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