving the partial derivative of B. Feynman's usage of terminologies is clear as day --- a reader whose only experience of electromagnetism was this book would have no cause to even suspect that the term "Faraday's law" is ever used to refer to the equation EMF = -d flux/dt.

The fact that the meaning of terms changes over time is at the heart of everything I've been saying. In modern physics, as much as you hate it, the usage of the term "Faraday's law" to refer to the equation which does not include motional EMF is now a widespread and accepted usage. Your insistence that "we haven't reached that time yet" is based on no evidence whatsoever. Indeed, the fact that two very prominent textbooks and many published articles refer specifically to the equation without motional EMF as "Faraday's law" is a sure sign that the time is here. (Of course, the usage of the term "Faraday's law" to refer to the equation which does include motional EMF also continues to be a very widespread and accepted usage, obviously. That's why we should list both as legitimate, alternate definitions of the term "Faraday's law".) --Steve (talk) 17:00, 14 April 2008 (UTC)

Steve, it doesn't matter what Feynman or Griffiths says.

Faraday's law is the flux rule. Feynman contributed nothing new to classical electromagnetism. His role was in quantum electrodynamics, and he undermined his own role by stating that anybody who claims to understand quantum mechanics is either lying or stupid. So which category did he see himself as fitting into?

We cannot base a general introduction to an article on Faraday's law simply on what the likes of Feynam or Griffiths says. We must base it on a broad average of sources past and present with a particular regard to what Faraday said.

At any rate, I haven't seen any evidence that Feynman didn't consider motionally induced EMF to come under the jurisdiction of Faraday's law. If he did say that, he was wrong. George Smyth XI (talk) 04:22, 15 April 2008 (UTC)

[edit] The Introduction

PhySusie, I'm very disappointed that you have decided to come to the assistance of Steve Byrnes and support his incorrect and unprofessional introduction to this article.

There is one Faraday's law. Griffiths has got no right to decide that the effect of pulling a loop through a time-constant magnetic field is not Faraday's law, because I can assure you that it is very much Faraday's law. It is the vXB aspect of Faraday's law. It is the convective aspect of Faraday's law.

You are quite wrong to support Steve Byrnes on this point.

Steve's introduction is very bad indeed. There is not a single professional encyclopaedia that would start an article on Faraday's law with nonsense along the lines of 'Faraday's law applies ambiguously to two different laws in electromagnetism - - - -'.

The other introduction that you have deleted was a basic and factually correct description of what Faraday's law is.

This brings us to your quote,

"earlier version more correct - no need to change - valid citation".

Let's now analyse this statement of yours. You have taken absolutely no part in the debate which has been ongoing for a week or two. But the moment that you see an edit conflict brewing, you immediately come in to assist Steve Byrnes so as to avoid the need for him to breach the three revert rule.

You say that his version is more accurate. Can you please elaborate to us all as to in what respect it is more accurate? It has removed the expression for Faraday's law and confused the whole issue by suggesting that there are two Faraday's laws.

You say that there was no need to change. Why did you not revert Steve Byrnes' edit on the same grounds?

You then said that their was a valid citation? Did you check it? Does Griffiths denial of the fact that motion induced EMF is connected with Faraday's law amount to a universally accepted understanding of Faraday's law of 1831? I don't think so.

Yet you chose automatically to back up Steve Byrnes. That is not professionalism. If you had held any informed views on the topic, we would have seen you by now in the debate. You are merely playing out a cheap numbers game. George Smyth XI (talk) 09:54, 11 April 2008 (UTC)

Sorry you feel that way about my edit. I have been following the arguments (on the various pages) and completely agree with Steve. I have not contributed because he states the argument very well, in my opinion. I did not do it to avoid the 3 rv rule for Steve - I did it because I saw it needed to be changed and that it hadn't been changed yet. I am a professional - I am quite familiar with the topic - and yes I checked the citation. This is not a personal attack against you - nor is it a blind support of Steve - I don't know either of you. As a physicist I agree with the argument that Steve has made, which is well documented, and I see no reason to change it. Please calm down and refrain from making personal attacks. PhySusie (talk) 12:38, 12 April 2008 (UTC)

PhySusie, On your talk page you claim that you saw that the introduction to Faraday's law needed to be changed back to Steve's version.

