# Magnetization of ferromagnetic material

lorenz0
Homework Statement:
Consider a ferromagnetic material and suppose we can approximate a portion of its hysteresis curve as a segment as shown in the figure. With this material we make a permanent magnet in the shape of a toroid with an overall length equal to ##D = 80 cm## including an air gap with a thickness of ##h = 5 mm##. Calculate the magnitude of the magnetization vector ##\vec{M}## of the magnet.
Relevant Equations:
##\vec{H}=\frac{\vec{B}}{\mu_0}-\vec{M}##
I have thought about the following
##\oint \vec{H}\cdot d\vec{l}=0\Leftrightarrow H_{int}(D-h)+H_{ext}h=0\Leftrightarrow (\frac{B}{\mu_0}-M)(D-h)+\frac{B}{\mu_0}h=0\Leftrightarrow M=\frac{D}{D-h}\frac{B}{\mu_0}## but (supposing what I have done is correct) I don't understand which value of ##B## I should take: perhaps the one at ##H=0\frac{kA}{m}##, namely ##B=0.5T##?

Thanks

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Homework Helper
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This one is not simple, but you got a good start, with ## \oint H \cdot dl=0 ##, so that ## H_{inside} \approx -H_{gap} h/D ##. To a good approximation, ## H_{gap}=M ##, from the pole model of magnetostatics, (compare to the computation of ## E ## with two planes of surface charge density ## \pm \sigma ##).
From the hysteresis curve, you need to find the linear relationship between ## M ## and ## H ##. Let me write it out for you: Using ## B=\mu_o H+\mu_o M ##, we have one point at ## M=.5/\mu_o=M_o ## and ## H=0 ##. The other point is at ## H=-2000 ## and ## M=+2000 ##. We can write ## (M-M_o)/(H-0)=(2000-M_o)/(-2000-0) ##, and solve for ## M ## in terms of ## H ##.
Combine this with ## H=-M h/D ## to solve for ## M ##.

Edit: alternatively to see that ## H_{gap}=M ##, in the material ## B \approx \mu_o M ##. The ## B ## must be continuous, so that ## B=\mu_o M=\mu_o H_{gap} ##.

Last edited:
• lorenz0 and Delta2
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for additional reading on the pole model vs. the surface current model of magnetostatics, especially post 1 of this link.
See https://www.physicsforums.com/insig...tostatics-and-solving-with-the-curl-operator/
for including the contribution to ## H ## from currents in conductors.
(Note: ## H ## in general comes from two contributions: magnetic poles and currents in conductors. For the above problem, to find ## H_{gap} ##, we just had magnetic poles to consider, as there were no currents in conductors).

Last edited:
• lorenz0
lorenz0
This one is not simple, but you got a good start, with ## \oint H \cdot dl=0 ##, so that ## H_{inside} \approx -H_{gap} h/D ##. To a good approximation, ## H_{gap}=M ##, from the pole model of magnetostatics, (compare to the computation of ## E ## with two planes of surface charge density ## \pm \sigma ##).
From the hysteresis curve, you need to find the linear relationship between ## M ## and ## H ##. Let me write it out for you: Using ## B=\mu_o H+\mu_o M ##, we have one point at ## M=.5/\mu_o=M_o ## and ## H=0 ##. The other point is at ## H=-2000 ## and ## M=+2000 ##. We can write ## (M-M_o)/(H-0)=(2000-M_o)/(-2000-0) ##, and solve for ## M ## in terms of ## H ##.
Combine this with ## H=-M h/D ## to solve for ## M ##.

Edit: alternatively to see that ## H_{gap}=M ##, in the material ## B \approx \mu_o M ##. The ## B ## must be continuous, so that ## B=\mu_o M=\mu_o H_{gap} ##.
You have been very helpful, thank you very much!

• Homework Helper
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and a follow-on: More useful than the value of ## M ## is the value of ## \mu_o M ##. I get ## \mu_o M=.22 ## T. Does this agree with what you obtained?

• lorenz0
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Perhaps a couple additional comments about the solution of this would be worthwhile. We are given a hysteresis curve that is linear, relating ## B ## to ## H ##. We can thereby write ## B=aH+b ## for constants ## a ## and ## b ##. With ## B=\mu_o H+\mu_o M ##, it should be apparent, without a rigorous proof, that ## M=a'H+b' ##, for constants ## a' ## and ## b' ##, i.e. there is a linear relationship between ## M ## and ## H ##. We made use of this in solving the problem above.

• lorenz0
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It may be of interest that changing the geometry from that of a toroid to a toroid with a gap made for the solution where we had the line ## M=-160 \, H ## intersecting the hysteresis curve, rather than having ## H=0 ##,
(## \mu_o M=.5 ## T), for the complete toroid where there aren't any poles. (Magnetic pole density ## \rho_m=-\mu_o \nabla \cdot M ## gives rise to a non-zero ## H ## in the magnetic pole model of doing the calculations).

Another case of interest is a spherical shape for the permanent magnet, where some advanced calculations show that ## H=-M/3 ##. The factor (1/3) is known as the demagnetizing factor ## D ## for the spherical geometry. Using the same hysteresis curve, the result would be that ## \mu_o M ## would be considerably less for the permanent spherical magnet than the toroid with the small gap. If I get a little extra time, I may compute this value. See https://encyclopedia2.thefreedictionary.com/Demagnetizing+Factor

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• lorenz0
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For the spherical geometry, using the same hysteresis data, I get that ## \mu_o M=.0074 ## T. It would be interesting to see some experimental data for this type of thing. They could also try gaps of different thicknesses with the toroid, and see how it affects the strength of the permanent magnet. The larger the gap, the smaller the magnetization ## \mu_o M ## of the permanent magnet.

• lorenz0
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One other item of interest worth mentioning in regards to these permanent magnets is what the ## H ## represents. The ## H ##, computed from the currents in conductors along with a contribution from magnetic poles in general does not represent an actual magnetic field, but is a useful mathematical construction that always obeys the pole model formula ## B=\mu_o H +\mu_o M ##.

In a permanent magnet, surprisingly, the ## H ## is found to point opposite the magnetization ## M ##. For the permanent magnet, it is the ## B ## that is the actual magnetic field, and it is found to point in the direction of ## M ##, as expected.

For linear materials such as the iron in transformers, where we can write ## B=\mu H ##, the ## H ## can be treated like an actual field, but that is not the case with the permanent magnet.

• lorenz0