# MMF/Flux density across air gap for a salient pole

EEstudent90
Hi

I am trying to understand the figure 2 I have attached to this post.

From text:
Due to the appropriate design of the pole shoe, the air-gap flux density varies cosinusoidally even though it is caused by the constant magnetic potential difference in the air gap.

Ok, I get that the air-gap flux density varies in a sinusoidal manner, because the air gap length is not constant i.e. non-constant reluctance and hence non-constant magnetic flux density.

What I do not understand is how we can draw the air gap MMF can be constant across the air gap?

I am not very good in magnetics and Ampere's law etc., therefore my guess to why the MMF is constant across the air gap is by using Ampere's law with the red Ampere loop as shown in Figure 1.

Now my problem is that the MMF should change sign at B and D (according to attached Figure 2), right? How will Ampere's law handle this?

Again, not very good at this, hope someone can help me.

Figure 1:

Figure 2:

#### Attachments

2.9 KB · Views: 632

Homework Helper
Gold Member
2020 Award
I can try to give you what I think may be correct, but @jim hardy knows this kind of thing better than I do. Perhaps he can give feedback on whether I got it right. ## \\ ## If I am interpreting it correctly, figure 1 is drawn poorly, (they drew it more accurately in figure 2), and needs to have the gap at point A be smallest going straight up on the center line. This will make the ## H ## in the ## \int H \cdot dl ## across this air gap needing to be the highest, giving the highest ## B =B_{\delta} ## at this point. This is the zero degree angle of the cosine distribution. The ## B_{\delta} ## is an approximate cosine shape. If the path of the integral goes slightly to the left or right before making contact with the stator, the path length (drawn poorly in figure 1) to the stator will be slightly longer, requiring a slightly lower ## H ## and ## B ## value than the path that goes straight up. (Remember the MMF around the loop is ## N I=\oint H \cdot dl ##, (## N ## =Number of turns), and in the magnetic material, the ## H ## is assumed to be near zero (it's very small). If you take a path with a longer air gap, the required ## H ## in the air gap is lower (## \int H_{air \, gap} \cdot dl=NI ##.(See below =we need to consider the air gap by point C in this integral as well). Meanwhile ## H ## over the path of ## \int H \cdot dl ## out of the top (by A) that stays a long distance in the air gap before making contact to the stator near point B will have a much lower ## H ## and ## B ## value because the air gap path is much longer and it has the same MMF as the shorter path straight up to A across the gap to the stator. ## \\ ## The path down by C is similar, and to be consistent, you want the return path across the air gap by C to take the same path (short or long) as you have for the outgoing path at A. That way, you only need to compute ## H ## at A and what happens at C will have the same contribution to the integral, essentially dividing the required ## H_{air \, gap} ## by two.

Last edited:
Asymptotic and EEstudent90
jim hardy
Gold Member
Dearly Missed
While @Charles Link got it , I think it can be stated simpler than that.

MMF is amp - turns. In the text from which fig 2 comes he calls it 'current linkages' Θ . see .https://www.physicsforums.com/attachments/current-linkage-ask-pdf.208916/
H is MMF per unit length

Walking around your red amperian loop and crossing the air gap anywhere on the pole face you'll enclose the same amp turns. None of them get expended in the iron, that's stated in the text as a simplifying assumption..
So mmf, amp-turns, is the same along a path that crosses the air gap anyplace on the pole piece.
But - the paths nearest the center traverse the shortest length of (almost non-permeable) air so MMF/Length, which is H, is highest there.
They shape the pole piece to make an air gap that results in cosine shaped flux.

Remember B = μμ0NI/Length where NI is MMF, NI/Length is H
giving the more familiar B = μH

Same reference :

print is large enough to read at that surrey.uk site

Any help ? Don't feel alone, most folks struggle with magnetic units. I sure did.

old jim

Asymptotic, EEstudent90, cnh1995 and 2 others
EEstudent90
Hi and thank you for the replies, I appreciate that you take the time and effort to help a newbie like me.

My first misconception was that I thought MMF was equal to H, but of course it is not, I could have seen that from Ampere's law even.
My second misconception (I think) was that my mental picture of MMF was field lines similar to the field lines of a magnet.

I think I am close to understanding this. I think I understand why the MMF is constant along the pole shoe as jim hardy said:
Walking around your red amperian loop and crossing the air gap anywhere on the pole face you'll enclose the same amp turns.

