Walter Lewin paradox

[mabilde]

Right answer for the Walter Lewin demo/paradox about electromagnetic induction in Lecture 16 ?

 

[Charles Link]

For the Professor Lewin video that I previously saw on the subject, I think professor Lewin had most of the information correct. In any case, the additional measurements that are made in this video are very interesting. See post 52 of this thread for the Professor Lewin video that I saw: https://www.physicsforums.com/threa...asure-a-voltage-across-inductor.880100/page-3 And yes, he does call that one "Lecture 16".$\\$ Edit: To the OP @mabilde I see you chose the same name as the author of this video. Are you the author? $\\$ Additional editing: In a conductor, even though there is an EMF, where EMF $\mathcal{E}=\int \vec{E}_{induced} \cdot d \vec{s}$, a reverse electrostatic field will necessarily develop=otherwise the current would necessarily be very large by the equation $\vec{J}=\sigma \, \vec{E}$ where $\sigma$ is very large for a conductor. In that sense, I don't know that obtaining a voltage measurement inside the coil necessarily proves anything. In any case, an interesting set of experiments, but I think further discussion is in order. $\\$ And my input on this: A voltmeter and/or an oscilloscope consists of connecting wires plus some resistance and measures the current in the resistor circuit. It becomes part of the circuit loop, and there is nothing magical about it. It really can't measure the electric field or an EMF between two points in the circuit. It becomes part of the circuit, and if the resistance is chosen properly and the leads are connected properly, ideally it doesn't disrupt the circuit being studied, but that does not mean that (changing) external magnetic fields won't effect the results that are obtained. The EMF that is used to analyze the circuit containing the voltmeter can usually be obtained by Faraday's law. IMO, there is no need to try to localize the source of the EMF's as is done in this video by Professor Cyriel Mabilde.

 

[sophiecentaur]

I often find this sort of thing confusing but it seems that it works like this. Every elemental length of wire will have an elemental emf induced in it.
Your measurement leads have an emf induced in them which will depend on their path through the field from the primary and this adds (vectorially) to the PDs across the two resistors and the emfs induced in the two sides of the secondary.
Trivial situation: If the measurement leads are short enough and connected straight across each resistor then the emfs will be the 'share' of 1V that each resistor gets - as a normal potential divider. This is because there are no other emfs involved and you could think of the circuit as a 1V (AC) battery (or a 0.5V battery in each leg of the loop.
More complicated situation: Connect the voltmeter diametrically across the mid points of the two semicircles. In my diagram, I have split the induced emf into four 0.25V contributions. The dashed line splits the circuit into two symmetrical parts and, treating it as a simple DC circuit, the total PD measured will be 0.4 X 2 = 0.8V, because you have, effectively a potential divider. (I hope I have my + signs right in the diagram)
Even more complicated: Move the measurement leads off centre or make them longer and tilt them. They will have their own emf induced into them, which can vary from 0.5V to -0.5V so the measured value can be anything between 0.1V and 0.9V, depending.

 

[mabilde]

Walking clockwise through the loop, then the emf's have the right polarity, but the resistor drops are opposed to this direction. The sum will be ZERO
Using two voltmeters (in series) as in your "split" circuit, will give 0.2 Volts on each voltmeter.
For example, considering the upper left : 0.25 Vemf -0.05 Vdrop = 0.2V
and for upper right: - 0.25Vemf +0.45 Vdrop = 0.2 V etc...So the sum of both voltmeters will be 0.4 Volts and not 0.8 Volts!
Note that in the first (left) case your voltmeter Hi is connected to the positive side of the emf, but in the right case the voltmeter Hi is pointing to the negative side of the emf.
Hope this will help to understand also the rest of my video and thanks for watching it.
Best greetings from Belgium
Cyriel Mabilde

 

[Charles Link]

Now here is how I would solve this voltmeter problem: Let the EMF from the changing magnetic field be clockwise and 1.0 volt. Let the resistor on the left be $R_1=100 \, \Omega$ and the resistor on the right be $R_2=900 \, \Omega$. Let the voltmeter across the middle be $R_3= 1 \, M \Omega$ . Let the clockwise current through $R_1$ be $i_1$, and the clockwise current through $R_2$ be $i_2$. The downward current through $R_3$ is $i_3$. $\\$ The 3 loop equations are: $\\$ 1) $.5=i_1R_1+i_3 R_3$ $\\$ 2) $.5=i_2 R_2-i_3 R_3$ $\\$ 3) $i_1=i_2+i_3$. $\\$ A little algebra gives (eliminating $i_2$) $\\$ $1.0=i_1R_1+(i_1-i_3)R_2$ $\\$ and $\\$ $0=i_1(R_1-R_2)+i_3(R_2+2R_3)$. $\\$ The $i_3 R_2$ term can be considered small in two places, so that $i_1=\frac{1.0}{R_1+R_2}$ and $\\$ the voltmeter reading is $i_3 R_3=-\frac{1}{2}\frac{(1.0)(R_1-R_2)}{R_1+R_2}=+.4 \, volts$. $\\$ @mabilde Thank you very much for your video. It was very interesting and instructive.

