Talk:Electromotive force/Archive 1

Emf
''Emf is often used as a synonym for potential difference, where the potential difference is induced by an energy source such as a battery. This usage is considered obsolete.''

I thought that it is not the same because of the internal resistance in the battery, and that therefore the concept of emf is still needed for a battery. - Patrick 12:09 Feb 4, 2003 (UTC)

What's obsolete?
I believe the expression electromotive force in words is obsolete, but the expression emf in letters is not. The problem is the word force &mdash; an emf is not a force, so it should not be called one. Thus I recommend moving this article to EMF (voltage), and writing "The expression EMF used to stand for electromotive force, but the term electromotive force is now obsolete, and EMF is a substitute for it."

Any comments?

Fg2 06:20, Sep 5, 2004 (UTC)


 * The term "electromotive force" is still used in the latest edition of Jackson's Classical Electrodynamics, and a literature search of the last 14 years finds at least 250 journal articles that use the term in their titles and/or abstracts. So, I don't think there is any evidence to support the argument that the term is "obsolete" in the sense of "no longer used" or even "no longer used by professionals".  Yes, it is a bit of a misnomer, but these things happen in language. &mdash;Steven G. Johnson 19:14, Sep 5, 2004 (UTC)

Further editing
I get confused with EMF -- is it the voltage based on the internal resistance of the cell, or is the sum total voltage which is required to support the circuit? Tanmay Altered the wording and also added a physical analogy for voltage as a force. Edit was done by me ElBarto as 203.79.121.174.


 * EMF is not based on internal resistance. It is equal in magnitude and opposite in direction to the open circuit potential difference of a source.--Light current 20:58, 27 December 2005 (UTC)

The electrons flow out of the positive terminal into the negative terminal. This is because Ben Franklin guessed wrong when researching this. So as a convention (to stay consistent), we say electrons flow from positive. This is Pasive Sign Convention.


 * I don't think that's quite right. Current flows out of the positive terminal and into the negative terminal, but current was defined by Franklin as the direction that positive particles of electricity would flow, before anyone knew that it was actually the electrons, or negative particles, that flow in metal conductors, whereas the protons, or positive particles, are at fixed locations.  So in fact, electrons flow out of the negative terminal and into the positive terminal, but since current is defined as positive charge flow, the direction of conventional current is just the opposite.


 * The mistake that Franklin made was not in choosing the direction of conventional current flow, it was in not understanding that it is negative charges that flow, whereas the positive charges are fixed in the solid structure of the metal. Of course, there was no way that he could have known at the time, since he lived and worked 100 years before Maxwell derived a complete theory of electromagnetism, 125 years before Thompson discovered the electron, and 150 years before Rutherford probed the structure of the atom and discovered the positively-charged nucleus.


 * But, in a twist of historical irony, Franklin's choice of showing current in the direction of positive electrical flow turned out, in the age of semiconductors, more correct than people often realize. In semiconductors, the total current results from the flow of both electrons and mobile holes, which represent the absence of an electron in the crystal lattice and which behave just like positive particles of electricity.  So the direction of current flow, which is the sum of the contributions from both electrons and holes, is the same direction as the flow of the holes, and opposite to the flow of the electrons.


 * We are actually very fortunate that Franklin made this "mistake" and defined current as flowing in the positive direction, because it makes electrical circuits much easier to understand than if he had chosen the opposite convention. It also eliminates the need for minus signs in all of the circuit equations and laws that govern electrical phenomena.  The only hard part is to learn and remember that the direction of current flow is not the direction of electron flow, but rather the direction of positive charge flow.


 * -- Metacomet 17:24, 25 February 2006 (UTC)

EMF or emf?
Silly people who abbreviate things in lowercase. :-) mph, rms, emf.  When will it end??

I know there's some convention for this, but it doesn't make sense to me. Is there anything in the manual of style about this? - Omegatron 15:38, August 9, 2005 (UTC)


 * Actually, it should be e.m.f. as in the old books!--Light current 00:00, 27 February 2006 (UTC)

Symbol
Moved here for Stevenj's talk page for a more general discussion: "Nothing, but there is no reason to put it in a large font. In any case, the most common symbol I've seen is a script \mathcal{E}, not a Roman-font E. —Steven G. Johnson". The symbol I added is a script E, and the Unicode character for EMF. If it's showing up as a normal E, then it's something to do with the fonts you have installed on your computer. porges 04:40, 12 October 2005 (UTC)

Can emf be measured directly? Or is it the pd you're measuring?
Can emf be measured directly, or must it always be inferred from the pd presented at the terminals? If it must be inferred, when people are taught about measuring various emfs with a voltmeter or potentiometer etc., they are being given slightly the wrong story. Yes?--Light current 02:50, 26 December 2005 (UTC)

Does nobody know the difference between emf and pd?--Light current 05:49, 25 February 2006 (UTC)


 * I am not 100 percent sure, but I think the only difference is a negative sign, and the fact that emf is a bit anachronistic as a terminology. I am not sure why it is necessary or useful, and IMHO, it is easier to use the more precise and well-defined concepts of voltage, potential difference, and electric potential depending on the context.  -- Metacomet 19:22, 25 February 2006 (UTC)

Unfortunately, many of the electromagnetic principles (ie Lenz law) use the term 'emf' but the people who discovered these principles could not have measured the emf they were talking about!--Light current 03:11, 26 February 2006 (UTC)


 * I am confused. What do you mean when you suggest the emf cannot be measured?  Isn't emf simply the voltage that shows up on my portable multi-tester when I place it across two terminals (without electrocuting myself!)?  Or actually, times -1 to change the sign?  Or is emf more complicated than that? -- Metacomet 04:52, 26 February 2006 (UTC)

Youre measuring the opposing pd with your meter- not the actual emf (theyre the same value of course but opposite polarity) See my first post under this heading.--Light current 04:55, 26 February 2006 (UTC)


 * And if I disconnect the two leads and connect them to the opposite terminals, my multitester will read the negative value of the pd, which is the... I am not trying to be difficult, but I just don't see any difference between emf and pd except for the minus sign, which is actually arbitrary in both cases anyway (since it depends on which way you decide to hook up the terminals, and which terminal you choose to call '+'.  -- Metacomet 05:00, 26 February 2006 (UTC)

Fortunately, the answer is in the article:

