# Homework Help: Question regarding a magnetic field path around a gapped core

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1. Apr 21, 2017

### doktorwho

1. The problem statement, all variables and given/known data

$l/l_0=100$
$S=3 (cm)^2$
$B_r=200μ_0H_c=2mT$
$R=10Ω$
No current runs through the feromagnetic core. Calculate the charge $q$ that goes through $R$ from the moment in the diagram to the moment it is removed from the core.
2. Relevant equations
3. The attempt at a solution

I just have one small lack of understanding concerning this problem.
From the ampere's law i get:
$Hl+H_0l_0=0$
and
$Hl+\frac{l_0}{μ_0}B=0$
Here the professor said that we from the diagram suppose that $H=-H_c$ and then continued to calculate $H$ to be $H=25/\pi [A/m]$
Why did he suppose that? And what would happen if i supposed $B$ to be $B_r$. Is there a thing that happens that tells me that my supposition was wrong and that i should do the other one? Could you explain this?

2. Apr 21, 2017

From the hysteresis curve (of $B$ vs. $H$), the magnetic field in the iron core is $B= B_r$ and is constant. (The iron core is a permanent magnet, and the magnetism $M$ is at a saturated level. Sometimes, the $B$ can vary slightly with $H$, but in the graph that is supplied, the $B$ is constant. The $H$ is zero for this case where there are no windings that carry a current wrapped around the iron core. This is an extra detail if you want to study the topic in more detail). In any case, Faraday's law $\varepsilon=- \frac{d \Phi}{dt}$ where $\Phi=B A$ , ($A=S$), is what you need to use to find the voltage and thereby the current that flows when the loop of wire is removed. You can then integrate over time to get the charge $Q=\int I \, dt$.

3. Apr 21, 2017

### TSny

Your equation $Hl + \frac{l_0}{\mu_0} B = 0$ gives one relation for the two unknowns $B$ and $H$. Another relation is given by the magnetization curve (hysteresis curve) as given in the problem.

You can find the values of $B$ and $H$ that satisfy both relations by plotting $Hl + \frac{l_0}{\mu_0} B = 0$ on the magnetization curve. Where does the line $Hl + \frac{l_0}{\mu_0} B = 0$ intersect the magnetization curve? What are the values of $B$ and $H$ at this point of intersection?

What if the gap size were reduced such that $\frac{l}{l_0} = 400$?

For background reading, see section 36.5 here
http://www.feynmanlectures.caltech.edu/II_36.html

4. Apr 22, 2017

I'm puzzled about Feynman's solution that assumes you have an $H_1$ in the material and an $H_2$ in the air gap. It seems you should be able to simply assume that $B$ is continuous across the gap. To first order, the only thing that a computation of $H$ will do is show that the $H$ from the magnetic poles, which have magnetic surface charge density (in the magnetic pole model) $\sigma_m=M \cdot \hat{n}$, will make it such that by using $B=\mu_o H+M$, that in the gap where $M=0$, the $H$ from these poles will make $B$ continuous across the gap with an increase in $H$ because of these poles (to precisely account for the $M=M_o$ in the material and $M=0$ in the gap). To first order, the $H$ in the material is left unchanged because of the opposite poles on the two faces. $\\$ I don't think the equation that Feynman presents for the air gap in this manner is correct. The exact solution for a substantially finite air gap is going to be a complex function of the geometry involving the diameter of the endfaces. To first order, for reasonably small gaps, the gap has infinite sheets of magnetic surface charge $+\sigma$ and $-\sigma$ on the two surfaces. Any computation of $H$ and $B$ using these infinite sheets of magnetic surface charge, (which is somewhat advanced for a first year course in E&M, but quite workable using Gauss's law ), will simply show that $B$ is continuous across the gap to a good approximation. $\\$ Editing. Additional comment: Because for static magnetic poles, there is the equation $\nabla \times H=0$, analagous to the electrostatic $\nabla \times E=0$ equation, you can write $\oint H \cdot \, ds=0$, but this really does very little to solve for the value of magnetic field $B$ in the gap. For this problem. I think it is incorrect to try to employ $\oint H \cdot ds=0$ to solve for the magnetic field $B$ in the gap or in the material. $\\$ Additional Editing... @vanhees71 You might find this problem of interest. I would enjoy your feedback on it.

Last edited: Apr 22, 2017
5. Apr 22, 2017

### TSny

For a very small gap, B in the gap is approximately the same as B in the material. Note the reasoning for this that Feynman gives just above and below figure 36-11. So, Feynman does take B to be continuous across the gap.

