Inducing EMF Through a Coil: Understanding Flux

[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. :)

 

[Charles Link]

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, Ec and Ei , 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.

Very good @alan123hk

For both the secondary coil around a changing magnetic field, and the other problem that I "linked" in post 97, separating the $E$ into electrostatic $E_s$ and induced $E_{induced}$ components seems to offer some insight into the physics...

We spent a lot of time in this thread dissecting the voltage that one can measure from a coil such as the secondary coil of a transformer. It might be worth mentioning though that the steady ac voltage that we get from our power lines is only one of a couple of things that occur in the operation of a transformer. The self-balancing that occurs so that $|N_p I_p| \approx |N_s I_s|$ so that the total magnetomotive force stays approximately the same in the formula $\oint H \cdot dl=\sum N_i I_i$, (when there is additional power/current in the coils), is rather remarkable.

In the above thread we analyzed the voltage in the coil and things like Professor Lewin's puzzle, but it may be very worthwhile at this time to review a thread that a bunch of us worked on a couple years ago regarding transformers. I think that thread was very much complete, where we all agreed to the solutions, but IMO it really pays to try to keep the things we learn at our fingertips, rather than forgetting some of the finer details. See

https://www.physicsforums.com/threa...former-homework-problem.1002895/#post-6491336

To me this is very relevant in analyzing the details of how these changing magnetic fields operate.
I plan to study this one again. Perhaps others will also find it useful. :)

 

[alan123hk]

I just tried simplifing Jefimenko's equations https://en.wikipedia.org/wiki/Jefimenko's_equations based on quasi-static conditions and got some interesting results.

Of course, Jefimenko's equations already provides a complete and accurate solution, and trying to simplify it seems pointless. But my purpose in doing this is really to show that there is some basis for decomposing the electric field into two parts ($~E_c ~,~ E_i$) in the approximate model, and it is not based on a reckless act of pure imagination. In addition, of course, the above simplified equations can also be directly derived based on Maxwell's equations and quasi-net field conditions. Finally, I would like to say that everyone is welcome to provide different opinions or point out errors.

 

[Charles Link]

See https://www.physicsforums.com/threa...we-apply-the-biot-savart.927681/#post-5979758
especially posts 16 and 17, that discuss the self-balancing of the transformer that almost magically supplies power to whatever degree is needed, with the magnetic field in the transformer simply operating at the keep-alive value where there is zero load. The magnetic field from any additional current in the secondary is completely balanced and offset by additional current in the primary.

@alan123hk I need to study your derivations more carefully, but I'm glad you are keeping an open mind to the $E_c$ and $E_i$. It works well enough in some cases that it seems there must be some real physics behind it.

Edit: Looking it over briefly, @alan123hk it looks like this may be just what we are needing. I think I would give your derivations an A+. :)

 

[Charles Link]

For completeness, I want to post one more "link" that I think may be relevant in the discussion:

See https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/

This helps explain some of the physics that occurs when we have a system where we have an iron core to enhance the magnetic field, as well as to show in some detail what the formula $B=\mu_o H+M$ is all about.

With the combination of 3 or 4 threads that we have on Physics Forums, I do think students could likely learn more about the subject of transformers here on PF than they could find in most of the textbooks on the subject.

IMO it is important for students to learn both the pole model and magnetic surface current model of magnetostatics. It is remarkable that both methods get the same result for the computed magnetic field $B$.

In this thread, we did spend a lot of time discussing whether we are justified in separating the $E$ from a conductive coil where there is a changing magnetic field into $E_c$ and $E_{induced}$ components, but to me it does provide one additional piece to the puzzle, and even if we don't have complete agreement on it, perhaps we have lessened the disagreement.

Right now our audience seems somewhat limited, as has been pointed out in another thread, https://www.physicsforums.com/threa...d-to-10-years-ago.1055276/page-3#post-6980034 , but perhaps there are still a good number that will read some of these posts. I do think the reader could find them rather educational.

 

[Charles Link]

As mentioned in post 106 above, I've been reviewing a couple of the Physics Forums threads on transformers. Another that is interesting is the following, where @Dale writes the coupled differential equations for the primary and secondary currents.

See https://www.physicsforums.com/threa...er-core-when-coupling-high-low-power.1001193/ see post 22 and 43.

A couple of things that come from these equations is $V_s/V_p=N_s/N_p$, and that the magnetic flux is the same for both the primary and secondary coil in complete coupling.

The current balancing $|N_p I_p| \approx |N_s I_s|$ could also be predicted mathematically from these equations, where the primary current responds to any increased secondary current, when the secondary resistance is lowered, so that the magnetic flux remains at the keep alive, no load value when there is a load.

I included the "link" above, as well as others in a couple of the above posts, because I wanted to make this thread into a handy reference for anyone else who also wants to review the transformer subject. I found it useful to review this material. IMO, the transformer topic is an important one, that involves some very useful physics. The transformer is a remarkable device and I think they would do well to give it increased emphasis in the E&M courses. We spent a lot of time in my college E&M classes learning how $B=\mu_o H+M$, and doing boundary value problems, but the transformer makes such good use of the E&M principles that it could really be at the core of the E&M studies. In any case, I'm hoping at least a couple people find the last couple of posts of interest.

