Inducing EMF Through a Coil: Understanding Flux

[Charles Link]

It might be of interest at this point to look more closely at the EMF that gets generated when a loop containing $N$ coils is introduced. We could just use Faraday's law with the flux being $N$ times as much, but I did notice there is a way to explain this using the $E_{induced}$. See https://www.feynmanlectures.caltech.edu/II_22.html
This one is subject to some interpretation, but right after equation (22.3) Feynman says there can be no electric fields inside an ideal conductor. I interpret that as saying there arises an $E_{s}$ in the conductor, where $E_s=-E_{induced}$. This $E_s$ will integrate along the path inside of the coil to give it the factor of $N$ times the EMF of a single loop. In addition, external to the $N$ loops, it must also give this same result, since $\nabla \times E_s =0$. This (the electrostatic component) is what I believe we measure with a voltmeter when we measure the EMF from the coil. See also post 27 above and the thread that is linked. In private discussions (PM) with others, I don't have complete agreement on this, but perhaps you @alan123hk might concur. The connections of the leads to the voltmeter can be close enough together that we clearly aren't measuring the $E_{induced}$, but the integral $\int E_s \, dl$ between two points must be path independent, and thereby gives the same result that it does going through the voltmeter and voltmeter leads, as it does through the coil, resulting in a very reliable voltage measurement.

Then there is also the case of a single loop that is open, the subject of the Walter Lewin paradox, where it matters which side of the localized changing magnetic field the voltmeter is placed, but that is a separate problem, and it has already been discussed in detail. For that case, the voltmeter is also seeing $E_{induced}$, and how much (integral of) $E_{induced}$ that is observed (depends on location of the meter) makes a difference in the measurement.

For $N$ in the above, the location of the meter could make it $N \pm 1$, etc., which usually doesn't have a significant effect in the voltage reading, provided $N$ is fairly large.

See also https://www.physicsforums.com/threa...-voltage-across-inductor.880100/#post-5529838 post 22.

 

[alan123hk]

I interpret that as saying there arises an Es in the conductor, where Es=−Einduced. This Es will integrate along the path inside of the coil to give it the factor of N times the EMF of a single loop. In addition, external to the N loops, it must also give this same result, since...=0 ... perhaps you @alan123hk might concur

Yes, I very much agree with your previous analysis, you explained very well why the induced emf is proportional to the number of turns of the coil.

The connections of the leads to the voltmeter can be close enough together that we clearly aren't measuring the Einduced, but the integral ∫Esdl between two points must be path independent, and thereby gives the same result that it does going through the voltmeter and voltmeter leads, as it does through the coil, resulting in a very reliable voltage measurement.

As described in Feynman's lectures https://www.feynmanlectures.caltech.edu/II_22.html
"we have assumed that there are no magnetic fields in the space outside of the “box,” this part of the integral is independent of the path chosen and we can define the potentials of the two terminals"

Just like the AC voltage provided by the power company, users obtain stable AC voltage independent of the transmission line path through the transmission cables outside the power company's substation facilities.

Please note that the user does not have the means or right to go into the power company facility and put extra turns of wire on the transformer to change the voltage.

 

[Charles Link]

Yes, I very much agree with your previous analysis, you explained very well why the induced emf is proportional to the number of turns of the coil.

Very good. :)

It should be noted that in some previous discussions on PF where $E_s=-E_{induced}$ in the conductor was introduced, (by another member), one major criticism was presented of
how can a non-conservative field ($E_{induced}$) be the opposite of and basically be represented by a conservative field ($E_s$)?
If $\nabla \times E_{induced}=-\dot{B}$, it should be impossible to have $E_s=-E_{induced}$ (in the conductor) with $\nabla \times E_s=0$.

It looked like it could be a valid objection, but then I later saw the explanation: When you go once around the loop (of the $N$ turn coil) $\oint E_{induced} \, dl=-\dot{\Phi}$. When you go once around the same path with $\oint E_s \, dl$, you get $\dot{\Phi}$ rather than zero, (even though $\nabla \times E_s=0$), because you are now in the adjacent ring of the coil.

