What is the relationship between voltage and flux in Faraday's Law?

[hutchphd]

Therefore, my own conclusion is that although this may seem confusing, it may simply be because everyone in different situations may have different perspectives and priorities when dealing with the same issue.

It is confusing, and that is why scientists agree on certain definitions. You can always think about electric fields as made of cotton candy...I could not care less. But you cannot define "the potential difference" in a non-conservative field. That definition is already taken for conservative fields. So when $\partial_t \vec B \neq 0$ we do not call it a voltage difference nor should you, regardless of your perspective.

 

[alan123hk]

It is confusing, and that is why scientists agree on certain definitions. You can always think about electric fields as made of cotton candy...I could not care less. But you cannot define "the potential difference" in a non-conservative field. That definition is already taken for conservative fields. So when ∂tB→≠0 we do not call it a voltage difference nor should you, regardless of your perspective.

OK, I understand and agree that I should not define "potential differences" in non-conservative fields. By definition there is no potential difference in non-conservative field.

But in actual transformer circuit engineering analysis, people usually call it "voltage drop". Of course, it does not mean that it is intended to say that there will be a potential difference in a non-conservative electric field. This may just be the name they use in circuit analysis, or just think of it as the voltage drop in the lumped circuit analysis equation.

Quote from https://www.electronics-tutorials.ws/transformer/voltage-regulation.html
" Thus voltage drops due to the windings internal resistance and its leakage reactance causes the output terminal voltage to change."

 

[Charles Link]

It might interest you that in measuring the voltage from an inductor, the result is $V=+\int E_{induced} \cdot ds$. If the source is purely electrostatic, the result is $V=-\int E_s \cdot ds$. (Notice the signs on these two integrals).
It is conceivable that what is being measured in most cases is the resulting electrostatic field that necessarily occurs which, as mentioned in post 25, is equal and opposite the $E_{induced}$ that is contained within the power source. The electrostatic field, being conservative, must have $\oint E_s \cdot ds=0$, and thereby we will get the same voltage for any path between the two terminals.
Professor Lewin's case (where you can get two different readings) is probably not applicable to most circuits that one encounters, but one lesson from it would be when taking a voltage reading (e.g. with an oscilloscope), be sure your leads aren't surrounding or near a current carrying line (which may be generating a changing magnetic field). To minimize stray signals, it can be helpful to have the two wires that are doing the measurement to be in the form of a coaxial cable, that has an outer conductive sleeve that is separated from the center conductor by an insulator. Coaxial cables are commonly used in the electronics lab when taking oscilloscope readings of a circuit, especially if the component being measured and the oscilloscope are not in close proximity.

 

[alan123hk]

I came up with another idea.
The situation is like this, the induced electric field will exert a force to the charge in the conductor, and then the current will flow through the circuit, and then according to Ohm’s law, a voltage will be generated across the resistor, so there are electric charges accumulation on both ends of the resistor. These charges accumulated in fixed positions will generate a conservative electric field, so actually there are two different electric fields superimposed. This statement seems more convincing and accurate, and doesn't contradict the definition. I personally think that it seems quite reasonable, at least for the time being.

 

[DaveE]

Their is a certain schadenfreude inherent but it also slightly disconcerting. This is a fundamental albeit subtle mistake for EE professors to screw up.

I can only speak for my experience with EE professors, but, I don't think they do screw this up, at least not at Caltech or Santa Clara Univ. What they do do is partition the problem into "circuit elements". Some like inductors and transformers have induced voltage, which is correctly explained, in my experience. Others have no significant flux linkage with changing fields. This is a nice way to model the world of circuits since, as a rule, this is how things are built (except, of course when they aren't). They know that a transformer winding can generate voltage in a magnitude and polarity different than you would expect from it's ohmic resistance, they know why, and they move on. Or they do detailed EM modeling and analysis of a distributed, perhaps complex structure. Any analog EE understands that to really model the workings of a transformer winding is a significant undertaking, that, frankly, you hope someone else did for you; or you just model the external salient features because that's all you needed.

