Inducing EMF Through a Coil: Understanding Flux

[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.

 

[Dale]

It is a separate puzzle why the current x voltage product works, and perhaps might also be worth addressing in detail.

See figure 11.3.3 and the section that contains it.

https://web.mit.edu/6.013_book/www/book.html

This is probably the main proof that shows that circuit theory is a valid simplification of Maxwell’s equations. Especially the “lumped element” aspect

 

[Charles Link]

See figure 11.3.3 and the section that contains it.

I wasn't able to access this section on my computer, but I think I have a pretty good idea of the route that it takes. I have to wonder if at times the physicist can over-analyze something=going the Poynting route, etc.

It still in any case is remarkable to me that the ideal transformer functions with its keep alive voltage, with just a small ac current $I_{po}$ in the primary coil, and then can deliver power with large (ac) currents and large (ac) voltages in the secondary coil, and offset any magnetic field from the secondary coil with $N_p I_p=-I_s N_s$, so that with $\oint H \cdot dl=\sum N_i I_I$, the power carrying $N_i I_i$ terms cancel each other.

It may be worth quantifying a couple of comments on the above when we talk about ohmic losses in a conductor. For a coil delivering one hundred volts with one ampere of current, I believe ballpark numbers might be somewhere around one watt of ohmic power loss and thereby about one volt of voltage drop for a good conductor. In any case, it seems to be a good approximation to assume $E_{total} \approx 0$ so that $E_c=-E_{induced}$ inside the conductor, although we don't have total agreement on that methodology.

Edit: It is perhaps fortuitous that a very skilled physicist can integrate the Poynting vector over an arbitrary extended surface and show the energy (computed from the fields) through it is conserved, but in the case of the ideal transformer, the more practical calculation seems to be that the power output of the primary coil, the $IV$ product, (with the phase term, etc.), agrees with the power delivered by the secondary coil using the $IV$ product.

 

[vanhees71]

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.

That seems to imply that you only consider the limit $\sigma \rightarrow \infty$. Then it's right.

 

[vanhees71]

See figure 11.3.3 and the section that contains it.

https://web.mit.edu/6.013_book/www/book.html

This is probably the main proof that shows that circuit theory is a valid simplification of Maxwell’s equations. Especially the “lumped element” aspect

Do you know, whether this book is available as a pdf? It seems to be excellent, i.e., one of the few sources where the current-conducting wires are completely treated.

 

[Charles Link]

@Dale It looks like a very good reference. I now figured out how to use the arrows on my keyboard to access the later chapters of it. Thanks. :)

 

[Frabjous]

Do you know, whether this book is available as a pdf? It seems to be excellent, i.e., one of the few sources where the current-conducting wires are completely treated.

You can download chapter by chapter here.
https://ocw.mit.edu/courses/res-6-001-electromagnetic-fields-and-energy-spring-2008/

There are a couple of other books by Melcher that are available
Electromechanical Dynamics
https://ocw.mit.edu/courses/res-6-003-electromechanical-dynamics-spring-2009/
Continuum Electromechanics
https://ocw.mit.edu/ans7870/resources/melcher/resized/cem_811.pdf

 

[vanhees71]

Why do they make it so difficult to simply download the entire book? But it looks so great, that I'll make an effort to get everything together :-).

 

[Frabjous]

Haus also wrote an interesting book with Penfield
Electrodynamics of Moving Media (1967)

 

[tech99]

That seems to imply that you only consider the limit $\sigma \rightarrow \infty$. Then it's right.

If we consider a very small segment of line yes, but at high frequencies the inductance of the line per unit length is important and gives the electron inertia.

 

[Charles Link]

Even though the MIT lectures are very good, I think the section 9 based on $H$ could use a little updating. Even Feynman says in his lectures that the $H$ turned out to be something different than what was initially thought.
(Let me see if I can find the paragraph...see https://www.feynmanlectures.caltech.edu/II_36.html right after (36.32):
"This purely algebraic correspondence has led to some confusion in the past. People tended to think that H was “the magnetic field.” But, as we have seen, B and E are physically the fundamental fields, and H is a derived idea. So although the equations are analogous, the physics is not analogous. However, that doesn’t need to stop us from using the principle that the same equations have the same solutions".).