Distinction between emf and p.d.
''If an external circuit is not connected to a source of emf, an electric current cannot exist. Thus, between the terminals of the source, there must exist an electric field that exactly cancels the generated emf. The source of this field is the electric charges separated by the mechanism generating the emf. For example, the chemical reaction in the battery proceeds only to the point that the electric field between the separated charges is strong enough to stop the reaction. This electric field between the terminals of the battery creates an electric potential difference that can be measured with a voltmeter. The polarity of this measured pd is always opposite to that of the generated emf. The value of the emf for the battery (or other source) is the value of this 'open circuit' voltage. emf itself cannot be measured directly.''
 * --Light current 05:04, 26 February 2006 (UTC)


 * I guess we will simply have to agree to disagree. If I connect the red wire of my voltmeter to the + terminal of a 9V battery, and the black wire to the the – terminal of the same battery, the meter will read +8.92 Volts.  But if I then connect the red wire to the – terminal and the black wire to the + terminal, the meter reads – 8.92 Volts.  So in one case, haven't I measured the pd, and in the other case the emf? -- Metacomet 05:20, 26 February 2006 (UTC)

Not so fast! ;-) If I were to say that emf is a cause and pd is an effect, does that help any? You always need a source of emf in any circuit, but you can only measure its effect as a pd - yes?--Light current 05:23, 26 February 2006 (UTC)


 * You make an interesting point, and I will have to think about it a bit. I am not ready to throw in the towel (or my multi-tester) just yet, but you have got me thinking, which is always a dangerous thing (especially when smoke starts to rise from the top of my head).  -- Metacomet 05:30, 26 February 2006 (UTC)

Good. I'll come back tomorrow when youve had some time!--Light current 05:31, 26 February 2006 (UTC)


 * ...and some sleep. Which is often when I do my best thinking.       ;-)        -- Metacomet 05:34, 26 February 2006 (UTC)


 * I've thought about it, and I've slept on it, and I still don't see the difference. Sorry, Light Current.  The cause-and-effect argument is interesting, but I don't think it really amounts to a difference.  Fundamentally, the source of the electric field is the charge distribution, and the electric potential is a measure of the potential energy per unit charge of the arrangement of charge density.  And the potential energy is simply a measure of the energy required to arrange the charge into its current distribution from some arbitrarily chosen reference distribution, often chosen such that all charge is at infinite distance.  The emf, based on how it is defined, is exactly the same as electric potential, except that the convention chosen introduces a minus sign.  -- Metacomet 18:26, 26 February 2006 (UTC)


 * Also, although you have made an interesting argument in terms of cause-and-effect, you have not really responded to my thought experiment involving the voltmeter and 9V battery. Doesn't that example show that it is possible to measure emf directly, and that it is nothing more than the negative of pd?  -- Metacomet 18:26, 26 February 2006 (UTC)

Well no I dont think so! emf is the fundamental cause of any voltage or pd becuase it is a source of energy. A battery generates an emf, but you can only measure its pd. A generator generates an emf, but we measure its pd. Dissimilar metals in contact cause emfs, but again we only measure their pd.

To create any voltage difference (pd) in a closed circuit, there MUST be a source of emf somewhere. Take a resistive circuit with a battery driving it. The battery is commonly called a source of emf. The voltages across the resistances that you measure are called pds. This is because a pd is not a source of energy and arise purely due to the currents flowing thro the resistors. The point is that the pd (which is what you're measuring) only appears as a result of the emf and the load resistance. If there were no pd developed (say because the resistor was zero) to balance the emf, infinite current would flow from the source of emf. Is any of this making sense?--Light current 18:48, 26 February 2006 (UTC)

THis is why I find it so interesting that he early scientists came up with this idea of emf when they must have only been able to measure pd.


 * I don't agree with your analysis. A few thougths:


 * 1. Go back to Maxwell's Equations.  Electrical circuits, batteries, and resistors all must obey Maxwell's Equations.  Kirchoff's Laws, Ohm's Law, and even Coulomb's Law can all be derived as special cases of Maxwell's Equations.  The "source" of emf is nothing more than a separation of positive and negative charges, which is an arrangement of charges in space.  This charge distribution causes an electric field to exist between the separated charges, and as long as there are no time-varying magnetic fields nearby, then the electric field can be represented by a scaler potential function, and a potential difference can be defined between any two points in space, including nodes in an electric circuit.  Whether you choose to call it emf or pd, it is still the same thing.

Whilst I agree with most of your logic here, but, because emf must be found by integrating around a closed loop, it cannot form a scalar field. I think what youre describing above is potential or potential difference. Maybe youre talking about electrostatic emf?--Light current 23:12, 26 February 2006 (UTC)


 * 2. A battery is not always a source of emf.  Some batteries are rechargeable.  So if I reverse the flow, now the source of emf is a consumer of emf.

Agreed. I didnt say it was always a net source of emf, but it will still have and emf in oppostion to the charging emf! I dont think theres any thing that consumes emf.--Light current 23:17, 26 February 2006 (UTC)


 * 3. Read the comments that Stevenj wrote in the new section he added below.  In some treatments, emf is used to describe something entirely different, which involves time-varying magnetic fields.

THis may be the source(!) of the problem in confusing electrostatic emf with that generated by mag fields etc! Prof Stephen J may well be correct again!--Light current 23:17, 26 February 2006 (UTC)


 * -- Metacomet 19:48, 26 February 2006 (UTC)


 * BTW, you have said that emf is a source of energy whereas pd is not a source of energy. Well, as you know, in a thuderstorm, there are large potential differences that arise between the thunderhead and the Earth.  If potential difference is not a source of energy, please try to explain that to anyone who has ever been struck by lightning!  -- Metacomet 19:56, 26 February 2006 (UTC)

No I thought of that one. Thats why I said closed circuit!! :-))--Light current 22:57, 26 February 2006 (UTC) The answer is of course that a thunder cloud is a charged capacitor and all charged capacitors are sources of emf!! No problemo! :-)--Light current 23:06, 26 February 2006 (UTC)

Maxwell's use of the term electromotive force

 * It is also interesting to note that, according to the WP article, Maxwell used the term electromotive force to refer to what we would today call the electric field, and not what we would call voltage or potential. Since the electric field is defined as the force per unit charge acting at a point in space on a positive test charge, then Maxwell's is actually using the term force more or less in the correct sense.  Electric potential, on the other hand, is really nothing but potential energy per unit charge, and so using the term electromotive force to represent potential energy is a bit of a misnomer.  -- Metacomet 19:28, 25 February 2006 (UTC)


 * Note that in terms of units of measure, electric field is volts per meter or equivalently newtons per coulomb. And from the Lorentz force law, we have:


 * $$ \mathbf{E} = q\mathbf{F} $$


 * So again, Maxwell's use of electromotive force to represent what we call the electric field makes good sense.