Further, I think every good analog or RF EE knows about induced voltages in wires. [generic insult directed to digital EE's deleted for civility]. As others have said, you don't have to spend too much time using oscilloscope probes to realize this happens. They may say something about RFI or antennas, but, you know, "a rose by any other name...".

I think this thread is the dual case of "why can current flow through a capacitor when there aren't any conductors across it." Good EEs are comfortable with the physics of both idealized inductors and capacitors. Frankly I think Lewin does a disservice to Kirchhoff's laws which are only intended to be applied in lumped element circuits, and which I think all the good EEs understand. In the lumped element world, this problem is "what L, and C really are", not in the circuits, not in the induced voltage in loops. Yes, we know L⋅(di/dt) is an induced voltage, we know why and then we use it as a circuit element just like we use C⋅(dv/dt); works great, less filling.

Honestly, I don't think it's that complicated unless people can't keep their contexts straight.

 

[hutchphd]

Honestly, I don't think it's that complicated unless people can't keep their contexts straight.

Well said. I can say that one of my best lifelong friends who is a very good analog EE had to be brought kicking and screaming to this truth. He is not an RF guy though.
And Lewin very clearly talks about colleagues who accused him of chicanery in his lecture (see 48:25 et seq) which I, too, found surprising. I believe something akin to "politically correct speech" may actually be important here.

.

 

[Delta2]

I came up with another idea.
The situation is like this, the induced electric field will exert a force to the charge in the conductor, and then the current will flow through the circuit, and then according to Ohm’s law, a voltage will be generated across the resistor, so there are electric charges accumulation on both ends of the resistor. These charges accumulated in fixed positions will generate a conservative electric field, so actually there are two different electric fields superimposed. This statement seems more convincing and accurate, and doesn't contradict the definition. I personally think that it seems quite reasonable, at least for the time being.

I totally agree with this view. The electric field at all times (including the case where we have time varying magnetic flux) is simply $$E=-\nabla V-\frac{\partial A}{\partial t}$$

with the conservative component being the $-\nabla V$ and the non-conservative component $-\frac{\partial A}{\partial t}$ (A the magnetic vector potential).

V is the scalar potential here which obeys the equation of post #20 (which is the time dependent generalization of Poisson's equation) , which @hutchphd insists not to call it potential difference but only call it so when the aforementioned non-conservative component $-\frac{\partial A}{\partial t}$is zero i.e when the magnetic field is constant in time.

 

[hutchphd]

You may call it what you like. I won't even object to calling it a voltage (it has the appropriate units). I do object to calling it a potential difference because the potential, in fact, does not exist and so the difference is similarly illusory.

 

[Delta2]

because the potential, in fact, does not exist

I don't seem to get it, are you saying that V exists only in the static case (and DC steady state perhaps?)
Then what is the V in my post #37 and #20?

 

[etotheipi]

I don't seem to get it, are you saying that V exists only in the static case (and DC steady state perhaps?)
Then what is the V in my post #37 and #20?

I think, whenever $\partial_t \boldsymbol{B} \neq \boldsymbol{0}$, it's probably fine to call the scalar field $V$ in $\boldsymbol{E} = -\nabla V - \partial_t \boldsymbol{A}$ the potential so long as you made this terminology clear to the reader, and that the potential on its own isn't a meaningful thing anymore.

 

[Delta2]

Ok I think I see now thanks.
EDIT: I now get the meaning of the image of post #38, the word potential is overloaded here...

