Inducing EMF Through a Coil: Understanding Flux

[SredniVashtar]

I am sorry, but I don't see any uncertainty at all in the voltage measured across the terminals of a multiturn or single turn coil.

 

[Charles Link]

I am sorry, but I don't see any uncertainty at all in the voltage measured across the terminals of a multiturn or single turn coil.

This one is simple=I don't think I even need to draw a picture:
Take a single conductive loop that is slightly open with a changing magnetic field inside of it of changing flux $\dot{\Phi}$. Let the open space in the ring be on the right side, and put a voltmeter across the right side. I think you would agree that you read $V=\dot{\Phi}$.

Next put the voltmeter on the left side of the ring, and have one voltmeter wire go above and over the ring to the voltmeter. Take the other voltmeter wire and run it under the ring to the voltmeter. I think you will agree that you read zero volts. This is basically a variation on what Professor Lewin does.

With $N$ turns you measure $N \dot{\Phi}$, but there will be a possibility that you get an "error" caused by the way you connect the voltmeter wires of $\pm \dot{\Phi}$. We did the case of $N=1$ above.

additional item: For the case above with the voltmeter on the left, attach the first wire to the lower part of the open ring on the right (and run it over the top to the voltmeter), and attach the second wire to the upper part of the ring on the right, (and run it under the ring to the voltmeter), and I believe you will then measure $V=2 \dot{\Phi}$.

 

[SredniVashtar]

This one is simple=I don't think I even need to draw a picture:
Take a single conductive loop that is slightly open with a changing magnetic field inside of it of changing flux $\dot{\Phi}$. Let the open space in the ring be on the right side, and put a voltmeter across the right side. I think you would agree that you read $V=\dot{\Phi}$.

Next put the voltmeter on the left side of the ring, and have one voltmeter wire go above and over the ring to the voltmeter. Take the other voltmeter wire and run it under the ring to the voltmeter. I think you will agree that you read zero volts. This is basically a variation on what Professor Lewin does.

With $N$ turns you measure $N \dot{\Phi}$, but there will be a possibility that you get an "error" caused by the way you connect the voltmeter wires of $\pm \dot{\Phi}$. We did the case of $N=1$ above.

additional item: For the case above with the voltmeter on the left, attach the first wire to the lower part of the open ring on the right (and run it over the top to the voltmeter), and attach the second wire to the upper part of the ring on the right, (and run it under the ring to the voltmeter), and I believe you will then measure $V=2 \dot{\Phi}$.

I do not see that as "indeterminacy" at all. Voltage is perfectly determined; it just depends on the path. Which is the whole point Lewin was trying to make.

 

[Charles Link]

I do not see that as "indeterminacy" at all. Voltage is perfectly determined; it just depends on the path. Which is the whole point Lewin was trying to make.

The EE who hooks his voltmeter up to a transformer coil can get a couple of answers. When $N$ is large, it then makes the answers nearly the same. It seems we are entering a "semantics" problem now, but I don't think it is correct to say voltage is perfectly determined. I think it would be better to discuss the different scenarios that we have and the answers we get, than to wind up disagreeing on what can be how well we say what we are trying to say.

 

[alan123hk]

In order to reduce possible confusion and increase mutual understanding, I am wondering whether Faraday's law of induction can be expressed in the following form in practical applications.

The magnetic flux of each loop from 1to n can be different. If the magnetic flux passing through the loop $\mathbf {\phi_i}$ creates an opposite induced electromotive force, its magnetic flux becomes negative. $$\oint \vec E \cdot d \vec l =- \frac{d}{dt} \sum_{i=1}^{n} \phi_i $$ I think this may be able to express some problems encountered in practice, such as the increase and decrease of one or several loops formed by the voltmeter and its two leads when measuring the voltage of the solenoid inductor.

