Inducing EMF Through a Coil: Understanding Flux

[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.