H-field and MMF in a magnetic circuit

• I
Homework Helper
Gold Member

Main Question or Discussion Point

I have been revising magnetic circuits concepts and I think I am missing something regarding H-field and mmf.

If the 100-turn primary of a transformer is energized with 1A dc, its mmf will be 100A-T. And all the line integrals H.dl enclosing the primary coil will equal 100A. As per magnetic circuit laws, H-field in the core is very small compared to that in the space between the limbs, which is correct considering the huge difference in permeabilities of the two media.

My question is, how is the H-field minimum inside the primary coil? The H field is created by electric currents and does not depend on the permeability of the material. How does it get redistributed in the magnetic circuit (little in the core, more in the air gaps etc)? Does magnetization of the core have an effect on this mechanism?

Any inputs are appreciated.

Related Other Physics Topics News on Phys.org
Homework Helper
Gold Member
I think I can answer your question. The MMF equation is actually a form of Ampere's law for magnetic materials. The derivation begins with using the equation $B=\mu_o(H+M)$, and taking the curl of both sides. $\nabla \times B=\mu_o \nabla \times H+\mu_o \nabla \times M$.
Maxwell's equation is then employed. $\nabla \times B=\mu_o J_{total}+\mu_o \epsilon_o \dot{E}$ which becomes $\nabla \times B=\mu_o J_{total}$ for steady state, where $J_{total}=J_{conductors}+J_m+J_p$.
The bound magnetic current $J_m=\nabla \times M$.
( Polarization current $J_p=\dot{P}$ is presented here for sake of completeness, but is often not of interest).
This leaves $\nabla \times H=J_{conductors}$. This is integrated over an area $dA$ and Stokes theorem is used. The number of turns gets included when counting the current. The result is $\oint H \cdot dl=NI$.
The $H$ is really just a mathematical construction, but a very useful one. It is assumed that the flux is continuous in these problems because $\nabla \cdot B=0$. The additional assumption is made that $B=\mu H$, for some $\mu$ that is known from the hysteresis curves. The function $H$ can undergo discontinuities that arise because of magnetic poles (magnetic surface charge $\sigma_m=\mu_o M \cdot \hat{n}$ at the surface boundaries, e.g. at the faces of the air gap. In general, magnetic pole density $\rho_m=-\mu_o \nabla \cdot M$).
In doing these calculations using the MMF equation, it is not necessary to compute the magnetic poles. Instead, all that is necessary is to make the assumption that $B$ is continuous, and that $H$ changes values from $H_1$ to $H_2$ as the material changes. (Were assuming constant cross-sectional area here). The result is $\oint H \cdot dl=H_1L_1+H_2 L_2$, etc. and all you need to do is write $B_1=\mu_1 H_1 =\mu_2 H_2=B_2=B_o$, and the complete solution results.
In the case of a transformer with an air gap, the term $\frac{ B_o}{\mu_{air} }L_{air}$ can often be much larger than the term $\frac{B_o}{\mu_{core} }L_{core}$, if $\mu_{core}$ is sufficiently large. $\\$ Additional item: The equation $\oint H \cdot dl=NI$ always applies, and because of the additional assumption of the continuity of $B$, it is really rather remarkable how much can be computed from it. The $H$ is really just a construction, and the actual magnetic field is the $B$.
One thing that is actually of prime importance in what goes on, regarding the magnetic field in the transformer, is the magnetic surface currents that are part of the $J_m$. These get totally removed from the $\oint H \cdot dl =NI$ equation. $\\$ Edit: It took me a while to figure out what the real puzzle may be here, but I think the next couple of paragraphs may make $H$ appear much more logical: $\\$The $B=\mu_o ( H +M)$ equation, actually also plays a prominent role here. The derivation of this equation is rather lengthy, and the $H$ that comes out of this is somewhat complex: Besides having currents in conductors as sources of $H$, (comes from Biot-Savart and/or Ampere's law), magnetic poles are also sources of this $H$. That is why we are justified in assuming $H$ can undergo discontinuities at the material boundaries, so that $H$ takes on a new value in the gap than it did in the adjacent core. $\\$ (If the only sources of $H$ were the currents in the coils, the behavior of this $H$ function would be really very difficult to explain. A detailed analysis with the magnetic pole model really makes the discontinuos behavior of the $H$ quite logical. You may also note that the $H$ from the coils that comes from Biot-Savart indeed has $\nabla \cdot H=0$ so that the portion of $H$ that comes from the currents in the coils will be continuous).$\\$ When we write $\oint H \cdot dl=NI$, the $H$ in this equation is a very precise $H$ that includes not only the $H$ from the currents in conductors, but also the $H$ from magnetic poles. Likewise, when the $H$ is computed in the air gap, the effect of the magnetic poles around the gap get taken into account=really in a very subtle way, by simply saying that the value of $H$ there is some unknown $H_{air}$ that is different from $H_{core}$. $\\$ (You might also find it of interest that for the $H$ from the poles, $\oint H_{poles} \cdot dl=0$. Just like the electric field from electrical charges , where $\nabla \times E=0$, the $H$ from the poles has $\nabla \times H_{poles}=0$. Note: The $H$ from the magnetic poles points in the opposite direction in the core than it does in the gap. In general it is also much weaker in the core, because the two faces of the gap contain opposite poles whose effects will mostly cancel in the core. Across the gap the opposite sign of the poles makes their contributions add). $\\$
The value $H_{air}$ that is used in these equations is assumed to be the result of the $H$ from the current in the coils along with the $H$ from the poles. Likewise, the $H_{core}$ is assumed to result from the $H$ from the currents in the coils along with anything that comes from the magnetic poles.$\\$ You also might find this thread of interest, although it is rather lengthy. I think we really dissected the problem of the transformer with an air gap in this thread: https://www.physicsforums.com/threads/magnetic-flux-is-the-same-if-we-apply-the-biot-savart.927681/