 

[Charles Link]

@alan123hk That looks like it could be a useful concept especially for those who want to look at the problem of $N$ loops in more detail. From what I have computed, the solution does occur in unit increments for the ideal case of the uniform and localized $\dot{\Phi}$ if the wires of the voltmeter stay in the same plane. This surprised me a little, but the wires are either on the right side of the magnetic field, and pick up the $\dot{\Phi}$ from Faraday's law, or they are on the left and don't. No doubt the solution is a continuous one, if you run the wires out of the plane when you start to go to the other side, in practical cases, in an actual measurement.

Edit: and also note that the digital result between left and right side is independent of how the voltmeter wires are strung or the points of contact on the coil, so long as the wires stay in the same plane, other than being $N$ positions up or down on the coil.

When you first begin the two wires can be on the same position on the coil horizontally, ($N$ turns apart),but as you move them along the coil to separate them horizontally, the voltmeter reading is unaffected. This is really a very interesting result, and it agrees with a Faraday law analysis, as well as by looking at the $E_{induced}$ and $E_s$.

Edit 2: I find the digital incremental result here very interesting.

(When I first looked at it, I was expecting an analog type result that it would make a difference where you placed the voltmeter wires, so that you could get a variable voltage source that varied continuously, but that appears to not be the case=the change happens in digital increments by going to the next turn.)

I haven't seen this problem in any textbook, but I'm hoping others concur with the solution that I got, when the voltmeter wires stay in the same plane in the ideal case presented above.

 

[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.

For the case of an alternating current in a coil, I think we see the successive building and collapse of the magnetic field. This involves energy flowing outwards then inwards. If we measure the phase delay against distance from the coil we it as constant. This is because the measured phase is the addition of the outgoing and incoming energy. We see a similar effect if we measure the phase of sound pressure in a tube containing standing waves. The phase/distance characteristic is flat (actually there is a sudden 180 degree switch every half wavelength). In the alternative case of radiated energy, both sound and EM, we see a progressive phase retardation with distance.

 

[Charles Link]

I would like to comment further where on my post 58:

" 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."

My one further comment here is that Feynman does seem to overlook one detail in that even though $\dot{B}$ may be zero in the region outside, it doesn't mean that $E_{induced}$ is necessarily zero. Feynman IMO was certainly on the right track, but he should have said what I have said above, that we for the most part measure the electrostatic $E_s$ portion, and that it, (i.e. the line integral of $E_s$ for the path through the voltmeter), will dominate over the $E_{induced}$ portion when $N$ is large. For small $N$ we have the case that Professor Lewin has presented, where we don't have one single voltage.

Edit: IMO Feynman was explaining to us why there is indeed a voltage that we can measure, and although he didn't explicitly introduce an $E_s$, I do believe that is what he was referring to when he said the result is independent of the path, and that is because the $\int E_s \, dl$ inside the conductor will be the same value over the same two points for any path outside the conductor through a voltmeter. I believe that is basically what he intended to say, but I leave it open for debate whether you agree or disagree.

Note: In the conductor $E_s=-E_{induced}$, and although Feynman did say the $E_{total}=0$ in the conductor, he did not go into detail and introduce $E_s$ and $E_{induced}$. I do think he would have done well to include this last detail, but again, I leave it open for debate.

One additional edit: Feynman says that since $\dot{B}$ is zero in the region outside, the integral $\int E \, dl$ is path independent. There is a little truth to that statement because if you keep the voltmeter wires and the voltmeter on the same side of the changing magnetic field region, you can string the wires any way you choose, and you will get the same result, (i.e. the $E_s$ part is clearly path independent, but also any non-zero $E_{induced}$ will give the same result for $\int E_{induced} \, dl$ if you keep the wires and voltmeter on the same side). The part that he may have overlooked here is that if you have the points of contact on the right side, but run the wires to the voltmeter on the left (no longer encompassing the changing magnetic flux, e.g. as Professor Lewin does) you get a different result. We can second-guess Feynman in a couple of places here, but once again, I think he could have presented it more carefully and in more detail by introducing the $E_s$, and saying that it is the $E_s$, when it is the dominant term, i.e. for large $N$, that is responsible for making the voltage reading (almost) path independent.