 

[sophiecentaur]

Walking clockwise through the loop, then the emf's have the right polarity, but the resistor drops are opposed to this direction. The sum will be ZERO
Using two voltmeters (in series) as in your "split" circuit, will give 0.2 Volts on each voltmeter.
For example, considering the upper left : 0.25 Vemf -0.05 Vdrop = 0.2V
and for upper right: - 0.25Vemf +0.45 Vdrop = 0.2 V etc...So the sum of both voltmeters will be 0.4 Volts and not 0.8 Volts!
Note that in the first (left) case your voltmeter Hi is connected to the positive side of the emf, but in the right case the voltmeter Hi is pointing to the negative side of the emf.
Hope this will help to understand also the rest of my video and thanks for watching it.
Best greetings from Belgium
Cyriel Mabilde

And greetings to you from sunny Essex UK.
Thanks for your comment. (The demo on the video is great, btw) I obviously got the signs in my calculation wrong and, because I am familiar with Potential Dividers, I can do it in an easy way . Take the top left corner as reference zero for the first half of the calculation. The two resistors are a potential divider with 10:1 ratio. The upper voltmeter should be showing the unbalanced volts between the two nodes of 0.2V (= 0.25-(0.5/10)) Likewise the other voltmeter will see -0.2V, giving 0.4V overall.
It's been a long time since I did this stuff and I'v looked at the problem again. I should have written some reference letters on the diagram and it would be easier to discuss but @Charles Link has solved the problem in yet another way (with Kirchoff 1) - and got the same answer!.

 

[Charles Link]

@mabilde I am very glad you posted this video. The experimental apparatus and results is very interesting. I am also very glad that it showed the results that confirms a combination of Faraday's law, and Kirchhoff's voltage law.

 

[alan123hk]

Right answer for the Walter Lewin demo/paradox about electromagnetic induction in Lecture 16 ?

This is a cool experiment, which demonstrates that the induced electric field $(E_i)$ is evenly distributed over every elemental length of the single-loop circuit. Thank you very much for your efforts and sharing.

Of course, in this case, we can define a scalar potential as $(V_{ab} = IR - E_i L)$ in the circuit analysis.

However, it is worth mentioning that the scalar potential should have the operational meaning only when the leads of the voltmeter connected to these two points do not receive additional net induced EMF. In this case, the voltmeter will display the value predicted by the equation, as shown in the video. But in any case, of course this scalar potential is real, and we only need to use it correctly, an example of this is the output of the transformer secondary.

I think that, generally speaking, two electric fields are superimposed in an electromagnetic induction circuit, one is the electrostatic field $(E_c)$ corresponding to the scalar potential, and the other is the induced electric field $(E_i)$ , and $~ E_{total} = E_c + E_i~$ . In a zero-resistance wire, $(E_c)$ and $(E_i)$ cancel each other in the direction of the wire. In the resistor, $~ E_c~$ may be larger or smaller than $~E_i ~$, or equal to each other, depending on the circuit.

In any case, the closed-loop line integral of the circuit's $~ \oint \vec E \cdot d \vec L ~$ is equal to the $~ total ~~EMF = -\frac {dΦ} {dt} ~$, which is not conservative and different from the zero value of the closed-loop line integral in the DC circuit (assuming the battery has an internal electric field, which points from positive terminal to the negative terminal). I believe this is the key point that the professor wants to explain. What he said does not contradict or deny the existence of the above-mentioned scalar potential and electrostatic field.

 

[rude man]

Hi @alan123hk,
good commentary, thank you.

My overall intention in the blog was to call attention to the existence of two E fields of separate provenance which, somewhat incredibly, many incl. Dr. Lewin seemed unaware of, and that that view provided an alternative explanation of the phenomena..

My criticism of Dr. Lewin's lecture was in essence his comment that "Kirchhoff was wrong". Kirchhoff's voltage law as applied to electric fields rather than potentials applies to what you call the $E_c$ field only. Lewin implicitly defined voltage as $\int (\bf {E_c + E_i}) \cdot \bf dl$. He thus seemed unaware of what exactly "voltage" is, mistaking it for $\int \bf E \cdot \bf dl$ where $\bf E = \bf E_i + \bf E_c$.

As you point out, the voltmeter readings nolens volens include the emf set up by the Faraday setup he has. He never explained that those readings are not actual voltages. Had he held up the voltmeter leads directly above the ring he would have read the correct voltage which is 0.4V and which would have canceled the emf of the ring itself with the emf induced by the flux cutting the voltmeter loop. But he didn't, alas.

Failure to identify the two electric fields of differing sources ($E_c$ from charges only and $E_i$ from induction is I feel a serious omission. For one thing, had he been asked if & where charges are produced he would have had no answer, or so I believe.

What he did succeed in doing is that if your only goal is to analytically determine voltmeter readings for the different lead positions by considering the total E field alone then yes he did that. What he did not do was to point out that those readings are not the voltages between the two lead connections to the ring except if the leads are held directly across the resistors.

You might look at the references I provide, especially MacDonald's. He too addresses the Lewin "conundrum" & properly defines scalar potential aka voltage.

Also, did you see the infamous C. Mabilde video? That got me started on this whole odyssey. His data were correct but what he really showed was a simulation of the voltages around the ring (by equating the flux rate of his voltmeter lead setup to the actual voltage). He also, by not explicitly identifying the sepate but coincident E fields, came up with some hocus-pocus transformer-like coupling mechanism for the voltmeter readings. Still, he was spot-on redressing Lewin for his Kirchhoff remark, synthesizing the actual voltages around the ring, and although I never checked into it, Lewin's self-contradictory statements about the fields in the wire. Taken with that grain of salt the video is a real eye-opener.

As for my analysis of the battery circuit, that has generated enough controversy that I recommend ignoring it. What is certain is that the sum of line-integrated $E_c$ fields around the circuit including the battery = 0 a la Kirchhoff.

 

[DaveE]

Lovely video.

