#562 – Electroboom!

[Sredni]

Which is another thing I think I understood now: charge particles that are inert referenced to the magnetic flux are not interacting with the magnetic field, even if it's changing.

Whoa! My spider senses are tingling all over.
Care to rethink this?

[thinkfat]

Which is another thing I think I understood now: charge particles that are inert referenced to the magnetic flux are not interacting with the magnetic field, even if it's changing.

Whoa! My spider senses are tingling all over.
Care to rethink this?

Yes, of course, but I'm pretty sure about that. It explains, IMHO, that Lorentz' force and the electromagnetic force are two independent effects and not just two sides of the same coin. It would also serve as an explanation for why, as @bsfeechannel said, a wire in time varying magnetic flux doesn't have an "inner electric field" while when it is moving in a (constant) magnetic flux, there is an "inner electric field". In the case of the time varying magnetic flux the charge particles interact not with the magnetic field but with the electric field that results from the magnetic flux change. That's how I understood the Maxwell-Faraday equation: change in magnetic flux causes a rotational electric field.

[bdunham7]

But you said that you are measuring the voltage across both the wire and the resistor with that configuration. Or did get it wrong?

I'm measuring from one point to another.  In this case, it a happens to be both across the resistor and across the partial turn, outside of any non-conservative field.

[bsfeechannel]

By irreconcilable I mean that they aren't going to be the same number in a non-conservative field.

It's because YOUR definition of voltage fails. The line integral of the electric field is telling you the right answer. And how do we know it is right? Because you can find absolutely no contradiction between what it predicts and what you measure.

The line integral of the electric field says you'll find different voltages between two points under a non-conservative field. You connect your meters there and BANG! that's what they measure.

Heck, it even gives you the numerical value of what the meters are going to show with absolute accuracy.

YOUR definition of voltage doesn't predict any of that and does not allow me to calculate what the meters are going to measure requiring all sorts of tricks and cheats both speculative and practical to make the experiment conform to your assumptions and it leaves a lot of loose ends that, when tried to be dealt with, lead to contradictions.

The general definition of voltage depicts reality. YOUR definition of voltage depicts the struggle to understand it.

You say "be it conservative or non-conservative", but I think perhaps the Magneto-Quasi-Static idea has fooled you into thinking that the concepts from conservative fields can be applied to non-conservative fields as long as they hold steady for some period of time.  I'm pretty sure that is untrue, but I'm struggling to demonstrate that.

It is the other way around. The concepts from the non-conservative field can be applied to the conservative field, which is just a special case of the former when you do not have a varying magnetic field. The quasi static is there to guarantee that the wavelengths associated with the time-variations are much greater than the dimensions of our geometry, otherwise things get really hairy.

[jesuscf]

I believe what many of us participating in the discussion initially didn't understand, me explicitly included, is that Dr. Lewins experiment must be seen as a whole, including his choice of probing the voltages. The whole idea about formulating a "law" in science is to be able to explain and predict the outcome of an experiment, and as I already said but probably nobody really took notice, the test instrument setup including geometrical arrangement of probe wires is inevitably part of the experiment. Dr. Lewin proposed a circuit and the challenge was to explain the measurements, and quite honestly, you cannot do that with Kirchhoff. It doesn't work.

I beg to disagree.  A proper experiment would had included many measurements from different geometric probing configurations.  Lewin  purposefully picked a configuration that he believed would eliminate the effect of the varying magnetic field in the measurement equipment while measuring the voltage between the top and the bottom of the ring.   He ended up cancelling the effect of the magnetic field both in the probes and in the ring!  That is equivalent to measure the voltage across the resistors directly.

Lewin then assumes that the voltage across any two points in the ring wire is zero volts, because he applies ohms law and the wire resistance is almost zero, and he goes AHA! KVL doesn't work!!!  The problem here is that Lewin forgot that a piece of wire in a circuit under the influence a varying magnetic field doesn't behave as zero ohm resistor but as non-ideal voltage source.  When you account for that extra little piece of information, all of a sudden KVL works perfectly, no matter the probing geometry, if you include the probes as part of your circuit.

[bdunham7]

The general definition of voltage depicts reality. YOUR definition of voltage depicts the struggle to understand it.