Last edited:
Homework Helper
Gold Member
Additional comment: One calculation that can be useful in these problems is once you have found the $B_o$ and the $M$ and the $H's$ in the two regions, is to compute $\sigma_m=\mu_o M \cdot \hat{n}$ from the results, (i.e. using $M=\frac{B}{\mu_o}- H$) and then compute $H_{air}=H_{poles} +H_{conductors}$. The $H_{poles}=\frac{\sigma_m}{\mu_o}$, (two infinite sheets of opposite magnetic surface charge for a narrow gap), and $H_{conductors}=\frac{NI}{L_{total}}$. $\\$ You will find, (assuming I didn't make any "typos"), that this $H_{air}$ is exactly what was computed from the $\oint H \cdot dl=NI$ equation with the assumption that there is an $H_1$ and an $H_2$. (We are assuming constant cross-sectional area in this problem), along with $B=\mu H$, and the assumption that $B$ is continuous around the loop. $\\$ Notice also that the simplified procedure using $\oint H \cdot dl=NI$ gets the results without needing to compute $\sigma_m$, etc. $\\$ The $H_{poles}$ term in the gap is quite large, and needs to be, in order to have $B$ be continuous as it goes from a region of magnetic material, (where the $H$ is quite small, but with large $\mu$), to a region of very small $\mu_{air}=\mu_o$. It is the magnetic poles= magnetic surface charges=(not to be confused with magnetic surface currents which this pole model ignores) that creates this large $H_{air}$.

Last edited:
Homework Helper
Gold Member
Thanks @Charles Link for the wonderful responses!
When we write ∮H⋅dl=NI∮H⋅dl=NI \oint H \cdot dl=NI , the HH H in this equation is a very precise HH H that includes not only the HH H from the currents in conductors, but also the HH H from magnetic poles
And these poles are created by the bound currents, right?
Note: The HH H from the magnetic poles points in the opposite direction in the core than it does in the gap. In general it is also much weaker in the core, because the two faces of the gap contain opposite poles whose effects will mostly cancel in the core. Across the gap the opposite sign of the poles makes their contributions add).
Could you please elaborate this point? Shouldn't the H-field of the poles be from N to S both inside and outside the core?