What remains for me is: How can I draw the MMF distribution from North-pole to south-pole (between the blue lines, see figure in next post), what happens with the MMF distribution when I leave the surface of the pole shoe and how will I draw the ampere loop to describe this?
View attachment 211011
are you in the same class as the fellow who started this thread back in August?
No, but I saw the thread, that was what inspired me to understand this properly.
Basically I am an electrician and just out of curiosity I try to learn "electrical engineering stuff" because I find it fun and interesting. What's holding me back is that I most of the time I don't have the necessary basic knowledge and lacking some math skills.

Again, thanks.

EEstudent90
The figure in my last post is not showing, so posting it here:

Homework Helper
Gold Member
2020 Award
With this improved figure, I think I have a better explanation of the ## cos(\theta) ## factor. If you call the distance between the two circles going straight across the gap as ## d ##,(## d ## is the difference in the radii of the two circles) the distance ## s ## of a line connecting these two has that goes at angle ## \theta ## across the gap is ## s= \frac{d}{\cos(\theta)} ## (the gap is small so the two radii can be considered to be parallel planar surfaces. The longer path makes for a smaller ## H ##: ## H(\theta)=H_o cos(\theta) ##. ## \\ ## Editing: See post #11. I believe this post (#7) is in error, and this result gets explained and corrected in post #11.

Last edited:
EEstudent90
jim hardy
Gold Member
Dearly Missed
My first misconception was that I thought MMF was equal to H,
Don't feel alone i still slip back into that trap when been away from the subject for too long.

My second misconception (I think) was that my mental picture of MMF was field lines similar to the field lines of a magnet.
"Line" is an antiquated unit for Flux , renamed to Maxwell.
One Maxwell per square cm is one Gauss , roughly the flux density of Earh's magnetic field .
Now they use Webers for flux and Teslas for flux density
a Weber is 108 lines or Maxwells and a Tesla is 104 Gauss.
104 lines/cm2 is a Tesla , and with 104 square cm in a square meter that makes 108 lines in a Weber. A Tesla is one Weber per square meter . It's quite alot of flux.
Basically I am an electrician and just out of curiosity I try to learn "electrical engineering stuff" because I find it fun and interesting. What's holding me back is that I most of the time I don't have the necessary basic knowledge and lacking some math skills.
It's hard for me too because the units are all named after people and i'm terrible with names. And as i progressed through school the textbook authors were changing unit systems
MMF i learned as Gilberts . That unit went away thankfully.
You'll still run across H in Oersteds , an Oersted is amp-turns per cm / 4pi
hopefully you won't see MMF as Gilberts anymore - amp turns X 4pi/10

aarrrgghhh no wonder i'm scrambled.

EEstudent90
Homework Helper
Gold Member
2020 Award
@EEstudent90 One thing that can be helpful in understanding these magnetic solutions with ## \oint H \cdot dl=NI ## is that because ## \nabla \cdot B =0 ##, flux lines of the magnetic field ## B ## are continuous. Meanwhile ## H ## changes as you enter different materials, and to a first approximation can be considered constant throughout a given material because ## B=\mu H ## is continuous, and ## H ## doesn't change much until ## \mu ## changes. Meanwhile, in magnetic materials, ## \mu ## is very large, making ## H ## very small, while ## B ## remains nearly the same as it goes from one material to the next.

Last edited:
EEstudent90
EEstudent90
Again thanks. I am very very close to understanding this, but I still have to understand how the MMF (NI/L) is distributed after we leave the pole shoe.

Please take a look at my crappy paint drawing below. I have tried to draw the MMF distribution across the air gap as a function of angle, where I start at the middle of top pole shoe. I don't understand how the distribution will be between the blue dots, and have therefore not drawn anything after I leave the pole-shoe, shown in the paint drawing below.

How can I draw my ampere loop on the picture below, to show how the MMF distribution is between the blue dots in my first paint drawing?

Sorry if you have explained this question I ask in this post already, but if that is the case I have not understood it.

Arg, this is destroying my sleep lol, but it is interesting.

Homework Helper
Gold Member
2020 Award
Again thanks. I am very very close to understanding this, but I still have to understand how the MMF (NI/L) is distributed after we leave the pole shoe.

Please take a look at my crappy paint drawing below. I have tried to draw the MMF distribution across the air gap as a function of angle, where I start at the middle of top pole shoe. I don't understand how the distribution will be between the blue dots, and have therefore not drawn anything after I leave the pole-shoe, shown in the paint drawing below.
View attachment 211031
How can I draw my ampere loop on the picture below, to show how the MMF distribution is between the blue dots in my first paint drawing?
View attachment 211033

Sorry if you have explained this question I ask in this post already, but if that is the case I have not understood it.