 

[Charles Link]

I think in the equation $E=-\nabla V-\frac{\partial{A}}{\partial{t}}$, the $-\nabla V=E_s$ is the electrostatic component that I previously mentioned, (posts 25 and 33), and $E_{induced}=-\frac{\partial{A}}{\partial{t}}$.
To apply it to the case of the voltage that we measure from an inductor or secondary coil of a transformer=outside of the component the magnetic field is essentially/(ideally) zero, and we then measure $V$ when we attach a large resistor (voltmeter/oscilloscope) across the component.
It becomes a more difficult problem when the magnetic field is changing across the components of interest. See https://www.physicsforums.com/threa...op-with-a-triangle.926206/page-8#post-6154120 We(a bunch of us) finally were all in agreement on the solution=see posts 192 to 197.

 

[DaveE]

Well said. I can say that one of my best lifelong friends who is a very good analog EE had to be brought kicking and screaming to this truth. He is not an RF guy though.
And Lewin very clearly talks about colleagues who accused him of chicanery in his lecture (see 48:25 et seq) which I, too, found surprising. I believe something akin to "politically correct speech" may actually be important here.

I think the basis of much of this confusion is that EEs and physicists don't speak the same language when they start drawing schematics. Lewin drew "lumped element" resistors connected with "wires". Then explains how those wires have induced voltages.

This can be confusing to EEs who assume that schematics are highly abstract descriptions of what the creator wanted to communicate about salient features of the circuit. In the EE mode, wires are always superconductors with no induced voltage, i.e. a topological connection between things that they think are worth paying attention to. To most EEs, if you think induced voltages are significant, then you need to draw an inductor to tell people "pay attention to magnetic effects". In some sense a schematic, to us, is a way of communicating all of the stuff that you need to pay attention AS WELL AS all of the complex stuff you are allowed to ignore (we tend to call these parasitic effects). My complaint with Lewin in these videos is that he was using mixed metaphors, if you will. If he had drawn this as a wire loop (perhaps with regions of different resistivity), and left out the "resistors", I think many EEs would "shift gears" and understand this is an EM problem, where Maxwell is more useful than Kirchhoff.

In defense of physicists, this abstract approach can often lead to oversimplification of real world behavior, and faulty assumptions about how the real world works. It is often the case that schematics can't really describe the true behavior of some analog/RF circuits. Problems also arise because in industry, schematics serve multiple roles: functional description for other engineers, data input to CAD SW, navigation data for people working with HW, data input to ERP/MIS systems (purchasing, etc.). IMO, there has never been a perfectly drawn schematic, ever; it's impossible.

 

[vanhees71]

I would like to express my views on the issue of non-conservative electric fields and potential differences.

Although the induced electric field is non-conservative, but for the fixed path and direction in the electric field, such as R1 and R2 shown in the figure below, there can indeed be a potential difference, because their voltage will respectively vary with their resistance, and they consume energy. But, of course, the placement and direction of the two voltmeters and their leads may affect the measured value.

View attachment 272957

Consider again the transformer shown in the figure, the secondary output voltage is also produced by the non-conservative electric field induced by changing magnetic field, therefore, its corresponding line integral of the electric field should only be along the path of the secondary coiled wire. If the line integral follows other paths, of course there may be different results.

In addition, We can see that the placement and direction of the secondary external output circuit loop has no (or extremely little) impact on the output voltage. The reason is obvious, because for the same magnetic flux , it only has one turn, but the number of turns inside the transformer is usually much larger than one turn.

This language is, from my experience, precisely what confuses students. If there is no potential, how can there be a potential difference? What we discuss here are precisely no potential differences but the non-vanishing "electromotive forces" (where "force" in this context is not force in the Newtonian sense but means an energy-like quantity). It's precisely NOT a potential difference, because it's the line integral of a vector field along a CLOSED loop. If the vector field were conservative, i.e., would have a potential, this EMF would be 0. Then and only then is the potential difference given by line integrals along an arbitrary open path connecting the two points where the potential difference is taken, and then and only then is this potential difference independent of the path chosen to connect the two points.