Homework Helper
Gold Member
And these poles are created by the bound currents, right?
These poles are created by the magnetization. By bound currents, it normally means magnetic surface currents, but also any bound currents that occur because $\nabla \times M=J_m$ is non-zero. Magnetic poles occur where $-\mu_o \nabla \cdot M=\rho_m$ is non-zero. Perhaps the best representation of these two concepts is a uniformly magnetized cylinder of magnetization $M$ in the + z direction. There will be a surface current per unit length $K_m =M \times \hat{n}$ on the outer surface, just like the current of a solenoid. This follows from the discontinuity in $M$ at the surface, and taking the curl there, with Stokes theorem. $\\$ It turns out that unless the cylinder is infinite in length, there will always be poles on the endfaces of magnetic surface charge density $\sigma_m=\mu_o M \cdot \hat{n}$, due to the discontinuity in $M$. (Notice because of the dot product, we will get both a "+" pole and a "-" pole).$\\$ In the magnetic surface current theory, poles are ignored, and the magnetic field both inside and outside the magnetized cylinder is calculated completely from the surface currents using Biot-Savart. The geometry is the same as that of a solenoid. $\\$ In the magnetic pole theory, the magnetic field $B$ is computed simply by computing the $H$ from the poles and using the inverse square law just like with $E$ but with $\epsilon_o$ replaced by $\mu_o$. The surface currents are ignored in the pole theory. The $B$ is computed everywhere using $B=\mu_o( H +M)$. Notice inside the cylinder, the $H$ actually points opposite the $M$. For a very long cylinder, far from the end faces, $B=\mu_o M$. $\\$ Surprisingly, Biot-Savart with the surface currents gets the exact same answer for $B$ everywhere as these "pole" calculations.
Could you please elaborate this point? Shouldn't the H-field of the poles be from N to S both inside and outside the core?
For the case of a gap in a transformer, there is a "+" (North) pole and a "-" (South) pole. The $H$ points from + to - in the gap. If we look at the corresponding electrostatic problem, and compute the $E$ for it, the gap is like a parallel plate capacitor. For finite plates, there will be a small electric field outside of the plates on both sides. In the transformer problem, this small $H$ that is outside the plates is assumed to stay inside the transformer core and loop around the core at its constant small value. It's value can readily be computed. As previously mentioned, $\nabla \times H_{poles}=0$ so that $\oint H \cdot dl=0$ by Stokes' theorem. That means $H_{core}$ points opposite $H_{gap}$ and also $H_{gap} L_{gap}+H_{core}L_{core}=0$. To compute $B_{gap}$, we simply use $B=\mu_o ( H+M)$ and $M=0$ in the gap, so that $B_{gap}=\mu_o H_{gap}$. $\\$ Right now, we were just considering the poles. There will also be an $H_{conductors}$ from the current in the coil given by $H_{conductors}=\frac{NI}{L_{total}}$. (We are assuming uniform cross-sectional area). $H_{total}=H_{conductors}+H_{poles}$. $\\$ Notice again, we can do these calculations without actually computing the value of the magnetic surface charge density=surface pole density $\sigma_m=\mu_o M \cdot \hat{n}$. $\\$ One additional note: There is a system of units that uses $B=\mu_o H+M'$ instead of $B=\mu_o(H+M)$. This just means that $M'=\mu_o M$, where $M'$ is the magnetization vector for this other system. A good portion of the older MKS literature uses $B=\mu_o H +M$.