Likewise if you go around $N$ times, you then have traversed the entire coil, and get $\int E_s \, dl=N \dot{\Phi}=EMF$ rather than zero. Thereby, we still have a conservative field $E_s$, so the path integral outside through the voltmeter between the two ends of the coil also gives the same EMF, but where the coil boundaries gave us the special property where $\oint E_s \, dl$ could be non-zero.

@alan123hk The above is a little extra work, but you may find it of interest.

Note: It took a little proof-reading and a couple edits, but I think I now have it reading like it should. :)

 

[alan123hk]

If ∇×Einduced=−B˙, it should be impossible to have Es=−Einduced (in the conductor) with ∇×Es=0.

above is a little extra work, but you may find it of interest.

I have also thought about this issue, here are my thoughts

There may not necessarily be an induced electric field with curl inside the wires of the coil. If the changing magnetic flux only concentrated at the center of the coil and does not pass through the coil wire, then the induced electric field in the wire should have no curl, at this time, the electric field generated by the accumulated charge can completely offset the induced electric field, so the electric field inside the wire becomes zero.

Even if the magnetic flux passes through the wire of the coil and a curly electric field appears inside the wire, the electric field generated by the accumulated charge cannot and does not necessarily completely cancel the curly electric field. All it has to do is make the line integral of the electric field along the coil wire zero. The effect should be very close to before. The voltage across the coil remains unchanged, complying with Faraday's law. However, strong eddy currents will be generated inside the coil wires, causing greater losses.

 

[Charles Link]

It may be worthwhile at this time to consider the very ideal case that $\dot{B}$ is uniform (in the z direction) inside a small circular cylindrical region of basically infinite length, (the ideal very long solenoid discussed above, with a changing current (changing at a constant rate) in its windings), and with $\dot{B}$ essentially zero everywhere else.

We do find that to have $\nabla \cdot B=0$,(magnetic flux is conserved), the integrals of $B$ and $\dot{B}$ over the infinite area outside of this region need to be finite, even though $B$ and $\dot{B}$ are essentially zero.

Outside of this circular region we have that $\nabla \times E=-\dot{B}=0$, but that does not mean that $E=0$. Instead we can use Stokes' law, along with symmetry and get that $\oint E_{induced} \cdot \, dl=-\dot{\Phi}=2 \pi r E_{induced}$, (vector direction should be apparent, but is omitted), so that we can accurately compute $E_{induced}$ as a function of $r$ for this case.

It's the very ideal case, and kind of a simple example, but it may be worthwhile taking a close look at it.

 

[Charles Link]

Even if the magnetic flux passes through the wire of the coil and a curly electric field appears inside the wire, the electric field generated by the charge cannot and does not necessarily completely cancel the curly electric field.

I think you followed my explanation in post 33, but I would like to say it slightly differently, with basically the same mathematical idea: Suppose the $E_{induced}$ is uniform around the coil at a given radius as we have in our above example. We integrate with a closed integral and get $\oint E_{induced} \, dl=-\dot{\Phi}$. The $E_s$ meanwhile is electrostatic so that we necessarily should have $\oint E_s \, dl=0$. This looks like we can then conclude that we can not have $E_s=-E_{induced}$, because that would imply $\oint E_s \, dl=\dot{\Phi}$ if we circle the loop.

$\oint E_s \, dl=\dot{\Phi}$ is in principle what we have though in traveling around a single ring in the conductor "coil"=we are physically one millimeter over at the finish of the integral if the wire is one millimeter in diameter (it would be easier to illustrate just by pointing to the coil and we would be in the adjacent wire from where we started=one strand over). We thereby have an electrostatic field with $E_s=-E_{induced}$ where we can go once around a loop (which is part of a coil) and have a non-zero value , $\dot{\Phi}$, for the path integral. For all practical purposes, the electrostatic $E_s$ does indeed cancel the $E_{induced}$ in the conductor coil.

The coil gives a mathematical twist to things, so that the conservative electrostatic field $E_s$ can in fact have the non-zero loop integral value that is necessary for it to cancel the $E_{induced}$.