 

[Charles Link]

The inductor and the secondary coil of a transformer that delivers the "voltage" to a resistive load is a problem that involves a changing magnetic field, but upon looking at this problem a few times over the last couple of years, I'm starting to believe more and more that what is measured at the output terminals is indeed a potential difference which is just the integral of the electrostatic field. The induced electric field $E_{induced}$ is most often contained within the voltage source. Whether the EMF is generated by a changing magnetic field, or is the result of a battery, or solar cell, the net result is that it behaves as if it were an electrostatic source, e.g. like that of a Van de Graaff generator. $\\$

For the inductor and transformer,we must have that $E_{total} \approx 0$ inside the coil, so that an electrostatic $E_s=-E_{induced}$ must be present, where $E_{total}=E_{induced}+E_s$. Since the electrostatic field is conservative, i.e. $\oint E_s \cdot ds =0$ for any and all closed paths, the integral $V=\int E_s \cdot ds$ is well defined outside the device, (it's absolute value is equal to $\mathcal{E}=\int E_{induced} \cdot ds$ inside the device), and is the potential difference.

For the case of the battery, the EMF is not from a changing magnetic field, but rather a voltage created by a chemical reaction, but the same conservative electrostatic field rules apply, external to the device.
It is this electrostatic field, i.e. $\int E_s \cdot ds$ that is measured by a voltmeter or oscilloscope, which can be modeled as a large resistor, with a small current passing through it. The voltage is the product of the resistance and the current.

Taking Professor Lewin's "paradox" into account, it would appear that with an oscilloscope or voltmeter we could get slightly different readings from an inductor type voltage source, depending on where we put the lead wires, but for a toroidal type transformer or similar geometry, the lead wires generally wouldn't encompass a region of changing magnetic flux, and the readings would be unambiguous. These readings, with minimal induced electric field component, would be readings of the electrostatic field, i.e. $V=\int E_s \cdot ds$ which is also equal in absolute value to the EMF $\mathcal{E}= \int E_{induced} \cdot ds$ inside the component.

 

[alan123hk]

This language is, from my experience, precisely what confuses students. If there is no potential, how can there be a potential difference? What we discuss here are precisely no potential differences but the non-vanishing "electromotive forces" (where "force" in this context is not force in the Newtonian sense but means an energy-like quantity). It's precisely NOT a potential difference, because it's the line integral of a vector field along a CLOSED loop. If the vector field were conservative, i.e., would have a potential, this EMF would be 0. Then and only then is the potential difference given by line integrals along an arbitrary open path connecting the two points where the potential difference is taken, and then and only then is this potential difference independent of the path chosen to connect the two points

I understand what you mean, according to definition and theory, it is true.

Therefore, I think we should probably not say that the output of the transformer is a potential difference.
For example, It is not quite correct to say "..potential difference induced across the secondary winding.." in this education material https://phys.libretexts.org/Bookshelves/University_Physics/Book:_University_Physics_(OpenStax)/Map:_University_Physics_II_-_Thermodynamics_Electricity_and_Magnetism_(OpenStax)/15:_Alternating-Current_Circuits/15.07:_Transformers

Even if the load is far away from the transformer and will not be interfered by the electromagnetic field of the transformer, it is always the one turn of the output winding of the transformer, and the close line integral of this turn along with the output winding is still not zero.

In addition, all AC circuit should not have a potential difference, because it involves an varying electromagnetic field. So many people like to say "AC potential difference" is not very correct.

I am not arguing anything about the definition and theory, I just think that they may be applying an equivalent model to solve practical problem under certain conditions, the model may not be perfect, but for them, this method is very useful and reasonable.

 

[Charles Link]

See https://www.physicsforums.com/threa...op-with-a-triangle.926206/page-8#post-6154120 especially posts 192-197 and the solution by @cnh1995 (post 192), and the subsequent posts explaining the solution.

When we do encounter a problem where the magnetic flux is changing inside the circuit loops, so that EMF's need to be included, it is apparent from this problem that there is still a (electrostatic) potential difference between A and B, given by $V_{AB}=\int\limits_{A}^{B} E_s \cdot ds$.