Arg, this is destroying my sleep lol, but it is interesting.
Read (and study) my post #7. (Editing: See below, and I believe post #7 is actually in error). Meanwhile the MMF comes from the current in the windings. It is constant in the equation ## NI= \oint H \cdot dl ##. The thing that you change is the path across the gap to find the ## H ## for a different direction across the gap. After calculating ## H ## in the gap, you compute ## B ##. This will also give you a good estimate is of ## B ## in the material, since the flux lines of ## B ## are continuous. See also: https://images.google.com/imgres?imgurl=https://cdn.miniphysics.com/wp-content/uploads/2012/07/solenoid-magnetic-field.jpg&imgrefurl=https://www.miniphysics.com/ss-magnetic-field-due-to-current-in-a-solenoid.html&docid=4BFH21z5e7KBgM&tbnid=9Ao6GVbwH4WGZM:&vet=1&w=1578&h=812&source=sh/x/im ## B ## from the shoe as a function of angle will be quite similar to this but the shoe helps spread it. Also, surprisingly, it is the air gap distance, along with ## NI ## that determines the strength of the magnetic field ## B ##. ## \\ ## Editing: I think I need to make a correction to my post #7. The ## H ## vector will in general point straight across the gap. If you take a path at an angle ## \theta ## in crossing the gap, the dot product ## H \cdot dl ## contains the ## cos(\theta) ## factor, balancing the longer ## dl = dl_o/\cos(\theta) ##, (where ## dl_o ## is straight across the gap), so that you already have the correct ## H ## pointing straight across the gap. For paths that go outside the shoe, towards points B and D, ## H ## will be much lower due to the longer gap distance. Thereby, I'm not sure that the book's assessment that ## B ## follows a precise cosine distribution with angle ## \theta ## is even completely accurate.

Last edited:
EEstudent90
Homework Helper
Gold Member
2020 Award
@EEstudent90 Please see my edited post #11 above. I think that might explain it.

jim hardy
Gold Member
Dearly Missed
I don't understand how the distribution will be between the blue dots,

i only see one blue dot - were's the 'between' ?

I don't understand how the distribution will be between the blue dots, and have therefore not drawn anything after I leave the pole-shoe, shown in the paint drawing below.

Maybe it'll help to apply the same mental trick we use for electric circuits.

Imagine yourself very small sitting in the air gap on that pole piece right in its middle
Think dirt simple in baby steps.......
(excuse MY crummy Paint sketch)

Looking horizontal you see the lines of flux streaming up out of the pole. Note they take a path resembling an Amperian loop - after all it's amp-turns that are making the flux flow.....
Looking down with X-Ray vision you see the amp-turns encircling the rotor. They make MMF equal to NI pushing flux UPward.
Right in the middle you take your "Acme Portable Gauss Meter"* and you measure flux density equal to μ0NI/Length where 'Length' is the distance from pole piece to stator at point of measurement. Note we're measuring flux density , not flux...

Now you move over a few millimeters and repeat your flux density measurement.
This time Length is different . SO Flux will be μ0NI/(new Length) , different from before.
So the same MMF (NI) applied to a different distancegives different H(NI/Length)

For the example in that textbook , the one that started all this,
It is incumbent on the machinist who fabricates that rotor pole piece to shape it such that Length(of air gap) is cosine( angle).

Note that MMF is constant across the pole but H is not.

Much confusion results from flipping between concepts of fluxΦ and flux density B , MMF in NI and H in NI/meter, we get mixed up.
In circuits we usually aren't troubled by amps per square meter and volts per meter
so it's easy for us practical guys to fall into those mental traps.

* see next post - jh

Last edited:
jim hardy
Gold Member
Dearly Missed
'Gauss Meter' is not a figment of my imagination

i found one of these in a junkshop . Had to give ten bucks, though .

I love it. Handy for checking if car alternator is charging the battery .

EEstudent90
i only see one blue dot - were's the 'between' ?

Jim hardy post#15: Now you move over a few millimeters and repeat your flux density measurement. This time Length is different . SO Flux will be μ0NI/(new Length) , different from before. So the same MMF (NI) applied to a different distancegives different H(NI/Length)

Maybe I have stated my question a bit poorly, but my problem is how the distribution of the MMF will look like between the blue dots in my paint drawing. Below are two examples.

Look like this?

Or like this?

How can I decide how the MMF distribution will be (picture 1 or picture 2, if any) using Ampere's law or something similar?