Edit: I asked myself looking at @Dale equations, if it would be possible to have a transformer with an air core, rather than an iron core, if you could wrap the wires of the primary and secondary in a torus shape, so that the coupling would be nearly complete?

The answer, which took me maybe a minute or two to figure out, is that the inductance $L=\Phi/i$ is proportional to the permeability $\mu$ of the core, and that without the iron, the (ac) keep-alive current would need to be very high (i.e. amperes or more) to get an appreciable voltage, and the result would be huge resistive power losses in the primary coil. I think I got that correct, but I welcome any feedback. The iron core also makes for better coupling, so that the primary and secondary windings can each be wrapped very locally, rather than spread out in a torus shape.

Edit 2: See https://www.britannica.com/technology/air-core-transformer

Looks like because of the $\mathcal{E}=-d \Phi/dt$, we can get sufficient voltages with an air core at higher frequencies.

 

[Charles Link]

For a little more on the transformer, I want to say a couple of things on something that seems to be unnecessarily taught like voodoo, see https://en.wikipedia.org/wiki/Magnetomotive_force

where the formula they use follows from a variation of one of Maxwell's equations $\nabla \times H=J_{conductors}$ when the $\dot{D}$ term can be ignored. For a derivation of the MMF (magnetomotive force) equation,
see https://www.physicsforums.com/insig...tostatics-and-solving-with-the-curl-operator/

The $\oint H \cdot dl=\sum N_i I_i$ can be very useful in computing the magnetic flux in cases where there is an air gap in the transformer. See https://www.physicsforums.com/threads/absolute-value-of-magnetization.915111/

It also is useful in analyzing the transformer mathematically with the current balance that occurs to see how the keep-alive value of $B$ is what results even when there are large currents in the secondary coil. These large currents are offset almost precisely by the additional current in the primary coil, with $N I_p=-NI_s$.

IMO they really unnecessarily complicate things when they teach it in the MMF (magnetomotive force) manner. The process that they do follows from the modified Maxwell equation mentioned above, and it really is much more straightforward, at least for a physicist, to see that it simply follows from Maxwell's equations.

 

[Charles Link]

One more item on the subject of transformers that really needs to be mentioned to make this set of "links" complete is the laminations that are used in the iron core. We had a rather thorough discussion of this in a previous Physics Forums thread.

See https://www.physicsforums.com/threads/i-dont-understand-transformers-how-to-apply-them.1002399/

@Baluncore pointed out in post 25 that the reason for the laminations is a skin depth problem, rather than simply to block eddy currents. See also his post 69 and the complete discussion from post 70 to post 82. I found his input very enlightening.

There are now also some transformer core materials where the laminations are not needed. See https://www.customcoils.com/blog/what-types-of-cores-used-in-the-toroidal-transformers/

Hopefully the above is useful for the student who is interested in learning some of the details behind transformer core materials.

I found it very useful for myself to review these various discussions (with the "links" I listed in the last few posts) and the mathematical solutions that were presented in solving these transformer problems.
Hopefully at least a couple of others also find it good reading. :)

 

[Charles Link]

With a short review of the transformer, back to @alan123hk 's post 107. This to me is the way the problem of the voltage of the coil needs to be taught=if we teach it simply as an EMF without a local $E_{induced}$ then we really need to find an explanation for how the EMF displaces itself from the coil and suddenly finds itself across the stretch of the voltmeter.

The methodology with $E_{total}=0$ in the conductor IMO gives a satisfactory explanation , where an $E_c=-E_{induced}$ arises, and we know that this (electrostatic) $E_c$ has $\nabla \times E_c =0$, so that this $E_c$ will persist on an external path, and thereby provide for what is usually the vast majority of what we read on the voltmeter or oscilloscope.

IMO @alan123hk 's calculations (see also his post 74) lend much credence to this concept, and I do think this does give additional aid to our understanding of the transformer. The concept of voltage is taken for granted, until we examine it more closely, with Professor Lewin's paradox in mind, as well as also really not having an explanation for how an EMF also gives rise to a voltage.

IMO, the separation of $E_{total}$ into $E_{induced}$ and $E_c$ offers a very satisfactory explanation. We can survive without it, and still have good explanations for most of the operation of the transformer, but to me this is one of the last pieces of the puzzle, and the explanations are more complete if it is included.

 

[vanhees71]

Of course in reality nothing is instantaneous. I haven't followed in detail all these old threads, but I guess that finally it got to the standard textbook treatment in the usual quasistationary approximations, where the displacement current is neglected (except in capacitors). This of course makes the solution instantaneous, i.e., it's neglecting retardation. From the field-point of view it's the near-field approximation, justified precisely if the spatial extension of the circuit is small compared to the typical wavelength of the em. field.

The separation of the fields in gauge-dependent parts never has any physical significance, precisely because it's gauge dependent.

