Inducing EMF Through a Coil: Understanding Flux

[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.

 

[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!

 

[Charles Link]

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

I mad the best case that I could for the idea of splitting the $E$ into an $E_s$ and an $E_{induced}$ for the problem of a transformer coil operating at 50 or 60 Hz. Inside the conductor of the coil, we have a sea of electrons, and it should very quickly respond, in what would be a quasi-static type of charge redistribution, to any $E_{induced}$ that would exist from a changing magnetic flux.

We have that $\oint \vec{E}_{induced} \cdot dl=-\dot{\Phi}$. For the type of symmetry we have in the problem with the long solenoid, with a uniform changing magnetic field over the circular area, I do believe we should be able to say that the amplitude of $E_{induced}$ is $\dot{\Phi}/(2 \pi r)$, and if you want to call it $E$ instead of $E_{induced}$ for the free-space anywhere outside of the core where the magnetic field is changing, I do think there is sufficient symmetry to justify its computation. If we are not justified for this as a starting point, then I think we are indeed left with simply working with closed loops and calculating EMF's for those closed loops.

If we do have the starting point of an $E_{induced}$ in the region where the conductive coil is present, the sea of electrons will give $\vec{E}_s =-\vec{E}_{induced}$ inside the conductor coil. This then gives rise to the (integral of the) electrostatic $E_s$ through any other external paths connecting the same two points, including through a voltmeter. The calculations are remarkably consistent, but it is really up to the individual whether it is deemed as good physics.

One item I did look to explain is how a voltmeter can measure what is an $E_{induced}$. If we are limited to writing it as an EMF in a closed loop, we certainly still have calculations that will get us the answer, (for what the voltmeter reads, etc.), but to me, introducing $E_s=-E_{induced}$ in the conductor sheds some light on the problem.

With this though, I think I rest my case=you can either see some merit in looking at it in such a manner, or stick to the EMF's with closed loops as being the method you much prefer.

 

[vanhees71]

Faraday's Law doesn't involve electrons at all. That's in the inhomogeneous Maxwell equations and the additional assumption that the force density on a charged fluid is (in SI units)
$$\vec{f}=\rho \vec{E} + \vec{j} \times \vec{B}=\rho (\vec{E}+\vec{v} \times \vec{B}).$$

 

[Charles Link]

I do think in many places in plasma physics and solid state physics, they do make attempts to calculate the electric field, and wherever possible, it can help to identify the sources. This transformer coil or inductor coil seems to me to be kind of a simple system to work with. I presented it the best I could, and it's ok with me that not everyone agreed to the solution. Maybe a couple years from now, we might see more agreement and/or someone else might be able to make use of it.

 

[vanhees71]

The electric field as a whole is of course a physical, observable field. The only thing I object to is the attempt to interpret the potentials and parts of the field somehow defined with them. That are always gauge-dependent quantities, and one cannot so easily interpret them physically, and they are not observable.

 

[Charles Link]

@vanhees71 Very good. I think I can agree with your latest post.

I would like to present one more scenario that I may have presented above, but I don't think I presented it in it's complete form:
Suppose we have a region of a uniform magnetic field in the z-direction that is changing by a constant rate. We might even have additional details that it is created by a long solenoid, possibly with an iron core, and we have that the current in the solenoid is changing at a constant rate, thereby causing the magnetic flux to change. We wish to construct a probe to measure the $\dot{\Phi}$ or $\dot{B}$ or ## $of the magnetic field, or even get a measurement of what the$ \dot{E} $is, assuming the probe doesn't significantly alter the$ \dot{E} $that exists without the probe. How might we do this in an optimal manner? From the discussions in this thread, it seems that could best be done by putting an open-circuited conductive coil of a large number of turns, e.g.$ N=100 $or more around the solenoid, and measure the voltage that is created with a voltmeter or oscilloscope. We essentially have a secondary coil around the original coil, and we have made it into a transformer. It should also be apparent why we would want to have a large$ N $, both for signal to noise reasons, as well as so that we don't encounter significant errors from the Professor Lewin problem that has been discussed above and in other threads. Note that we do have one stumbling block in that the$ E $inside the probe's conducting coil will be nearly zero, so that the probe does alter, perhaps in a predictable manner, the$ E $field around the original coil. I believe we have presented a satisfactory analysis for this probe in the above thread, with the result that we can have accurate experimental results for$ \dot{\Phi} $,$ \dot{B} $, and$ \dot{E} ## that exist without the probe.

For any readers who are first learning about the Faraday effect and inductors and transformers, I think they may find this somewhat educational.