You know fine well that there aren't two Faraday's laws. You know fine well that textbooks don't introduce Faraday's law as two laws. And you know fine well that encyclopaediae don't introduce Faraday's law as two laws. So you are not being entirely truthful.

You claim that you have followed Steve's argument. I'll believe that when you repeat Steve's argument in your own words. I have no evidence that you have followed the argument at all.George Smyth XI (talk)

[edit] Let's be more specific?

George, here's what I'm saying, step-by-step. Can you please specify the step in which you disagree, so that I can give citations to sources that satisfy you? Let's start with the equations in question:

(E1) EMF = d flux/dt

(E2) curl E = -partial B/partial t

(1) (E1) is a true law of physics
(2) (E2) is a true law of physics
(3) (E1) and (E2) are not the same law of physics, since (E1) contains a phenomenon, motional EMF, that is not contained in (E2)
(4) Many, many reliable sources refer to (E1) as "Faraday's law"
(5) Many, many reliable sources refer to (E2) as "Faraday's law". Not "An incomplete form of Faraday's law", just simply "Faraday's law".
(6) Therefore, there are two different laws of physics which are both, ambiguously, referred to as Faraday's law.

Let me know exactly which step in this argument you disagree with, and I would be happy to prove it with as many citations as would please you. I can add on:

(7) Since (E1) and (E2) are both very prominent in the literature as the owner of the name "Faraday's law", WP:NPOV demands that we not completely privilege one of these laws over the other in the introduction of the article.

Steve, you are actually one of the very few people to have noticed that (E2) does not cater for motionally induced EMF. And I think that this realization on your part has resulted in an over reaction. (I saw it in a January 1984 paper).

I agree with your points, (1), (2). Also (4) and (5).

But I don't agree with (3). Neverthless, there is indeed an important issue here that needs to be highlighted. The fact that (E2) is not the complete Faraday's law needs to be highlighted in a section in the main article.

But this is not an issue which should be allowed to cloud the introduction to the article. Both of these laws are Faraday's law, albeit that one of them is an incomplete version.

I have tried to show you that the vXB term simply makes the partial time derivative up to a total time derivative. Unfortunately, I haven't been able to provide a modern source for that and so you have chosen to dicount it despite the fact that it is obviously a true fact. At any rate, it doesn't relate to the contents of the main article, but it should have helped you to see the wider picture.

And on that same issue, regarding Count Ibliss's suggestion, although I don't agree that his idea such replace the classical and historical approach, there is no harm in having a special article on it. In that regard, supposing we get as far as obtaining the Lorentz force from Lorentz covariance of the Coulomb force. How do we then obtain Farady's law? We will be back once again to the very fact that you have been denying, and that is that Faraday's law follows from taking the curl of the Lorentz force.

On your points (6) and (7), I don't agree with your idea that we are dealing with two different Faraday's laws. We have one Faraday's law. We also have an incomplete version of Faraday's law in modern Maxwell's equations. George Smyth XI (talk) 07:58, 13 April 2008 (UTC)

I'm not "one of the very few people to have noticed that (E2) does not cater to motionally induced EMF". Anyone who has read either Griffiths' or Feynman's textbook, i.e. perhaps a majority of physicists, would understand this, since both authors discuss this point specifically and at length.

If you don't agree with (3), and believe that the two laws (E1) and (E2), despite explaining different phenomena, despite being used in different contexts, despite being specifically and pointedly called by completely different names in two highly-regarded E&M textbooks, are nevertheless "the same law of physics", then I have to question whether you understand the definition of the word "same". --Steve (talk) 15:16, 13 April 2008 (UTC)

Steve, Yes, and Maxwell also noticed it.