It is perhaps a very good thing that Professor Lewin has attracted some attention to this problem, but it looks as if we now have satisfactory solutions to it, and a good interpretation of it, along with suitable definitions to resolve the "paradox".

 

[vanhees71]

Yes, BUT $V_{AB}$ depends on the path you use to connect $A$ and $B$, i.e., you have to write
$$V_{AB}=\int_{C} \mathrm{d} \vec{r} \cdot \vec{E}.$$
It depends on the path, because if you have two different paths $C_1$ and $C_2$, both connecting $A$ with $B$ you get, for the closed path $C'=C_1-C_2$ (i.e., you go from $A$ to $B$ along path $C_1$ and then back from $B$ to $A$ along path $C_2$),
$$\int_{C'} \mathrm{d} \vec{r} \cdot \vec{E} = \int_{C_1} \mathrm{d} \vec{r} \cdot \vec{E} - \int_{C_2} \mathrm{d}\vec{r} \cdot \vec{E} = -\int_A \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B} \neq 0.$$

 

[Charles Link]

Yes, BUT $V_{AB}$ depends on the path you use to connect $A$ and $B$, i.e., you have to write
$$V_{AB}=\int_{C} \mathrm{d} \vec{r} \cdot \vec{E}.$$
It depends on the path, because if you have two different paths $C_1$ and $C_2$, both connecting $A$ with $B$ you get, for the closed path $C'=C_1-C_2$ (i.e., you go from $A$ to $B$ along path $C_1$ and then back from $B$ to $A$ along path $C_2$),
$$\int_{C'} \mathrm{d} \vec{r} \cdot \vec{E} = \int_{C_1} \mathrm{d} \vec{r} \cdot \vec{E} - \int_{C_2} \mathrm{d}\vec{r} \cdot \vec{E} = -\int_A \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B} \neq 0.$$

In the notation I used, $E_s$ is the electrostatic component of the electric field. This integral will be path independent. If you study the solution by @cnh1995 in detail,(post 192 of the "link"), I think you will find it to be a very good one. (To add more detail, I solved the problem for the currents using loop equations, essentially 6 equations and 6 unknowns. By introducing the electrostatic integrals, e.g $V_{AB}$, @cnh1995 came up with a much simpler solution, the details of which I explained in post 193).

See also post 42. I think we are justified in making a distinction between the electrostatic $E_s$ and the induced $E_{induced}$, with the result that $E_{total}=E_s+E_{induced}$.

 

[alan123hk]

In short, there is charge accumulation at both ends of the output port of the inductor or transformer, thereby forming a voltage. In the actual working environment, if there is no changing magnetic field in the external space, for example, there is no leakage flux from the inductor or transformer, the output voltage is basically the same no matter where the external connection wires and load are placed. In this case, as long as we know the reasons and the constraints behind it, then I guess the voltage can be approximated as a potential difference, but of course there are exceptions, that is, we cannot cross the two output connection wires and form another closed loop around the magnetic field of the inductor or transformer, because according to Faraday's Law, there will be an induced voltage, and then the output voltage will change.

 

[vanhees71]

In the notation I used, $E_s$ is the electrostatic component of the electric field. This integral will be path independent. If you study the solution by @cnh1995 in detail,(post 192 of the "link"), I think you will find it to be a very good one. (To add more detail, I solved the problem for the currents using loop equations, essentially 6 equations and 6 unknowns. By introducing the electrostatic integrals, e.g $V_{AB}$, @cnh1995 came up with a much simpler solution, the details of which I explained in post 193).

See also post 42. I think we are justified in making a distinction between the electrostatic $E_s$ and the induced $E_{induced}$, with the result that $E_{total}=E_s+E_{induced}$.

I forgot that we have this strange splittings of the em. field. I still think it's more confusing than helpful to talk about such (usually gauge dependent) splittings particularly in this context of Faraday's Law with $\dot{\vec{B}} \neq 0$, where the electric field is not conservative.