I did some calculations on this topic, (see the link below), showing/proving that without currents in conductors, and using $J_m=\nabla \times M/\mu_o$ with Biot-Savart, that $B=\mu_o H +M$. This is basically the pole model formula of magnetostatics. When magnetic surface currents are designated to be the sources, Biot-Savart is used, and the poles with pole(=magnetic charge) density $\rho_m =-\nabla \cdot M$ are ignored in the calculation of $B$. We can define an $H=(B-M)/\mu_o$, but $H$ then has a less significant role.
The $H$ has magnetic pole density $\rho_m=-\nabla \cdot M$ as sources, with the inverse square law.

See (also linked in post 109 above) https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/
posts 25-28.The $H$ is then redefined to add currents in conductors with a Biot-Savart type contribution (divided by $\mu_o$ ) and then added to both sides of the formula, so that we continue to have $B=\mu_o H+M$.

Note: Some books use $B=\mu_o(H+M)$. I prefer to use $B=\mu_o H+M$.

 

[Charles Link]

and a follow-on to the above: I think I just came up with a good qualitative explanation of what is going on with the collection of magnetic dipoles vs. the collection of electric dipoles, and why we get such similar results when they are added together, but just slightly different:

Edit: (Note: The MIT notes =section 9 got me thinking about this, because they begin with the $B =H$ from a microscopic magnetic dipole has the same geometry as the $E$ from an electric dipole. It puzzled me why this doesn't carry over completely macroscopically with a collection of these dipoles=it almost does, but with one correction term for the interior).

The building block of the electric dipole can be thought of as a brick with a + surface on one end and a - surface on the other. When we stack them together, the charges on the adjoining faces cancel, and we have a large rectangle that has a + surface at the top and a - surface at the bottom, and all the other charges cancel.

For the magnetic dipole, we can think of the building block as a rectangular current that runs around the outside of the block. The result when we stack them together is that all the currents cancel, except for a surface current that runs around the large rectangle that we build.

For the electric dipoles, we calculate the field everywhere for the large rectangle using Coulomb's law, and for the large rectangle made from magnetic dipoles, we use Biot-Savart's law on the surface currents to compute the magnetic field $B$.

The results of this are a little surprising, but very good in a way: We don't get the expected $B=H$ everywhere, where for $H$, we would calculate using magnetic poles or charges, and is analogous to $E$, computed from the poles $P$. We do get the result $B=H$ everywhere outside the bricks, but inside we pick up an additional term, which depends on the units we use, but it is of the form $B=H+M$.

I did the calculation of this about twelve years ago using a cylindrical shape (as opposed to a rectangle), with magnetization $M_z=M$ and made it a cylinder of semi-infinite length, so there was only one endface that contained the pole. The results surprised me=I was surprised to find that the surface current method gives the same answer as the pole method, other than the additional $M$ term inside the cylinder. I'll give a "link" to those calculations momentarily...see https://www.overleaf.com/project/5ca6af5c8dccb27da809813e
You will find today's date on the article, but the article dates back to around 2012.
I used cgs units where $B=H + 4 \pi M$.
I computed the $B$ in the plane of the single endface using both the surface current and pole methods. The result is that just above the single endface, for $r<a$, we have $B_z=2 \pi M$. The result is that if you make a cylinder of infinite length from two semi-infinite cylinders, where both have magnetization $M$ in the z direction, it is clear that you will get the $4 \pi M$ result inside the material. Meanwhile, the cylinder of infinite length has no poles, so $H=0$. Thereby we don't have that $B=H$, but rather $B=H+4 \pi M$.

In the plane of the endface, for $r> a$ the pole method gives $B_z =0$ by inspection. I was surprised to find the surface current integral gave this same result, (but gave $B_z=2 \pi M$ for $r<a$, which also agrees with what the pole method gives, which is as it would appear to be adjacent to an infinite sheet of magnetic surface charge density with $\sigma_m=M \cdot \hat{n}=M$ on the endface).
Note: Magnetic surface current per unit length $\vec{K}_m=c \, \vec{M} \times \hat{n}$.

I also showed both methods get the same result for $B_r$ in the plane of the endface.