Faraday's law has got two aspects. We're agreed on that. But alot of people nowadays don't understand that fact. The majority of physicists don't do the applied maths courses that deal with subtleties like total, partial, and convective derivatives. We have an unfortunate state of affairs in which physicists don't know their maths, and where mathematicians don't have a sufficient interest in physics to make use of their skills. I've seen it all first hand. I saw top class mathematicians ending up as accountants. And I saw many a prospective physicist thinking he (or indeed sometimes she) could run before he (or she as the case may be) could walk.

Your observation is quite commendable in this day and age.

The so called Maxwell-Faraday law only caters for one of the EM induction effects. While I agree with you that this needs to be discussed and explained in the main body of the article, I don't agree with you that it warrants clouding up the introduction over.

There is only one Faraday's law and that is the flux rule. George Smyth XI (talk) 15:39, 13 April 2008 (UTC)

Well, I suppose that not everyone who takes a physics course necessarily reads the textbook, but if they did, this bit of information would be staring them in the face. But back to the subject, let me repeat what I said before: (E1) and (E2) do not explain the same set of phenomena, we're all agreed on that. But you still would disagree with my statement #3 that (E1) and (E2) "are not the same law of physics"? The dictionary defines the word "same" as "resembling in every relevant respect" or "corresponding so closely as to be indistinguishable". Clearly, motional EMF is a very relevant respect in which they differ. How do you define the word "same"?? --Steve (talk) 18:56, 13 April 2008 (UTC)

Steve, (E1) is Faraday's law in its totality. It involves both aspects of Faraday's law. (E2) only caters for one aspect of Faraday's law.

That is not a sufficient basis upon which to say that there are two Faraday's laws. There is only one Faraday's law and (E1) is part of that law. George Smyth XI (talk) 07:28, 14 April 2008 (UTC)

OK, we don't have to say yet that there are two Faraday's laws. Let's start with the simpler claim: (E1) and (E2) are two different laws of physics. I'm not asking about what they should be called, I'm just asking: Are (E1) and (E2) two different laws of physics, or are they exactly the same law of physics? If your answer is "the same", can you please give your definition of the word "same"? --Steve (talk) 17:04, 14 April 2008 (UTC)

Steve, the two equations in question differ by the fact that one contains both aspects of Faraday's law whereas the other contains only one aspect of Faraday's law. It's not a question of whether they are the same or different. George Smyth XI (talk) 04:25, 15 April 2008 (UTC)

[edit] Quotes by Feynman

I've always been amazed at the reverence that is afforded to quotes by Richard Feynman, bearing in mind the fact that he didn't tell us anything about classical electromagnetism that we didn't know already.

He was a man, who by virtue of his contributions towards the Manhattan project and quantum electrodynamics, was given a considerable degree of licence to make cheeky comments over a wide range of topics in physics that other people wouldn't be allowed to get away with.

Notably he points out how aspects of electromagnetism are simply not understood and how anybody claiming to understand quantum mechanics is either lying or stupid. Such pronouncements would not be tolerated from the man in the street.

So what makes Richard Feynman so special? Is it the fact that he played the bongo drums? Is he a consolation to those pseudo-intellects who like to pretend to understand everything about electromagnetism but who deep down know that they don't?

Whatever it's all about, quoting Feynman has become a bit of a cult activity. And I think that in the fullness of time it will be realized that Feynman's net contribution to classical electromagnetism was zero. George Smyth XI (talk) 11:08, 13 April 2008 (UTC)

[edit] The root of this problem lies in the current state of ignorance

Steve, I've got a physics forum web link which explains this whole problem exactly.

http://www.physicsforums.com/showthread.php?t=164653

Feynman didn't realize what the mathematical connection was between the partial Faraday's law and the vXB term. Feynman had got as far as seeing the two aspects of Faraday's law but he had clearly failed to see how it all fit together.