 * -- Metacomet 19:40, 25 February 2006 (UTC)

disputed
I don't think that the sign convention has anything to do with the difference between electomotive force and potential difference. Potential difference is, strictly, the line integral of electric field, $$\int \mathbf{E}\cdot d\mathbf{s}$$. This is not well-defined in a circuit with a time-varying magnetic field, because it is path-dependent. Emf, as I understand the term as it is used nowadays, is simply a generalization of potential difference to handle such cases, by including a "fictitious" potential drop over things like ideal inductors, in exactly the right amount to make Kirchhoff's voltage law valid.

I'll double-check a reference book when I get a chance, but meanwhile I'm adding the disputed tag because I'm dubious about the current text. Even if it is commonly used simply to mean the driving voltage(s) in a circuit, which is what the current text seems to suggest, I don't recall seeing a consistent difference in the sign convention.

—Steven G. Johnson 18:22, 26 February 2006 (UTC)

(PS. Sorry about adding the disputed tag yesterday without explanation. WP gave me an error message and I thought that the edit hadn't occurred.)


 * That's what I was afraid of. If the difference between pd and emf were nothing more than sign convention, then my argument is that there is no difference at all.  But if, as you point out, the difference relates to the presence or lack of time-varying magnetic fields, then there is a very major difference indeed.  Thank you for jumping into the fray, so to speak.   -- Metacomet 18:46, 26 February 2006 (UTC)


 * I quote from Principles of Applied Electricity by AM Howatson, pub Chapman and Hall 1969:


 * ''The line integral of E around a closed path (which is zero as far as electrostaic E is concerned) is the electromotive force (e.m.f) in that path.
 * This is true when it is due to chemical action, as in a battery, or thermal, as in a thermocouple, or any other. In all cases, the e.m.f. is that agent which enables current to flow in a closed circuit. Without it there can be no steady current. We can write in general,


 * $$ \mathcal{E} = \oint_C \mathbf{E} \cdot d\mathbf{l} $$ (line int around a closed path)


 * where e is any source of emf.


 * E.m.f. being a line integral of E, is measured in volts but unlike V it cannot be given a value at any point and so cannot represent a scalar field.''


 * It should be noted that potential difference is a scalar quantity, wheras emf is not. Herin lies another major difference between emf and pd.--Light current 20:37, 26 February 2006 (UTC)

Formal definitions of emf and electric potential
A couple of things:

1. The line integral of a vector field is by definition a scalar. The issue is that you cannot define a unique scalar field at all points in space because the line integral of the electric field is not in general path independent. But the integral itself is definitely a scalar. It's just not a unique function of space.

2. I think we need to agree on the definitions of both emf and electric potential before we proceed.

3. According to the reference from which you quoted, the definition of emf is:


 * $$ \mathcal{E} = \oint_C \mathbf{E} \cdot d\mathbf{l} $$

Furthermore, this definition is valid under all circumstances, regardless of whether there is a time-varying magnetic field or not.


 * Comment: this definition cannot be right.  If it is, that means that the emf in most electrical circuits, in the absence of a time-varying magnetic field, is zero.


 * This is only the emf due to varying mag fields, and fails to take account of the electrostatic e.m.f. from batteries etc! In my book, it implies that e.m.f is only created by induction. In the case of static fields, emf as such is not really mentioned.


 * Here any voltage difference is given by
 * $$ \mathcal{V}_a^b = -\int_a^b\mathbf{E} \cdot d\mathbf{l} $$

--Light current 00:22, 27 February 2006 (UTC)

That's fine, as long as you understand that a voltage difference, even in an electrical circuit, is identical to a potential difference between the two points in space which coincide with the two nodes in the circuit. In other words,

$$ \mathcal{V}_a^b = \mathcal{V}_b - \mathcal{V}_a = \Phi(b) - \Phi(a) = \Phi_a^b $$

In fact, that is arguably a pretty good definition for the voltage drop between two nodes in a circuit. Again, just because we are talking about an electric circuit doesn't mean we cannot use Maxwell's Equations. In fact, Maxwell's Equations are completely general, and apply in any situation. Electrical circuis are simply a special case.


 * -- Metacomet 02:01, 27 February 2006 (UTC)


 * Yes, well that seems sensible to me unless Im missing something important. We have to be careful in trying to combine circuit theory and field theory tho'! Gotta watch difference between dc and ac!--Light current 02:41, 27 February 2006 (UTC)

4. Electric potential, on the other hand, is defined if and only if there is no time-varying magnetic field passing through the surface enclosed by Contour C, or if the time rate of change of the magnetic field is small enough so that its impact is insignificant. Under these limited conditions, we can say that the curl of the electric field is zero:


 * $$ \nabla \times \mathbf{E} = 0 \qquad \qquad {\partial \mathbf{B} \over \partial t} = 0$$
 * OR
 * $$ \nabla \times \mathbf{E} \approx 0 \qquad \qquad \bigg| {\partial \mathbf{B} \over \partial t} \bigg| << \bigg| { \mathbf{B} \over \tau } \bigg| $$


 * where $$ \tau = {d \over c} $$ and d is the maximum distance that a signal would have to propagate in the physical structure. These restrictions are equivalent to saying:


 * $$ \omega << {1 \over \tau} $$  or $$d << { \lambda \over 2 \pi } $$


 * Under these conditions, and only under these conditions, we can define a unique scalar function, called the electric potential, such that the electric field is the negative of the gradient of the electric potential:


 * $$\mathbf{E} = - \nabla \Phi $$

5. Using vector calculus, the last equation is equivalent to:


 * $$ \Phi(b) - \Phi(a) = - \int_a^b \mathbf{E} \cdot d\mathbf{l} $$


 * for any two points a and b. But again, this electric potential difference is defined if and only if there is no time-varying magnetic field passing through the region of space containing points a and b.

6. This last equation also provides the basis for Kirchoff's voltage law:


 * $$\oint_C \mathbf{E} \cdot d \mathbf{l} = 0 $$


 * with the condition, again, that there are no time-varying magnetic fields passing through the surface enclosed by Contour C, or that any such magnetic field is varying slowly enough that its effect is insignificant.