Feynman IMO had it right when he began with the idea that the total electric field is zero in the conductor, but I do think this one needs to have the step that $E_s=-E_{induced}$ in the conductor, and then observe that $\int E_s \, dl$ in the conductor will have the same result when the path is through the voltmeter, regardless of how the voltmeter wires are routed. This is the case because $\nabla \times E_s=0$ in all cases.

 

[Charles Link]

With my post 68 above, I would like to make an additional comment or two: Even the case of a single open conducting coil surrounding a changing magnetic field is very interesting. With the opening on the right and the voltmeter attached to the right, you will read $V=\dot{\Phi}$ every time, regardless of how you let the voltmeter wires go, i.e. they can be close together or separated, and even the points of contact aren't at all fussy=it is a very stable result=you could almost call it a voltage...

But now if you have the voltmeter on the left side, you read zero volts every time=also a very stable result.

In both cases, we could attach multiple voltmeters with different paths between the two points. Since there is zero $\dot{\Phi}$ enclosed by two separate voltmeters, they both read the same. Perhaps this is what Feynman was referring to with the integral he mentioned as path independent.

If you take a voltmeter on the left, and one on the right, (connecting the same two points) they do enclose the $\dot{\Phi}$, so they will read differently. Professor Lewin demonstrated this to us, but we can only wonder if Feynman may also have seen the same thing, but simply chose not to elaborate on it.

 

[vanhees71]

I do not see that as "indeterminacy" at all. Voltage is perfectly determined; it just depends on the path. Which is the whole point Lewin was trying to make.

That's a contradiction in adjecto: voltage is a potential difference. If the electric field had a potential, the voltage difference cannot depend on the path in the line integral connecting the two points. The whole point is that for time-varying fields the electric field is NOT a potential field but obeys Faraday's Law, $\vec{\nabla} \times \vec{E}=-\dot{\vec{B}}$. What's measured in Lewin's experiment is not a voltage but an EMF along the wire look containing the volt meter.

 

[Charles Link]

Perhaps one scenario that illustrates the absence of a voltage in its simplest form is the single conductive open loop around the changing magnetic field, with the opening to the right. If we try to measure a voltage with contacts at top and bottom, (putting the voltmeter on either left or right side), we don''t know whether we will read $\dot{\Phi}$ or zero, unless we know where the opening is at. If the opening is on the left, everything is reversed. Professor Lewin usually had a couple resistors around the ring. The scenario with the open conductive loop is a variation on what Professor Lewin did.

It should be noted if we make an open coil of $N$ loops, with $N$ large, we will read something close to $V=N \dot{\Phi}$ regardless of how we hook up the voltmeter wires, basically with an uncertainty of $\pm \dot{\Phi}$. This is where a EE (electrical engineer) can say he is basically measuring a voltage, even though it arises from the changing magnetic field. I (and others) have also proposed that this $V=N \dot{\Phi}$ from the conductive coil is mostly of an electrostatic nature, and that is one reason behind why the measurement is very nearly path independent for large $N$.

Edit: Very good explanation @vanhees71 in post 70.

additional edit: Since we are on a new page here, I'll reiterate a couple of comments about Feynman's discussion of the scenario: IMO Feynman could have been a little more thorough=he did start out saying the electric field was nearly zero in the conductor (the coil), but he really needed to present it as Professor Lewin did, that the integral is not path independent if you run one of the voltmeter wires on the other side of the changing magnetic field. I welcome any feedback others may have on Feynman's discussion following his equation (22.3). See https://www.feynmanlectures.caltech.edu/II_22.html

 

[SredniVashtar]

That's a contradiction in adjecto: voltage is a potential difference. If the electric field had a potential, the voltage difference cannot depend on the path in the line integral connecting the two points. The whole point is that for time-varying fields the electric field is NOT a potential field but obeys Faraday's Law, $\vec{\nabla} \times \vec{E}=-\dot{\vec{B}}$. What's measured in Lewin's experiment is not a voltage but an EMF along the wire look containing the volt meter.