I'll just add that this isn't just an academic exercise. If you ever find yourself in an EE lab trying to make measurements, for example of a high current power supply switching transistor, you will absolutely have to deal with these effects. Oscilloscopes are one of the more difficult instruments to use well, if you need accuracy. It's all about the probes, which novice EEs tend to ignore. That long ground lead that seems so convenient at first is a huge problem in high frequency circuits.

This is a great example of why you have to do "sanity checks". Don't just make a measurement once, with one setup. Move stuff around and see if the data changes unexpectedly.

 

[DaveE]

Still, he was spot-on redressing Lewin for his Kirchhoff remark, synthesizing the actual voltages around the ring

Yes! I'll spare y'all the diatribe I wrote in a related post a while ago. But just point out that Kirchhoff's voltage and current laws are for LUMPED ELEMENT CIRCUITS. If you want to use them in the presence of induced emf, you must put some inductors in your schematic. This, of course, is why lumped element models don't work sometimes.

Why does Lewin say most physicists get this wrong? It's because he is asking a ill-formed question. I, for one, don't think of Maxwell's equations if you show me a schematic diagram. That format is so highly stylized that it lacks the required context for EM problems.

 

[alan123hk]

Failure to identify the two electric fields of differing sources (Ec from charges only and Ei from induction is I feel a serious omission. For one thing, had he been asked if & where charges are produced...

Let's take a look at how the distinguished Professor Lewin explained the build up of charges in the circuit and the role of these charges in the overall electric field. I think this is a good explanation.

This video is relatively long, maybe we can watch it from 34 minutes.

 

[rude man]

Let's take a look at how the distinguished Professor Lewin explained the build up of charges in the circuit and the role of these charges in the overall electric field. I think this is a good explanation.

This video is relatively long, maybe we can watch it from 34 minutes.

Thank you for that! I am happy to agree that with this video Dr Lewin has fully and correctly explained split E fields and the necessary buildup of charge in a resistively asymmetrical ring. I am in total agreement (except see below).

He's given several lectures on this topic (Faraday induction). It appears that the one I saw and responded to in my blog (my 1st blog btw; I discuss this theme in more detail and perhaps a bit better in the next blog, "How to Recognize Split E Fields" (link below). I think everything I said about that video (lecture 16 I guess) in my 1st blog is correct. He never mentioned split fields or charge buildup in that lecture and in fact admitted as much in this video.

OK, it's a bit advanced but telling students that the voltage between A and B depends on how the meter is hooked up is just plain wrong. As Mabilde correctly said, there is one and only voltage between A and B and that is 0.4V (in lecture 16). It's not advanced physics to define potential difference correctly IMO.

My problem with him then remains defining potential difference as the line integral of the electric field. This is in general incorrect. "Potential" eo ipso implies a conservative E field: . For his integral to signify potential difference, the E must be replaced by what you called $E_c$. This was really my only criticism of his lecture 16 (along with his corollary "Kirchhoff was wrong").

Anyway, watching Lewin is pure pleasure. I haven't laughed so hard in years! And I wondered then and do now: where are all those dummies who don't understand, or even believe in, his lectures? Omitting charge buildup is one thing but disbelieving his voltmeter readings is ludicrous. Hope they aren't MIT students!

https://www.physicsforums.com/threads/how-to-recognize-split-electric-fields-comments.984872/

Again I suggest ignoring the battery part. Too controversial! Also, you may be amused or bemused by all the comments, or you might just ignore them.

https://www.physicsforums.com/insig...ectric-vector-potential-and-its-applications/
(this may be of interest as proof that the Em field is invariant with whatever is placed in it, like a ring or chunk of metal or anything else. The Em field is solely defined by the magnetic field.)

 

[weirdoguy]

or you might just ignore them.

Or read them and try to understand what was the criticism.

 

[Charles Link]

Just my two cents: In all of this I think @rude man has presented some reasonably good physics. I am glad to see that in this thread it is starting to see a much better reception. :)

 

[alan123hk]

signify potential difference, the E must be replaced by what you called Ec.

Yes, this is the definition of potential difference.
http://www.electropedia.org/iev/iev.nsf/display?openform&ievref=121-11-26

When considering the charge build up on the surface of the conductor, a picture of energy flow inadvertently appeared in my mind, so I sketched it out very roughly.
https://en.wikipedia.org/wiki/Poynting_vector

 

[tedward]

Hi @alan123hk,
good commentary, thank you.

My overall intention in the blog was to call attention to the existence of two E fields of separate provenance which, somewhat incredibly, many incl. Dr. Lewin seemed unaware of, and that that view provided an alternative explanation of the phenomena..

My criticism of Dr. Lewin's lecture was in essence his comment that "Kirchhoff was wrong". Kirchhoff's voltage law as applied to electric fields rather than potentials applies to what you call the $E_c$ field only. Lewin implicitly defined voltage as $\int (\bf {E_c + E_i}) \cdot \bf dl$. He thus seemed unaware of what exactly "voltage" is, mistaking it for $\int \bf E \cdot \bf dl$ where $\bf E = \bf E_i + \bf E_c$.

As you point out, the voltmeter readings nolens volens include the emf set up by the Faraday setup he has. He never explained that those readings are not actual voltages. Had he held up the voltmeter leads directly above the ring he would have read the correct voltage which is 0.4V and which would have canceled the emf of the ring itself with the emf induced by the flux cutting the voltmeter loop. But he didn't, alas.

Failure to identify the two electric fields of differing sources ($E_c$ from charges only and $E_i$ from induction is I feel a serious omission. For one thing, had he been asked if & where charges are produced he would have had no answer, or so I believe.