Note that we could even put a voltmeter across the adjacent rings of the coil, and we would measure a voltage $V=\dot{\Phi}$. It just occurred to me that this implies there is an $E_s$ in the z direction outside the wires, as well as other directions. We clearly then don't have $\vec{E}_s=-\vec{E}_{induced}$ everywhere, but inside the conductor, this is close to being the case. It should also be noted, we only need $E_s=-E_{induced}$ in the conductor. It necessarily will not be the case outside the conductor.

 

[alan123hk]

I want to use a simple diagram to express. The situations of multi-turn spiral winding structure coils and single-turn spiral structure coils are actually the same. Let’s take a single turn as an example.

Since $~\oint E_{induced} \, dl=-\dot{\Phi}$ and $~\oint E_s \, dl=0~~$, how can they cancel each other out?

The reason is that we do not need a complete closed loop line integration of the electric field produced by the charge to cancel the EMF (closed loop line integration of the induced electric field) of the single turn coil as shown below.

Since the closed-loop line integral of the induced electric field has a vertical line segment assumed to be the zero valve. Therefore, the line integral of the electric field produced by the charge need not include this line segment. In fact, there may be many other ways to explain it.

 

[vanhees71]

Why do you think the EMF were 0? For time-dependent fields the exterior of the long solenoid is not field-free anymore!

 

[alan123hk]

There is an error in the text on this image. I'm not saying that the induced electric field is zero. What I actually mean is that the line integral of the induced electric field on this vertical line segment is zero, just like the text below the picture describes.
This is because the direction of the vector integration path is perpendicular to the direction of the induced electric field.

 

[vanhees71]

Hm, but if the magnetic flux through any surface with the loop as a boundary the electric field cannot vanish everywhere along this loop:
$$\mathrm{d}_t \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}=-\mathcal{E}=-\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E},$$
where I assumed that the surface, including it's boundary is at rest (in the calculational frame of reference).

 

[Charles Link]

With the surface that we are considering, the single turn, it needs to include the short stretch of air space from the one turn to the other to be complete. It is here that we pick up the non-zero electric field=electrostatic field integral that will indeed give us the necessary $-\dot{\Phi}$ so that @vanhees71 is also correct, and everything is satisfied. Remember that the electrostatic integral will give the same result between the two endpoints, independent of the path.

Note the integral of the total electric field around/through the conductor is zero, because the total electric field inside the conductor is zero everywhere.

Edit: Note that the line integral around the (open) loop of $\int E_s \, dl=\dot{\Phi}$. One could almost write it, (as I did above in posts 33 and 36), as $\oint E_s \, dl$, especially when comparing it to $\oint E_{induced} \, dl$. Note also that $\int E_s \, dl$ taking the alternate path through the air, across the same two endpoints, gives the same $\dot{\Phi}$ result.

@alan123hk I think you might find this latest posting of interest. I was very glad you drew the diagram for us in post 37. :)

One other item worth mentioning is that with this single open coil, it will matter for the voltage reading which side of the page the voltmeter is attached. This is the subject of the Walter Lewin "paradox" which we have successfully resolved. That can be analyzed simply by looking at the EMF's in the circuit loops, (Faraday's law), but calculations using $E_s$ and $E_{induced}$ are in complete agreement with the EMF method.

 

[Charles Link]

See: https://www.physicsforums.com/threa...duction-lecture-16.948122/page-6#post-6857043
You might find this thread of interest=look at in particular posts 187, 188, 189, and 194. There doesn't seem to be complete agreement in the Physics Forums regarding this topic, but you still may find this of interest. It is up to the individual to ultimately determine what is good physics and what isn't.

Note: I got the necessary edits in to these posts before the thread was closed. I stick by my conclusions on this. We don't have complete agreement with everyone, but that is ok.