-- Metacomet 22:52, 26 February 2006 (UTC)

A more general approach
From Gauss's law for magnetism, we have:


 * $$ \nabla \cdot \mathbf{B} = 0 $$

From vector calculus, there is an identity which states:


 * The divergence of the curl of any vector field A is always zero:
 * $$\nabla \cdot ( \nabla \times \mathbf{A} ) = 0 $$

Thus, we can define a vector field that is uniquely defined at all points in space such that the curl of this so-called vector potential A is equal to the magnetic flux density B:


 * $$ \mathbf{B} = \nabla \times \mathbf{A}$$

or equivalently,


 * $$\int_S \mathbf{B} \cdot d \mathbf{a} =   \oint_C \mathbf{A} \cdot d \mathbf{l} $$

From Faraday's law, we have:


 * $$\nabla \times \mathbf{E} = - {\partial \mathbf{B} \over \partial t}$$

Combining the last equation with the differential form of the prior equation, and simplifying, we have:


 * $$\mathbf{E} = - \nabla \Phi - {\partial \mathbf{A} \over \partial t}$$

where &Phi; is the electric potential, as before, except that we are now allowing time-varying magnetic fields in this region.

If we now take the definition of emf, as cited above, we have
 * $$ \mathcal{E} = \oint_C \mathbf{E} \cdot d\mathbf{l} = - \oint_C { \partial \mathbf{A} \over \partial t } \cdot d\mathbf{l} = - {d \over dt } \int_S \mathbf{B} \cdot d \mathbf{a} = - { d \Phi_B \over dt }$$


 * Comment: So here again, we have a problem, because this last equation suggests that the emf is always zero in the absence of a time-varying magnetic field.

-- Metacomet 23:31, 26 February 2006 (UTC)


 * Just goes to show you cant always believe the predictions of vector calculus! I find common sense is far more reliable.--Light current 00:04, 27 February 2006 (UTC)

Just as long as common sense doesn't become conventional wisdom, and then evolves into common nonesense! When so-called common sense fails to provide a meaningful answer, I always approach the problem from two perspectives: rigorous analysis based on first principles and valid mathematics, and physical intuition. Both of which are telling me that there is a problem with the way you are defining emf. -- Metacomet 00:09, 27 February 2006 (UTC)

I believe that a better and more general definition of emf would be:


 * $$ \mathcal{E}(b) - \mathcal{E}(a) = \int_a^b \mathbf{E} \cdot d\mathbf{l} $$

-- Metacomet 00:13, 27 February 2006 (UTC)
 * Yes I think you've just defined potential difference here (but not emf)--Light current 00:18, 27 February 2006 (UTC)


 * I must admit my 'definitions' were confined originally to circuit theory. If we are going into field theory then things may be more complex.--Light current 00:16, 27 February 2006 (UTC)

Bottom-line
The way I see it, it all boils down to just three issues:

1. What is the definition of electric potential?


 * The potential difference between two points is $$\int \mathbf{E} \cdot d\mathbf{s}$$ between those two points; this is path-dependent if you have time-varying magnetic fields. This is somewhat distinct from the definition of scalar potential &phi; &mdash; when you have time-varying fields, you can still define a &phi; but it depends on the gauge choice for the vector potential A and is decoupled from the potential difference (odd as that sounds): $$\int_a^b \mathbf{E} \cdot d\mathbf{s} \neq \phi(a) - \phi(b)$$ in electrodynamics, where $$\mathbf{E} = -\nabla\phi-\partial\mathbf{A}/\partial t$$. I don't think this stuff is clearly described in Wikipedia, probably because you rarely see vector potentials in undergraduate-level courses. —Steven G. Johnson

2. What is the definition of emf?


 * See below.

3. Are we assuming that the system is dynamic, electro-quasi-static (EQS), or fully electrostatic (FES)?


 * emf is defined in dynamic circuits as well as static circuits, see below.

After we resolve those three issues, then we will need to address two others:

4. Are emf and electric potential the same thing or different things?


 * Different, but related; see below.

5. If different, what is the relationship, if any, between emf and electric potential?

-- Metacomet 02:16, 27 February 2006 (UTC)


 * I totally agree!--Light current 02:25, 27 February 2006 (UTC)

Awesome! I wasn't sure that we were ever going to agree about anything. ;-)   -- Metacomet 02:31, 27 February 2006 (UTC)


 * This has been extremely minor MC. You should see some of my other arguments! (They go on for weeks) The only problem is : you've landed yourself with all these tasks now! :-)--Light current 02:34, 27 February 2006 (UTC)

You don't need to tell me. I have been involved in a few beauties myself. But I am trying very hard to maintain composure and not let things get out of hand. So far, this exchange has been a pleasure, and I think that you and I will reach a good consensus eventually.

I don't mind doing some work, as long as I am learning and WP is improving. Then it is more than worthwhile.

I have to call it a day for now. I will take a stab at the first three tasks over the next several days. You will, of course, have ample opportunity to comment and revise.

-- Metacomet 02:39, 27 February 2006 (UTC)


 * I look forward to cooperating and trying to help!--Light current 02:43, 27 February 2006 (UTC)

emf
I looked in a couple of standard reference books. In Jackson, Classical Electrodynamics, he only considers the induced emf in circuits without chemical batteries. In this case,


 * $$ \mathcal{E} = \oint_C \mathbf{E} \cdot d\mathbf{l} $$

(as l.c. quoted above). i.e. it is the sum of the potential drops around the circuit, which is non-zero when you have time-varying magnetic fields and a nonzero inductance. In circuit theory, this emf is usually included as a fictitious "potential drop" in the circuit to make Kirchhoff's voltage law valid.

In Landau and Lifshitz, Electrodynamics of Continuous Media, they also consider the emf in electrostatic situations where you have a chemical battery. In this case he defines the emf as the sum of the contact potentials around the circuit. The contact potential is the discontinuity in the potential as you go from one conductor to the next, due to the difference in work functions for the materials (the work required for a thermodynamically reversible removal of a charged particle through the surface of a conductor). This is not the same as total potential difference around the circuit, which is zero for electrostatics, because it does not include the ohmic potential drop IR within the conductors. When you have a battery, he writes: "the circuit includes conductors in which the current is carried by different means (e.g. metals and solutions of electrolytes). Because the work function is different for different charged particles (electrons and ions), the total contact potential in the circuit is not zero even when the conductors at each end are similar."

In both cases, the emf is essentially the driving voltage in the circuit. It is related to a sum of potential differences $$\int \mathbf{E} \cdot d\mathbf{s}$$ around the circuit but is not the same. (There is no sign difference as long as we are talking about potential drops.)