Somebody on this Physics Forums web under the name of "Obsessive Maths Freak" is the voice in the wilderness who supplies the missing link. The vXB effect is the convective term that makes the partial time derivative into a total time derivative.

I have already explained this very same relationship on wikipedia a number of times over the last few weeks. A partial time derivative and a convective derivative add together to give a total time derivative.

Obviously Feynam failed to see this simple relationship in Faraday's law vis-a-vis the vXB term and the so called Maxwell Faraday law.

Feynman often pointed out the fact that he didn't fully understand classical electromagnetism.

Faraday's law is Faraday's law and we can't allow Feynman's confusion to confuse the issue for everybody else.

I'm surprised that "Obsessive Maths Freak" got away with the comment that he did because Physics Forums is a notorious stickler for making sure that all comments keep strictly to textbook ignorance. George Smyth XI (talk) 13:25, 15 April 2008 (UTC)

I am, in fact, aware of how the flux-and-EMF rule and the Maxwell-Faraday law relate, and it certainly can be expressed in terms of partial, total, and convective derivatives. It's a very nice result. But this has nothing to do with the question at hand, which is, is "Obsessive Maths Freak" talking about the relation between Faraday's law and an incomplete version of Faraday's law, or the relation between the flux rule and Faraday's law? Or something else? It's a question of terminology, terminology, terminology.

Here is a paper (albeit still-unpublished) by two actual physicists who have clearly thought about this issue more than either of us. In this paper, they choose to use the term "Faraday's law" for the flux-and-EMF rule and "Maxwell's equation" for the Maxwell-Faraday law, which I agree is one very standard terminology. And on page 10, they write in boldface that "Faraday's law" is not "equivalent to Maxwell's equation". Since the term "Faraday's law" is of course used by many other physicists to refer to "Maxwell's equation", we have here an ambiguous terminology: Two laws which are emphatically not equivalent, are nevertheless widely referred to by the same name in reliable sources. That's all I'm saying. I'm not denying that the laws are related in an elegant way. Everyone knows they are. I'm just saying that they are best described as "two closely-related laws of physics". --Steve (talk) 17:50, 15 April 2008 (UTC)

To beat this horse to death: you need both the Maxwell-Faraday equation and the Lorentz force law to come up with Faraday's law of induction because you cannot relate EMF (which is work) to the fields without a force law. Therefore, regardless of any of these arguments, the Maxwell-Faraday equation by itself cannot possibly be the same as Faraday's law of induction. As support for this simple logic, see the Feyman quote in Faraday paradox where Feynman says specifically that we need both the Lorentz force law and the Maxwell-Faraday equation.

With this logical basis in place, George's argument could be supported to say that "There is only one Faraday's law" and supplemented with "and the Maxwell-Faraday equation is not it."

Assuming all that can be placed behind us, George has some additional discussion about the importance of the magnetic force component of the Lorentz force law in combination with the Maxwell-Faraday equation when one sets the goal of deriving Faraday's law. That discussion seems separate from the terminology issue, as Steve suggests, because there is no debate over Faraday's law being distinct from the Maxwell-Faraday equation, in just the same way that a whole is different from each one of its parts.

Brews ohare (talk) 22:41, 15 April 2008 (UTC)

[edit] Revision of Intro

Supposing we were to add a bit to the introduction for the purposes of clarity. After having explained both the aspects of EM induction that are inherent in Faraday's law, supposing we then introduce the Maxwell-Faraday law and draw attention to the fact that this well known law from Maxwell's equations, which is also referred to as Faraday's law, only actually caters for one aspect of Faradays law. It does not cater for the convective vXB aspect.

That would be a way of presenting all the important facts in the introduction in a coherent manner. George Smyth XI (talk) 06:49, 16 April 2008 (UTC)

Hi George: I don't like the idea of putting a controversy up front with inadequate space in the introductory few lines to straighten things out. To explain about v × B in the intro is just too complicated, and would drive the reader away muttering about "damn technical gobbledygook".