I did a repeat of post 27 of this thread here, because the discussion in the thread that is linked is closely related to what we discussed in the last several posts. I think for the most part, we have discussed Professor Lewin's paradox more than enough, but for completeness I thought I would include this one more time. I'm not sure if anyone might want to comment further, but for a couple of the items, posts 187, 188, 189, and 194 in the linked thread, I would welcome any feedback. The case of the coil with several or many turns is slightly different than Professor Lewin's single open ring, and it appears we do have some agreement now on what the results would be.

Edit: It should be noted that the thread that is linked above was closed, mostly due to some large disagreement that there was at the time. It looks to me though that some of the issues are finally getting resolved, and we are seeing more agreement.

additional comment: Using $\int E_s \, dl$, we now IMO have a more complete explanation for what is going on with the inductor, than somewhat logically, but seemingly an incomplete explanation of multiplying the changing flux by the number of turns $N$ to get the emf. It left open what to do in the case that there is e.g. $N=2.5$ turns. With the introduction of $E_s$ and $E_{induced}$ we can more accurately predict and calculate what will result in some of these cases.

@vanhees71 @alan123hk I added a few things, edits, etc., in both posts 41 and 42 that might be of interest that you may not have seen yet. I welcome any feedback you may have, but I am already pleased with the feedback you have provided so far. :)

Edit: One more item worth mentioning is that for the path consisting of the voltmeter resistor and the voltmeter wires, we have Ohm's law: $\int E_{total} \, dl=IR=V_{measured}$.

 

[alan123hk]

additional comment: Using ∫Esdl, we now IMO have a more complete explanation for what is going on with the inductor, than somewhat logically, but seemingly an incomplete explanation of multiplying the changing flux by the number of turns N to get the emf. It left open what to do in the case that there is e.g. N=2.5 turns. With the introduction of Es and Einduced we can more accurately predict and calculate what will result in some of these cases.

In most practical applications, unless special exceptions are made, inductor voltage is measured in units of integer turns. In a special case, such as 2.5 turns, it most likely means that the magnetic flux through one turn of the coil is only half of the normal case, in which case we can simply substitute N = 2.5 into the equation to calculate the emf.

 

[Charles Link]

In most practical applications, unless special exceptions are made, inductor voltage is measured in units of integer turns. In a special case, such as 2.5 turns, it most likely means that the magnetic flux through one turn of the coil is only half of the normal case, in which case we can simply substitute N = 2.5 into the equation to calculate the emf.

When I worked out the details using $E_s$ and $E_{induced}$, I indeed found integer increments if the voltmeter leads stayed in the same plane. This is even the case with a single open ring, (with a changing magnetic flux inside of it)=it doesn't matter where the voltmeter leads are placed on the ring, but it does matter which side the voltmeter is on. In the one case the voltage is $\dot{\Phi}$ regardless of where you attach the leads, but if the voltmeter is on the other side you read zero volts.

 

[Charles Link]

Just to add to the above, you can take a coil with $N$ turns and place the voltmeter leads across just a couple turns and, (if I analyzed it correctly), you get an integer result, even if you move the voltmeter leads laterally apart from each other, rather than having just the vertical separation of the leads on the separate rings of the coil. It surprised me to find that you don't get any fractional change in the measured voltage, i.e. it stays the same integer number times $\dot{\Phi}$, as you move the leads laterally apart on the coil.

Moving one of the leads vertically to the next wire over gives the integer increment.

Edit: It should be noted that if you put the voltmeter leads together on the same ring and then start to separate them laterally, you will continue to read zero volts on the meter. The result seems to make sense and is consistent with an analysis using $E_s$ and $E_{induced}$ and the path integrals. The voltage comes in increments by moving one of the leads vertically to an adjacent ring.

This item did come up around post 104 in the thread about Professor Lewin's paradox (see the link in post 42 above), and I think it could some day make for the topic of a good video, if someone would add to what Professor Lewin did with a single ring, to show the case of a coil with multiple rings,(performing measurements with a voltmeter/oscilloscope). It could be interesting to see the experimental confirmation of the result posted above. The coil could even contain an iron core, as in the case of a transformer, etc. One additional note, for the coil of $N$ turns the wire would need to be insulated, so provisions would have to be made for some way of getting the voltmeter leads to contact the conductor, but that is for the experimentalist to figure out. @alan123hk and @vanhees71 , I welcome your feedback once again on these latest inputs, but I think we are starting to get some agreement on what the results will be.