—Steven G. Johnson 19:43, 27 February 2006 (UTC)


 * On a quick reading of the above I think I concur with these satements. How to express this simply on the page is now the question. Let's try to keep it simple for simple folk like me to understand after I ve forgotten about it again!(ie in a couple of years time)--Light current 22:12, 27 February 2006 (UTC)


 * I don't understand much of what you are talking about. So I am out of the discussion.  Have fun.  -- Metacomet 00:38, 28 February 2006 (UTC)


 * Hey! you've done much more than I could have (esp on vector calculus) and have presented a very good explanatory math. You have helped to define the fact that there is a difference between dc emf and induced emf which was not really obvious to us (me). So please dont feel sidelined or inferior in any way-- youre not! Please comment further -- we welcome your sensible suggestions! :-)
 * --Light current 00:56, 28 February 2006 (UTC)

LC -- It's fine. That's only part of the issue. It's mainly that I really don't understand a lot of what Steven is talking about. I don't have the horsepower to keep up, and I have already gotten in over my head once before (not to mix metaphors too much). I am not going to make that mistake again. Once you guys sort it out, I will take a look at it, and see if I can help with the editing if necessary. I can be your "guinea pig" in terms of making it simple enough to understand. -- Metacomet 01:09, 28 February 2006 (UTC)


 * OK see you soon. Stephen J is as Im sure you know Asst Professor of something at MIT, but even he can make mistakes or forget things sometimes. I know what you mean tho'- I got in very deep once but just managed to escape with my integrity intact. Its a dangerous place (WP) if you dont keep your wits about you. Anyway thats why I like it- it keeps my mind sharp! Please look in on the page from time to time to see if we've made any silly errors-- Im sure we will -bye for now!--Light current 01:17, 28 February 2006 (UTC)

The discussion revived!

 * I wish I had put this article on my watch list as participating in this conversation would have been fun. Why is the accuracy tag still present?  Did you guys give up before this was settled or did you all agree to disagree?  Alfred Centauri 00:53, 4 April 2006 (UTC)


 * I deferred (or is it demurred) to the superior knowledge of the Asst Prof.--Light current 01:00, 4 April 2006 (UTC)


 * So, what then is still in dispute? From a quick read of the discussions on this page, It appears that there was some type of agreement although I'm not really sure. Alfred Centauri 01:17, 4 April 2006 (UTC)


 * The rest, as they say, is in the edit history (and here). This is all there is! Sad!--Light current 01:19, 4 April 2006 (UTC)


 * The actual definition of emf is disputed. Perhaps you could help?--Light current 01:28, 4 April 2006 (UTC)

Surely emf is simply a measure of the ability of the reaction to occur? I know it can be related to many other things but the main thing i think about when faced with problems including emf is the Gibbs free energy of the system and its ability to actually do the electrical work.


 * Consider an isolated chemical battery. We know that there is an electric field outside this battery due to the different charge densities on the terminals of the battery.  Further, we know that this external field is conservative.  That is, the work associated with a unit test charge moving from the positive terminal to the negative terminal of the battery is independent of the path taken as long as the whole of that path is external to the battery.  Equivalently, the work associated with a unit test charge moving in a closed path is zero as long as the whole of the path is external to the battery.  The value of work per unit charge in moving form the positive to the negative terminal is called the potential difference between the positive and negative terminals.
 * Now, consider the region inside the battery where, within the electrolyte, there are free charge carriers. Free charge carriers accelerate in the presence of an electric field yet, inside the battery, there is no current.  Apparently, in the region internal to the battery, there is no electric field.  Yet, we know that between the terminals of the battery, there is an electric field so we can only conclude that, inside the battery, there is also an equal but opposite electric field between the terminals.
 * Integrate the net electric field along a closed path where part of that path is from the negative to the positive terminal through the inside of the battery. This part of the path contributes nothing to the integral since the electric field is zero inside the battery.  Thus, the only part of the path that contributes to the integral is the external path but we already know what the value is for that part of the path - it is precisely the potential difference between the terminals of the battery.
 * Now, for a different perspective, let us decompose this electric field into a conservative part and a non-conservative part. The conservative part is due soley to the charge densities on the terminals of the battery and the non-conservative part exists only in the region inside the battery.  The integral of the conservative componenent of the field along any path from the positive to the negative terminal is defined as the potential difference between the terminals.  If we integrate the non-conservative part of the field along a closed path, there is no contribution from any part of the path outside the battery.  The only contribution to the integral comes from that part of the path inside the battery.  Thus, we can equate the line integral inside the battery (from the negative to the positive terminal of the battery) of the non-conservative part of the field to the line integral inside the battery (from the positive terminal to the negative terminal of the battery) of the conservative part of the field.  Or put another way, the line integral inside the battery of the conservative part of the field from positive to negative is equal and opposite to the line integral inside the battery of the non-conservative part of the field from positive to negative.
 * I'll have some more thoughts later... Alfred Centauri 16:31, 4 April 2006 (UTC)

Charge densities on cell terminals
Youre not on vacation again are you? Im going to go thro your text para by para. In your first para, you state that the charge densities on each terminal of the battery are different. Did you mean to say they were the same but of opposite sign? After all, is this not a case of charge separation as in a caopacitor? ;-). I Dont see how they could be different-- unless you know better! I see no reason to disagree with the rest of your para 1. But one question I have is: Why are you allowed to exclude the interior of the battery when doing the line integral? --Light current 16:44, 4 April 2006 (UTC)


 * Equal but opposite means equal in magnitude but opposite in sign which, quite clearly, is different. Charge density can be positive or negative.