Last edited:
Homework Helper
Gold Member
One additional point may be worth mentioning at this point: (I don't want to get too far off-topic, but this item really needs to be addressed if we are going to accept the above results of the $H$ that we were working with). $\\$ It is the case of a permanent cylindrical magnet where the $H$ from the poles points from "+" (North) to "-" (South) inside the magnet. The $H$, i.e. $\mu_o H$, is actually just a correction factor that results in $B$ inside the magnet to $B=\mu_o M$ for the case of a magnet of finite length. Surprisingly this $\mu_o H$ computed from the poles using the inverse square law is precisely the exact correction that is needed, so that we have $B=\mu_o (H+M)$. $\\$ This $H$, (which can be small far from the end faces of a long permanent magnet), in any case points opposite the magnetization. It clearly is not doing anything to keep the magnet magnetized. (This is one of the reasons I have said that the $H$ is really just a mathematical construction. The actual magnetic field is the $B$). The pole method is not normally taught in the standard curriculum anymore. Instead they simply teach the permanent magnet as having its magnetization $M$ maintained by the $B$ from its own magnetic surface currents. (See Introduction to Electrodynamics by by D. Griffiths). $\\$ The $B$ calculated from these magnetic surface currents using Biot-Savart gives a result that is identical for the finite cylinder as that of the "pole" method. In the surface current method, there is no computation of any "poles". Instead, Biot-Savart gets a smaller answer inside the magnet near the end faces because of the "missing" nearby surface currents because the cylinder is of finite length. $\\$ =============================================================================$\\$ And to connect it to the MMF, there is also an $H$ from the currents in the conductors. This $H$ when applied to the transformer material is found to be able to cause magnetization $M$ to occur. For linear materials, we can write $M=\chi H$, so $B=\mu_o (H+M)=\mu_o(1+\chi)H=\mu H$.
In addition, the $H$ here also includes any $H$ in the opposite direction from the poles of the gap. $\\$ This seems to pose a slight dilemma from the case of the permanent magnet above, but I think that is answered by the fact that the permanent magnet is in a saturated state of magnetization, and doesn't respond linearly. $\\$ For the linear operation of the transformer, the driver of the magnetization can be assumed to be either the $H$ or the $B$, and looking at either one as the driver is appropriate, because they are linearly related with $B=\mu H$. If we write $M=\chi' B$ for a linear response, using $B=\mu_o(H+M)$, we can do a little algebra that will give the result $M=\chi H$, where $\chi=\frac{\mu_o \chi'}{1-\mu_o \chi'}$. $\\$ In some ways, it may seem odd to treat the $H$ from the poles on equal footing with the $H$ that occurs from the currents in a coil, but some lengthy derivations are able to show that the equation $B=\mu_o (H+M)$ is on sound footing, where $H$ includes the not only the $H$ from the MMF, but also the $H$ from magnetic poles. This $B=\mu_o(H+M)$ equation applies in general, even when the materials are not linear in their response. $\\$ In the permanent magnet, the driver is the $B$, and is not closely, and certainly not linearly, related to the $H$, which is completely different from a transformer in its linear region of operation.

Last edited:
Homework Helper
Gold Member
Consider the following diagram.

Because of the current in the coil, the core is magnetized and N-S poles are created near the air gaps. Is this correct? The B-field is directed from S to N inside the coil and from N to S outside the coil.
The H-field in the air gap is shown in blue. Is its direction correct? As per my understanding, the blue Hair is equal to Hcoil+Hpoles, where both are directed downward in the blue air gap.
But "inside" the coils, Hcoil is upward and Hpoles is still downward, so they oppose each other.
Is that correct?

Attachments

• 22.1 KB Views: 175
Homework Helper
Gold Member
The blue $H_{air}$ is incorrect. In the transformer piece to the right, you will have another south pole on top, and a north pole on the bottom. This will result in two very short gaps in this configuration. You will get very little flux emerging from the square path that has two air gaps. It all stays in the square path. The magnetization $M$ and the $B$ can be assumed to be the same in both pieces. $\\$ Meanwhile, if you removed the piece at the right, then your $H_{air}$ would be approximately correct, but the MMF equations do assume a continuous flux and really work best with small gaps. $\\$ With the right piece in position as in the above diagram, the $H_{air}$ will go from north pole to south pole across the short gaps in both places.

Last edited:
Homework Helper
Gold Member
The blue $H_{air}$ is incorrect. In the transformer piece to the right, you will have another south pole on top, and a north pole on the bottom. This will result in two very short gaps in this configuration. You ill get very little flex emerging from the square path that has two air gaps. $\\$ Meanwhile, if you removed the piece at the right, then your $H_{air}$ would be approximately correct, but the MMF equations do assume a continuous flux and really work best with small gaps.
Yes, you are right!

But if the core was continuous (without the two small air gaps), would the direction of the blue Hair be correct?