It's not my definition of voltage, it is a definition that was taught earlier in Lewin's course, among a zillion other places that it pops up.

If you posit that a conductor is an equipotential even in a quasi-static but rotational, non-conservative field then you are saying that my example with the torus, winding and straight rod results in an equipotential over the entire surface of that conductive rod.  Is that an accurate assessment of your position?

[bsfeechannel]

But you said that you are measuring the voltage across both the wire and the resistor with that configuration. Or did I get it wrong?

I'm measuring from one point to another.  In this case, it a happens to be both across the resistor and across the partial turn, outside of any non-conservative field.

You are just measuring a voltage due to a conservative field that is present in the resistor and between the terminals of the wire. Assuming the wire has no resistance, you can slide the meter all the way down to the left, just before you cross the lines of the mag field. You're measuring nothing else. You're not measuring the voltage due to the field in the wire. For that you need to place your meter anywhwere three-dimensionally speaking to the left of the field.

[Sredni]

Which is another thing I think I understood now: charge particles that are inert referenced to the magnetic flux are not interacting with the magnetic field, even if it's changing.

Whoa! My spider senses are tingling all over.
Care to rethink this?

Yes, of course, but I'm pretty sure about that. It explains, IMHO, that Lorentz' force and the electromagnetic force are two independent effects and not just two sides of the same coin.

Well,

F = q ( E + v x B)

if B = 0 or v = 0 (or v//B) we kiss goodbye to Lorentz's force. We are left with E.
And a variable magnetic field in the core will be associated with a circulating Eind field around the core. So, a charge nearby will feel the force and will accelerate in the tangential direction. The changing magnetic field makes the particle move. I can call that 'an interaction'.
They built accelerators based on this principles: the betatrons.
See, for example, http://web.mit.edu/course/22/22.09/ClassHandouts/Charged%20Particle%20Accel/CHAP11.PDF

"A toroidal vacuum chamber encircles the core of a large magnet. The magnetic field is produced by pulsed coils; the magnetic flux inside the radius of the vacuum chamber changes with time. Increasing flux generates an azimuthal electric field which accelerates electrons in the chamber.
In the absence of an air gap, there is little magnetic flux outside the core."

I think you should reconsider.
(Also magnetic and electric effects are the two faces of the same coin. A magnetic field is but an electric field in another frame of reference. But I'm not gonna touch that in here. But if you are interested there is a new book with a nonconventional approach

Kjell Prytz
Electrodynamics: The Field-Free Approach:
Electrostatics, Magnetism, Induction, Relativity and Field Theory
Undergraduate Lecture Notes in Physics
2015, Spinger

He does not even use fields. Just forces between charges. It needs a bit of editing in my opinion - looks like a work in progress, with some of the theory discussed in the exercises and apparent gaps in the text but it makes a very interesting read. I am pretty sure bsfeechannel and rsfeec will find it interesting)

[bsfeechannel]

It's not my definition of voltage, it is a definition that was taught earlier in Lewin's course, among a zillion other places that it pops up.

Care to provide a reference to it in Lewin's course, please?

[bdunham7]

You are just measuring a voltage due to a conservative field that is present in the resistor and between the terminals of the wire. Assuming the wire has no resistance, you can slide the meter all the way down to the left, just before you cross the lines of the mag field. You're measuring nothing else. You're not measuring the voltage due to the field in the wire. For that you need to place your meter anywhwere three-dimensionally speaking to the left of the field.

How do you define the term 'equipotential'?

Let's assume I'm measuring between any two points with an analog voltmeter, just to make things a bit clearer.  This would be a galvanometer in series with a resistor.  So if I connect such a voltmeter to two points, current will not flow if they are equipotential, but will if there is a potential there.  You can hide the entire apparatus from me, but as long as my meter and test leads are free from any external fields, B or E, I will get a certain result that will indicate whether those two points are equipotential or not.  Free of external fields, there is no dispute over what voltage is.  I don't need to have any knowledge of the source or cause of the voltage, or lack thereof.  Or do you disagree with that as well?

[bdunham7]

Care to provide a reference to it in Lewin's course, please?

Maybe I can look a bit later, but which part do you doubt?   C = 4 * pi * E₀ * r  or Q = CV?