It may take a little bit of work to read through the calculations of the "link".
Just maybe a couple readers will find it of interest.

 

[vanhees71]

Even though the MIT lectures are very good, I think the section 9 based on $H$ could use a little updating. Even Feynman says in his lectures that the $H$ turned out to be something different than what was initially thought.
(Let me see if I can find the paragraph...see https://www.feynmanlectures.caltech.edu/II_36.html right after (36.32):
"This purely algebraic correspondence has led to some confusion in the past. People tended to think that H was “the magnetic field.” But, as we have seen, B and E are physically the fundamental fields, and H is a derived idea. So although the equations are analogous, the physics is not analogous. However, that doesn’t need to stop us from using the principle that the same equations have the same solutions".).

I did some calculations on this topic, (see the link below), showing/proving that without currents in conductors, and using $J_m=\nabla \times M/\mu_o$ with Biot-Savart, that $B=\mu_o H +M$. This is basically the pole model formula of magnetostatics. When magnetic surface currents are designated to be the sources, Biot-Savart is used, and the poles with pole(=magnetic charge) density $\rho_m =-\nabla \cdot M$ are ignored in the calculation of $B$. We can define an $H=(B-M)/\mu_o$, but $H$ then has a less significant role.
The $H$ has magnetic pole density $\rho_m=-\nabla \cdot M$ as sources, with the inverse square law.

See (also linked in post 109 above) https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/
posts 25-28.The $H$ is then redefined to add currents in conductors with a Biot-Savart type contribution (divided by $\mu_o$ ) and then added to both sides of the formula, so that we continue to have $B=\mu_o H+M$.

Note: Some books use $B=\mu_o(H+M)$. I prefer to use $B=\mu_o H+M$.

I never understood this issue with "currents" vs. "pole" models. They should be equivalent, precisely because you can substitute any magnetization by the corresponding current density, $\vec{J}_m$ you quote above. Haus et al. write about this in their preface.

Of course I completely agree with Feynman: the physical fields are $\vec{E}$ and $\vec{B}$, while $\vec{H}$ and $\vec{D}$ are auxilliary fields, and there's always the question, how you split the sources into free parts and/or lump them into the field parts (polarization and magnetization). At the end both versions should lead to the same results for the physical fields (within the approximations made).

 

[Charles Link]

I never understood this issue with "currents" vs. "pole" models. They should be equivalent, precisely because you can substitute any magnetization by the corresponding current density, J→m you quote above. Haus et al. write about this in their preface.

Please see the edit I just added a few minutes ago about the MIT notes=in parenthesis=second paragraph=that's what is the stumbling block=the interior does not give the identical result that one might expect.

The reason may be that with the microscopic electric dipole, we don't consider what happens to the space between the charges...

Thereby, where $B$ is the real physical field everywhere, and $H$ is calculated from the poles, (ignore currents in conductors for now), we have instead of $B=H$, we need to modify it to $B=H +4 \pi M$.

It is interesting that when we take the case of a finite cylinder, rather than of infinite length, we then have poles on the endfaces, and the $H$ from them is exactly the correction that is needed, so that $B=H+ 4 \pi M$ also works for the case of the interior of the cylinder when we don't have $H=0$.

 

[vanhees71]

I'd say the only difference between the electric polarization and the magnetization is that in the electric case in addition to the polarization charge density, $\rho_{\text{pol}}=-\vec{\nabla} \cdot \vec{P}$ you can have also "free-charge" densities $\rho_{\text{free}}$, which you can't have in the magnetic case, because there are no magnetic monopoles.

From a very fundamental point of view (QED) you have of course currents as the sources of the magnetic field, including the "spin part" of the matter fields (Dirac fields for electrons).

 

[Charles Link]

I'd say the only difference between the electric polarization and the magnetization is that in the electric case in addition to the polarization charge density, $\rho_{\text{pol}}=-\vec{\nabla} \cdot \vec{P}$ you can have also "free-charge" densities $\rho_{\text{free}}$, which you can't have in the magnetic case, because there are no magnetic monopoles.