Basically, is the following reasoning correct?
Assuming no other air gaps in the core (just like in a normal transformer core), if we make a round trip from S to N "inside" the coil and N to S outside the coil along the blue air-gap line, we will encounter two H-fields, Hcoil and Hpoles.
Hpole will always be from N to S, but Hcoil (due to Biot-Savart) will be from S to N "inside" the coil and from N to S "outside" the coil. Hence, inside the coil, both H fields oppose and outside the coil (in the blue air gap), they add.

Homework Helper
Gold Member
If the core were continuous, there would be no poles. The magnetization $M$ would be uniform in the clockwise direction. Poles occur as a result of a non-zero $\nabla \cdot M$. ($\rho_m=- \mu_o \nabla \cdot M$ ).Basically, if $M$ is uniform, you don't have any poles. $\\$ You do not get any poles from the coil itself. It has a direction and so you can label one end north and one end south, but these are not poles. You only get poles if you have a gap, e.g. if you remove the piece on the right, or have small gaps when the piece on the right has air gaps.

Homework Helper
Gold Member
One additional input: From Gauss' law, you can compute the poles: the equation you have is $-\mu_o \nabla \cdot M=\rho_m$. Applying Gauss's law with a pillbox at an end face: $\int M \cdot \hat{n} \, dA=-\frac{Q_m}{\mu_o}$, where $Q_m$ is the enclosed (fictitious) magnetic charge. $\\$ The magnetization $M$ is zero outside the material. You should be able to show quite readily that magnetic surface charge density= surface magnetic pole density $\sigma_m=\mu_o \, M \cdot \hat{n}$.

Homework Helper
Gold Member
You only get poles if you have a gap, e.g. if you remove the right piece.
Right again!

So if the core is continuous, wouldn't the bound currents create a downward H-field just like the poles would if I removed the right piece?

Hence, inside the coil, both H fields oppose and outside the coil (in the blue air gap), they add.
Is this true in any case?

Homework Helper
Gold Member
The bound currents (and now we are doing a magnetic surface current analysis=any poles of the pole model are ignored), the equation $\nabla \times M =J_m$ gets us the surface currents, so that surface current per unit length $K_m=M \times \hat{n}$. If you look closely at this equation, it gives a cross-product. These surface currents will make our sample look like a toroidal solenoid. (I'm assuming continuous toroidal $M$ with no gaps). The $B$ is computed from Biot-Savart. In the surface current model, you compute the $H$ from $B=\mu_o(H+M )$. If you do that with this toroid of magnetization $M$ with the $K_m$ as above, (its much easier to compute for a long solenoid), you will find $H=0$. $\\$ Unlike currents in conductors that by definition have an $H$ associated with them, these bound currents that come from magnetization $M$ have a $B$, but there is no implicit $H$ from them. An $H$ might arise when $B=\mu_o (H+M)$ is computed, but they are not given an implicit $H$. This may seem very odd, but it all comes our of the assembling of the $B=\mu_o(H+M )$ equation, where currents in conductors are included into the definition of $H$ as an add-on, to an equation that is derived for magnetic systems with magnetic poles (surface currents are ignored) in the absence of currents in conductors. $\\$ It should be noted, the derivation of $B=\mu_o(H+M)$ is not found in most E&M textbooks. A couple of years ago, I managed to derive it for a system that had magnetization $M$, but had no currents in conductors. It then became apparent, how, if this same equation was going to hold when the currents in a conductor were turned on, that the currents in the conductors must also be considered to be sources of $H$, because they created a $B$. How do we keep that same equation $B=\mu_o(H+M)$ which works when $H$ is only composed of pole contributions, but the left side is getting bigger? It is necessary to have the corresponding $H$ from the currents in conductors be part of the $H$ so that the equation stays balanced. (That was always the convention when they taught us the pole theory back in 1975, but noone really knew why that needed to be the convention). $\\$ Additional item: The surface current model allows for a direct computation of $B$. It does not allow for a direct computation of $H$. That comes from what is the pole model equation $B=\mu_o(H+M)$. The $H$ is well defined in the pole model where both $\rho_m=-\mu_o \nabla \cdot M$ and currents in conductors are sources of $H$. The poles uses the inverse square law to compute $H$, and the currents in conductors uses $H= \frac{B}{\mu_o}$, where $B$ comes from Biot-Savart. The pole model ignores bound surface currents. It simply looks at the poles, along with currents in conductors. $\\$ And additional item: In the surface current model, the poles are completely ignored. The calculation for $B$ proceeds entirely from Biot-Savart on both the current in the conductors and the surface currents. Once $B$ is known in this model, $H$ can be found using $B=\mu H$. The $H$ that is determined this way will be the same $H$ as the pole model calculates, but the surface current model doesn't look at the poles to get this result. $\\$ And if you computed this (using the surface current model) for the toroidal system above with current in the coils that supplies an MMF, you will then get magnetization $M$ and surface currents, and the $B$ is the result of the current in the coils plus the surface currents. The $H$ that you compute from $B=\mu H$ will (perhaps surprisingly) be equal to the $H$ that is generated from the currents in the coil. $( \,H=\frac{NI}{2 \pi R} \, )$. There is no additional ( or subtractive ) $H$ because the pole calculations show there are no poles. ( $- \mu_o \nabla \cdot M=0$ ).