Following Popovic, I separate the concepts of voltage (line integral of the total electric field) and scalar potential (which is related to the line integral of the conservative part of the electric field). Using different symbols, like V for the former and phi for the latter.

This allows me to avoid any kind of confusion. I have put a few formulas in this post on EE SE

https://electronics.stackexchange.c...drop-exist-in-a-current-carrying-loop-of-wire

It's the answer by Sredni, with green background for pictures and formulae)

(Mind you, it was still on its draft form when I decided not to contribute any more, but the formulae are there)

To me "voltage" is the generally path dependent line integral of the total electric field and as such it does not in general admit a potential function. It can be decomposed into the line integrals of the irrotational and solenoidal components of E. The former is path independent and admit the potential function phi. When Etot only has the conservative part, voltage reduces to potential difference.

 

[vanhees71]

That's the trouble with using some slang. For me a voltage is a potential difference. One should of course make things clear by using math rather than unsharp everyday language.

 

[alan123hk]

According to the definition of magnetic vector potential,
$$ \mathbf {B} = ∇\times \mathbf{A}~~~~~~~~~~~~~~~~\mathbf{E}=-∇ \phi-\frac {\partial \mathbf{A}} {\partial t} ~~~(1) $$Since $$ ∇\times \mathbf{ E_i}=-\frac {\partial \mathbf{B}}{ \partial t } = -∇ \times \frac {\partial \mathbf{A}} { \partial t} $$
Compare the leftmost term with the rightmost term in above equation, we know that the induced electric field $$ \mathbf{E_i}= -\frac {\partial \mathbf A } { \partial t} ~~~(2) $$ Combing equation (1) and (2), we get the conclusion
$$ \mathbf{E} = \mathbf{E_s} +\mathbf{E_i} $$ where $\mathbf {E_s}$ = scalar electric field, $\mathbf {E_i}$ = induced electric field. Therefore, there should be only one possibility for $\mathbf {E}=0~$ which is $~\mathbf{E_s} +\mathbf{E_i}=0$

Obviously the potential difference can be written as $$ V_a-V_b = \int_a^b \left( \mathbf E_s\right) \cdot d \mathbf {r} = \int_a^b \left( \mathbf E -\mathbf E_i\right) \cdot d \mathbf {r} = \int_a^b \left( \mathbf E +\frac {\partial \mathbf{A}} { \partial t}\right) \cdot d \mathbf {r} $$ But since $\mathbf E$ is generally a non-conservative field, this approximately pure potential difference will only appear in space when $\mathbf {E_s}$ is much larger then $\mathbf{E_i} ~~$ or $\mathbf E \cong \mathbf E_s ~~~$

Then the question comes, $\mathbf E_i$ is a non-conservative electric field and $\mathbf E_s$ is a conservative electric field, so they cannot cancel each other out...

 

[vanhees71]

There's no physics in the artificial split of $\vec{E}$ into $\vec{E}_s$ and $\vec{E}_i$, because it's gauge dependent. Only the combination $\vec{E}$ is a physical gauge-invariant observable. It's completed by $\vec{B}$ to build a relativistic field, most easily relized as the antisymmetric Faraday tensor, $F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}$.

 

[weirdoguy]

There's no physics in the artificial split

I guess people from outside of physics would rather overcomplicate their life instead of just simply learning physics the way it is...

 

[Charles Link]

Then the question comes, Ei is a non-conservative electric field and Es is a conservative electric field, so they cannot cancel each other out...

That puzzled me too, but that part I was able to resolve=when you take the path along a loop of the coil, the integral $\int E_s \, dl$ is non-zero. In principle it is almost $\oint E_s \, dl$, but it is conservative, and once around the loop puts you in the adjacent ring. You find the voltage in comparing adjacent rings with a voltmeter is indeed $V=\dot{\Phi}$.

Note that $E_s$ will have a z-component to it between the rings, so that $E_s$ is only really the opposite of $E_{induced}$ for the tangential component inside the coil.