What he did succeed in doing is that if your only goal is to analytically determine voltmeter readings for the different lead positions by considering the total E field alone then yes he did that. What he did not do was to point out that those readings are not the voltages between the two lead connections to the ring except if the leads are held directly across the resistors.

You might look at the references I provide, especially MacDonald's. He too addresses the Lewin "conundrum" & properly defines scalar potential aka voltage.

Also, did you see the infamous C. Mabilde video? That got me started on this whole odyssey. His data were correct but what he really showed was a simulation of the voltages around the ring (by equating the flux rate of his voltmeter lead setup to the actual voltage). He also, by not explicitly identifying the sepate but coincident E fields, came up with some hocus-pocus transformer-like coupling mechanism for the voltmeter readings. Still, he was spot-on redressing Lewin for his Kirchhoff remark, synthesizing the actual voltages around the ring, and although I never checked into it, Lewin's self-contradictory statements about the fields in the wire. Taken with that grain of salt the video is a real eye-opener.

As for my analysis of the battery circuit, that has generated enough controversy that I recommend ignoring it. What is certain is that the sum of line-integrated $E_c$ fields around the circuit including the battery = 0 a la Kirchhoff.

I'm late to the party on this, but I've been obsessed with this problem recently and all the misconceptions surrounding it.

It seems that you accept that Lewin's voltmeters measure the Int(E.dl) along different paths, and give different measurements. Your issue seems to be a different meaning of 'voltage' as the sum of ONLY the electrostatic E-field, as McDonald does in his paper. The sum of these static produced fields indeed is zero around any loop, but what use does it serve? The consequence is that you can define different 'potentials' at different points in the same conducting wire. This leaves you with potentials that 1) are not measurable by a voltmeter, 2) vary in a region with ZERO net E- field , and 3) have no connection to the work done on a test charge as it moves from one point to another.

In what possible way does such a measurement correlate to what we usually think of potential? This is like moving an object horizontally in a gravitational field, which requires no work, and saying its potential energy has changed. It's a mathematical abstraction, nothing more, and has no physical basis.

I found it interesting that you picked out C. Mabildes' mistake in interpreting his results. He thinks he's measuring McDonald's 'scalar potential', when he's actually just measuring the flux through the pie-slice shaped area in his leads, which is proportional to the angle. You obviously see the problem, but generously say that he is 'simulating' measurement of the scalar potential, rather than saying that he's completely misinterpreting his results.

While I agree that it is instructive to consider both the induced E-field (Ei) and the electrostatic E-field (Ec) to analyze the problem, the fact is that it is only the combined field that does work on a test charge, and that field is non-conservative. Hence different measurements for different paths. Neither measurement is wrong, they're measuring different things. There is no 'true' potential difference between the two points, because you can not define potential for a non-conservative field.

 

[Averagesupernova]

I've never understood the controversy with Lewin's paradox. Why is it accepted that the conductors used in a measurement are somehow not supposed to behave as conductors of the circuit? It's ALL about knowing what introduces errors in what is trying to be measured.
-
Think of it this way: Set your voltmeter to AC, short the probes together and make a single loop around a solenoid setup similar to and driven as the one in the @mabilde video. The fact that a voltage will be read on the voltmeter should give a clue that a significant error will be introduced considering the voltmeter leads very much resemble the setup. There are many ways to 'blow your mind' with this but it always comes back to the same thing.

 

[tedward]

Not sure exactly what your take is, but... I totally agree with you that voltmeter leads form a circuit when connected and should be analyzed as such. And I agree with you that 'shorted' voltmeter leads around a solenoid will read the emf produced - it has to, by Faraday's law. If there's flux penetrating the measurement loop, it will show up in the reading.

But by the same token, if there is no flux in the measurement loop, then there is is no error - you read exactly what you intend to measure, again by Faraday's law. If I want to measure the voltage across a section of wire with a resistor, and I put my voltmeter leads on either side of the resistor, as long as there's no flux in the new loop (like the voltmeters on either side of Lewin's circuit), it's accurate.

Part of the 'bad thinking' around this topic is that people assume that a solenoid will affect any nearby wire regardless of placement. That's not really true - all that matter whether flux penetrates the surface bounded by the loop formed by the voltmeter leads and what you're trying to measure. In the @mabilde video, he most definitely has wide open measurement loops with flux passing through them. He thinks he's measuring 'scalar potential', but all he really did was create a very precise flux meter. Bottom line - emf affects a loop. You can't discuss the 'emf' in a section of a wire without considering what the loop is.

 

[DaveE]

Somehow, I'm reminded of a comment about legislators speaking on the floor; 'everything's been said, but not everyone has said it yet'. You can read, or you can type, but you probably won't do both in this thread.

 

[tedward]

I read this whole thread - there are several points that seem to have reached a consensus but are just plain wrong or misleading.

 

[alan123hk]

In an electromagnetic induction system, when current flows through a impedance, a voltage and corresponding electric field is generated. As for this voltage, or the approximate potential difference when the frequency is very low and tends to zero, it is difficult to say whether it can be measured with a voltmeter. I think it depends on the situation. In some cases, it can also be said that everyone's point of view may be different.

For example, when using a voltmeter to measure the voltage at the output load of an actual common transformer, one would think that it is measuring the transformer's EMF output, which is of course true. But some people may assumed that this is the voltage drop in the load impedance, because the measured voltage value is constant regardless of the position of the voltmeter itself and its leads, so this voltage is due to the electric field without curl, and this electric field should be generated by the electric charge of the system.