Last edited:
Homework Helper
Gold Member
And one error I would like to correct in some of my explanations above: In the case of electrical currents in the coils, the method of computation of the $H$ that they generate, e.g. in the pole model of calculation, is to use Ampere's law rather than Biot-Savart. Biot-Savart works fine for a long cylindrical geometry, (especially where the windings extend the length of the coil), but when calculating a toroidal geometry, and moreover when the windings are very localized, Ampere's law in the form $\oint H \cdot dl =NI$ is used to compute $H$.$\\$ It's not that Biot-Savart is wrong. Instead, what will actually occur is that the magnetic currents will redistribute themselves slightly to cause the path of the magnetization $M$ and the $B$ to follow around the toroid in a very uniform manner, so that the final result will be as if the windings were spread out around the toroid, with the magnetic surface currents also completely uniform. The windings can be localized in the case of the toroid, but in the MMF calculations, this has no effect, because of the way the magnetic flux behaves. All that matters is the number of turns $N$ and the current $I$. $\\$ The Ampere's law can either be employed as a path of a ring around the whole inside of the toroid, or a small short rectangular path can be used, just like with a solenoid, where the $H$ or $B$ is taken to be zero outside of the solenoid or toroid. The windings are treated as if they were uniformly spread out around the toroid. And once the windings are treated as if they were wrapped around the whole toroid, the $H$ or $B$ could then alternatively be computed by Biot-Savart rather than Ampere's law.

Last edited:
Homework Helper
Gold Member
And here's a question that comes to mind in regards to post 14:$\\$ If we take two identical iron toroids, and wrap them with 100 windings, (both with the same current), with one of them where the windings are only over a small region, and the other where the windings extend around the toroid, is the resulting magnetization $M$ the same in both cases? $\\$ Our MMF equations tell us that the magnetization is exactly the same. Any differences between these two cases is beyond the capabilities of what our MMF equations are able to calculate.

Homework Helper
Gold Member
Thanks, I think I've got the answer to my question in the OP (about air gap H field etc).
If the core were continuous, there would be no poles. The magnetization MM M would be uniform in the clockwise direction. Poles occur as a result of a non-zero ∇⋅M∇⋅M \nabla \cdot M . (ρm=−μo∇⋅Mρm=−μo∇⋅M \rho_m=- \mu_o \nabla \cdot M ).Basically, if MM M is uniform, you don't have any poles. \\
The reason I started this thread is I am interested in knowing the H-field in the region between the limbs. I have been reading this paper on power flow in a transformer using Poynting vector. (Full pdf is attached below).

On the second page (orange marks), it says "most of the H-field appears across the horizontal limbs."
What is the source of this H-field?