 

[SredniVashtar]

Oh boy, where do I begin?
I have debated this topic to the death many times on YouTube and on the Eevblog forum (where I posted numerous figures, and maybe if I find the link when I'm on my PC I'll post it here).
Oh, I have found one of my answers on EE Stack Exchange where I have a few figures (about middle post, the two open coils in a conservative and circulating electric field): https://electronics.stackexchange.c...how-different-values-circui?noredirect=1&lq=1
(I can't create a formatted link on this phone)

The key to understand the apparent conundrum is that, since the total electric field is not conservative, voltage is path dependent and the path integral ALONG the coil is different from the path integral ACROSS the terminals (or points on the coil) in the space around the coil.
Basically, Lewin's 'paradox' applies to all coils, not just his ring with two resistors. The circulating induced electric field associated with a time-varying magnetic field displaces the charges on the surface of the coil in such a way as to cancel it (completely in a perfect conductor or in open circuit, and almost completely in a real conductor when current is flowing ) inside it.

A voltmeter measures the path integral of the total electric field and therefore the voltage is in general dependent on the path.

I am writing this post just to add a reference that IIRC considers the case of the multiturn coil: Haus and Melcher, "Electric Fields and Energy" it's freely available on the MIT OCW site.

I find interesting the side question about the propagation of the effect of changing the magnetic field rate of change, tho. I believe it can be resolved considering that coils and transformers operate in a quasi-static setting, so the propagation delay is considered negligible in all space where the effects are non-negligible. But I'd love to hear other points of view about this.

 

[Charles Link]

See also from post 158 of the linked thread of post 42 above:
( https://www.physicsforums.com/threa...duction-lecture-16.948122/page-6#post-6857043 )

There is a very interesting paper in The American Journal of Physics [50, 1089 (1982)] by Robert Romer with the title "What do voltmeters measure?: Faraday's law in a multiply connected region". It is very relevant to the "Lewin Paradox". Lewin himself has referred to this paper.

It does appear we are finally getting considerable agreement on this topic. Thank you @SredniVashtar for your inputs. I do believe we did make some progress about 6-8 months ago in the thread linked in post 42 before it was closed, but now more and more it does appear we are getting a few people on the same page, and so far we don't have anyone in disagreement=perhaps I shouldn't speak too soon, but I am pleased with the progress. :)

 

[Charles Link]

A voltmeter measures the path integral of the total electric field and therefore the voltage is in general dependent on the path.

It is worth mentioning that for a large number of turns $N$ in the coil, the electrostatic (conservative) part of the electric field will normally be the dominant component for the path across the voltmeter leads, so that we indeed get a fairly reliable voltage reading, without needing to pay a lot of extra attention to how the meter is hooked up. See also post 31 of this thread.

 

[vanhees71]

A voltmeter measures the path integral of the total electric field and therefore the voltage is in general dependent on the path.

More generally, the voltmeter measures the line integral (it's not a path integral of course) defining the EMF, i.e., if the area with the line as a boundary is time-dependent you have
$$\dot{\Phi}=\frac{\mathrm{d}}{\mathrm{d} t} \int_{A} \mathrm{d}^2 \vec{f} \cdot \vec{B}=-\int_{\partial A} \mathrm{d} \vec{x} \cdot (\vec{E}+\vec{v} \times \vec{B}),$$
where $\vec{v}(t,\vec{x})$ is the velocity of the boundary curve. Of course, in Lewin's setup $\vec{v}=0$.

If you run multiple times along the boundary of the surface, e.g., by using a wire winding $N$ times around the surface, $\Phi$ gets an additional factor of $N$.

I am writing this post just to add a reference that IIRC considers the case of the multiturn coil: Haus and Melcher, "Electric Fields and Energy" it's freely available on the MIT OCW site.