Another example, in real life the power company will say what is the voltage of the electricity it is supplying to the customer, not what the EMF is because from the consumer's point of view it is just a voltage. Regarding the work done when a charge is moved, of course all the energy comes from electromagnetic induction. But for consumers, I think it's acceptable in the context of their application to think of energy transfer as a voltage mechanism or even roughly a potential difference, since the difference is so small that it can be ignored.

 

[Averagesupernova]

I read this whole thread - there are several points that seem to have reached a consensus but are just plain wrong or misleading.

I am curious about which consensus you think that would be. I'm also wondering if you are saying that it is impossible to position voltmeter leads in a way to obtain a measurement without having the emf affect the reading.

 

[tedward]

The consensus I was referring to was the existence of a measurable, scalar potential that exists in a conducting wire. While you can certainly separate the static E-field (Ec) and electrically induced E-field (Ei), as McDonald does in his paper, this potential is just a mathematical abstraction. It's not physical, because a change in this potential does no work on a charge. It's not measurable, because a voltmeter only measures the integral of total E-field along a path. @mabilde believes he's measuring this potential, but as pointed out by both myself and @rude man, he's simply measuring the changing flux through passing through the loop formed by his leads (if you watch his video, it's really easy to see the pie-slice shaped loops the flux is passing through). This scalar voltage doesn't even conform to Ohm's law for any resistors whose length is not negligible when subjected to and induced field. So this definition of voltage forces you to bend over backwards just in order to have something that always sums up to zero around a loop.

If you define voltage as E.dl, which does agree with Ohm's law and measures the work done on a test charge, it does not sum to zero around a loop when you have a changing magnetic flux, in accordance with Faraday's law, and gives rise to path dependence (hence the two voltmeter measurements).

Regarding the voltmeter measurements, voltmeter leads can certainly be positioned without being affected by the emf. There's a very very simple test to determine whether your measurement is being affected by an emf of not. Look at the circuit loop formed by the voltmeter, the leads, and the portion of the circuit you are intending to measure. If there is a changing magnetic flux passing through this loop (i.e. the imaginary surface that your loop bounds), then that total flux term , -d(phi)/dt, will be included in your measurement - per Faraday's law - and you must account for it. If there is no magnetic flux passing through this loop, then your voltmeter is exactly in parallel with whatever you are measuring, and will show the intended voltage. This is still in keeping with Faraday's law, but as the flux is zero, you can think of it as Kirchoff's law.

The tendency by many to talk about the induced emf in a 'section' of a wire is very misleading. Emf doesn't act on a wire, it acts on a loop. You can certainly talk about the induced electric field in a wire, but that E-field adds vectorially with any other E-fields present (like Ec), and they sum to zero in a conductor.

 

[tedward]

In an electromagnetic induction system, when current flows through a impedance, a voltage and corresponding electric field is generated. As for this voltage, or the approximate potential difference when the frequency is very low and tends to zero, it is difficult to say whether it can be measured with a voltmeter. I think it depends on the situation. In some cases, it can also be said that everyone's point of view may be different.

For example, when using a voltmeter to measure the voltage at the output load of an actual common transformer, one would think that it is measuring the transformer's EMF output, which is of course true. But some people may assumed that this is the voltage drop in the load impedance, because the measured voltage value is constant regardless of the position of the voltmeter itself and its leads, so this voltage is due to the electric field without curl, and this electric field should be generated by the electric charge of the system.

Another example, in real life the power company will say what is the voltage of the electricity it is supplying to the customer, not what the EMF is because from the consumer's point of view it is just a voltage. Regarding the work done when a charge is moved, of course all the energy comes from electromagnetic induction. But for consumers, I think it's acceptable in the context of their application to think of energy transfer as a voltage mechanism or even roughly a potential difference, since the difference is so small that it can be ignored.

There is a simple resolution to whether or not a transformer has a 'voltage' across it. Let's just assume it's a coil of wire (solenoid) with an external B-field changing through it. If you want to measure the voltage between the terminals, there are two different paths you can consider. One path extends between the terminals on the outside of the solenoid, through free space. If you attach a voltmeter to these points, the outside path your are measuring IS the path of the voltmeter leads. There is definitely an electric field along this path, and the integral of this path is non-zero. Indeed, it is equal to the induced emf. But as it is a real electric field, it counts as a voltage ALONG that path. This is exactly how Feynman defines the voltage of an inductor in his E/M lectures.

The second path lies between the terminals and follows the wiring. It corkscrews along the solenoid. This is cumbersome to measure as you would have to attach voltmeter leads at nearby points, and add up the readings as you move the voltmeter leads around the solenoid one section at time, working your way around from one end to the other in a corkscrew pattern. If you did, you would find that this sum is zero, which is keeping with the fact that the solenoid wire as negligible resistance - it's a conductor. In the same lecture, Feynman explicitly states that Int(E.dl) around this path is zero.

The discrepancy is no mystery. The electric field formed by the changing flux is non-conservative, a fact that comes directly from the Maxwell version of Faraday's law: curl E = -d(phi)/dt. There for the int(E.dl) between two points is path dependent. In a circuit, any loop containing the outside path will be conservative, the emf is counted as a voltage. In a loop containing the path through the wires, the path integral will be non-zero as there is no voltage along the wire. BUT there is a changing flux through this loop, so that has to be accounted for via direct application of Faraday's law. Now the emf is on the other side of the equation, as it's not part of the loop integral. You get the same equation, but the emf term is just that, an induced emf.