 

[vanhees71]

That's fine, because here you argue with entirely physical observable quantities and not gauge-dependent ones. I guess whay you mean is not the "Faraday effect" but "Faraday's Law of Induction", i.e., one of Maxwell's equations, $\dot{\vec{B}}+\vec{\nabla} \times \vec{E}=0$.

The Faraday effect is the rotation of the polarization vector of light going through a medium under the influence of a magnetic field:

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

 

[Charles Link]

Before I give in completely to agreeing that the separation of the electric field into $E_s$ and $E_{induced}$ constituents is perhaps unsound, I think it would be worthwhile to take another look at a seemingly difficult problem that was solved rather readily by using this concept.

See https://www.physicsforums.com/threads/faradays-law-circular-loop-with-a-triangle.926206/page-6
posts 192 and 193.

@cnh1995 solved it in a couple of steps, where it is a much more difficult process to write loop equations with EMF's and Faraday's law for the various loops.

I have to believe that this one was probably solved in what would be the Coulomb gauge, but in any case, being able to generate the correct result lends considerable credibility to the methodology. It seems to be a very useful computational tool to introduce an "electrostatic" potential at each node, and also to assume that an $E_{induced}$ can be computed from the changing magnetic flux when there is sufficient symmetry. I do think this second method is worth further study. I welcome any feedback.

Note in this solution, it was not necessary to solve for the electrostatic potential everywhere, but simply to find the value of it at two nodes, assigning a value of zero to the 3rd node. It made the assumption that there is a value for the electrostatic potential at each node, and worked from there.

Edit: In some ways I see Faraday's law for the closed loop as a simplification, where we can ignore any $E_s$ that arises, because $\nabla \times E_s=0$ so that $\oint E_s \cdot dl=0$, so that the EMF $\mathcal{E}=\oint E \cdot dl=\oint E_{induced} \cdot dl$. That one is open for debate. There may be the more correct way of looking at it, but regardless of how much we know or don't know, we still will never have all the answers. :)

@alan123hk I welcome your feedback on this one.

 

[vanhees71]

In Faraday's Law there's always only a closed loop. The fundamental equation is of course the local Maxwell equation,
$$\vec{\nabla} \times \vec{E}+\partial_t \vec{B}=0.$$
The complete integral form for the most general case of an arbitrarily moving (i.e., time-dependent) area $A$ with boundary $\partial A$ and the velocity $\vec{v}(t,\vec{x})$ along the boundary reads
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B})=-\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
This is of course gauge independent.

 

[alan123hk]

Sorry, I didn't read this https://www.physicsforums.com/threads/faradays-law-circular-loop-with-a-triangle.926206/page-6 in detail because I thought it was quite long and complicated.

But allow me once again to offer some personal thoughts on this long-standing and much-criticized issue. I never said that decomposing the electric field into two parts, $E_c$ and $E_i ~$, describes a precise physical phenomenon. It is only an approximation or calculation method. Everyone may have a different opinion as to whether it works or not, But I'd actually be a little surprised if a lot of people think this is a really bad approach. We just use Helmholtz's theorem to decompose the total E into two parts, $$ E= E_c + E_i$$where $E_c$ is generated by the instantaneous value of the charge density and $E_i$ is generated by the instantaneous value of the current density. Note that The two sources are not independent because the continuity condition relates the instantaneous value of charge density and current density, $$∇ \cdot j +\frac {\mathbf{ \rho}} {\partial t} = 0$$ I think that under the condition that the quasi-static field is satisfied, we can calculate a $E_c$nd $E_i$ separately and then add them to get the total $E$. Although this method is not very accurate, in practical applications it is sometimes simple and intuitive. This method of operation is just a process, not the goal

 

[Charles Link]

Very good @alan123hk If you get some extra time, I think you might find it worthwhile to look at the solution in post 192 by @cnh1995 of the thread that I linked, where in post 193 I detail the logic behind it. The members of PF did a lot of spinning their wheels on this one, so it doesn't pay to read the whole thing. I think I solved it around post 152 by using 6 equations and 6 unknowns, (with Faraday's law and loop equations), but @cnh1995 solved it with a much simpler solution. Try solving it yourself, and then see how @cnh1995 solved it. I think you will like his solution. He really took something that looked like a lot of work, and made it look easy.

I looked at it again myself, and to help you follow post 192, it helps to see the original post that says $r_1=(3/2) r_2$. He just uses simple circuit theory=Ohm's law, where the current in each segment is the (induced EMF plus the electrostatic potential difference ) divided by the resistance, and the current into each node is the current out of the node. He then gets two relations between $V_a$ and $V_c$, and solves for them.

The assumption that there is an electrostatic potential at each node, and thereby an additional electrostatic potential difference across each resistor in addition to the induced EMF was key to his simple solution.