I find interesting the side question about the propagation of the effect of changing the magnetic field rate of change, tho. I believe it can be resolved considering that coils and transformers operate in a quasi-static setting, so the propagation delay is considered negligible in all space where the effects are non-negligible. But I'd love to hear other points of view about this.

Of course, AC theory treats "compact" setups, where the wavelength of the em. waves is very large compared to the extensions of the circuit. Then all fields can be approximated by the near-field approximation, i.e., quasistationary approximations:

https://itp.uni-frankfurt.de/~hees/pf-faq/quasi-stationary-edyn.pdf

 

[Charles Link]

More generally, the voltmeter measures the line integral (it's not a path integral of course) defining the EMF

From post 49, @vanhees71 , might you elaborate on this a little more please. It's an integral along a path, but , correct me if I don't have it correct, I think you are wanting the term "path integral" to be reserved for a particular type of quantum mechanical integral that is known as a "path integral".

 

[vanhees71]

It's an integral along a path, but not a path integral. A path integral is a functional integral used in quantum mechanics, QFT, and statistical physics. It's an entirely different mathematical notion than a line integral in classical vector calculus. In German it's always distinguished clearly: the "path integral" is a "Pfadintegral" and the "line integral" is a "Kurvenintegral". I hope there are no English textbooks calling a "line integral" a "path integral". At least I'm not aware of any one ;-)).

 

[SredniVashtar]

It's an integral along a path, but not a path integral. A path integral is a functional integral used in quantum mechanics, QFT, and statistical physics. It's an entirely different mathematical notion than a line integral in classical vector calculus. In German it's always distinguished clearly: the "path integral" is a "Pfadintegral" and the "line integral" is a "Kurvenintegral". I hope there are no English textbooks calling a "line integral" a "path integral". At least I'm not aware of any one ;-)).

As a non native speaker I wasn't aware of this strict distinction. In my own language we use versions of "line integral" and "curvilinear integral".

But I have just checked that on wikipedia path integral directs here

https://en.m.wikipedia.org/wiki/Path_integral

So it seems it's a bit of a gray zone, maybe due to sloppy use by non native speakers...

 

[tech99]

I find interesting the side question about the propagation of the effect of changing the magnetic field rate of change, tho. I believe it can be resolved considering that coils and transformers operate in a quasi-static setting, so the propagation delay is considered negligible in all space where the effects are non-negligible. But I'd love to hear other points of view about this.

It appears to me that, if the energy is transported from primary to secondary by means of an EM wave then it will travel at velocity c, but if it is transported by induction then it will travel instantaneously. If we have a lossless coil and voltmeter and take no energy from the fields then I suppose the transfer is instantaneous.

 

[alan123hk]

My personal thought is that since nothing can travel faster than the speed of light, electromagnetic induction cannot produce instantaneous effects beyond spatial.

Although according to the integral form of Faraday's law of electromagnetic induction, it seems that an induced electric field can appear instantaneously in distant space, I think this is just a misunderstanding, and the integral form of Faraday's law of electromagnetic induction does not mean that instantaneous effects beyond spatial.

For example, the sudden appearance of vertical magnetic flux generated by a small solenoid coil at a certain point in space cannot cause an instantaneous change in magnetic flux in a large closed circular path on the horizontal plane.

This is because it takes time for electromagnetic waves caused by changes in magnetic flux to propagate to the boundaries of large closed circular path. There will be no net magnetic flux through the closed circular path until the span of the electromagnetic wave exceeds the closed circular path.

However, we usually ignore this transmission process in simplified models because the distance is much smaller than the wavelength.

 

[tech99]

My personal thought is that since nothing can travel faster than the speed of light, electromagnetic induction cannot produce instantaneous effects beyond spatial.

Although according to the integral form of Faraday's law of electromagnetic induction, it seems that an induced electric field can appear instantaneously in distant space, I think this is just a misunderstanding, and the integral form of Faraday's law of electromagnetic induction does not mean that instantaneous effects beyond spatial.

View attachment 336256​

For example, the sudden appearance of vertical magnetic flux generated by a small solenoid coil at a certain point in space cannot cause an instantaneous change in magnetic flux in a large closed circular path on the horizontal plane.