 

[alan123hk]

The methodology with Etotal=0 in the conductor IMO gives a satisfactory explanation , where an Ec=−Einduced arises, and we know that this (electrostatic) Ec has ∇×Ec=0, so that this Ec will persist on an external path, and thereby provide for what is usually the vast majority of what we read on the voltmeter or oscilloscope.

IMO @alan123hk 's calculations (see also his post 74) lend much credence to this concept, and I do think this does give additional aid to our understanding of the transformer. The concept of voltage is taken for granted, until we examine it more closely, with Professor Lewin's paradox in mind, as well as also really not having an explanation for how an EMF also gives rise to a voltage.

I agree with your idea.

This is how I understand it. The magnitude and direction of the magnetic field and induced electric field generated by the changing current are fixed in space. So how do the voltage and current output by the transformer, that is, the power flow based on the Poynting vector, be transmitted to the voltmeter along any path of the wire?

These voltages and currents, or correspondingly the redistributed E and B fields transmitted along the wire according to the Poynting vector, are clearly generated by the charges and currents on the wire.

Of course, quasi-static fields can only approximate its operation, and it only provides a simplified concept to describe it, rather than a complete, rigorous and accurate theory. I think just stating that this is an approximate approach can avoid misunderstandings.

 

[vanhees71]

Power flow (or better energy flow) is of course not transmitted along the wire but through the electromagnetic field. It's a common misconception due to the very unfortunate popularity of the "water-pipe analogy" of circuits advertised by physics didactics for high school. It's, however, not an analogy at all but a severe misconception!

It's sufficient to look at the example of a DC-carrying coaxial cable to see that the energy transport along the cable is (almost) only in the free space between the conductors, i.e., only there is a (non-negligible) component of the Poynting vector along the cable. In the interior the Poynting vector is (almost) directed radially inwards. For a thorough discussion, see A. Sommerfeld, Lectures on Theoretical Physics, vol. 3.

 

[alan123hk]

Maybe my English expression is not good. The electric field inside a conductor is zero, so of course energy cannot be transferred.

I guess I should say that magnetic and electric fields guided by transmission lines transfer energy. In a coaxial line, these energies of course travel in the space between the two coaxial structures. On a transmission line consisting of just two separate wires, these energy-carrying electric and magnetic fields will of course expand into all surrounding space as they propagate and transfer energy.

My main argument is that the energy traveling along any path outside the transformer is primarily the electric and magnetic fields guided by the wires or transmission lines, which can be considered a different source than the electromagnetic fields (Poynting vector with fixed magnitude and direction) coming from inside the transformer.

 

[Charles Link]

So how do the voltage and current output by the transformer, that is, the power flow based on the Poynting vector, be transmitted to the voltmeter along any path of the wire?

The power flow of the transformer is a separate item, and the Poynting vector description no doubt offers the most accurate explanation. It is a good thing for the engineer that the power can be computed by the current x voltage product. It is a separate puzzle why the current x voltage product works, and perhaps might also be worth addressing in detail. I find it of interest that with the current balance $N_p I_p=-N_s I_s$, we don't see any increase in the magnetic field in the core in the process of more energy being relayed from the primary to the secondary coil. The process is rather remarkable. Perhaps we would do well to examine it in more detail.

In any case, so far I have only tried to give an explanation for the voltage on the voltmeter (and not the power flow). It would take a separate meter, with low resistance and placed in series, to measure the current. We now seem to have a little more agreement on the details of what the voltage is than we first had, even if it is not complete agreement.

Edit: With the power flow, this is presently just a not completely thought out guess, but I wonder if with our quasi-static approximation, that the voltage can be thought of as acting electrostatically on the current, and thereby making for the power transfer, (but scratch that=it lacks good physics). Perhaps that is how many EE's think of it, rather than trying to do an analysis of the electromagnetic fields. It doesn't explain how the power gets from the primary part of the circuit to the secondary, but I have yet to solve that with the Poynting vector method.

 

[vanhees71]

Maybe my English expression is not good. The electric field inside a conductor is zero, so of course energy cannot be transferred.

In a current-conducting wire the electric field inside is not zero, but it's in very good approximation along the wire to drive the conduction electrons through it against the friction of these electrons. That's what's described by Ohm's Law. In non-relativistic approximation and for homogeneous, isotropic conductors it reads $\vec{j}=\sigma \vec{E}$.

 

[alan123hk]

In a current-conducting wire the electric field inside is not zero, but it's in very good approximation along the wire to drive the conduction electrons through it against the friction of these electrons. That's what's described by Ohm's Law. In non-relativistic approximation and for homogeneous, isotropic conductors it reads j→=σE→.

I just want to describe that in the general transmission line model, it is usually assumed that due to the skin effect, the alternating current flows concentratedly through the conductor surface, and the electric field inside the conductor is approximately zero relative to the outside, corresponding to the energy mainly propagated by the external electromagnetic field, rather than losses within the conductor. Of course, there is still an electric field close to the conductor surface, but the ohmic losses produced in this case are still typically much smaller than the effective power transfer provided by the transmission line.