I added a couple of things to the above post, but please study this carefully. We are not considering free charges or currents in conductors. It is almost a remarkable coincidence that we get the formula that we do: $B=H+ 4 \pi M$, where the $4 \pi M$ is the correction for the interior, and it gets added to the $H$, (where $H$ is calculated from the poles, i.e. $\rho_m=- \nabla \cdot M$, etc. , even for the finite cylinder in the interior).

 

[vanhees71]

Obviously I don't understand your point. $\vec{B}=\vec{H}+4 \pi \vec{M}$ is the standard definition (in non-rationalized Gaussian units). I don't see, which difference it should make whether I consider $\vec{M}$ as a source term or the equivalent current density $\vec{\nabla} \times \vec{M}$. The physical fields come out the same with both assumptions.

The most simple example I can come up with is the homogeneously magnetized sphere ("hard ferromagnet model"). There the equivalent current is, of course, rather a surface current. It's in my lectures notes (sorry, they are in German, but I think the "formula density" is high enough, so that you can understand it):

https://itp.uni-frankfurt.de/~hees/publ/theo2-l3.pdf (Sect. 3.3.3).

 

[Charles Link]

In my paper that I "linked" from twelve years ago, I constructed a finite cylinder from a couple of semi-infinite ones with opposite magnetization, that overlapped but leaving one section that still had $M_z=M$ of length $L$. The result was that $B=H+4 \pi M$ works for this as well, with this one now having an $H$ contribution from both endfaces, where the infinite cylinder has $H=0$.

It should be somewhat clear though that for the infinite cylinder the correction for the interior is $B=4 \pi M$. ($H=0$).That is what we get from the surface currents=you can even use ampere's law instead of Biot-Savart to compute it.

 

[Charles Link]

Obviously I don't understand your point. B→=H→+4πM→ is the standard definition

What we are explaining here is why we don't simply have $B=H$ everywhere. We do have $B=H$ everywhere outside the macroscopic collection. The $M$ or $4 \pi M$ turns out to be the correction to the $H$ for the interior, but without working the Biot-Savart on it, or doing some other detailed vector calculus identities, I don't think you would guess that the correction is indeed this $4 \pi M$.

They teach it as a definition, and some even handwave that it comes from $D=E+4 \pi P$, but I don't think either of those is a satisfactory answer.

It is something that needs to be computed by doing Biot-Savart on the finite cylinder case, and seeing what the correction for the interior is. Otherwise, I later did a calculation where vector identities also get you the result that $B=H+ 4 \pi M$ . See https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/ posts 25-28.

 

[vanhees71]

It's of course only an approximation. You can also take into account higher multipole expansions starting from microscopic (quantum) electrodynamics. What we discuss here are only the leading-order linear-response approximations for the interacting medium-field system. I think what you do in the old thread is the usual derivation of exactly this, i.e., using the dipole approximation for the magnetostatic case. It's pretty similar, what I do in my manuscript.

On the other hand one should of course do this also fully relativistically, where there seems to be quite some confusion concerning the macroscopic classical electrodynamics. Of course an entirely classical "electron theory" is anyway not the true thing, and you must use QED, and that's a pretty delicate issue too. The usual line of attack is to start with the non-equilibrium many-body formalism of QFT and derive the Kadanoff-Baym equations with all the obstacles added to QFT by the fact that QED is a gauge theory. Then you do some gradient expansion to derive semi-classical transport models and finally (magneto-)hydrodynamical descriptions. It's a pretty complicated thing and imho not really fully solved. One already "classical" paper in this direction is, e.g., this:

https://doi.org/10.1016/0003-4916(72)90329-6

 

[Charles Link]

I find even the classical model of this rather interesting. When, as I saw in the MIT notes, that the $B=H$ for the microscopic magnetic dipole, at least the terms of interest, has the same form as the $E$ for the electric dipole, it is understandable how some of the early researchers thought that $H$ was indeed the magnetic field, and why they even based much of the magnetostatics instruction on the pole model. I do think I came up with a good input though on why the formula for the magnetic field needs to read, even in a classical sense, $B=H+4 \pi M$. (cgs units), instead of $B=H$.