This is because it takes time for electromagnetic waves caused by changes in magnetic flux to propagate to the boundaries of large closed circular path. There will be no net magnetic flux through the closed circular path until the span of the electromagnetic wave exceeds the closed circular path.

However, we usually ignore this transmission process in simplified models because the distance is much smaller than the wavelength.

Hertz carried out what amounts to a race between a wave travelling along a wire and waves in free space, and found that close to the source the propagation seemed to have a velocity of either zero or infinity.
"On finite velocity of electromagnetic actions" Feb 2, 1888.

 

[vanhees71]

My personal thought is that since nothing can travel faster than the speed of light, electromagnetic induction cannot produce instantaneous effects beyond spatial.

Although according to the integral form of Faraday's law of electromagnetic induction, it seems that an induced electric field can appear instantaneously in distant space, I think this is just a misunderstanding, and the integral form of Faraday's law of electromagnetic induction does not mean that instantaneous effects beyond spatial.

View attachment 336256​

For example, the sudden appearance of vertical magnetic flux generated by a small solenoid coil at a certain point in space cannot cause an instantaneous change in magnetic flux in a large closed circular path on the horizontal plane.

This is because it takes time for electromagnetic waves caused by changes in magnetic flux to propagate to the boundaries of large closed circular path. There will be no net magnetic flux through the closed circular path until the span of the electromagnetic wave exceeds the closed circular path.

However, we usually ignore this transmission process in simplified models because the distance is much smaller than the wavelength.

Yes, that's the quasistationary approximation, applied to circuit theory, where the spatial extent of all involved elements of the circuit is much smaller than the typical wavelength of the electromagnetic waves involved in its operation.

It's also clear from a mathematical point of view that the homogeneous Maxwell equations are constraints on the fields and don't describe a cause-effect relation between them. The sources of the electromagnetic field are the charge-current distributions, and the solution of the corresponding wave equations for $(\vec{E},\vec{B})$ obeys Einstein causalitys as is evidenced by the use of the retarded propagator in the "Jefimenko equations".

 

[Charles Link]

See https://www.hellovaia.com/explanations/physics/electromagnetism/retarded-potential/
which discusses the Lienard -Wiechert retarded potentials.

 

[Charles Link]

I would like to go back to the discussion around post 46, and see if @SredniVashtar might agree with me on this point: When we have just a couple of loops in the coil, Professor Lewin's considerations certainly apply, but when we have many loops such as a transformer coil or solenoid surrounding a changing magnetic field, we can, with a good deal of accuracy, measure a voltage (and get an oscilloscope reading) because we are seeing what in effect is the electrostatic line integral of the path through the conductive coil, with the electrostatic field being opposite the induced electric field.

The electrostatic line integral is the same value over the path with the same two endpoints, so we get the same result going through the voltmeter. The factor of $N$ that the line integral gets is important here, because we still can get as much as a single loop of $E_{induced}$ affecting the voltmeter reading, but so long as we don't make a coil with the voltmeter leads of multiple loops around the changing magnetic field, then we get a reliable voltage reading that for most practical purposes is measuring the line integral of $E_s$, which is the same in absolute value to the line integral of $E_{induced}$ through the path of the coil.

The measurement of the EMF could be said to be an indirect measurement of the $E_{induced}$, (we actually are observing the $E_{electrostatic}=E_s$,and by having a coil with $N$ turns, we are amplifying the $E_s$ reading, (from that of a single coil by $N$), and although it might appear that way, the $E_{induced}$, (the induced electric field value) is not affected by the coil.

[Edit: (Note: The voltmeter reading is $V= N \dot{\Phi} \pm \dot{\Phi}$, where the second term comes, (basically with a minus sign), from Professor Lewin's EMF, and may be absent, depending on which side the voltmeter leads are attached). For the most part, the voltmeter is reading the line integral of $\int E_s \, dl= N \dot{\Phi}$ inside the coil, which is the opposite that of $\int E_{induced} \, dl$ inside the coil. For a single loop or just a couple of loops, as Professor Lewin demonstrates, it can be very important how the voltmeter is placed, but for large $N$, it is no longer so important, and we can get a reasonably accurate number for the voltage].