As the intro stands, it does a good job of presenting what you yourself would call "Faraday's law". Later, under the "Maxwell-Faraday equation" the point you want to raise is mentioned. If there is tinkering to be done about the terminology issue, this is where I'd see it done. Brews ohare (talk) 15:38, 16 April 2008 (UTC)

I tried a compromise. George's preferred use of the term "Faraday's law" is the one that's privileged right at the top as the standard usage. I still think this violates WP:NPOV to some extent, but I can live with it. The Maxwell-Faraday equation is given at the end of the introductory section, where it's a described as a law that many physicists call "Faraday's law". It's also stated that the Maxwell-Faraday equation doesn't attempt to explain motional EMF. So it's still easy enough for a reader to learn that the term "Faraday's law", as it's used by physicists, has some ambiguity, but they have to read the whole introduction, not just the first sentence...which again I'm not a huge fan of, but I can live with it. It's also, I think, not overly technical. So, everyone happy? --Steve (talk) 17:31, 16 April 2008 (UTC)

I think it's quite good now. I was originally supporting the point that Brews has just made that we shouldn't put too much controversy in the introduction.

But on the other hand, as in the case of Maxwell's equations, when slight variations in format do exist, and do generate alot of discussion, then there is no harm in having a clarifying paragraph at the end of the introduction.

There is only one Faraday's law, but Maxwell derived it in a unique way in his 1861 paper with the curl function. (Maxwell of course used a cumbersome cartesian notation).

The curl presentation is to the best of my knowledge Maxwell's own.

Heaviside then uses it but specifically with partial time derivatives and hence precludes the vXB effect. I know that Brews didn't want a discussion of the vXB effect in the introduction but I think that Steve has got around that by simply stating in English that the Maxwell-Faraday equation doesn't cater for motionally induced EMF.

It's a further matter of interest to what extent Maxwell himself was treating his equation (54) as a partial time derivative. He actually uses the total time derivative symbols and the preceding equations indicate that he is fully aware of the implications. But the evidence tends to suggest that he is only interested in the partial time derivative effect.

If we move on down towards equation (77) where he develops the vXB effect, we note that he specifically mentions that the dA/dt terms are to refer to a fixed point in space. Hence, although he uses the total time derivative symbols in equation (77), he clearly means them to represent partial time derivatives. As for equation (54), this is not so clear.

The only thing we know for sure is that Maxwell developed Faraday's law into the curl form, and Heaviside unequivocally precluded the vXB effect by specifically making it a partial time derivative.

So on reflection, I am now in favour of a special mention of this Maxwell-Heaviside- Faraday law at the end of the introduction.

What I was opposed to was the idea of introducing Faraday's law as a term ambiguously applied to two different laws in electromagnetism. George Smyth XI (talk) 01:18, 17 April 2008 (UTC)

Well George, I would go with Steve's last version, but do not like your recent changes. They are of historical interest, but in the intro they are simply confusing, as a reader is likely to think the straight derivative form you attribute to Maxwell is still in use as accepted practice. That is misleading. It should be removed. You should follow the adage "half a loaf is better than no bread" and go back to Steve's version. 4.240.72.26 (talk) 05:28, 17 April 2008 (UTC)

[edit] The equation $\nabla \times \mathbf{E} = - {{d\mathbf{B}} \over dt}$

...as a law of physics that incorporates motional EMF, is original research by George. I've certainly never seen it in any textbook. It comes, I think from the fact that he's been reading his Maxwell, and gotten used to Maxwell's outdated definition of E. To be specific, Maxwell wrote the equation

(E3) $\mathbf{E} = \mathbf{v} \times \mathbf{B} - \frac{\partial\mathbf{A}}{\partial t}-\nabla \phi$

which would presumably be complemented by

(E4) $\mathbf{F} = q\mathbf{E}$

whereas modern writers would write

(E5) $\mathbf{E} = - \frac{\partial\mathbf{A}}{\partial t}-\nabla \phi$

(E6) $\mathbf{F} = q\mathbf{E} + \mathbf{v}\times\mathbf{B}$

Obviously, (E5) and (E3) cannot both be simultaneously true, and nor can (E4) and (E6). But while every textbook writes (E5) and (E6), the equation (E3) is in no modern textbook, because it's false according to modern definitions; likewise with (E4) for nonzero velocity.