 

[Charles Link]

I forgot that we have this strange splittings of the em. field. I still think it's more confusing than helpful to talk about such (usually gauge dependent) splittings particularly in this context of Faraday's Law with B→˙≠0, where the electric field is not conservative.

For me, I find it useful to explain the concept of "voltage", that a EE measures, that is virtually identical, whether the source is an inductor or transformer, or battery, or solar cell.

Meanwhile, another concept that I have found to offer considerable insight into how magnetism and a transformer works is "magnetic surface currents". See https://www.physicsforums.com/insights/permanent-magnets-ferromagnetism-magnetic-surface-currents/

One of the arguments against presenting the magnetic surface currents is that there is no actual charge transport, so the currents are thereby a mathematical construction. That there is no charge transport turns out to be a very important feature in the construction of a transformer. From Faraday's law, there are necessarily reverse EMF's in the core of the transformer that result in (real) eddy currents with reverse magnetic fields that would render the transformer useless.

The solution here is actually very simple=laminations are used in a transformer with plastic insulation between the iron layers. The eddy currents are thereby almost completely blocked, while the magnetic surface currents that are at least mathematically responsible for generating the magnetic field are unaffected by the laminations, and the transformer works just as it should, where the magnetism is computed from the magnetic surface currents using Biot-Savart.

It might interest the reader to see a discussion I had about these transformer laminations with the late jim hardy=see https://www.physicsforums.com/threads/magnetic-flux-is-the-same-if-we-apply-the-biot-savart.927681/ posts 19-20.

In any case, since the subject of transformers and Faraday's law came up in this discussion, I think the reader might find this last "tidbit" of transformer laminations of much interest.

 

[vanhees71]

The surface current is fine though. It's not a current in the sense of charge transport but mathematically equivalent to magnetization $\vec{j}_{\text{M}}=\vec{\nabla} \times \vec{M}$. If you assume homogeneous magnetization throughout the material, all you get is a surface current along the boundary of the magnet.

 

[Charles Link]

The surface current is fine though. It's not a current in the sense of charge transport but mathematically equivalent to magnetization $\vec{j}_{\text{M}}=\vec{\nabla} \times \vec{M}$. If you assume homogeneous magnetization throughout the material, all you get is a surface current along the boundary of the magnet.

Very good. Just to provide some detail to the reader, magnetic current density $j_M=\nabla \times \vec{M}$ arises, (at least mathematically), in any region where the magnetization is non-uniform. The result is that at a surface boundary we have magnetic surface current per unit length $\vec{K}_m=\vec{M} \times \hat{n}$. This is useful in understanding the above post 52, in the discussion of transformer laminations.
(Note: Using units where $B=\mu_o(H+M)$ ).

 

[alan123hk]

A gentleman made a short film to explain Lewin's paradox.

This short film is titled "Lewin's Paradox Explanation: Kirchhoff's Voltage Law (kvl) doesn't work on DC either!", But of course he didn't mean that "Kirchhoff's Voltage Law (kvl) doesn't work on DC either". Another gentleman replied that he had the same idea, but according to his narrative, the professor said that he didn't understand it, so obviously the professor did not agrees with this method of explanation, but I think this method of explanation does not seem to be completely nonsense, so I really don’t know.

 

[Charles Link]

A gentleman made a short film to explain Lewin's paradox.

This short film is titled "Lewin's Paradox Explanation: Kirchhoff's Voltage Law (kvl) doesn't work on DC either!", But of course he didn't mean that "Kirchhoff's Voltage Law (kvl) doesn't work on DC either". Another gentleman replied that he had the same idea, but according to his narrative, the professor said that he didn't understand it, so obviously the professor did not agrees with this method of explanation, but I think this method of explanation does not seem to be completely nonsense, so I really don’t know.

The two resistors in the configuration are different. Meanwhile, the voltmeters are attached to completely different points in the circuit. IMO, this video is poor physics, and is not on par with that of Professor Lewin's.

