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

1. Sep 12, 2017

### EEstudent90

Hi

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

From text:
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:

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2. Sep 12, 2017

### jim hardy

3. Sep 12, 2017

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: Sep 12, 2017
4. Sep 12, 2017

### jim hardy

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

5. Sep 13, 2017

### 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:
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
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.

6. Sep 13, 2017

### EEstudent90

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

7. Sep 13, 2017

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: Sep 13, 2017
8. Sep 13, 2017

### jim hardy

Don't feel alone i still slip back into that trap when been away from the subject for too long.

"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.
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.

9. Sep 13, 2017

@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: Sep 13, 2017
10. Sep 13, 2017

### 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.

11. Sep 13, 2017

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: Sep 13, 2017
12. Sep 13, 2017

@EEstudent90 Please see my edited post #11 above. I think that might explain it.

13. Sep 13, 2017

### jim hardy

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

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: Sep 13, 2017
14. Sep 13, 2017

### jim hardy

'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 .

15. Sep 13, 2017

### EEstudent90

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?

16. Sep 13, 2017

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: Sep 13, 2017
17. Sep 13, 2017

@EEstudent90 Please see my last comment of post #16. I'm hoping it is helpful. :)

18. Sep 13, 2017

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: Sep 13, 2017
19. Sep 13, 2017

### 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.

20. Sep 13, 2017