 * I exclude the region inside the battery in paragraph one for the reason given in the following paragraphs. Within the battery, there is a non-conservative field that is superimposed on the conservative field originating from the charge on the terminals of the battery. Outside the battery, the electric field is conservative. To be more precise, in this idealized example, the curl of the electric field is non-zero (in fact, infinite) at the boundaries of the battery. Alfred Centauri 17:37, 4 April 2006 (UTC)

a)Is the magnitude of the charge density equal on both terminals?

b) You must not exclude the interior of the battery. Add an opposing field if you like inside, but I cant see how exclusion is legitimate. A conservative and non conservative field can sum in a region of space, can they not? --Light current 22:37, 4 April 2006 (UTC)

a) There is no need for the charge density to be equal on both sides just as long as there is more electrostatic repulsion on one side then the other. Physics require an unbalanced force for movement not an absence of oposing forces.69.213.70.93 19:43, 8 September 2006 (UTC)

Current inside batteries (cells)
With regard to your paragraph two, I have to take issue with your statement that there is no current inside the battery You have stated that there are free charge carriers - correct. You have also stated that in the presence of an electric field, charge carriers will accelerate (correct). You have stated there is an electric field within the battery (correct). Flow of charge carriers is defined as electric current. Ergo, there must be a current inside the battery!! Im still considering the interesting ramifications of the rest of your para 2. Watch this space!--Light current 17:09, 4 April 2006 (UTC)


 * Uh, I stated that "Apparently, in the region internal to the battery, there is no electric field." I justify this conclusion precisely because there is no current in the battery. Alfred Centauri 17:37, 4 April 2006 (UTC)

You have not shown that there is no electric field inside the cell. You must justify your statement that there is no current in the the battery explicitly by demolishing my above argument. --Light current 22:39, 4 April 2006 (UTC)


 * There is no need to demolish any argument. That there is no current circulating inside a battery is an observed fact.  If you believe otherwise, please point me to a reference that supports your belief.
 * But, if you insist, I will demolish your argument. I never stated that there is an electric field within the battery. Alfred Centauri 00:20, 5 April 2006 (UTC)

Ah now you seem to be changing your statement. What do you mean by 'circulate'. Do you mean a flow wholly contained within the confines of the cell, like coffee circulating in a cup after stirring it?--Light current 00:52, 5 April 2006 (UTC)


 * The things that move inside the cell are the electrons, which the negative plate shoves off through its electronic circuit to the negative cell terminal from which it goes to the load circuit, and does its work. Then the electron returns to the cell positive terminal, where it moves through the positive plate electronic circuit to the positive plate, where it causes the positive plate material to shove the electrons into the electrolyte which transports them back to the negative plate.WFPM (talk) 16:09, 1 September 2009 (UTC)The driving power in all of this is the emf potential of the chemical reaction at the negative plate/electrolyte interface, which forces the electron through the circuitry and overpowers the chemical reduction action in the positive plate.WFPM (talk) 04:01, 2 September 2009 (UTC)

The electric field inside a cell
When you say apparently there is no electric field inside a cell, on what evidence or thinking do you base this statement? You seem to have generated a circular argument with no justification. --Light current 22:52, 4 April 2006 (UTC)

Para two, second comment. You say:we know that between the terminals of the battery, there is an electric field Correct. It is generated by the internal drift of charge carriers in the cell due to the chemical action. Wheres the problem? I think you may have mentioned this problem before--Light current 22:57, 4 April 2006 (UTC)

Para 2. Another question. Are you saying this:

''The internals of a cell constitute a generator of emf. As such, you cant actually measure it becuase its generated by or involves a non-conservative field,- just like an electromagnetically induced emf. '' --Light current 23:14, 4 April 2006 (UTC)


 * There is no circular argument, LC. The argument goes like this:


 * (1) There is a conductive medium inside the battery.


 * (2) An electric field within a conductive medium causes a current.


 * (3) There is no current inside an isolated battery (excepting some minute leakage current that depends on the physics of the battery). If you disagree with this, then please point me to a supporting reference.


 * (4) Thus, there is no (net) electric field inside the battery.


 * It would seem to me that (1) and (2) are statements of fact. Also, it would seem to me that (3) is as statement of fact but if you can show me otherwise, I'll recant.  The conclusion (4) is a logical consequence of (1) (2) and (3).  There is no begging the question fallacy in this argument.


 * The static electric field is not generated by the internal drift of charge carriers (that is typically called a current). The static electric field is due to the accumulation of electrons on one terminal and by the removal of electrons from the other.  This accumulation and removal of electrons occurs because of the chemical reaction between the electrolyte and the plates.


 * Regarding you final question. How do you propose to measure emf?  Alfred Centauri 00:29, 5 April 2006 (UTC)


 * What's this about non-conservative electric fields? All static electric fields are conservative!  There must be an electric field inside the battery.  Pfalstad 00:40, 5 April 2006 (UTC)


 * Hmmm... where exactly did I claim that a static field is not conservative? In fact, I said that the electric field due to the charge on the battery terminals is conservative.  However, inside the (isolated) battery there is no electric field.  I'm not making this up - allow me to cite a source:


 * "These charges [on the battery terminals] give rise to an electrostatic field intensity E both outside and inside the battery. Inside the battery, E must be equal in magnitude and opposite in direction to the nonconservative Ei produced by chemical action, since no current flows in the open-circuited battery and the net force acting on the charge carriers must vanish.".  This is from pg 206 of Field and Wave Electromagnetics, 2nd Edition, David K.Cheng


 * I will cite additional sources if required. Alfred Centauri 03:21, 5 April 2006 (UTC)


 * Unfortunately I don't know much about batteries and I don't have that source. But if there's no electric field inside the battery, then the field is not conservative, because in a conservative field, all closed-loop line integrals are zero.  If you integrate through the battery and then outside to the other end, you will get a nonzero answer if there is no field inside the battery.  Pfalstad 20:17, 5 April 2006 (UTC)


 * (1) is wrong, or misleading. A battery does not act like a conductor.  Otherwise, there couldn't ever be a voltage difference across the two ends of it.  Pfalstad 00:43, 5 April 2006 (UTC)

Ah yes, but two equal and opposing voltages may exist across a conductor, resulting in zero current. I suspect this is what Alfred is driving at.--Light current 00:56, 5 April 2006 (UTC)


 * Your not thinking right, Paul. Recall that a length of wire disconnected from any external circuit and immersed in an appropriate changing magnetic field will have a measurable potential difference across the ends of this conductor.  Nonetheless, there will be no (net) electric field inside the wire.  How? Within the wire, the induced electric field exactly cancels the static electric field due to the charge at the ends of the wire.  Similarly, a battery disconnected from any external circuit will have a measurable potential difference across the terminals but there will be no (net) electric field inside the battery. Alfred Centauri 03:21, 5 April 2006 (UTC)


 * If there's a changing magnetic field, it's not electrostatics. But an isolated battery, with no current flowing, is an electrostatic case.  So the electric field must be conservative..  Inside and outside the battery.  Pfalstad 20:17, 5 April 2006 (UTC)