Attachments

• 22.1 KB Views: 192
• 39.9 KB Views: 197
• 39.9 KB Views: 169
• 78.1 KB Views: 179
• 312.4 KB Views: 111
Homework Helper
Gold Member
Meanwhile I need to go to the drawing board to try to figure out how to work the Poynting vector for this case. If I'm not mistaken, contained in those posts is a simple calculation showing the power balance based on $P=I \mathcal{E}$, but your present question is more difficult. It remains an open question to me whether the authors of the paper have made a correct analysis regarding the Poynting vector. $\\$ With these transformers, I am still on somewhat of a learning curve. The topic really contains a lot of finer detail. $\\$ And an additional comment or two: The Poynting vector, if I understand it correctly, is about power transfer particularly in electromagnetic waves. There may also be DC type applications of it, but my experience with the Poynting vector is somewhat limited. In any case, in the simplest approach, the power transfer will be governed by equations of the form $P=I \mathcal{E}$. I think a Poynting vector approach could be considerably more difficult.

Last edited:
Homework Helper
Gold Member
I think a Poynting vector approach could be considerably more difficult.
I agree. I'm just reading that paper out of curiosity!

In this wikipedia article, it is said that there are two components of the H-field in a magnetic material, one is from free currents and the other is from bound currents' demagnetizing field.
https://en.wikipedia.org/wiki/Magnetic_field#H-field_and_magnetic_materials.
Inside the magnetic material, this demagnetizing H-field is in the opposite direction to that of the magnetization. But in the above diagram, if the core is continuous, there will be no poles. So where will this demagnetizing H-field start and end?

Could you explain this last line of that article?
"The magnetic H-field, therefore, re-factors the bound current in terms of "magnetic charges". The H field lines loop only around 'free current' and, unlike the magnetic B field, begins and ends near magnetic poles as well."

Homework Helper
Gold Member
The imperfect geometry of the rectangular form, (as opposed to a more regular torus), makes for much more possibility to have regions where the magnetic pole density $- \mu_o \nabla \cdot M=\rho_m$ is non-zero. Unlike a toroidal geometry, there necessarily will be regions where magnetic poles appear. $\\$ I was able to determine with a proof of the equation $B=\mu_o(H+M)$, that the $H$ as defined as previously discussed, consisting of the contributions from poles as well as from currents in conductors is quite exact, but I have yet to do a similar type of derivation for the $H$ in the Poynting vector equation. If indeed that $H$ is one and the same, the imperfect rectangular geometry will give some additional regions of significant $H$. This is where I need to study it further though. Until the Poynting vector is given a more thorough study, it is uncertain for me whether the field of interest is $H$ or $B$. In addition, the transformer case is more of a DC and/or longitudinal type electromagnetic behavior, where the wavelength at the frequency that transformers operate is extremely long. The problem is interesting, but it really is beyond my present level of what I can readily compute. $\\$ Additional input: I think it might be worthwhile to work with a toroidal geometry, and see if the same conclusions can be reached. I think the energy flow would perhaps be even more efficient for the toroidal case, but it might put a damper on the conclusions that the authors have reached. $\\$ And another additional item: I haven't studied the paper in full detail, but I would imagine the electric field $E$ from the Faraday EMF is the source of the $E$ for the Poynting vector. It remains to be seen what exactly the $H$ is that needs to be employed here. If a toroidal geometry is used, I think the $H$ from the coils in principle covers a path that goes around the whole toroid. Meanwhile, the power transfer can be quite significant, and I would be rather surprised to find that a "leakage" flux is responsible for any significant amount of power transfer. It might be the case that it is responsible for significant power losses, but I don't think it could ever be concluded that the power transfer depends on this. $\\$ One detail I haven't ironed out yet is that $N_p I_P=N_s I_s$ , but I believe the $H's$ from these are basically equal and opposite. Somehow, I'm expecting this is where the power transfer occurs, but I need further study of the problem. One possibility is it is the same type of calculation involving two electrical charges, where the electric field of interest for a single charge is the one that comes from the other charge, and it doesn't experience a force from its own electric field. $\\$ @vanhees71 My experience with the Poynting vector is somewhat limited. Might we have your input on this paper of post 16 ? I am starting to believe that the authors are incorrect in determining that a "leakage" flux is responsible for the power transfer. Perhaps the Poynting vector analysis is rather straightforward. Perhaps you can show us that this is the case. $\\$ And I didn't have time to read this paper yet, but perhaps it is more accurate than the one of post 16. Just a quick reading of it, it appears it may have the answer: http://www.hep.princeton.edu/~mcdonald/examples/transformer.pdf