Usually we don't think about the path difference due to the difficulties of measuring the corkscrew voltage. But if you have a single turn loop, the circuit is now planar. You can easily attach a voltmeter to either side of the loop (the open side connecting the terminals one one side, the loop section on the other side). When you analyze the voltmeter circuit loops, one will contain the flux, one will not. The loop that that contains the flux is just measuring the electric field on the outside path, or the flux it self, however you want to count it. This is the 'voltage' of the inductor as Feynman defines it. The other voltmeter loop does not contain a flux. There is not electrical field to measure, so it reads zero. This is equivalent to Lewin's two volmeters. He just has another resistor, and has opened up the loop so it is formed by the whole circuit.

 

[Averagesupernova]

If there is no magnetic flux passing through this loop, then your voltmeter is exactly in parallel with whatever you are measuring, and will show the intended voltage.

There is no magnetic flux passing through the pie shaped conductors that radiate out from the center of the solenoid and make contact with the single loop conductor in the @mabilde video. The flux lines cut lengthwise along those conductors.

 

[alan123hk]

Your issue seems to be a different meaning of 'voltage' as the sum of ONLY the electrostatic E-field, as McDonald does in his paper. The sum of these static produced fields indeed is zero around any loop, but what use does it serve? The consequence is that you can define different 'potentials' at different points in the same conducting wire. This leaves you with potentials that 1) are not measurable by a voltmeter

Lewin demo/paradox and general ordinary transformers are electromagnetic induction systems, so the working principle is basically the same.
Haven't the common transformers we use every day already shown that a voltmeter can be used to measure scalar potential in such systems?
Even though it is difficult to measure with a voltmeter in some cases like the Lewin demo/paradox, is there really no other way?

2) vary in a region with ZERO net E- field

I guess you mean that the electric field inside a perfect wire is 0. If the electric field in a region is zero, then of course there is no change in potential in that region. But note that this scale potential still acts on the wire surface and outer space

3) have no connection to the work done on a test charge as it moves from one point to another.

So isn't the voltage output by the transformer doing work on the load resistance?

 

[tedward]

There is no magnetic flux passing through the pie shaped conductors that radiate out from the center of the solenoid and make contact with the single loop conductor in the @mabilde video. The flux lines cut lengthwise along those conductors.

I completely disagree. The pie-wedge shaped surface of the voltmeter loops are parallel to the ground. The magnetic field of the solenoid points straight up, at least in the interior of the cylinder he is measuring. The magnetic field therefore penetrates this surface at a perfect 90 degree angle, straight through. All he's measuring is the rate-of-change of this flux on his voltmeter.

 

[tedward]

Lewin demo/paradox and general ordinary transformers are electromagnetic induction systems, so the working principle is basically the same.
Haven't the common transformers we use every day already shown that a voltmeter can be used to measure scalar potential in such systems?
Even though it is difficult to measure with a voltmeter in some cases like the Lewin demo/paradox, is there really no other way?


I guess you mean that the electric field inside a perfect wire is 0. If the electric field in a region is zero, then of course there is no change in potential in that region. But note that this scale potential still acts on the wire surface and outer space


So isn't the voltage output by the transformer doing work on the load resistance?

View attachment 321777​

When you measure the output voltage of a transformer (taking a voltmeter as the load that completes the circuit), you can analyze it two ways. If you draw a loop that passes through the interior of the conducting wire in the solenoid, The only voltage drop is across the load. The emf is provided by the changing magnetic field which penetrates the 'loop'. The loop is a bit hard to visualize as it consists of a corkscrew, but imagine a piece of rotini pasta. The field lines penetrate this surface once every turn, giving the net emf. In this analysis, there is no voltage across the transformer, just an induced emf, using Faraday's law to analyze the loop.

The other analysis loop consists of the voltmeter and a path that connects the terminals of the solenoid OUTSIDE the windings (hence outside the circuit). There is no flux which penetrates this surface, but there is an electric field that runs outside the solenoid, whose integral is equivalent to the induced emf. Using this loop, you could fairly refer to that electric field integral as a voltage, and use Kirchoff's law to analyze the loop.

Either way, on the load side you measure a voltage. You can think of this as the induced emf from the magnetic field, or a voltage between points directly outside the solenoid. The former is more physically correct I would say as it is the fundamental principle involved.

The scalar potential ignores the induced emf completely. It bases it's definition of potential solely on the electric field generated by the electrostatic distribution of charge at the ends of the resistor. So if the voltmeter is the only load, it has charge build up on the ends of it's internal resistor. This charge distribution creates a static field that cancels out the induced field in the wire, but concentrates in the resistor itself. The sum of this field ONLY adds up to zero around the loop, and as such you can assign a number, or 'potential' , to any point in the loop. But as this 'potential' ignores the induced emf completely, it has no relation to work on a test charge as it moves around in the solenoid windings. It costs nothing to move a charge from higher to lower 'potential' or vice versa along the solenoid path, and thus renders the concept of potential meaningless (mot to mention, this definition of potential or voltage does not agree with Ohm's law unless the resistor in question has zero length - in a circular loop for instance, if a resistor took up a quarter of the circles length, the static or scalar potential across it would differ from I*R significantly).

However, it DOES require to work to move a test charge through the load resistor. Hence it takes different amounts of work to move a test charge from one terminal of the solenoid to the other depending on which path you take. So if you define voltage between two points in the traditional way, as the amount of work-per-unit-charge required to move a test charge between the points - or equivalently Int(E.dl)- you find the voltage depends on the path taken, i.e. path dependence. This is what a voltmeter measures and why Lewin gets different readings on his voltmeters.