[bsfeechannel]

I'll tell you what my crystal ball forecasts: that your method to measure voltage across the rod will force you to move the sensing instrument on a path that, together with the rod, will form a closed path around the variable magnetic region.

No, your insistence on a closed path is your downfall.  You only insist on it because it makes your math work.

The insistence in the closed path is because we are talking about circuits, which by definition are closed paths.

[bdunham7]

The insistence in the closed path is because we are talking about circuits, which by definition are closed paths.

Your arguments are growing feeble.  I'm disappointed!

Obviously a debate over KVL necessarily refers to a closed path, but what we seem to be debating is related to specific parts of that path and I see no problem in breaking them down.

[bsfeechannel]

How do you define the term 'equipotential'?

Let's assume I'm measuring between any two points with an analog voltmeter, just to make things a bit clearer.  This would be a galvanometer in series with a resistor.  So if I connect such a voltmeter to two points, current will not flow if they are equipotential, but will if there is a potential there.  You can hide the entire apparatus from me, but as long as my meter and test leads are free from any external fields, B or E, I will get a certain result that will indicate whether those two points are equipotential or not.  Free of external fields, there is no dispute over what voltage is.  I don't need to have any knowledge of the source or cause of the voltage, or lack thereof.  Or do you disagree with that as well?

Suppose I agree. Please, go on.

Care to provide a reference to it in Lewin's course, please?

Maybe I can look a bit later, but which part do you doubt?   C = 4 * pi * E₀ * r  or Q = CV?

Where, when, what did Lewin exactly say? You need to support your claims with evidence, otherwise you'll have a feeble argument. As for your electrostatic equations, please enlighten me as to what this has to do with a rotational electric field along the path of a circuit. I read your posts, obviously, but nothing in them gave me a clue about the fact that you'll not have undefined voltages between two points under a varying magnetic field.

[Sredni]

It should measure zero, simply because it needs to. 

Emphasis mine.

Ok, I have re-read the description of your concept of 'absolute voltage', the description of 'the machine' and how it should work in a nonconservative and in a conservative field. I... I...

I need a shot of tequila.

<pause>

I don't know where to begin. Let me say that your instrument appears to run on a lot of wishful thinking. You use the formula for self-capacitance to infer voltage? But is your sphere alone in the vacuum of space? I don't think so. First off, your machine requires two spheres, so you will have to account for the field lines going from one sphere to the other. Then you have the object you want to measure the charge of that will interact with those field lines, and the source of the field, and the whole planet beneath your feet. You will have a full fledged matrix of coefficients of capacitance to handle - and that alone would kill the independence of your instrument.

And that thing about the field lines being perpendicular to the path of the spheres? Wishful thinking at its best. The moment the field induces charges on your rod, they will distort the field - not only inside the body to create the zero field, but also outside. And the distortion will be significant because the field lines needs to be perpendicular to the surface of the conductor.
And then your spheres will have induced charge themselves, that will disturb the field even more and change the charge distribution of the rod as you move near and away from it.
And why should the charge displaced on the rod jump on the spheres (which will have their own induced charge already?)

And what do you want to do in the nonconservative case? "extend the ends out to the point where the fields are negligible" What fields? Not the B field, it was never outside the core to begin with. So it must be the electric field. It can't be Ecoul, since it is generated by the charges themselves. Must be the induced field Eind, then. And you expect to see the displaced charge where there is no Eind field? If there is no more Eind field, what keeps the charges segregated at the extremes of the rod? All that charges of the same sign crammed together at one extreme? Wouldn't you think they'd repel each other?

And then you bring the alleged charge transferred on the spheres back together in the same place where you can measure it? My crystal ball was right. You are closing the path, after all (but let's not talk about this now). Even admitting you can magically duplicate (this could actually work) or take the charge of one extreme of the rod and the charge on the other extreme and then infer the voltage between them - you would see a voltage between them even in the electrostatic case. (because you have removed the external field Eext that causes the electrostatic induction of the rod). So, should we also say that there is a voltage between two points of a conductor in an electrostatic field?

Tequila is my lady, tonight.