@SredniVashtar I welcome your feedback. It seems we may be starting to get some agreement on this topic, but it previously has not been completely accepted. There has been some opposition to "splitting" the electric field into $E_{induced}$ and $E_s$ components. I do think introducing the two components can have its merits in some cases.

See https://www.feynmanlectures.caltech.edu/II_22.html
right after equation (22.3). He does say that the electric field is basically zero in the ideal conductor, and he seems to imply IMO that we are measuring the electrostatic component with a voltmeter, (i.e. normally in most cases, with of course taking into account any additional $\int E_{induced} \, dl$ that may appear in the lead wires of the voltmeter, which is basically the source of Professor Lewin's puzzle) , but he doesn't elaborate on it. I welcome your feedback.

One additional note: When the lead wires are kept next to each other, the contribution of $\int E_{induced}$ in one wire usually cancels that of the other wire, but if they are separated and go around the source of changing magnetic field their combined $\int E_{induced} \, dl$ will be $-\dot{\Phi}$. Further inspection of this seems to indicate that if they are wrapped around in the other direction they will pick up a contribution of $+\dot{\Phi}$. (Professor Lewin never looped the wires of his voltmeter to wrap around to pick up this $+\dot{\Phi}$, but if he had, it seems clear that this is the result he would get. You could even wrap them multiple times, and then pick up multiples of $\dot{\Phi}$. Note with the low currents in the voltmeter wires, attaching the voltmeter does not change the physical system being measured to any significance.

 

[SredniVashtar]

@Charles Link I just gave a quick cursory look at your last post but... Can you clarify why you think that the multiple turn coil would behave differently from the single loop one?
Sure, there will be more interaction between the lateral surface charge, but the gist is the same: the surface charge redistributes in order to reduce the total field inside the copper to the value expected by ohm's law: exactly zero when the coil is open, and the small value E = j / sigma when a current is flowing in the load attached to the coil.

One loop will require a small distribution of charge because the induced electric field to neutralize only makes one turn around the the dB/dt; N loops will require more charge because the line integral is N times bigger. The voltage you see along paths in the space (outside of the dB/dt region) between the terminals is N times the emf of a single loop. The voltage you see along a path that follows the interior of the conductor is either zero or the negligible ohmoc loss due to the time resistivity of copper.

And if you place your probes on two points on different turns (with your voltmeter on the outside of the coil) you will only measure an integer number of the single turn emf (minus a small ohmic loss).

It's easier to see with a picture but I need to find them and it won't be easy these days.

 

[Charles Link]

Can you clarify why you think that the multiple turn coil would behave differently from the single loop one?

To see the argument in its entirety, you need to make a distinction between the electric field $E_{induced}$ caused by the changing magnetic field, and the electrostatic field $E_s$ that has its origins in the conductor, and the component of it that is most important is equal and opposite the $E_{induced}$ in the conductor. Once you follow $E_s$ along and through the coil, in however many loops you choose, you can then take the same two endpoints, but go through a voltmeter. The reason the electrical engineer (EE) is able to get a reliable voltage reading is that for many loops $N$, he measures $V=N \dot{\Phi}$, with an uncertainty of $\pm \dot{\Phi}$ that Professor Lewin has brought to the forefront. The voltage reading of $V=N \dot{\Phi}$ is the electrostatic part, which is independent of the path.

Professor Lewin put resistors in various places in his single closed ring, so the mathematics is slightly different, but I'm really not doing anything very new here with the coil of $N$ turns. However, I am introducing an $E_s$ and $E_{induced}$, which has been used previously to solve a related problem on PF in a very simple manner. I'll give a "link" to it momentarily: See https://www.physicsforums.com/threa...duction-lecture-16.948122/page-6#post-6857043 post 187. The last paragraph there gives a "link" to the problem I just mentioned, but the whole post 187 is relevant here. This "split" electric fields has created a lot of disagreement on the Physics Forums, but I think it can be a very useful concept at times.