 

[vanhees71]

Note that the EMF is of course also gauge invariant, but that's the integral along a closed (!) path:
$$\mathcal{E}=\int_C \mathrm{d} \vec{r} \cdot \vec{E}=-\int_C \mathrm{d} \vec{r} \partial_t \vec{A}.$$
Now assume for simplicity that the integration path is at rest. Then you can write
$$\mathcal{E}=-\frac{\mathrm{d}}{\mathrm{d} t} \int_C \mathrm{d} \vec{r} \cdot \vec{A}.$$
Now in another gauge
$$\vec{A}'=\vec{A}-\vec{\nabla} \chi$$
with an arbitrary field, $\chi$, but the gradient field integrated over the closed path is 0, and thus
$$\mathcal{E}=-\frac{\mathrm{d}}{\mathrm{d} t} \int_C \mathrm{d} \vec{r} \cdot \vec{A}'.$$
That's also clear from Faraday's Law, because (again for time-independent surfaces $A$ and its boundary $\partial_A =C$)
$$\mathcal{E}=\int_A \mathrm{d}^2 \vec{f} \cdot \vec{\nabla} \times \vec{E}=-\int_{A} \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B},$$
and $\vec{B}=\vec{\nabla} \times \vec{A}$ is of course also gauge-independent.

You get also of course the same result as above, because
$$\int_{A} \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}=\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d}^2 \vec{f} \cdot (\vec{\nabla} \times \vec{A}) = \frac{\mathrm{d}}{\mathrm{d} t} \int_{C} \mathrm{d} \vec{r} \cdot \vec{A}.$$

 

[Charles Link]

Now in another gauge

Perhaps it is very much a simplification, but perhaps we can specify that we are working in the Coulomb gauge, $\nabla \cdot A=0$. In general, the "split" may not be the best physics=it is no doubt an approximation when saying $E_s=-E_{induced}$, but it is one that seems to work in this case to explain what is going on.

Otherwise, we are left with a bit of a puzzle to explain how we measure an EMF with a voltmeter,(or oscilloscope), with the sign of $E_{induced}$ going opposite what the voltage reading is, and using a voltmeter to measure electrostatic voltages such as from a van de Graaf generator, or a battery. To me the answer is that the voltage from the inductor coil is for the most part (quasi) electrostatic.

 

[vanhees71]

In the Coulomb gauge $\vec{E}_s$ is non-local, as is $\vec{E}_i$. In the sum the non-local terms cancel, i.e., they reduce to the retarded Jefimenko solutions with the charge and current distributions as causal sources.

 

[Charles Link]

In the Coulomb gauge $\vec{E}_s$ is non-local, as is $\vec{E}_i$. In the sum the non-local terms cancel, i.e., they reduce to the retarded Jefimenko solutions with the charge and current distributions as causal sources.

I would have to study it further, but that could be what we are looking for=to have the $E_s$ come from the charge distributions, (basically from inside the conductor coil), and the $E_{induced}$ to come from the (changing) current distributions.

Note if $\nabla \cdot A =0$, that should imply that $\nabla \cdot E_{induced}=0$.

In some cases, such as a circuit with no changing magnetic fields in the loop, we are far better at simply looking at Faraday's law with $\mathcal{E} =0$, than to try to compute $E_{induced}$ from external changing magnetic fields that would over-complicate the calculations. In the case of the inductor coil, I do think it makes for good physics to introduce an $E_{induced}$ and $E_s$.

Edit: It should be noted for this inductor or transformer coil, we are for the most part interested in the low frequency 50 or 60 Hz case, and we are not attempting to solve the more general case or cases involving rf etc.

additional edit: It may be worth emphasizing=assuming there is an $E_s=-E_{induced}$ along the path of the coil, because $\nabla \times E_s=0$, the integral $\int E_s \, dl$ between the same endpoints for the path outside the coil will have the same value of $V=N \dot{\Phi}$. For large $N$, this should be much larger than any $\int E_{induced} \, dl$ over this outer path through the voltmeter.

IMO this offers a very good explanation to how we can all have ac power in our homes where the voltage delivered is a reliable number without needing to pay much attention to which side of the transformer the electrical wires happen to be running.