In some ways, I'm probably somewhat behind the times. The latest calculations that you @vanhees71 mention are far beyond anything that I have computed. From the viewpoint of other students though, they do have to learn to walk and run before they can fly=they might find what I have computed on the topic of interest. Cheers. :)

Edit: Just an additional comment or two: The magnetic field $B$ computed from Biot-Savart seems to be fairly reliable because it is the steady-state solution of $\nabla \times B=\mu_o J+\mu_o \epsilon_o \dot{E}$, ignoring the $\mu_o \epsilon_o \dot{E}$ term.

I find it interesting that the magnetic pole model, where it is considered to be plus and minus magnetic charges instead of a current loop, works as well as it does. We find comparing with a Biot-Savart analysis, that instead of $B= \mu_o H$ everywhere, we need to use $B=\mu_o H +M$ (MKS units), so that in the material there is a correction term of $+M$ that gets added on in computing the magnetic field $B$ with the pole model.

I do encourage students to learn how to work magnetostatics problems using both the Biot-Savart surface current method, and the pole method.

 

[vanhees71]

$B=H$ holds in vacuo, where $M=0$ (in Gaussian or Heaviside-Lorentz units). What else should be $M$ without matter present?

The magnetic-pole model works, because it's equivalent to assuming the current density to be $\propto \vec{\nabla} \times \vec{M}$, including possible surface-current densities like in my example of the homogeneously magnetized sphere, which is the most simple example due to symmetry.

In SI units the constitutive equation is $\vec{B}=\mu_0 (\vec{H}+\vec{M})$. I also mess up where the $\epsilon_0$ and $\mu_0$ should go. The mnemonics is that the material sources like magnetization belong to $\vec{H}$ and then you need $\mu_0$ to get $\vec{B}$ dimensionally correct ;-). By definition of the SI units the macroscopic Maxwell equations read
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{E}+\partial_t \vec{B}=0$$
and
$$\vec{\nabla} \times \vec{H} -\partial_t \vec{D}=\vec{j}, \quad \vec{\nabla} \cdot \vec{D}=\rho,$$
with $\rho$ and $\vec{j}$ the "free charges and currents", including polarization charge densities $\rho_{\text{pol}}=\vec{\nabla} \cdot \vec{P}$ and magnetization currents $\vec{j}_{\text{mag}}=\vec{\nabla} \times \vec{M}$ with possible surface-charge densities and surface-current densities at boundaries between different media (or a medium and vacuum).

 

[Charles Link]

Note in the above post 131, that the current loop running around the brick is a more accurate description of what is going on with the building block of the magnetic moment than a plus magnetic charge at one face and a minus at the other. Thereby, at least in this classical description, we need to compute what $B$ gives in the material using this other building block, and compare it to what the plus and minus block gives. The result we find is that $B=H+4 \pi M$,(cgs units) , with the $4 \pi M$ being the correction term.

I'm repeating some of what I previously said, but this new part is something I didn't put in the original explanation. Even the current loop is a classical description=it may not be the perfect description, but it is deemed to be closer to the complete description than a (fictitious) plus and minus magnetic charge.

The plus and minus charge pole model says that $B=H$ everywhere, but the current loop model says that in the material $B=H + 4 \pi M$, so we add the $4 \pi M$ to the $H$ and consider that to be a description that is much more mathematically accurate. It still is an approximation, as @vanhees71 mentions in post 139, but over the years it has been accepted as being a fairly good one.

 

[vanhees71]

The $4 \pi M$ is not "a correction term" but a source term due to the presence of a medium. It's describing the response of the medium to the external electromagnetic fields, i.e., the rearrangement of the charge and magnetic-moments distribution of the particles making up the medium.

I don't understand why you think that $\vec{B}=\vec{H}$ everywhere in the magnetic-charge-dipole model. Of course, what you call $\vec{H}$ is indeed model-dependent, i.e., in which arbitrary way you split the sources into "free/external" and "internal" parts. You can arbitrarily shuffle sources to fields and vice versa. The physical results are the same, particularly $\vec{E}$ and $\vec{B}$, which are the observable fields.