According to Maxwell's definition of E, George's equation and the accompanying text make perfect sense. According to the modern one, it's manifestly incorrect. curl E is a vector field, independent of v, so cannot equal both the partial derivative of the vector field B and the convective total derivative of B simultaneously.

And once again, George, please, please read WP:NOR. Wikipedia is emphatically clear that original research is forbidden, no matter how obvious you think it is. --Steve (talk) 03:18, 17 April 2008 (UTC)

Steve, the equation $\nabla \times \mathbf{E} = - {{d\mathbf{B}} \over dt}$ is not original research. It is equation (54) in Maxwell's 1861 paper. It is the curl of equation (77) ( $\mathbf{E} = \mathbf{v} \times \mathbf{B} - \frac{\partial\mathbf{A}}{\partial t}-\nabla \phi$ ) in the same paper. It doesn't contradict anything in the modern textbooks.

You have been reading too much Feynman and not enough Maxwell.

OK, so how do we resolve the legitimate problem that the anonymous has raised? He has a point. Maxwell's equation (54) is not seen in modern textbooks in total time derivative format, and it may confuse the readers.

We could of course put in the partial time derivative version by Heaviside and explain it all. But that gets back to the original issue of not putting too much detail in the introduction.

We can either drop the subject in the introduction, or we can try and use the better known Heaviside partial time derivative in the introduction alongside a coherent, simple and factually correct explanation.

I'll try to do that now and see how it goes. George Smyth XI (talk) 09:44, 17 April 2008 (UTC)

[edit] Is there something missing in Faraday's law or just in Wikipedia's explanation?

Faraday's law of induction is confusing because for one matter it is incomplete. For another the way “magnetic flux linked to the circuit” is undefined. For a third it is by no means responsible for the working of all electrical generators. Aren’t thermocouples or some piezo crystals electrical generators?

A permanent magnet could move all day in the vicinity of a stationary conductor without inducing the slightest EMF in that stationary conductor. Inducing EMF in a conductor depends on the relative direction of that movement. The following experiment should demonstrate this:

Consider two permanent magnets the north pole of one parallel to and facing the south pole of the other. The magnetic flux existing between these two surfaces is among other constant parameters dependent on the distance between these two surfaces. Reducing this distance would certainly change the magnetic flux thus satisfying the requirements of Faraday's law. Should there be a stationary conductor just in the middle of this changing field no EMF is induced in it thus disproving to Faraday’s law.

The contradiction is probably due to the second point of confusion namely just how is the “magnetic flux linked to the circuit”.

AdrianAbel (talk) 08:12, 29 April 2008 (UTC)

I would say "magnetic flux through the circuit", which is perfectly unambiguous in the case that the circuit is a loop of thin wire, which is really the only case that (this version of) Faraday's law should be said to apply to. I agree that "linking" isn't the right word. You're right about "electrical generators"...is there a more specific term for "electrical generators involving spinning things and magnets and wires"? Or, it could just say "most electrical generators".

through the circuit" is correct.

AdrianAbel (talk) 17:52, 1 May 2008 (UTC)

In the "experiment", could you specify what you mean by "conductor" (A loop of wire? A flat planar conducting sheet?) and how you know that "no EMF is induced"? I'd suspect there would be an EMF, but I couldn't say for sure unless you're more specific. Thanks!! :-) --Steve (talk) 16:35, 29 April 2008 (UTC)

A loop of electrical conducting wire such as a copper wire connected at its ends through a resistor of finite resistance. The loop fits through a hole just in the middle of the magnets. Any EMF generated in this wire while moving the magnets closer and farther away from each other must be measurable across the resistor with a voltmeter. Is that specific enough or would you rather see a drawing?