 

[Charles Link]

@vanhees71, @alan123hk Please see a couple edits including the "additional edit" in the above post.

 

[Charles Link]

See also https://www.physicsforums.com/threads/what-is-the-coulomb-gauge.763088/

and

https://en.wikipedia.org/wiki/Helmholtz_decomposition

See also
https://www.physicsforums.com/threads/why-does-a-voltmeter-measure-a-voltage-across-inductor.880100/
especially post 18,
but even for much of the discussion, we make the case for why the voltage that we see from the inductor is not from the $E_{induced}$ from inside the inductor, but rather the electrostatic field $E_s$ that arises from it, which is also present externally to it.

 

[vanhees71]

This contradicts the math provided in #78!

 

[Charles Link]

This contradicts the math provided in #78!

I do think this one is worth some detailed discussions=
For post 78, one formula that is given is that the EMF around the loop is given by Faraday's law. Plenty of other math there as well...

But do look over the thread:
https://www.physicsforums.com/threads/why-does-a-voltmeter-measure-a-voltage-across-inductor.880100/

I don't think I contradict any of the well-established math or physics with an $E_s$ that has the properties that I have been assigning to it. I need a little more feedback to respond to this one though, but patience is in order=I've read through the Physics Forums post on the Coulomb gauge that I "linked" in post 83.

I also went back and looked over very carefully the discussion with the voltmeter and the inductor=I was on a learning curve then, but I did correctly figure out that the voltage from the inductor is $V=+\int E_{induced} \, dl$ while the voltage from electrostatic sources is $V=-\int E \, dl$.
Putting a driving (electrostatic) voltage across the inductor (post 18) showed that we indeed do not see the $E_{induced}$ (the Faraday E) with a voltmeter. Instead, we simply see the electrostatic driving voltage. (Whether we treat the inductor as a passive element by driving it with a voltage across it, or as an active element, by giving it a $dI/dt$ and measuring the voltage, its properties are still the same=the total electric field is basically zero inside the inductor).

I need additional feedback though please, to address the issues in as much detail as I can. I hope to present a good enough case for it, that the merits of $E_s$ and $E_{induced}$ for this inductor/transformer coil become clear... I thought the Coulomb gauge was exactly what I needed to put the math on solid footing...

I also thought @alan123hk had it right in post 74. Looking at post 75, I thought introducing the Coulomb gauge might resolve that difficulty...

One thing that I read in post 78 has come up with the EMF before=that it seems to exist only around a closed path, at least by one school of thought. That could really place its limitations on what we can say about the electric field, and especially $E_{induced}$. I do like to think that by symmetry, etc., that in many cases we can solve for $E_{induced}$. Even though some of the things have been expressed with EMF's over a closed loop, it shouldn't prevent us from doing physics with partial loops and calculating electric fields, if we can...

 

[vanhees71]

But by definition earlier in the thread $\vec{E}_s=-\vec{\nabla} \Phi$. It's contribution to the EMF is 0 (let's not discuss the more complicated case of regions that are not simply connected here, because that's complicating things even further).

I don't know, whether I understand your issue with the Coulomb gauge right. Using Faraday's Law leads to gauge-independent answers, because it only involves $\vec{E}$ and $\vec{B}$ that are gauge-independent.

In the static case, there's of course no induction, and electric and magnetic fields decouple completely.

 

[Charles Link]

In the static case, there's of course no induction, and electric and magnetic fields decouple completely.

I like to thing that it is possible to compute $E_{induced}$ in a coil as a tangential vector with $E_{induced}=\dot{\Phi}/(2 \pi r )$.