Homework Helper
Gold Member
2020 Award
The MMF comes from the number of turns and the current and runs upward through the iron core surrounded by the current windings. ## MMF=NI ##. The magnetic "voltage drops" occur across the air gaps and the resistance is proportional to the distance. The "magnetic voltage drop" is measured by ## \int H \cdot dl ## across the gap. ## \\ ## Meanwhile ## \int H \cdot dl ## is very near zero in the magnetic material, because ## H ## is small in the magnetic material. (If ## B ## is finite and ## \mu ## is very large, that makes ## H=\frac{B}{\mu} ## very small). ## \\ ## The ## H_{gaps} ## can be computed because ## \int H_{gaps} \cdot dl= H_{gaps}L_{gaps}=NI ##. Then the magnetic field ## B ## can be computed: ## B_{gaps}=\mu_o H_{gaps} \approx B_{everywhere} ##. (This last "approximate" expression is just an approximation to illustrate a concept. The flux lines of ## B ## are continuous so if the flux lines spread out ## B ## will of course decrease.) ## \\ ## And to explain your question in the post above-the MMF is similar to an electrical battery or EMF being in the circuit. The voltage increases in going through the battery. Just like in a floating electrical circuit, the exact location of the ground is really arbitrary. The upper shoe has a "magnetic voltage" that is ## NI ## higher than the lower shoe.

Last edited:
EEstudent90
Homework Helper
Gold Member
2020 Award
View attachment 211041

Maybe I have stated my question a bit poorly, but my problem is how the distribution of the MMF will look like between the blue dots in my paint drawing. Below are two examples.

Look like this?
View attachment 211042
Or like this?
View attachment 211043

How can I decide how the MMF distribution will be (picture 1 or picture 2, if any) using Ampere's law or something similar?
@EEstudent90 Please see my last comment of post #16. I'm hoping it is helpful. :)

Homework Helper
Gold Member
2020 Award
Perhaps one thing that can be a little confusing in all of this is that usually a solenoid, with its changing magnetic flux when driven by an ac current, is used to create an EMF. Here instead, the solenoid is used as a source of MMF, and a similar equation to ohm's law (the magnetic analog of the electrical ohm's law) is used to compute the resulting ## H ## and ## B ##. ## \\ ## The equation ## \oint H \cdot dl =NI ## is the result of applying ampere's law ## \oint B \cdot dl=\mu_o I_{total} ## to loops of integration that contain magnetic materials. (For an extra detail, ## I_{total} ## consists of currents in conductors ## I ## along with magnetic surface currents ## I_m ## from the magnetization ## M ## that occurs in the material. The derivation to get the equation in the form ## \oint H \cdot dl=NI ## from the equation ## \oint B \cdot dl=\mu_o(I+I_m) ##, essentially eliminating the ## I_m ## term, is somewhat advanced, but the result is an equation with ## H ## that can be readily utilized).

Last edited:
EEstudent90
EEstudent90
Ok I think I got it, please see if what I write makes any sense, @Charles Link and @jim hardy

Below I have drawn three scenarios.

Scenario 1 (left):
MMF across airgap is at its maximum as maximum of net current enclosed by my Ampereian loop is maximum.

Scenario 2 (middle):
MMF is lower across the air gap where my Amperian loop is crossing, because net current enclosed is less as we have included another conductor, but this one is in opposite direction and therefore subtract and not add.

Scenario 3 (right):
MMF is zero across the air gap where my Amperian loop is crossing, because net current enclosed is now zero (equal amount of current going into and out of my surface enclosed by my Amperian loop.

Homework Helper
Gold Member
2020 Award
@EEstudent90 I think you got it !! :) I am finding this, including your latest post, very educational, because, unlike @jim hardy , I haven't had much previous experience in working problems involving the MMF. :)

EEstudent90
EEstudent90
@EEstudent90 I think you got it !! :)

If that is the case, I quote Fred Flintstone and say "yabba dabba doo!" :)