Yes charge separation occurs due to the chemical reaction. Drift was the wrong word to use. I withdraw it. Alfred, it is up to you to provide a reference against the widely held belief and evidence that current flows within a cell.(KCL). Im talking about when its on load - are you? Ive just seen your term 'isolated' Do you mean not connected to anything at all?? I do not propose to measure the emf. If you look at my previous posts, you will see that I have argued the case that it is not possible to do so directly. All one can do is measure a pd across the battery terminals. Maybe one could measure the pd inside the cell as well (Wet cell). Any way,you have not answered my question about whether you are stating that the internal field is non conservative and that is why it cannot be measured. Please do so.--Light current 00:48, 5 April 2006 (UTC)

Correction
If Alfred is talking about an isolated battery sitting alone on a shelf with nothing connected to its terminals, then, in this case:


 * 1) there is no external current
 * 2) therefore there is no internal current from one terminal to the other (not sure about circulating currents inside)
 * 3) there is an external field.
 * 4) this must be balanced by an internal field of opposite sense

Is THAT what you are saying? If so I agree.--Light current 01:06, 5 April 2006 (UTC)


 * Ah! Now I understand where our disconnect was.  Yes, I was referring to a battery that is not connected to an external circuit.  Alfred Centauri 02:16, 5 April 2006 (UTC)

Your last para
Ive just read this carefully, and I think I agree with what I think it says (I think!). In fact, Im sure I said something similar myself (but not as well put) further up the page. i.e

e.m.f= -pd

We must now move on!--Light current 01:22, 5 April 2006 (UTC)


 * We agree. The sign of the emf is opposite of the potential difference.  That is what I had hoped to show with my battery example.  So, does professor Steve disagree with us on this?  Alfred Centauri 02:21, 5 April 2006 (UTC)

Dunno! He says the emf is the sum of the potential drops in the (closed) cct and then the signs are all correct. But he doesnt give an explanation for his conclusion. So he seems effectively to be saying the same as we. --Light current 03:47, 5 April 2006 (UTC)

It does now seem likely that the same defn can be used for electrochemically generated emf and induced emfs. Would you agree?--Light current 04:25, 5 April 2006 (UTC)

Early scientists and emf
Did the early scientists appreciate this subtlety of distinction between emf and pd when they were bandying around this 'emf' term ? especially as they could only measure pds with potentiometers. Or did they just ignore the sign difference? I think these experiments would have been before the days of Maxwell, and so these scientists may not have had the advantage(?) of vector calculus to help them. If they did appreciate the opposite signs, they appear to have been much more perceptive than todays electrical engineers (and certainly wiser than I!). Yet the voltages they measured they always referred to as emfs! So maybe they were using a shortcut or maybe they just referred to the magnitude and didnt give a hoot about the sense? Does anyone have any comments on this as it may be good to put in the history para of the article.

This subtle difference betweeen emf and pd is in danger of being forgotten unless we get this article right and as Alfred Centauri has pointed out in the past, the concept of emf is essential to understanding the fundaments of electrical engineering--Light current 16:05, 5 April 2006 (UTC)

Where is the magnetic field?
I believe that there is a serious problem with my analysis of the isolated battery.

First, I do believe that there cannot be a net static electric field within the electrolyte as there are mobile ions that, like the electrons in a conductor, will redistribute themselves in such away as to cancel any applied field. However this alone cannot create a non-conservative electric field!

Second, I argued that this lack of an internal electric field leads to a non-zero closed path integral of the electric field if part of the path was through the battery. But, if this is true, then by Stokes' theorem, the flux of the curl of the electric field through the surface bounded by the path must be non-zero. But then, according to Maxwell, there must be changing magnetic flux through this surface. This cannot be physical. I am not aware of anyone measuring a constantly changing magnetic field in the vicinity of a charged battery sitting on a shelf!

Thus, despite the citation I gave, I do not see how there can be a non-conservative electric field inside the battery. Reading Professor Steves' comments again, the word contact potential jumped out at me. I feel certain that when this contact potential is properly taken into account, the closed line integral of E will always be zero even when the path is through the battery. More thoughts later. Alfred Centauri 23:58, 5 April 2006 (UTC)




 * OK, restore stupid question: :-)
 * So, you are thinking that any source of emf has to be produced by a non-conservative field. Am I correct in this assumption?--Light current 00:04, 6 April 2006 (UTC)

Well, if we define emf by


 * $$ \mathcal{E} = \oint_C \mathbf{E} \cdot d\mathbf{l} $$

then yes, emf is due to a non-conservative field. We know that a non-conservative field can drive (deliver energy to) an electric charge moving along a closed path whereas a conservative field cannot.

However, if by emf, we mean a measure of the strength of 'something' that can 'pump' electric charge along a closed path, then maybe a non-conservative field is not absolutely required. For example, can we not think of a (rechargable) battery as a analogous to a capacitor where electrical energy is stored in chemical bonds (which are, after all, electrical) rather than the electric field between the plates? Alfred Centauri 00:49, 6 April 2006 (UTC)

Contact Potential - I see that there is not an article here on Wikipedia but there shall soon be one. The contact potential is a "discontinuous jump of the electrostatic potential φ at the junction". A discontinuous jump in the potential implies an impulse in the electric field. Thus, as we integrate E along a path through the battery, the value of the integral suddenly jumps at the first terminal/electrolyte interface and then again at the other electroyte/terminal interface. In between, while we're in the electrolyte, the integral doesn't change. These jumps are in the same direction and must add up to the opposite of the potential difference between the terminals in order for the line integral to be zero. There's the 'emf' and it is indeed equal to -p.d. Alfred Centauri 01:05, 6 April 2006 (UTC)


 * As we cannot have infinitely thin interfaces, then the integral function will not be discontinuous. This is a minor point I think you will agree with that does not affect the argument.--Light current 14:21, 6 April 2006 (UTC)


 * Yes that article will be a welcome one as I have already refered to contact potential under 'other sources of emf'--Light current 01:08, 6 April 2006 (UTC)

Here's why we can have a conductive medium (electrolyte) inside the battery yet not discharge the charge stored on the plates of the battery: Contact_electrification. Alfred Centauri 01:10, 6 April 2006 (UTC)


 * Because theres no pd or emf across the electrolyte and all the emf is generated in infinitely (or very) thin layers at each electrode? Neat!--Light current 01:21, 6 April 2006 (UTC)

I agree. At the interface, the integral (potential) changes very rapidly over a very small distance compared to the overall distance along the path. Alfred Centauri 14:38, 6 April 2006 (UTC) OK. good!--Light current 16:35, 6 April 2006 (UTC)

Status of disputed tag
LC, we still have a disputed tag on this article but the comments by the person who put it here are now archived. I think we should have a summary of the point(s) that are disputed, address them, remove the disputed tag, and then archive all of the comments regarding the dispute in a separate archive. What think? Alfred Centauri 14:01, 6 April 2006 (UTC)