I also like to believe that this electric field doesn't suddenly move out and show up in another part of a circuit loop on its own, but rather what does occur, does indeed occur with some logic:

If $E_{induced}$ is inside an ideal conductor coil, there will be an electrostatic field $E_s=-E_{induced}$ that arises to make $E_{total}=0$ in the conductor. This electrostatic field does have the property though that it will display itself in such a manner, even to points outside the coil so that the integral $\int E_s \, dl$ across any path connecting the same two points is the same.
This is how the $E_{induced}$ appears to show up in another part of the circuit loop, (such as the path through the voltmeter), where we can even arrange by symmetry, keep wires close together, etc., that we can't possibly be seeing $E_{induced}$ in that part of the reading of the voltmeter. What we instead see though is the integral of $E_s$ over the external path, where $E_s$ arose directly from $E_{induced}$.

and I think @alan123hk had it right in post 74, with his definitions of $E_s$ and $E_{induced}$ from the scalar and vector potentials.

 

[Charles Link]

and a follow-on: I followed the part $E_s=-\nabla \Phi$ , since the curl of a gradient is zero, $E_s$ can't be part of the closed loop EMF.

It is that same property though $\nabla \times E_s=0$ is exactly why (with the path across the voltmeter wires=not a closed loop) $\int E_s \, dl$ is what we see in the voltmeter, even though we might think we are seeing $E_{induced}$.

Inside the coil $\int E_s \, dl=-\int E_{induced} \, dl$.

Outside the coil, $E_{induced}$ can be zero, but we still see $\int E_s \,dl$ giving the value of $\int E_s \, dl=-\int E_{induced} \, dl$ that they had inside the coil.

It's almost amusing that your reason for why we don't need to pay attention to $E_s$ when it comes to considering EMF's ($\nabla \times E_s=0$), is exactly why the $E_s$ does what it does, and makes the whole thing work. :)

 

[vanhees71]

I like to thing that it is possible to compute $E_{induced}$ in a coil as a tangential vector with $E_{induced}=\dot{\Phi}/(2 \pi r )$.

This doesn't make sense. On the left-hand side you have a vector, on the right-hand side a scalar. The sources of the em. field are $\rho$ and $\vec{j}$. It doesn't make sense to interpret $-\partial_t \vec{B}$ in some sense as the source of (parts of) $\vec{E}$.

I also like to believe that this electric field doesn't suddenly move out and show up in another part of a circuit loop on its own, but rather what does occur, does indeed occur with some logic:

If $E_{induced}$ is inside an ideal conductor coil, there will be an electrostatic field $E_s=-E_{induced}$ that arises to make $E_{total}=0$ in the conductor. This electrostatic field does have the property though that it will display itself in such a manner, even to points outside the coil so that the integral $\int E_s \, dl$ across any path connecting the same two points is the same.
This is how the $E_{induced}$ appears to show up in another part of the circuit loop, (such as the path through the voltmeter), where we can even arrange by symmetry, keep wires close together, etc., that we can't possibly be seeing $E_{induced}$ in that part of the reading of the voltmeter. What we instead see though is the integral of $E_s$ over the external path, where $E_s$ arose directly from $E_{induced}$.

and I think @alan123hk had it right in post 74, with his definitions of $E_s$ and $E_{induced}$ from the scalar and vector potentials.

As I tried to explain already above, a physical interpretation of this artificial split of $\vec{E}$ is impossible, because it's gauge dependent.

 

[vanhees71]

and a follow-on: I followed the part $E_s=-\nabla \Phi$ , since the curl of a gradient is zero, $E_s$ can't be part of the closed loop EMF.

It is that same property though $\nabla \times E_s=0$ is exactly why (with the path across the voltmeter wires=not a closed loop) $\int E_s \, dl$ is what we see in the voltmeter, even though we might think we are seeing $E_{induced}$.

Inside the coil $\int E_s \, dl=-\int E_{induced} \, dl$.

Outside the coil, $E_{induced}$ can be zero, but we still see $\int E_s \,dl$ giving the value of $\int E_s \, dl=-\int E_{induced} \, dl$ that they had inside the coil.

It's almost amusing that your reason for why we don't need to pay attention to $E_s$ when it comes to considering EMF's ($\nabla \times E_s=0$), is exactly why the $E_s$ does what it does, and makes the whole thing work. :)

Again: you cannot interpret these two parts of the electric fields physically since this split of the electric field is gauge dependent!