Inducing EMF Through a Coil: Understanding Flux

[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.

 

[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.