 

[Charles Link]

The 4πM is not "a correction term" but a source term due to the presence of a medium

What I am using as a source term is the surface current per unit length term $K_m= c \, M \times \hat{n}$ that even Griffiths has in his computation of the vector potential $A$ in section 6 of his book. There is no $M$ source term. (The other source term is a $J_m=c \, \nabla \times M$). It is surprising that the surface current term calculation gives $4 \pi M$ as the correction, so it looks like a source term of $4 \pi M$, but it comes from a Biot-Savart or amperes law computation of $B$ from the surface current per unit length $K_m= c \, M \times \hat{n}$ on a long cylinder, where we get $B=4 \pi M$.

If the cylinder is chosen to be of finite length, then surprisingly enough, the result of the Biot-Savart computation of the surface currents of the magnetized cylinder is precisely $B=H + 4 \pi M$.

If you look this one over very carefully, I think you might come to agree with me on this.

Edit:

The $4 \pi M$ then gets added into the pole model description as a source term for $B$, where we have discovered that the pole model result that $B=H$ everywhere isn't completely accurate, by using our improved description of the current loop as a building block. Thereby the pole model then uses the $4 \pi M$ as a source term, but the pole model itself, at least as far as I can tell, would give no information on what this correction term, if any, might need to be.

If we stick with our fictitious plus and minus charge model as the building block, we would conclude that $B=H$ everywhere, (where $H$ is computed from the poles, i.e. $\rho_m=-\nabla \cdot M$ and $\sigma_m=M \cdot \hat{n}$). and note the $\sigma_m=M \cdot \hat{n}$ is a magnetic surface charge=on the endfaces=it is not a volume source term. When we compute $B$ for a long magnetized cylinder, using the pole model, the $\sigma_m$ on the endfaces are assumed to have little effect, so that $H =0$ for the long cylinder, even though it has $\sigma_m=\pm M$ on the endfaces. We then throw in our correction term (that we discovered from our surface current calculations) with the result that $B=4 \pi M$ for the long cylinder.
(and note, although I may be stating the obvious, the endfaces for a long cylinder are not infinite sheets of charge in which case we would have $H=-4 \pi M$. Instead their contribution is minimal, and considered to be zero).

Yes, the pole model does use $4 \pi M$ as a source term for $B$, but I believe it comes as the result of a more complete study of the surface current/current loop description. The pole model, with its plus and minus magnetic charges, by itself would have no information on what the correction term needed to be. The pole model uses $H$ and $4 \pi M$ as sources, but only the $H$ itself comes from the magnetic charge description. Had the fictitious plus and minus magnetic charge model been deemed to be completely accurate, then there would be no $4 \pi M$ correction term in the calculation of the magnetic field $B$.

 

[vanhees71]

Ok course the surface current is a source term. What else should it be? I've shown explicitly the equivalence of both descriptions in this example of the homogeneously magnetized sphere. It's of course valid for magnets of any shape and any polarization.

 

[Charles Link]

@vanhees71 Please read the last three paragraphs in the above post 144 that I just added (=below Edit). I think that might help tie it all together.

The pole model with $H$ from the poles (the plus and minus magnetic charges) was thought to represent the magnetic field at one time. Then the surface current calculation (with the microscopic current loops as building blocks) showed that the $B$ in the material was not just $H$, but was found to be $B=H+4 \pi M$, but that $B=H$ was indeed correct outside the material.

Note that using plus and minus charges as building blocks does give the correct $E$ in the material when we consider the $P$ in dielectric materials, (note that we do not ever need an additional $4 \pi P$ to get the fundamental physical field $E$), but the $H$ that is computed from magnetic charges does not represent the $B$ that is calculated from current loops in the material.

This so far does not include currents in conductors. Finally $H$ is then redefined, (as is necessary to keep the formula $B=H+4 \pi M$ intact), to include currents in conductors as sources of $H$ using Biot-Savart and basically adding it to both sides of the formula, so that $H$ is now at the level of a mathematical construction.

With the introduction of $B=H+4 \pi M$ in the material, $H$ was already determined to not represent the actual field $B$ in all cases. Had the $H$ not been redefined to include the currents in conductors, the formula $B=H+4 \pi M$ would not be valid when currents in conductors are present.