H in the material and H in the gap are very different. In doctorwho’s problem, there is no electric current in any windings around the iron. In this case, H actually reverses direction as you move from the material into the gap. See point d in Feynman’s figure 36-12 where H is of opposite sign to B. Also, H undergoes a dramatic change in magnitude. In doctorwho’s problem, the magnitude of H in the gap is 100 times stronger than in the material.

I don’t see anything wrong with Feynman’s argument where he uses $\oint H \cdot \, ds=NI$. (In doctorwho's problem $I=0$, so $\oint H \cdot \, ds= 0$.) It is similar to treatments that I have seen in other texts. Can you point to a specific step in Feynman’s derivation that you think is incorrect?

6. Apr 22, 2017

The equation that Feynman is using is basically Ampere's law where it is often written as $\nabla \times B=\mu_o J_{total}$where $J_{total}=J_{free}+J_m$ and since $B=\mu_o H+M$ and also $\nabla \times M=J_m$, Ampere's law takes the form $\nabla \times H=J_{free}$. (We're assuming steady state.) With Stokes law, this becomes $\oint H \cdot ds=I_{free}$. For the case at hand $I_{free}=0$ (there's no solenoid type windings), but I disagree with the applicability of this equation for the case at hand. $\\$ Instead, using $B=\mu_o H+M$ with $H=0$ you can compute $M$ to do the calculation in its most complete form. You can then compute magnetic surface charge density $\sigma_m=M \cdot \hat{n}$ on the endfaces. $H$ and $B$ across the gap can be computed from Gauss law. The calculation is straightforward and uses $\mu_o \nabla \cdot H=-\nabla \cdot M$. (Comes from $\nabla \cdot B=0$ along with $B=\mu_o H+M$). For the single endface of $+ \sigma_m$, $H=+ M/(2 \mu_o )$. Accounting for the $- \sigma_m$ endface, we find $H_{gap}=M/\mu_o$, and thereby using $B=\mu_o H+M$, with $M=0$ in the gap, $B_{gap}=M$ , which is precisely what it is in the material, with $M=B_r$. A much more refined calculation with a much larger gap would use the same $\sigma_m$, but instead of using Gauss' law for an infinite sheet of magnetic surface charge, essentially Coulombs law (with $H$ and $\mu_o$ replacing $E$ and $\epsilon_o$) would be used to compute the $H$ in the gap,(quite an elaborate integral and the magnetic field in the gap is not going to differ significantly from the very narrow gap case unless the gap becomes quite large). $\\$ The equation of $\oint H \cdot \, ds=0$ is of little or no use for this application, and does not give the correct answer for the $H$ or $B$ in the gap. (Perhaps Feynman's mistake is that he can compute $H$ in the material. Does he compute this $H_2=B_r/\mu_o$?. I'm going to need to look more closely at his derivation, but $H_2=B_r/\mu_o$ is incorrect. Editing.. upon closer inspection, I think Feynman is in error by assuming $H_2$ is constant, but I don't see him get an incorrect answer otherwise...). $H_2$ (in the material) is quite small but it will be affected by the $H_{gap}$ (Feynman's $H_1$), i.e. by the geometry of the gap, rather than being able to use its known value (which isn't known) to compute $H_{gap}$. $\\$ My answer is that $B_{gap}=+M=B_r$ to a very good approximation, for the most part independent of the size of the gap until the gap gets quite large. If you want to compute $H_{gap}$, (It is not necessary for the problem at hand), $H_{gap}=+M/\mu_o=B_r/\mu_o$, and is also to a good approximation independent of the size of the gap. (Note: Feynman even says in his lecture that $B_1=B_2$, that $B$ is essentially continuous across the gap.)

Last edited: Apr 22, 2017
7. Apr 22, 2017

### TSny

OK

Why are you setting $H = 0$? Neither $H_1$ (in the gap) nor $H_2$ (in the material) is equal to zero.

In Feynman's figure 36-11, the right surface of the gap will have the positive fictitious pole density. Choosing vector components to the left as positive components, then Gauss's law for the right surface of the gap would give $H_1 - H_2 = \frac{M}{\mu_0}$, or $H_1 = H_2 + \frac{M}{\mu_0}$. This shows the relation between $H_1$, $H_2$, and $M$.