 * I believe the correct procedure would be to summarise Steves comments from the archive page and post the summary here, addess the points, and then, if no further comments, remove the tag. Yes.--Light current 14:08, 6 April 2006 (UTC)


 * Oh boy... yes, this article still needs a lot of work. I just found it after splitting Voltage ("pd" in this context) off from Volt. I think our first task ought to be to collect definitions with cited sources; without them, all this talk is just talk. Melchoir 00:28, 12 April 2006 (UTC)


 * I think you'll find most of the article is correct. But please go ahead and find some refs!--Light current 00:39, 12 April 2006 (UTC)

Copied from archive page by --Light current 08:20, 2 May 2006 (UTC)
disputed I don't think that the sign convention has anything to do with the difference between electomotive force and potential difference. Potential difference is, strictly, the line integral of electric field, $$\int \mathbf{E}\cdot d\mathbf{s}$$. This is not well-defined in a circuit with a time-varying magnetic field, because it is path-dependent. Emf, as I understand the term as it is used nowadays, is simply a generalization of potential difference to handle such cases, by including a "fictitious" potential drop over things like ideal inductors, in exactly the right amount to make Kirchhoff's voltage law valid. I'll double-check a reference book when I get a chance, but meanwhile I'm adding the disputed tag because I'm dubious about the current text. Even if it is commonly used simply to mean the driving voltage(s) in a circuit, which is what the current text seems to suggest, I don't recall seeing a consistent difference in the sign convention. —Steven G. Johnson 18:22, 26 February 2006 (UTC) PS. Sorry about adding the disputed tag yesterday without explanation. WP gave me an error message and I thought that the edit hadn't occurred.)''

emf

I looked in a couple of standard reference books. In Jackson, Classical Electrodynamics, he only considers the induced emf in circuits without chemical batteries. In this case,

$$\mathcal{E} = \oint_C \mathbf{E} \cdot d\mathbf{l}$$

(as l.c. quoted above). i.e. it is the sum of the potential drops around the circuit, which is non-zero when you have time-varying magnetic fields and a nonzero inductance. In circuit theory, this emf is usually included as a fictitious "potential drop" in the circuit to make Kirchhoff's voltage law valid.

In Landau and Lifshitz, Electrodynamics of Continuous Media, they also consider the emf in electrostatic situations where you have a chemical battery. In this case he defines the emf as the sum of the contact potentials around the circuit. The contact potential is the discontinuity in the potential as you go from one conductor to the next, due to the difference in work functions for the materials (the work required for a thermodynamically reversible removal of a charged particle through the surface of a conductor). This is not the same as total potential difference around the circuit, which is zero for electrostatics, because it does not include the ohmic potential drop IR within the conductors. When you have a battery, he writes: "the circuit includes conductors in which the current is carried by different means (e.g. metals and solutions of electrolytes). Because the work function is different for different charged particles (electrons and ions), the total contact potential in the circuit is not zero even when the conductors at each end are similar."

In both cases, the emf is essentially the driving voltage in the circuit. It is related to a sum of potential differences $$\int \mathbf{E} \cdot d\mathbf{s}$$ around the circuit but is not the same. (There is no sign difference as long as we are talking about potential drops.)

—Steven G. Johnson 19:43, 27 February 2006 (UTC)

I dont really see any conflict here between what Steven is saying and what the article says. Can anyone? --Light current 08:26, 2 May 2006 (UTC)


 * The article seems to claim that the only difference between emf and potential difference is the sign. This is false as far I can tell; the sign convention is a minor thing compared to the underlying physical distinctions. —Steven G. Johnson 00:19, 3 May 2006 (UTC)

Im not an expert on this by any means, but it seems that Alfred Cetauri is saying that emf can only be generated by a non conservative field whereas pd is not so restricted. If this is correct, then maybe the article needs to say it.8-)--Light current 00:34, 3 May 2006 (UTC)


 * emf is defined on a loop; pd is defined between two points. They are generally incomparable. Melchoir 00:38, 3 May 2006 (UTC)

Wher is the loop in a chemical cell? 8-|--Light current 00:46, 3 May 2006 (UTC)


 * It passes through the cell and around the vacuum outside. Since the force on charge carriers outside the cell is uniformly zero, it doesn't matter which outside path you use, which is why one speaks of the emf of a cell. Melchoir 00:53, 3 May 2006 (UTC)
 * Well, not uniformly zero, sorry. Outside the cell, the force on a charge carrier is due only to the electric field, which is conservative, and that's why the exact path doesn't matter. D'oh! Melchoir 00:55, 3 May 2006 (UTC)

Ah no, outside the cell (between the cell terminals in fact) there exists a pd (not an emf). An emf is not directly measurable.--Light current 00:59, 3 May 2006 (UTC)
 * What do you mean, exactly, by "there exists"? Melchoir 01:07, 3 May 2006 (UTC)

You can measure it. Its there! 8-)--Light current 01:11, 3 May 2006 (UTC)
 * But, see, without more information that's a content-free statement. You can speak of the emf around a given loop or the pd between two given points. Neither emf nor pd is even defined at any given point in space (that's the potential's job). So to say there exists or does not exist an emf or a pd at a point or region of space is just undefined. Melchoir 01:18, 3 May 2006 (UTC)

No sorry, I mean that a pd can be measured between the terminals of the cell with a voltmeter.8-|--Light current 01:22, 3 May 2006 (UTC)
 * Well, yes, certainly. Melchoir 01:49, 3 May 2006 (UTC)


 * It seems that there are two distinct 'emf' definitions here. There is the usual closed contour integral definition (emf is the change in energy per unit charge when moved along a closed path).  That is, a non-conservative electric field can drive electric charge along a closed path whereas a conservative field cannot.  Then, there is the broader definition of emf which is simply the measure of the strength of a some kind of charge 'pump' such as a chemical battery.  In this case, there is no non-conservative field.  Instead, we have a physical apparatus that can separate charge onto conductive terminals where this charge can then participate in a current in an external circuit.  As we have discussed above, the closed contour integral in this case is zero as long as the contact potentials are taken into account.
 * I agree that emf differs from pd by more than just sign however, the sign of the emf is opposite that of the pd. It must be in order to cancel the pd in the contour integral.  Does Steve really think that the article implies that the only difference between emd and pd is sign?  Alfred Centauri 15:55, 4 May 2006 (UTC)