[bdunham7]

Yes, of course, but I'm pretty sure about that. It explains, IMHO, that Lorentz' force and the electromagnetic force are two independent effects and not just two sides of the same coin. It would also serve as an explanation for why, as @bsfeechannel said, a wire in time varying magnetic flux doesn't have an "inner electric field" while when it is moving in a (constant) magnetic flux, there is an "inner electric field". In the case of the time varying magnetic flux the charge particles interact not with the magnetic field but with the electric field that results from the magnetic flux change. That's how I understood the Maxwell-Faraday equation: change in magnetic flux causes a rotational electric field.

I can't prove it at this time,  but I think that doesn't work.  It posits two different types of EMF resulting from non-conservative forces (energy can be extracted) one with a so-called inner E-field in a conductor and presumably a potential gradient and the other with no inner E-field and an equipotential conductor.  Both EMF forces push on charges, what is different about them that would allow that result?  Remember, in both cases the charges are being pushed out and around to potentially (!) do work, whereas in the static, irrotational conservative field, that can't happen.

[bdunham7]

First off, your machine requires two spheres, so you will have to account for the field lines going from one sphere to the other. Then you have the object you want to measure the charge of that will interact with those field lines, and the source of the field, and the whole planet beneath your feet.

Actually, it only requires one sphere.  The 'machine' will measure Q precisely by adding or subtracting electrons until the sphere is exactly neutral.  The two-sphere model allows simultaneous measurements, so I can equate it to an ideal voltmeter in the right conditions.  As for the difficulties, experimental physics isn't simple!  Look at what Millikan did 100+ years ago.

And that thing about the field lines being perpendicular to the path of the spheres? Wishful thinking at its best. The moment the field induces charges on your rod, they will distort the field - not only inside the body to create the zero field, but also outside. And the distortion will be significant because the field lines needs to be perpendicular to the surface of the conductor.

Yes, idea requires me to move the spheres in contact and then back out to a field-free zone by only crossing field lines at right angles, or IOW, along an equipotential.  Difficult, but I don't think theoretically impossible.  Or if it is technically theoretically impossible due to the perpendicular field issue, at least possible to get infinitesmally close.

And then your spheres will have induced charge themselves, that will disturb the field even more and change the charge distribution of the rod as you move near and away from it.
And why should the charge displaced on the rod jump on the spheres (which will have their own induced charge already?)

Well there's the thing--I think the charges won't jump on the sphere because they're happy where they are.   Just like a the voltmeter that I drew above the static field won't show anything.  But I couldn't see an easy way to prove that.  And as you pointed out, I'd need to prove that in the general case.

And what do you want to do in the nonconservative case? "extend the ends out to the point where the fields are negligible" What fields? Not the B field, it was never outside the core to begin with. So it must be the electric field. It can't be Ecoul, since it is generated by the charges themselves. Must be the induced field Eind, then. And you expect to see the displaced charge where there is no Eind field? If there is no more Eind field, what keeps the charges segregated at the extremes of the rod? All that charges of the same sign crammed together at one extreme? Wouldn't you think they'd repel each other?

Yes, they will repel each other until they are evenly distributed in the no-field zone.  And yes, I mean E_ind, the rotational E-field around the flux.  If I have, for example, a U-shaped rod that goes partly around the flux and then extends out a ways, I would expect to see a gradient in the area subject to the rotational E-field and then two long ends with evenly distributed but not neutral charge.  Thus when I test the ends of that rod, I will see a potential between them regardless of what method I use because there's no external field to worry about.

And then you bring the alleged charge transferred on the spheres back together in the same place where you can measure it? My crystal ball was right. You are closing the path, after all (but let's not talk about this now). Even admitting you can magically duplicate (this could actually work) or take the charge of one extreme of the rod and the charge on the other extreme and then infer the voltage between them - you would see a voltage between them even in the electrostatic case. (because you have removed the external field Eext that causes the electrostatic induction of the rod). So, should we also say that there is a voltage between two points of a conductor in an electrostatic field?

See, I can predict your thinking as well--the closed path.  That's why I went to the extra effort of measuring each balls absolute charge independently of the other.  I can actually not close the path.  But IMO it doesn't matter, closing the path far away from the action has negligible effect and you shouldn't rely on technicalities, infinitesmals or magic to make a practical theory work. 

Tequila is my lady, tonight.

Scotch for me!

[bsfeechannel]

As for the difficulties, experimental physics isn't simple!