Applying Gauss law to the left face of the gap (with the negative pole density), you again find $H_1 = H_2 + \frac{M}{\mu_0}$. This is nothing new, as it expresses the same result as for the right face. (You would not add the results together for the left face and the right face. Each result separately expresses the correct relation between $H_1$, $H_2$, and $M$.)

The correct relation is $B_{gap} = B_1 = \mu_0 H_1 = \mu_0 \left( H_2 + \frac{M}{\mu_0} \right) = \mu_0 H_2 + M$. However, since $\mu_0 H_2 << M$ in our situation, we do have approximately $B_{gap} \approx M$.

I don't understand this remark. My understanding of the meaning of $B_r$ is that it represents the value of $B$ when $H_2 = 0$. (See doktorwho's magnetization curve.) But we do not have $H_2 = 0$ in our situation.

Feynman does not get $H_2=B_r/\mu_o$. He gets $H_2$ as the value of $H$ at point $d$ in his figure 36-12. (Note that point $d$ lies on the straight line corresponding to $I = 0$.) Likewise, $B_2 = B_1 = B_{gap}$ is the value of $B$ at point $d$. This value of $B$ does not equal $B_r$.

Also, note that $H_2$ is negative, so $H_2$ points to the right inside the material just to the right of the right surface of the gap. That is, $H_2$ points in the opposite direction to $B_2$ in the material.

For doktorwho's example, I think the answer is $B_{gap}=B_r / 2$ and this result depends on the fact that the gap distance is such that $l_2/l_1 = l/l_0 = 100$. For a different gap distance you would get a different result for $B_{gap}$. (The slope of the straight line corresponding to $I = 0$ in fig. 36-12 depends on the gap distance.)

I don't agree with this statement. Although it would be true that $H_{gap} \approx +M/\mu_o$, it is not true that $M = B_r$. Also, $H_{gap}$ is not approximately independent of the gap distance.

Yes, as noted in post #5.

Last edited: Apr 22, 2017
8. Apr 22, 2017

@TSny Some very good questions that I think can perhaps be best answered by looking at a similar but slightly simpler problem that appeared on Physics Forums about a year ago. Going from the case of a straight magnetized cylinder with uniform magnetization to one that wraps around on itself with an air gap should be a simple extension of the same principles. Please take a look at https://www.physicsforums/threads/magnetic-field-of-a-ferromagnetic-cylinder.863066/ [Broken] These magnetic problems are most easily done using the "pole" method, but from what I have been told from an E&M professor who currently teaches the subject, they normally show the "magnetic surface current" method (e.g. Griffiths text) and they don't present the "pole" model until a graduate course. This is different (but I think better) than how it was when I was in college, where we were taught the "pole" model, but it came with a lot of handwaving, and they only presented the "equivalent surface current" very briefly as a theory and didn't connect the two quantitatively. I believe I compared the two methods in detail in my responses to the problem of a year ago (see the above link.) I'm hoping this helps to answer a couple of the questions that you had. For some reason the "link" didn't work. Let me try again: https://www.physicsforums/threads/magnetic-field-of-a-ferromagnetic-cylinder.863066/ [Broken] And I don't know why the "link" isn't working because I was able to find it in a search of "ferromagnetic cylinder" and it comes up. It was posted by "cdummie" on March 21, 2016 in the Computer Science and Engineering homework section. The "link" has always worked previously. Perhaps you can do a "search" and review it. $\\$ @Greg Bernhardt I can't get the "link" to work. Might I be doing something different this time? I have referenced this "link" on several other occasions, I think the very same way, but for some reason it fails to work presently... Problem solved=see below. $\\$ Editing: Problem solved: A copy and Paste of the "link" worked. Please "link" to: Magnetic field of a ferromagnetic cylinder Hopefully this is helpful.

Last edited by a moderator: May 8, 2017
9. May 19, 2017

@TSny Editing... I see that Feynman's solution is accurate for certain transformer problems that have appeared (even recently) on PF where the transformer has an air gap. I don't think it is applicable to this problem though which involves a permanent magnet and no linear assumption can be made that $B_2=\mu \mu_o H_2$. The only $H_2$ in this problem is from the air gap, and it is small. There is no linear relationship between $B_2$ and $H_2$ here, unlike that in a working transformer where the magnetization $M$ is unsaturated. $\\$ Editing... I'm assuming from the $B$ vs. $H$ curve that was presented in the above problem that this is indeed a permanent magnet.... $\\$ Editing... Feynman's solution does shed much light on the transformer problem with an air gap, so I have now found the "link" very useful in any case. Thank you. :) :)

Last edited: May 19, 2017