The circuit analysis of this is kind of simple, and most of the current flow is in the inner loop. With the different resistors, the connection points to the outer loop are not at the same voltage. Thereby, the voltmeters in the outer loop will also read differently.

Edit: If you assume the internal resistance of the batteries is on the order of one ohm or less,( a google says Rinternal≈80mΩ), the circuit analysis is rather simple: The inner loop functions virtually independent of the outer loop. You get a 4.5 volt drop across the 150 ohm resistor, and a 1.5 voltage drop across the 47 ohm resistor. Picking a point as the 0 volt ground, you can assign voltages to the contact points of the outer loop. The voltmeter readings are just as expected. This is all very basic physics=perhaps I shouldn't call it poor physics,(the circuit analysis is somewhat interesting), but it seems to lack relevance to Professor Lewin's concepts. Edit 2: Perhaps it was made as kind of a joke=I'm not sure.

Edit 3: It took me a while, but I think I see what the fellow is trying to show with his video: If we do have EMF's in the line, represented by batteries, we will of course get different voltages on the two voltmeters if the resistors are different. Futhermore the batteries are all placed to be in the same direction as the EMF's in Professor Lewin's video. Notice also the opposite polarity on the voltmeters, as Professor Lewin also got. It's really a very clever video, but he doesn't expound much on it=it takes a little thinking to see the analogies. Thank you @alan123hk for sharing. :)

 

[Charles Link]

@alan123hk Please see the "Edits" in post 56.

 

[alan123hk]

It took me a while, but I think I see what the fellow is trying to show with his video: If we do have EMF's in the line, represented by batteries, we will of course get different voltages on the two voltmeters if the resistors are different. Futhermore the batteries are all placed to be in the same direction as the EMF's in Professor Lewin's video. Notice also the opposite polarity on the voltmeters, as Professor Lewin also got. It's really a very clever video, but he doesn't expound much on it=it takes a little thinking to see the analogies

It seems that he spent a lot of time making the circuit and the video, but what he wants to express is actually very simple.

He wants to point out that even assuming that the voltage between the upper and lower points of the middle structure is a constant value, but because of the connecting wires of the two voltmeters will have an induced voltage generated by the changing magnetic field, so the readings of the two voltmeters on the left and right will be different. Note that the circuit he showed has two batteries at the top and bottom. These batteries are used to simulate the voltage induced on the wires of the two voltmeters.

I believe the theory behind his idea also seems reasonable. Please note that the two voltmeters and their connecting wires do form a closed loop or two loops, respectively, and there is a magnetic field change in the middle of this closed loop. According to Faraday's Law, there is an induced voltage on this loop, and, of course, the final readings of the two voltmeters are produced by combining the voltage generated by the central structure.

In any case, I admire his passion for science, and his video is also helpful for stimulating thinking.

 

[Charles Link]

and there is a magnetic field change in the middle of this closed loop.

Just to make sure this part is clear, the fellow represented the changing magnetic field of Professor Lewin's apparatus with batteries as the EMF's. There is a significant DC current in the fellow's center loop, but this DC current makes only a small and constant magnetic field. The difference in the two voltmeters is due to the EMF's and resistors, directly analogous to Professor Lewin's changing magnetic field. Your explanation is very good, other than I think it would be much better and very exact if you deleted this phrase. It only can refer to Professor Lewin's closed loops.

 

[hutchphd]

I am sorry but this video is silly. It sets up a "straw man" man and knocks it down. Prof. `Lewin was showing that two ordinary voltmeters connected to the same two points in a circuit can give wildly differing readings. This does not require some "Uri Geller" rigging of stage props as presented here.
Lewin presented a cautionary tale because such circumstances can often occur in real circuits particularly if any appreciable currents exist. Slavish devotion to Kirchhoff's laws and the lumped element model can leave you scratching your head muttering "WTF?". This is particularly true if there are ground loops. The fact that this became controversial is simply a measure of lack of understanding of the physics involved by folks who should know better.