Homework Helper
Gold Member
2020 Award
@EEstudent90 One comment about the diagrams of post #19: I do think the statement is accurate that the MMF runs upward up the line going through the solenoid in the middle. If we choose a path for the integral that is a circle around the entire transformer so that in the plane of the circle we have as many + current sources as - ones, the MMF of ## NI =0 ## that we put in our equation ## NI=\oint H \cdot dl ##, but that doesn't mean that ## H ## is zero everywhere on the path of the integral. In this case of going counterclockwise around the circle, if the path is in the air, ## H \cdot dl ## may be negative going around the right side of the circle and positive on the left side of the circle with the results canceling as they need to because ## NI=0 ##. ## \\ ## If instead, the path of the integral is a counterclockwise semi-circle, with a straight line upward through the center, you get the correct answer for ## MMF=\oint H \cdot dl ## if you assume the ## MMF ## points upward in the middle. The case of using a semi-circle on the right side instead of the left is interesting: In the case of the right semi-circle, the ## MMF=-NI ## that is used in the equation even though it points upward in the middle with amplitude ## NI ##, because it points opposite the counterclockwise direction of the path of integration. (Alternatively you could look at the "-" signs on the current sources in the diagram to the right of center to determine this). On the right semi circle on the counterclockwise path , ## H \cdot dl ## picks up a minus sign on the dot product (because the vector ## H ## points clockwise there), balancing the minus sign from the ## -NI ##. It's an extra detail, but I think you might find it of interest. :)

Last edited:
EEstudent90
jim hardy
Gold Member
Dearly Missed
my alleged brain is still replaying it in my typical "Shoulda done better" routine.

https://www.physicsforums.com/threa...ld-from-a-balanced-three-phase-system.923994/

Observe how the rotor resembles a soelnoid.
Perhaps this thread has focused too much on the integral form of ampere's law.
That closed loop integral gives you just a number for magnitude(not direction) of total mmf around an amperian loop.and you have to traverse the whole loop to get your result. We're lucky that for the inside of a solenoid it works out to a nice easy formula and that's why it's in all the textbooks.
You have no idea what is its value at any single point outside the solenoid..

When you move your point of observation from inside to outside the solenoid, it becomes a new problem.
I believe the way to attack it is by superposition as we did in that three phase thread above.

To get the MMF at some point outside the solenoid we could use another form of Ampere's law
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html
(which is a kissing cousin to something known as Biot-Savart law)
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

It gives you a magnitude but not a direction.

Since in air B = μH and we're in air we can write μ0H = μ0HI/2πr
divide both sides by μ0 and you're left with H = I/2πr where r is distance from a wire.
But you have to figure out the direction from geometry.

At any point the mmf will be the sum of the mmf's from each individual turn, calculated individually with direction and added up.

These guys did that and explain their method better than i could
https://arxiv.org/pdf/1610.07876.pdf

They figured B not H but in air they're proportional
C1 and C4 are two of the turns in the solenoid
P is some point outside the solenoid
the curves are the leakage flux going outside the solenoid
their equation is complex , see last couple pages of their paper.

it's no wonder that textbook author didn't go into flux outside the pole pieces.

NASA printed a paper in 1960 on same subject with graphs to help avoid complex math
it's here

Sorry i don't have a simple answer. It's one of those things you find after you get out of school - your textbook authors avoided the tough questions.

old jim

Homework Helper
Gold Member
2020 Award
@jim hardy and @EEstudent90 As I have recently learned, (actually here on Physics Forums), the simplest way of finding the magnetic field of a transformer or other magnetic system with an air gap is to employ ampere's law in the form ## \oint H \cdot dl =NI ## around a loop. ## \\ ## The 3 phase system we worked last week was to calculate the magnetic field from what was individually 3 long current-carrying cables, and it contained no magnetic materials. There are two ways to calculate the magnetic field ## B ## from a single long current carrying wire:
1) Biot-Savart's law
and
2)Ampere's law ## \oint B \cdot dl=\mu_o I ## ## \\ ##
The derivation of this form of ampere's law ( ## \oint B \cdot dl=\mu_o I ##) is much simpler than the derivation for the form of ampere's law that is used in the air-gap MMF problem: ## \oint H \cdot dl=NI ##. (This second equation which is used in cases with magnetic materials is fairly simple, but the derivation of it gets somewhat detailed).## \\ ##The air-gap MMF problem is really a much more advanced problem than the problem of computing the magnetic field ## B ## from a long wire. In fact, the first time I saw the MMF problem with an air gap was here on Physics Forums about 6 months ago. I'm still on a learning curve with the MMF case. Even though the algebra of the MMF case is not real difficult, it takes some rather advanced E&M to show how the equation with magnetic materials ## \oint H \cdot dl=NI ## originates. :) :) ## \\ ## One link that treats the air gap problem is the Feynman lectures: equation (36.26), etc. http://www.feynmanlectures.caltech.edu/II_36.html ## \\ ## One Physics Forums posting that treats the problem of a transformer with an air gap is the following: https://www.physicsforums.com/threads/absolute-value-of-magnetization.915111/#post-5767374

Last edited:
Asymptotic and EEstudent90
EEstudent90
Thank you both. I will read through this thread again and study it good, and i'll look at the links you two posted in the last two posts.