Difficult, but I don't think theoretically impossible.

But I couldn't see an easy way to prove that.

But IMO it doesn't matter, closing the path far away from the action has negligible effect and you shouldn't rely on technicalities, infinitesmals or magic to make a practical theory work. 

So let me get this straight. You can't prove none of your claims and you want to reinvent the wheel with a "practical" theory that doesn't rely on "technicalities, infinitesimals or magic" that again you can't even demonstrate in practice because "experimental physics isn't simple".

You said you're not a KVLer, but you talk and think like one. Don't get me wrong, but your posts look like a bunch of delusional speculations.

On the other hand, Faraday's law of induction, which renders KVL useless for varying magnetic fields, is not only easily provable, but also easily demonstrated by anyone with a spool of wire, a battery and a meter. That's how Faraday himself discovered the phenomenon. You don't need a fancy physics lab.

As for the practical side of it, Maxwell's equation for Faraday's law is what is used everyday in the world to calculate the number of turns of the primary of any transformer. For a linear transformer, for example the peak voltage per turn is calculated as follows:

Vpeak = 2πf*Bmax*A

You can easily recognize it. Vpeak is the line integral of E around the turn. 2πf is the result of the time derivative of cos(2πf), Bmax is the peak magnetic field just before saturation and A is the transversal area of the core. Define the peak voltage on the primary and divide it by Vpeak and you have the number of turns. That simple. No need for brass balls.

There are billions of them in the world today. There's nothing more practical and consistent than this.

[bdunham7]

So let me get this straight. You can't prove none of your claims and you want to reinvent the wheel with a "practical" theory that doesn't rely on "technicalities, infinitesimals or magic" that again you can't even demonstrate in practice because "experimental physics isn't simple".

Sir, this is EEVBlog.

On the other hand, Faraday's law of induction, which renders KVL useless for varying magnetic fields, is not only easily provable, but also easily demonstrated by anyone with a spool of wire, a battery and a meter. That's how Faraday himself discovered the phenomenon. You don't need a fancy physics lab.

I can't imagine how you could possibly read a denial of Faraday's Law in anything I've written.  If some of it seems too far out there, address a  few simpler ideas.

Start with the long U-shaped conductor.  Is there potential across the ends or not?  If so, how goes your claim of equipotentiality in non-conservative fields?  If not, why not? 

As I was responding to Sredni and thinkfat it occured to me that there's another issue with the equipotentiality argument.  In the static, conservative, irrotational field, there isn't an E-field inside the conductor because the charges instantly rearrange themselves to oppose it.  Thus there is no net force on any charge within the conductor.  In the case of a charge moving in a magnetic field, you accept that there is an non-conservative field inside the conductor acting on the charges, right?  But then we get to the MQS system, and even though there is clearly a local force acting on charges inside the conductor--which is the EMF--and those forces continually push charges through the conductor, that the conductor is nonetheless equipotential in the same way as in the static case, and for the same reason--there's no electric field in a conductor.  But what I'm showing you with my U-shaped conductor example is that while the charges will rearrange themselves to counter any net electric field in the conductor, the result may not be an equipotential surface.  Instead, the charges are pumped one way or another, with the result that one end has a different potential than the other. 

Here is a non-closed path example without any as-of-yet uninvented machines, except I'm not sure they have electroscopes this good.  What happens when the current is turned on?

[Sredni]

Start with the long U-shaped conductor.  Is there potential across the ends or not?

You are asking the wrong question. It's probably my fault because in a previous post I used the term 'equipotential' in quotes to mean 'the voltage difference between two points is zero' (inside a certain cylinder), so let me explain again.
You can no longer talk about potential function for the total electric field. It has no longer sense because the path integral of the total electric field depends on the path joining two points.
Potential no mas.

So your question has no sense. You should ask, instead: is there a voltage across the ends?
And the answer is:

IT DEPENDS ON THE PATH

What path? The path along which you want voltage, which is a function of the endpoints AND THE PATH, to be evaluated.
For all paths that together with the conductor (let me rephrase: together with any path inside the conductor that joins point A and B - it does not matter which because the electric field inside is exactly zero if there is a gap)--- let me repeat: for all paths that together with the conductor form a closed path that DOES NOT go around the dB/dt region, the voltage difference is ZERO.
For all paths that, together with the conductor, DO GO around the dB/dt region, the voltage difference is 1 EMF (or -1 depending on how we go around, or a multiple of the EMF if we go around several times).