Last edited:
Homework Helper
Gold Member
And I didn't have time to read this paper yet, but perhaps it is more accurate than the one of post 16. Just a quick reading of it, it appears it may have the answer:
I read the paper. Accoring to the author, the energy flow from primary coil to secondary coil depends on two things:
1) The E-field in the central air gap region of the toriod due to the changing B-field in the toroidal core.
2)The H-field in the same central air gap region of the toriod due to the "conduction currents".
This ExH gives the Poynting Vector carrying the load power.

Now, in case of an ideal toroid, all the magnetic field is contained "inside" the core. But here, there exists an H-field in the central air gap of the toroid due to the "conduction currents". Doesn't this suggest that this is the "leakage" flux that is used in the power transfer?
I may be wrong here.

Homework Helper
Gold Member
I read the paper. Accoring to the author, the energy flow from primary coil to secondary coil depends on two things:
1) The E-field in the central air gap region of the toriod due to the changing B-field in the toroidal core.
2)The H-field in the same central air gap region of the toriod due to the "conduction currents".
This ExH gives the Poynting Vector carrying the load power.

Now, in case of an ideal toroid, all the magnetic field is contained "inside" the core. But here, there exists an H-field in the central air gap of the toroid due to the "conduction currents". Doesn't this suggest that this is the "leakage" flux that is used in the power transfer?
I may be wrong here.
I think the conduction currents can be minimized with laminations. See posts 19 and 20 of this thread, especially post 20 where @jim hardy gives an important input: https://www.physicsforums.com/threads/magnetic-flux-is-the-same-if-we-apply-the-biot-savart.927681/

Homework Helper
Gold Member
I think the conduction currents can be minimized with laminations
I am not sure we mean the same thing by "conduction currents". In the paper, conduction currents are the primary and secondary currents. And Jim's response in that thread is about how "eddy-currents" can be minimized with the laminations.

Homework Helper
Gold Member
I now have read the paper of post 19, although still somewhat quickly. The conduction currents are those of the $I_p$ and $I_s$ including the windings as well as any additional leads. The displacement current $\frac{\partial{D}}{\partial{t}}$ is being considered because Maxwell's (MMF) equation/Ampere's law actually reads $\nabla \times H =J_{conductors}+\frac{\partial{D}}{\partial{t}}$ for the non steady state case. The term $\frac{\partial{D}}{\partial{t}}$ is known as the displacement current. $\\$ It does appear that the Poynting analysis determines the energy travels through the air across the core. $\\$ It's somewhat interesting, but whether this result has any practical application remains to be seen. When you consider simple DC or AC circuits, the usual approach is to consider the power transfer as happening inside the conducting wires. As a graduate student, I once had a fellow graduate student make the argument that the electromagnetic energy actually travels around and outside the wires. He may have been right, but I normally find the simpler explanations as being more useful for practical applications.

Homework Helper
Gold Member
Additional comment: This "leakage flux" is interesting. In posts 14 and 15 I made the argument that if Biot-Savart is used in the computation, the windings need to be re-routed to extend around the entire toroid. For this analysis, perhaps Biot-Savart should be used in the computation of $H$ without artificially changing the geometry. It's an interesting question, but I don't have all of the answers. Some of the authors of these papers have put considerable effort into this Poynting vector explanation. It appears to be a very non-trivial problem.

Homework Helper
Gold Member
It's somewhat interesting, but whether this result has any practical application remains to be seen.
Yes, analysing energy transfer using Poynting vector is not very practical, even for the simplest of circuits. For all practical purposes, circuit theory approach is sufficient. Maybe this is why many college-level engineering and physics books do not address this approach for analysing circuits and electrical machinery.

Thanks a lot for your patient and articulate responses! You are a true expert in magnetism!