AdrianAbel (talk) 17:52, 1 May 2008 (UTC)

It is a lot to ask that a short definition cover all nuances of "linking". The section Flux through a surface and EMF around a loop attempts to fill in some details. Later examples, I believe, make the notion very clear. The treatment of surfaces that vary in time, as happens in the drum generator or the Faraday disc, require some extended discussion. I'd suggest that the definition is correct, but it does require the later discussion.

As for generators: any device that converts one form of energy to another is a generator. I've revised this phrase - take a look. Brews ohare (talk) 18:15, 29 April 2008 (UTC)

The current "many forms of electrical generators" is also an improvement. And for a definition the above suggested "through" the circuit is considerably more appropriate than a word like "linking" with so many nuances.

AdrianAbel (talk) 17:52, 1 May 2008 (UTC)

Hi Adrian: The word "through" is in common use with a huge variety of meanings. I suspect it sounds better just because the reader can imagine whatever meaning they like best, while the unusual word "linking" forces the reader to ask "What does that mean?" Unfortunately, the concept is not straightforward, so the latter reaction is closer to appropriate, I believe. In any case, a lot of examples are necessary before it sinks in, whatever word you choose. Brews ohare (talk) 02:33, 2 May 2008 (UTC)

Probably so. But I consider changing the original wording from Faraday from through to linking as a crime. The perpetrator is probably Saiduko. I would appreciate your correcting it in Wikipedia. If it's not done within a few weeks I'll change it myself and give more reliable sources.

AdrianAbel (talk) 11:42, 2 May 2008 (UTC)

Hi Adrian: I'm not at all clear about the "crime" involved here. Searching the term flux linkage with the qualifier Faraday produces 3330 hits (303 on Google books). However flux through + Faraday while requesting no linkage or link produces 13,000 hits (629 on Google books). So linkage is in the minority, but it's pretty common, especially with authors/educators.

It’s a crime because Wikipedia's article is about Faraday's Law of Induction and Farady never used such an expression as “linking”. Also what Wikipedia proclaims the law states, Faraday never stated. Depending upon the interpretation of the word “linked” it could even negate his law entirely as my above experiment would show.

If you’re still not convinced consider the following. If I would give you two open bicycle chains and a master link and ask you to link them together you could probably manage it in a jiffy. Then I would hand you a permanent magnet and ask you to now link the bicycle chains to the magnetic lines of force emanating from it. After first deciding that I’m not crazy you’d probably wonder where to start. In other words "Linking" in the classical sense means to or within a chain with identical or similar members. Internet links providers, for example.

Google.com found about 64 hits on the key “Faraday’s law of induction”. Many were repeats or childish. A single one used the term “linking” who’s author I assume got it from Saiduko or vice versa. If Saiduku’s work was written originally in Japanese the error could well be in the translation to English. If his book wasn’t so expensive I would check that. I could give you about 10 others references found by Google.com that use exclusively the term “through”. But it’s not really necessary. Farady’s original findings in “Experimental researches in electricity” are available in copy for 50$ or 35£. If you should have access to the Library of Congress there’s a copy in their rare books department in the Franklin street That library has not yet put it on line. Excerpts of it are available from other sources on line. In reading them you’ll find how meticulously Faryday recorded his findings. He uses the word “through” and "cutting" often as well as “introduce” and “insert”. It’s so clear that anyone could repeat it at any time.

My suggestion is to exchange the word “linked” to “cutting through”. Then add the sentence: This law is derived from Faraday’s observations first reported in 1831 in his “Experimental researches in electricity”. If you want to leave a reference to Henry, no objections. But please remove the link to Saiduko so that that chain of confusion gets cut off.