Compare this to a regular DC circuit with a battery and a resistor. Moving a test charge from one terminal of the battery takes the same amount of work whether you move around the circuit through the resistor, or on the other path directly through the battery - i.e. path independence.

The bottom line is that a charge only 'feels' electric field. It doesn't care whether a field is induced or electrostatic. Our standard definitions of voltage are based on force and work, and should therefore depend only on how a charge is affected, not on mathematical abstractions that have no physical meaning.

 

[alan123hk]

@tedward
Thanks to #29 for the informative and valuable reply, a great comment. I agree with most of the points. But there are still some points that I don't fully agree with.

We are now dealing with an electromagnetic induction system, and we assume that its total electric field is the result of the superposition of the induced electric field and the scalar electric field produced by the distributed charge. But the total E field of this system is actually a non-conservative field with no potential difference. Since every part of the overall system is interrelated and affects each other, I don't think that "The scalar potential ignores the induced emf completely".

But as this 'potential' ignores the induced emf completely, it has no relation to work on a test charge as it moves around in the solenoid windings. It costs nothing to move a charge from higher to lower 'potential' or vice versa along the solenoid path, and thus renders the concept of potential meaningless

I think no matter anywhere in the system, including in the solenoid winding or in the external load circuit, whenever there is current passing through the resistor, there will be a relative scalar potential difference and corresponding energy loss, so this scalar potential difference provides energy for the resistor (The scalar potential around the solenoid winding wire is more complicated, it may not be equal to the current times the resistance of the wire due to the induced electric field), which is one of its meanings in the system. It works even if the net electric field in the solenoid zero resistance winding is zero. when current flows through the solenoid winding, the actual work of this scalar field is to accept the energy provided by the induced electric field, how can it be said that its existence is meaningless?

When you measure the output voltage of a transformer (taking a voltmeter as the load that completes the circuit), you can analyze it two ways. If you draw a loop that passes through the interior of the conducting wire in the solenoid, The only voltage drop is across the load. The emf is provided by the changing magnetic field which penetrates the 'loop'. The loop is a bit hard to visualize as it consists of a corkscrew, but imagine a piece of rotini pasta. The field lines penetrate this surface once every turn, giving the net emf. In this analysis, there is no voltage across the transformer, just an induced emf, using Faraday's law to analyze the loop.

The other analysis loop consists of the voltmeter and a path that connects the terminals of the solenoid OUTSIDE the windings (hence outside the circuit). There is no flux which penetrates this surface, but there is an electric field that runs outside the solenoid, whose integral is equivalent to the induced emf. Using this loop, you could fairly refer to that electric field integral as a voltage, and use Kirchoff's law to analyze the loop.

Please note that the output port of the solenoid is defined by ourselves. In fact, any two endpoints on any wire segment can define an output, and even though the solenoid has a zero-resistance winding, there is a corresponding scalar potential.

Transformer with multi-tap output on secondary winding​

this definition of potential or voltage does not agree with Ohm's law unless the resistor in question has zero length

I don't understand why you say that the definition of this scalar potential does not conform to Ohm's law.
Please refer to the link below #6,
https://www.physicsforums.com/threa...erms-of-current-density.1048001/#post-6830095

 

[Averagesupernova]

I completely disagree. The pie-wedge shaped surface of the voltmeter loops are parallel to the ground.

Relative to the ground is not relevant. Coil relative to the pie shaped is relevant.

The magnetic field of the solenoid points straight up, at least in the interior of the cylinder he is measuring. The magnetic field therefore penetrates this surface at a perfect 90 degree angle, straight through.

Yes the flux lines are precisely parallel with bore of the solenoid, or as you call them, straight up. But, as the field grows and shrinks with the current in the solenoid the vertical flux lines move in and out from the exact center of the bore. This does NOT cause the flux lines to cut across the short wires that make up the pie shape at an angle that is non parallel to the to those short pie shaped conductors. The flux lines simply travel parallel with those conductors. This does NOT cause a current to flow.
-
Think of the flux lines as actual strings or wires. The only way current is induced is if the flux 'strings' are to move in such a way as to tangle with the wire. Sliding parallel like a violin bow along the strings is a good example to compare to. The bow moving across the strings or the strings sliding across the bow will get a sound out of a violin but the same comparison with induction will not yield a current.

All he's measuring is the rate-of-change of this flux on his voltmeter.

Well, naturally that is reflected by the voltage generated but I'm not sure this is how you meant it.

 

[alan123hk]

I would like to introduce the relevant situation further. Does not violate Ohm's law. It is precisely because the definition of the scalar potential must be obeyed, and Ohm's law must be obeyed at the same time, so the calculation equation of the scalar potential must be $V_1-V_2=IR_{12}-\varepsilon _{12}~~~\Rightarrow ~V_1-V_2=IR_{12}-\int_1^2~dl\cdot E^{'}~~~~~~$

 

[tedward]

What I meant by saying the scalar potential ignores the induced emf, is that this definition of voltage focuses on the scalar potential and disregards the induced emf as a voltage, even though yes, the scalar potential is influenced by the induced field. Let's imagine this experiment: A good diagram would be to think of a circuit as a clock, with single resistor in a circular loop with a magnetic flux at the center, so the loop IS the transformer secondary, similar to the Lewin circuit. We place the tiny resistor at the top of the loop, at the 12 o'clock position. Say the emf is 12V, and the resistor is 12 ohms, giving a current of 1A. You draw the net E-field (induced plus electrostatic) along the circuit. It's zero everywhere except for inside the resistor (which is negligible length for now), and it points to the right. This field is large, and when multiplied by the length of the resistor gives 12V (same as the emf).