Tomorrow I will add a couple of pictures that hopefully will clarify what I wrote.

As I was responding to Sredni and thinkfat it occured to me that there's another issue with the equipotentiality argument.  In the static, conservative, irrotational field, there isn't an E-field inside the conductor because the charges instantly rearrange themselves to oppose it.  Thus there is no net force on any charge within the conductor.  In the case of a charge moving in a magnetic field, you accept that there is an non-conservative field inside the conductor acting on the charges, right?  But then we get to the MQS system, and even though there is clearly a local force acting on charges inside the conductor--which is the EMF--

Stop. This is wrong.
It is NOT the EMF. We always end up here: forgetting the role of the displaced charge. The EMF -  or better, the induced electric field Eind, is compensated in the conductor by the coloumbian field Ecoul. If there is a gap and no current is flowing, this is always exactly true.
If there is a current flowing (because we closed the circuit) then it is zero only if the conductor is perfect (zero resistivity, infinite conductivity). Charges move without a field. In practice, with real conductors, the compensation is not perfect and there is a residual field E in the conductor, directed along the conductor and whose value complies with the local form of Ohm's law: E = j / sigma_copper.

How does it follow the conductor? Answer: surface charge. (Read here, if you are interested https://electronics.stackexchange.com/questions/532541/is-the-electric-field-in-a-wire-constant/532550#532550 )

So, it is not Eind that is acting on the carriers. It is the combined effect of the EMF (represented by Eind) and the displaced charge (represented by Ecoul).

and those forces continually push charges through the conductor, that the conductor is nonetheless equipotential in the same way as in the static case, and for the same reason--there's no electric field in a conductor.

and this is again (EDIT: in general) wrong for the reason above. But, if you consider a region of space that contains all of your piece of conductor and none of the magnetic flux, like a cylinder around the rod in the middle of the torus, in that region of space all paths you can imagine will never enclose the magnetic flux region. in this sense you can consider it as equipotential. Shrink the surface to include only the conductor if you want. But the moment you consider a path the forms, with the conductor, a closed path that cuts the flux - NO MORE.

(I have a picture to visualize this as well - but I drank too much tequila to operate a scanner)

[thinkfat]

I believe what many of us participating in the discussion initially didn't understand, me explicitly included, is that Dr. Lewins experiment must be seen as a whole, including his choice of probing the voltages. The whole idea about formulating a "law" in science is to be able to explain and predict the outcome of an experiment, and as I already said but probably nobody really took notice, the test instrument setup including geometrical arrangement of probe wires is inevitably part of the experiment. Dr. Lewin proposed a circuit and the challenge was to explain the measurements, and quite honestly, you cannot do that with Kirchhoff. It doesn't work.

I beg to disagree.  A proper experiment would had included many measurements from different geometric probing configurations. Lewin purposefully picked a configuration that he believed would eliminate the effect of the varying magnetic field in the measurement equipment while measuring the voltage between the top and the bottom of the ring.   He ended up cancelling the effect of the magnetic field both in the probes and in the ring! That is equivalent to measure the voltage across the resistors directly.

Lewin then assumes that the voltage across any two points in the ring wire is zero volts, because he applies ohms law and the wire resistance is almost zero, and he goes AHA! KVL doesn't work!!!  The problem here is that Lewin forgot that a piece of wire in a circuit under the influence a varying magnetic field doesn't behave as zero ohm resistor but as non-ideal voltage source.  When you account for that extra little piece of information, all of a sudden KVL works perfectly, no matter the probing geometry, if you include the probes as part of your circuit.

I suggest watching that other MIT video, https://youtu.be/u6ud7JD0fV4. The thing is, the measurement loops C₁ and C₂ are not influenced by any magnetic flux. The demonstration even uses a toroid core to make sure that there is no flux outside of the core. There's nothing you'd need to cancel or compensate for. And the setup is not even very peculiar, the probe wires are just laying there on the bench with no particular care taken to fixate them anywhere. They're just "flopping around in the breeze".

But anyway, would you be able to explain what the oscilloscope shows in that video with just KVL?