AdrianAbel (talk) 17:30, 3 May 2008 (UTC)

In this connection, I've found reference to "Neumann's law" of 1845 (apparently Faraday's law often is referred to as the Faraday-Neumann law): When the magnetic flux linked with a coil (or circuit) is changed in any manner, then an emf is set up in the circuit such that it (the emf) is proportional to the rate of change of the flux-linkage with the circuit. Electro-Magnetism: Theory and Applications -If this statement is historically accurate, there appears to be a long tradition of "flux linkage", eh? Care to advance why you think use of linkage is a "crime" and an "error"? ;-)

Hello Brews. The use of "linkage" in Neumann's law in 1845 surprises me. It could well have been written originally in German and mistakenly translated. I see a lot of such translation errors. But in any case he takes a paragraph and a drawing to define exactly what he means. Wikipedia does not.

AdrianAbel (talk) 17:30, 3 May 2008 (UTC)

Do you have a suitable exact quote stating the law that can go in the quotation box with a "more reliable" attribution? Brews ohare (talk) 13:36, 2 May 2008 (UTC)

No. Not even Faraday has one. He just reported his observations. I didn't even see an equation in his report. But if you just change it as I suggested it would be a big improvement.

AdrianAbel (talk) 17:30, 3 May 2008 (UTC)

OK, I've eliminated linking. Brews ohare (talk) 20:02, 3 May 2008 (UTC)

Thanks. AdrianAbel (talk) 16:29, 6 May 2008 (UTC)

[edit] What is EMF?

The stated law says: "The induced electromagnetic force or EMF in any closed circuit ..." Shouldn't this be "electromotive force" (i.e. volts, not newtons)? So the link is wrong too.

I wanted someone knowledgeable to agree and change it so I'm leaving it to you guys, the maintainers of the page. Dan Morris, 14:28, 2 May 2008 (UTC)

Quite a gaff, eh? Thanks. Brews ohare (talk) 20:01, 3 May 2008 (UTC)

[edit] Resistive Force

Since the article only addresses closed circuits, I was wondering what the equation for the resistive force of the eddy currents would be for just a piece of metal moving through a magnetic field (like an electromagnetic brake). Any ideas?

86.140.130.100 (talk) 20:11, 7 May 2008 (UTC) James T.

I'm not sure that there's a simple and exact formula; my impression was that the microscopic flows of current could be arbitrarily complicated. I don't know for sure. Anyway, whatever formulas exist (and I'm sure they do, at least approximately), would presumably belong not here, but rather in the article: Eddy current. (At least primarily.) :-) --Steve (talk) 22:53, 7 May 2008 (UTC)

Talk:Faraday's law of induction

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Contents

[edit] a good writer

[edit] ____

[edit] Change to derivative

[edit] English

[edit] Use of English

[edit] Image copyvio?

[edit] help with science classwork

[edit] + or -

[edit] One or Two Faraday's Laws?

[edit] Faraday's law debate

[edit] Partial derivatives

[edit] Derivations

[edit] A different example

[edit] Moving wire-loop example

[edit] Lorentz force law method

[edit] Faraday's law method

[edit] Moving capacitor example

[edit] Lorentz force law method

[edit] Faraday's law method

[edit] Discussion

[edit] Relevance of relativity

[edit] The Two Forms of Faraday's law

[edit] The Introduction

[edit] Faraday's law as two different phenomena

[edit] One Faraday's law, two effects

[edit] RfC: One Faraday's law or two?

[edit] The Introduction

[edit] Let's be more specific?

[edit] Quotes by Feynman

[edit] The root of this problem lies in the current state of ignorance

[edit] Revision of Intro

[edit] The equation

[edit] Is there something missing in Faraday's law or just in Wikipedia's explanation?

[edit] What is EMF?

[edit] Resistive Force

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[edit] The equation $\nabla \times \mathbf{E} = - {{d\mathbf{B}} \over dt}$