With these modifications to the pole model, the surface current method and the pole model give identical results for the computed magnetic field $B$ for virtually any magnetization function $M$ and any distribution of currents in conductors. (Note that $B=H+4 \pi M$ is basically a pole model formula. The surface current method does not recognize poles as sources. For the pole model, you first compute $H$ and then you get compute $B$ as $B=H +4 \pi M$ ).

@alan123hk I would also welcome your feedback on this. I think you might find it interesting, starting around post 130.

 

[alan123hk]

@alan123hk I would also welcome your feedback on this. I think you might find it interesting, starting around post 130.

I am really interested in this topic, but obviously my ability in this area is not enough, so what I can do at this stage is to slowly learn and then understand.

 

[Charles Link]

I am really interested in this topic, but obviously my ability in this area is not enough, so what I can do at this stage is to slowly learn and then understand.

@alan123hk Perhaps a good place to start would be the following thread that I started. I tried to keep things simple in this thread:

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

 

[Charles Link]

I would like to point out one item that might help to emphasize it=when the magnetization current runs around the outside of the rectangular building block, the result for bricks stacked together is that the current between the bricks cancels, but we get a surface current flowing (e.g. counterclockwise with $M$ in the $+\hat{z}$ direction=note $K_m=c \, M \times \hat{n}$ where $\hat{n}$ is the outer pointing unit normal vector to the surface) around the outer surface of the bricks. We then use Biot-Savart or ampere's law to compute the magnetic field $B$ both in the material and outside the bricks. If we make the bricks into the shape of a long cylinder, we find $B=4 \pi M$ is the result we get for $B$ inside the material in cgs units, with surface current per unit length $\vec{K}_m=c \, \vec{M} \times \hat{n}$.

This surface current is a well-known result that Griffiths comes up with in section 6 of his E&M text in the derivation of the vector potential $A$. There is no charge transport in these surface currents, so one can argue whether they are completely real or the result of the mathematics of the magnetic moment.

Edit: Note also that for the plus and minus magnetic charges on each block as building blocks, the charges butted up against each other from adjacent bricks in the z direction cancel each other, and we are left with a magnetic surface charge density $\sigma_m=+M$ on the +z end, and a magnetic surface charge density of $\sigma_m=-M$ on the -z end.

When we do the Coulomb's law (magnetic form ) for these bricks, we get the same result as the current loop bricks (using Biot-Savart) external to the material, (i.e. $B=H$), but inside the material we find $B=H+4 \pi M$ (cgs units), where $H$ is the Coulomb's law result, and $B$ is the Biot-Savart result.

 

[Charles Link]

The question arises, why do we even use an $H$ anymore if it is found to not represent the magnetic field $B$? The answer is that it is one of the more useful mathematical constructions. It comes in very handy with a transformer where we have a version of Maxwell's equations $\nabla \times H =J_{conductors}+\dot{D}$, which in the steady state with Stokes' theorem becomes $\oint H \cdot dl=NI$. (MKS units).

It also comes in very handy for computing the magnetic flux when a transformer has an air gap, where we get magnetic poles on the two faces at the air gap. $H$ is then assumed to take on two different values=one in the material and another in the air gap, and the magnetic material is assumed to be linear. We then have with $B$ being continuous that $B=\mu H_{material}=\mu_o H_{gap}$ and magnetic flux $\Phi=BA$. With the integral around the complete inside path of the transformer we have $\oint H \cdot dl=NI$, so that we have two equations to solve for the unknowns $H_{material}$ and $H_{gap}$. Feynman must have considered this one to be important as well, because he did a write-up on it. See https://www.feynmanlectures.caltech.edu/II_36.html right around equation (36.26).The $H$ is something that comes out of the pole model of magnetization in materials. It really doesn't show up in a surface current presentation. It is also used in hysteresis curves of $M$ vs. $H$, where the $H$ typically comes from the current in a conductor of a solenoid around a long sample of cylindrical shape. Note that as mentioned in post 146, besides coming from contributions from the poles, which are insignificant in the case of a long cylinder, $H$ is defined to also include the currents in conductors as sources, using Biot-Savart.

For a Physics Forums thread of a transformer with an air gap see https://www.physicsforums.com/threads/absolute-value-of-magnetization.915111/ also referenced in post 111 of this thread.