You then separate the diagram into two components - one for the induced field only, one for the electrostatic field. From symmetry, the induced field diagram is uniform around the clock, pointing clockwise, just as if it would be in free space, from 0 (midnight) to 12 noon. Now the static field diagram must cancel out the induced field diagram everywhere but in the resistor, so this diagram is also uniformly spread in the COUNTER-clockwise direction - starting at 12 noon and going back to 0, or midnight. The other difference about the static field is that it now just in the resistor, it points to the right, or clockwise.

Now since the static field sums up to zero around the clock, we can define a number that we'll call scalar potential. We disregard the induced field diagram and only focus on the potentials on the static field diagram. Taking midnight to be ground (0V), the scalar potential increases in the clockwise direction, against the static field. One hour is one volt, so going clockwise to noon gives you your 12V. But in the tiny space between noon and midnight (say this is one-second on the clock), the potential suddenly drops back down to zero.

Now overlay the actual clock circuit over the diagram, and write down the potential values around the diagram (same as the clock numbers in volts). Pick any two points, say 3:00 and 5:00. The scalar potential predicts a difference of 2V. Hook up a voltmeter to these points, and you measure zero. This is because the voltmeter only measures integral of NET E-field, it has no way of discerning static from induced. Now, pick up a small test charge and place it in the loop (this part is just a thought experiment of course). As you move this charge between 3:00 and 5:00, you will find it takes no work to move it in either direction, no matter the size of the charge, even though the scalar potential difference is 2V.

Now move the test charge from one side of the resistor to the other, through the resistor. If you go left from just after midnight to noon, it will take you 12V times your charge worth of work, as the scalar potential predicts. (If you go the other direction, the field pushes your charge for you and the work you do is -12V * q.) But now push your charge from say 2:00 to 10:00 through the resistor. The scalar potential difference is 8V. But the work you do is still 12V*q, just as when you moved it from midnight to noon.

What we're seeing is that you cannot do work calculations using the scalar potential only. This is what I mean when I say that scalar potential is unrelated to the work done on a charge. Without including the induced field, the notion of potential difference as work - something that is foundational for thinking about voltage - no longer holds. The best you can say, as you mentioned, is that the TOTAL of the scalar potential in the solenoid is equal to the amount of work done on the load. The individual values along the path are meaningless. (Edit: see below)

What about Ohm's Law? As long as the resistor is negligible length, there's no problem. But now say our 12 ohm resistor extends uniformly from 6:00 to 12 noon clockwise, so on the left side of the clock. Emf and current are still the same, 12V and 1A respectively. So by Ohm's law, the Int(E.dl) voltage (I'll call this path voltage) across this resistor is 12V. But considering the induced emf ONLY, by symmetry, this must be 6V, spread uniformly. Therefore, the electrostatic potential across the resistor is also 6V, as they add up to 12V. But the scalar potential, 6V, doesn't give you IR which is 12v. This is what I mean when I say scalar potential doesn't conform to Ohm's law. You can still apply Ohm's law, but not without including the induced field, per the equation you linked to.

It seems to me that using this definition of voltage is bending over backwards to satisfy a single requirement: "Sum of voltage around a loop is always zero." But what are we giving up? You have to redefine Ohm's law to include a separate term, the integral of the induced field. You give up the notion of potential difference as work, unless you include that same term. You can't measure it with a voltmeter. And it does nothing to change the physics - the fact that the net E-field is still non-conservative and therefore path dependent.

On top of that, imagine a circuit that has no symmetry, just a messy loop with an irregular field. Trying to calculate the static potential at every point means you have to somehow integrate the induced field up to any point. It's a damn near impossible task except for the most symmetric circuit. And the minute you do work out all the scalar potentials, as soon you move wires around everything is different. The net E-field voltage doesn't have this problem.

It's far more useful to define 'path voltage' as the integral of E.dl, as it is the same voltage that V=IR uses, is measurable by a voltmeter, and naturally corresponds to work done on a test-charge along a given path. Yes, you now have to give up path independence and the idea of a fixed potential, but that's a direct consequence of the Maxwell-Faraday equation, and should not be surprising.

(In your transformer diagram, all choosing different taps does is change the amount of emf in your circuit, equivalent to increasing the number of windings or strength of d(phi)/dt.) EDIT: Ok I understand why you included this diagram, the scalar potential markings indicate the amount of emf you'll get. But that still only applies if you use a path between the terminal points that flux penetrates. If you add up the E.dl voltages along the wire itself, between ANY two points, or as in my clock circuit above, you get zero.

 

[tedward]

I would like to introduce the relevant situation further. Does not violate Ohm's law. It is precisely because the definition of the scalar potential must be obeyed, and Ohm's law must be obeyed at the same time, so the calculation equation of the scalar potential must be $V_1-V_2=IR_{12}-\varepsilon _{12}~~~\Rightarrow ~V_1-V_2=IR_{12}-\int_1^2~dl\cdot E^{'}~~~~~~$

Right - this is the new definition of Ohm's law that accommodates scalar potential. But the good old V = IR that everyone actually uses and is measured by a voltmeter is the path voltage, Int(E.dl).

 

[hutchphd]

It is precisely because the definition of the scalar potential must be obeyed, and Ohm's law must be obeyed at the same tim

The only law that must be obeyed in physics territory is Maxwell. Period.

(And tax Law)