I actually like that setup, because the core is a solid loop, there is no point in arguing "stray magnetic fields" and no way to contort the setup to "eliminate bad probing".

[thinkfat]

"A toroidal vacuum chamber encircles the core of a large magnet. The magnetic field is produced by pulsed coils; the magnetic flux inside the radius of the vacuum chamber changes with time. Increasing flux generates an azimuthal electric field which accelerates electrons in the chamber.
In the absence of an air gap, there is little magnetic flux outside the core."

Emphasis mine, this is what I was referring to. It's the electric field the electrons interact with.

[jesuscf]

I suggest watching that other MIT video, https://youtu.be/u6ud7JD0fV4. The thing is, the measurement loops C₁ and C₂ are not influenced by any magnetic flux. The demonstration even uses a toroid core to make sure that there is no flux outside of the core. There's nothing you'd need to cancel or compensate for. And the setup is not even very peculiar, the probe wires are just laying there on the bench with no particular care taken to fixate them anywhere. They're just "flopping around in the breeze".

But anyway, would you be able to explain what the oscilloscope shows in that video with just KVL?

I actually like that setup, because the core is a solid loop, there is no point in arguing "stray magnetic fields" and no way to contort the setup to "eliminate bad probing".

It is the same experiment and they are making exactly the same mistake Lewin made!!!  What induces the voltage in the wires is the rate of change of magnetic flux through the loops, even if the magnetic flux is confined in the core.   Adding a core to confine the flux inside the loops increases the efficiency of the transformer as no flux is wasted outside, but it doesn't change  the experiment.  They are still assuming that the wires in the secondary behave as short circuits with zero volts across them, while under the influence of external varying magnetic field.  That assumption is not correct.  Under the influence of an external magnetic field those secondary wires behave as a non-ideal voltage sources (or put in other words, the wire has both an 'inductance' and 'resistance').  The resistors in the loops also behaving as non-ideal voltage sources (they also have 'inductance' and 'resistance').

Check this video from 'fromjesse' with the very same experiment where he explains much better than me what is going on.  The title of the video says it all: "The Lewin loop inside an iron core - KVL still holds":



Also, I am attaching a couple of figures.  The first figure shows the circuit Lewin think he was dealing with.  The second figure shows the actual circuit Lewin should had used, based on his original experiment.  When you prove correctly with the oscilloscope you get rid of the induced voltages in series with the proves in the second figure, but the induced voltages in the wires connecting the resistors remain.

[thinkfat]

Check this video from 'fromjesse' with the very same experiment where he explains much better than me what is going on.  The title of the video says it all: "The Lewin loop inside an iron core - KVL still holds":

I'm afraid this won't make your argument stronger. You really need to take geometry and the fields into account and where the magnetic flux "flows".

The core is split, it's two toroids which both carry about half of the total flux each. His "Lewin loop" is around the center column where the two toroids join, same as the yellow reference loop, they both experience the total magnetic flux. However, when he measures the voltage across the "red" parts, his probe leads and the red wire form a loop around one of the two toroids, encompassing half of the magnetic flux. So what he actually measures is not the voltage across that red wire, it is the EMF in his measurement loop, which is, since it experiences about half of the total flux, half of the voltage induced in his reference loop.

PS: this would be a brilliant experiment: Use a split core with different permeability in each of the toroids, so that the flux is no longer evenly split. I wonder if the voltages would still add up

[bsfeechannel]

Sir, this is EEVBlog.

I didn't mean you cannot discuss such things here. What I meant is to call your attention to the fact that you need to provide evidence and proof to your claims.

I can't imagine how you could possibly read a denial of Faraday's Law in anything I've written.  If some of it seems too far out there, address a  few simpler ideas.

I'm not saying you're denying Faraday's law. Perhaps you don't understand the full implications of it yet. But that's not your fault.

Here is a non-closed path example without any as-of-yet uninvented machines, except I'm not sure they have electroscopes this good.  What happens when the current is turned on?

OK. Let's suppose that you produce a sawtooth magnetic field pulse in the torus, which your drawing seems to indicate. We will have a square pulse voltage between the terminals of the wire. While the square pulse exists, let's suppose your electroscopes indicate the displacement of charges at the terminals of the wire.

Where do we go from here?