Inducing EMF Through a Coil: Understanding Flux

[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.

 

[Charles Link]

It might be worthwhile as an exercise to go back to post 74 from @alan123hk , where $E_{induced}=-\frac{\partial{A}}{\partial{t}}$, and compute $E_{induced}$ from the $A$. I normally like to simply compute $E_{induced}$ from the EMF using the symmetry with the path distance being $2 \pi r$.

The vector potential $A$ is known for a uniform $B$ to be $A=(B \times \vec{r})/2$.

I think it is also possible to compute $A$ from the currents of a long solenoid that make up the uniform field, with $A(x)=\frac{\mu_o}{4 \pi} \int \frac{J(x')}{|x-x'|} \, d^3 x'$, but perhaps not a simple integral. @vanhees71 Might you have worked through this more difficult computation?

The answer we are looking for for $E_{induced}$ from using the EMF is $E_{induced}= \dot{B}(\pi r^2)/(2 \pi r)=\dot{B} r/2$, (pointing clockwise),

(Note for $r>a$, we should get $E_{induced}=\dot{B} a^2/(2r)$, but we don't have an expression for $A$ for $r>a$).

and that's exactly what we get using $E_{induced}=-\frac{\partial{A}}{\partial{t}}$ with $A=(B \times \vec{r})/2$.

It's nice to see that everything is consistent.

(Note: I omit most vector symbols for brevity=it's difficult to write vectors all the time in Latex).

 

[vanhees71]

Your formula for $\vec{A}$ is of course correct. It's in the Coulomb gauge for static fields (i.e., you cannot use it to calculate it for time-dependent fields, even in Coulomb gauge, because there you need to solve a wave equation rather than a Poisson equation).

If you have a long solenoid and you can use the quasistationary approximation, then your argument is also fine, and you don't calculate this unphysical field $\vec{E}_{\text{induced}}$ but simply the electric field sincd in this approximation you have $\vec{B}=B(t) \Theta(a-R) \vec{e}_3$ (with $\vec{e}_3$ the direction of the cylinder axis and $a$ the radius).

Then using Faraday's Law in integral form and the cylindrical symmetry you get $\vec{E}=E_{\varphi} (R) \vec{e}_{\varphi}$, and with this ansatz you immediately find your solution, but it's not this strange part of the electric field but the electric field itself and you don't even need the potentials, i.e., you work with gauge-invariant quantities througout!

 

[Charles Link]

@vanhees71
Might you comment on the computation of the vector potential $A$ from a long solenoid. (See post 101). For the long solenoid, several years ago I computed the magnetic field $B$ using Biot-Savart, (for the general case, and not simply on-axis), and found it to be uniform, just as it should be, but that integral wasn't easy either.

Has the $A$ for the long solenoid been computed from the current density in the coil? I think it might be a fun exercise, but not an easy one. :)

See https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Electricity_and_Magnetism_(Tatum)/09:_Magnetic_Potential/9.04:_Long_Solenoid

On this one they used Stokes' theorem, rather than doing a very complicated integral.

See also https://physics.stackexchange.com/q...-potential-of-a-solenoid-in-the-coulomb-gauge

Looks like on this one, they cranked it out. :)

 

[vanhees71]

I'd use the Coulomb gauge, i.e.,
$$\Delta \vec{A}=-\vec{j}.$$
In the approximation that
$$\vec{j}= n I \delta(R-a) \vec{e}_{\varphi},$$
which neglects that in reality the wire is wound in a helix and not as circles with $n$ the number of windings per unit length.

Then instead of using Biot-Savart's Law it's simpler to use the symmetry of the problem and make the Ansatz
$$\vec{A}=A(R) \vec{e}_{\varphi}.$$
First of all we note that
$$\vec{\nabla} \cdot \vec{A}=0,$$
i.e., the ansatz automatically fulfills the Coulomb-gauge condition.

Now in cylindrical coordinates we must write, making use of this,
$$-\Delta \vec{A}= \nabla \times (\nabla \times \vec{A}) = -\vec{e}_{\varphi} \frac{\mathrm{d}}{\mathrm{d} R} \left (\frac{(R A)'}{R} \right).$$
So we get
$$\frac{\mathrm{d}}{\mathrm{d} R} \left (\frac{(R A)'}{R} \right)=-n I \delta(R-a). \qquad (*)$$
First we solve the differential equation for $R \neq a$, where the right-hand side is $0$:
$$\left (\frac{(R A)'}{R} \right)'=0 \; \Rightarrow \; (R A)'=2 C_1 R \; \Rightarrow \; A=C_1 R + \frac{C_2}{R}.$$
Now we must find the solution such that the $\delta$-function singularity comes out right. In $R=a$ shouldn't be a singularity, which means that
$$A(R)=C_1 R \quad \text{for} \quad R<a.$$
Also for $R \rightarrow \infty$ the potential should go to $0$, i.e.,
$$A(R)=\frac{C_2}{R} \quad \text{for} \quad R>a.$$
We have to find $C_1$ and $C_2$. Now the highest derivative must lead to the $\delta$-distribution singularity and thus $A'(R)$ must have a jump at $R=a$ and $A$ itself must be continuous there. The latter condition gives
$$C_1 a = \frac{C_2}{a} \; \Rightarrow \; C_2=C_1 a^2.$$
Integrating (*) over an infinitesimal interval $(a-\epsilon,a+\epsilon)$ and then letting $\epsilon \rightarrow 0^+$ one gets
$$\left (\frac{(R A)'}{R} \right)_{R=a+0^+}-\left (\frac{(R A)'}{R} \right)_{R=a-0^+}=-n I.$$
For $R>a$ we have $R A=C_2=\text{const}$, i.e., this contribution vanishes. For $R<a$:
$$[(R A)'/R]_{R=a}=2 C_1=n I \; \Rightarrow \; C_1=n I/2.$$
This solves the problem.
$$A(R)=\begin{cases} n I R/2 & \text{for} \quad R<a, \\ n I a^2/(2R) & \text{for} \quad R \geq a.\end{cases}$$
Now
$$\vec{B} =\vec{\nabla} \times (A \vec{e}_\varphi)=\begin{cases} n I &\text{for} \quad R<a, \\ 0 & \text{for} \quad R>a. \end{cases}$$

 

[Charles Link]

Thank you @vanhees71

For the boundary condition, with the surface current flowing, I think you are essentially taking $\int \nabla \times (\nabla \times A) \cdot dS =\oint \nabla \times A \cdot \, dl=\oint B \cdot dl=-(B^- -B^+)L=\int J \cdot dS=n \mu_o I L$ around a narrow rectangular strip (of length $L$) with one long edge just inside the solenoid and the other just outside.

(I had to look up/google the curl in cylindrical coordinates).

Your derivation made for very good reading. Thank you. :)