# I Magnetization currents of Magnets with μ_r close to 1

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1. Nov 20, 2016

### particlezoo

Neodymium-Iron-Boron magnets have a relative magnetic permeability of 1.05.

According to the formula:

B = μ0 * (H + M)

Both the magnetizing field H and the magnetization M contribute to the magnetic field B. The magnetization M has sources and sinks, so therefore it follows from Maxwell's equations that H also has sources and sinks.

Since Neodymium magnets have magnetic permeability close to unity, then its value of M must be small relative to the value of H. The magnetization current is equal to ∇ × M. So does that mean that Neodymium Magnets have less magnetization current than soft magnetic materials possessing the same strength? That doesn't make sense to me. Why should magnetization current of a Neodymium magnet be less than that of, say, high permeability silicon steel of the same B field intensity? Is this "magnetization current" the actual magnetization current, or not? Or does it make sense to say that H fields of Neodymium magnets are due to "free electrical currents" somehow sustained internally inside the Neodymium magnets?

2. Nov 20, 2016

I googled the Neodynium iron boron magnet and also saw this relative permeability =1.05 value. It seems somewhat unclear what this is referring to. One of the problems that I think occurs in the area of magnetism is too often even the people making these data sheets either have limited expertise and/or they are not specific enough for it to be readily interpreted. The strength of the magnetic fields listed for these magnets (in the 1T range) are very typical of high quality magnets. Meanwhile for a permanent magnet, the magnetization is nearly saturated and the H (geometry dependent) is often near zero. (Otherwise, in a permanent magnet the H is actually negative=contribution to H comes from the magnetic poles and points opposite the magnetization M.) The term "relative permeability" is in my opinion, rather meaningless, unless they are telling you the magnetization is saturated and that little additional magnetization occurs if the magnet is put in a solenoid and an H is applied. $\\$ editing... And it is the magnetization $M$ that causes the magnetic field inside and outside of a permanent magnet, due to the magnetic surface currents. For a previous posting regarding magnetic surface currents and permanent magnets, see https://www.physicsforums.com/threads/magnetic-field-of-a-ferromagnetic-cylinder.863066/ Note: This post uses an alternative definition for $M$ where $B=\mu_o H+M$.

Last edited: Nov 20, 2016
3. Nov 20, 2016

### particlezoo

This makes H in opposition to B in the case for a permanent magnet, whereas for a solenoid current H is aligned with B. Correct?

If that's correct, then doesn't this create conflict for two different expressions for magnetic field energy density? If we compare (1/2)B2/μ and (1/2)(BH), then the former would be positive only and the latter would be negative for permanent magnets and positive for solenoid currents. Isn't B = μH essentially false when considering cases where there is magnetization (due to the non-divergence of B contrasted with the divergence of H)? It would therefore seem that only one of these two expressions can be correct. I think the second one is more fundamental, but this would mean that the potential energy of the magnetic field of an isolated permanent magnet (due to its own B and H fields) is actually negative. The first one depends on the meaningfulness of μ. As you point out:

It appears to me that this is what they meant. But when you speak of "additional" magnetization, that appears to be a reference to differential permeability. It makes me wonder then that the magnetic permeability is only real as a differential property.

4. Nov 20, 2016

The permanent magnet is indeed a case where the electromagnetic energy is negative and thereby an energetically favorable state. I believe your expression energy density $U=(1/2)B \cdot H$ is correct where $H=(B-M)/\mu_o$ (using your definition for $M$.) What this also shows is it is necessary to have sufficient magnetization/magnetic susceptibility in order to obtain an energetically favorable permanent magnet state. For additional reading, you may find of interest something I wrote up on the magnetism subject. The American Journal of Physics declined interest in publishing it, with their claim that much of the contents of this paper is already somewhat common knowledge. In any case, you might find it of interest. It's written in c.g.s units, but the basic quantities $B, H,$ and $M$ are all proportional regardless of the units chosen. https://www.overleaf.com/read/kdhnbkpypxfk

5. Nov 20, 2016

Just one additional comment on the energy: For a pure magnetic field $U=B^2/(2 \mu_o)$. For the case of a magnetic moment in a magnetic field $E=-m \cdot B$. For the case of the magnetization, $U=-M \cdot B$ would apply for an external field, but for the case of a permanent magnet, the magnetization is responsible for making the surface currents that make the magnetic field $B$ that causes more magnetization. The result is $dU=-B \cdot dM$ with $M= \chi B$.$\\$ $\chi$ in general is not a constant, but in any case, I'm not sure that for this case the simplistic $U=(1/2)B \cdot H$ is applicable. No doubt someone has already done a thorough thermodynamic analysis of this, but I do not have a good reference for it.

6. Nov 20, 2016

### particlezoo

If I recall correctly, for an ordinary "soft" magnetic material, the external H field accounts for all the B field outside of it, but needs a correction term inside, which it gets from M. However, it seems that in the case of a "hard" permanent magnet, you are saying M is the cause of the of the field inside and outside the magnet, and H serves as the correction term.

7. Nov 20, 2016

This gets a little complicated by the definition of $H$ which includes both contributions from currents in conductors (Biot-Savart), and contributions from (fictitious) magnetic poles. The pole model is mathematically correct, but does not explain the underlying physics. $\\$ In the surface current model of magnetization, (which is really the preferred way of proceeding=the underlying physics is completely explained by this method), the magnetic fields (for uniform magnetization $M$ ) occur as a result of the fields from magnetic surface currents (computed by Biot-Savart), along with the fields from currents in conductors (also computed using Biot-Savart). In this model, the magnetization $M$ does not create a local $B$. Any $B$ is the result of surface currents or currents in conductors. The $H$ really is not used in the surface current method, although the magnetic field from the current in conductors can be designated by the letter $H$. $\\$ The pole method, alternatively works with the $H$ and uses the equation $B=\mu_o H+M$ to get what appears to be a $B$ from the $M$ at that location, with a correction from $H$ (pole) terms. This method mathematically gets the same answer for $B$ as the surface current method, but it is really, in some ways, obsolete, because it can provide some very incorrect interpretations of the underlying physics. (e.g. the magnetization $M$ does not cause a local contribution to $B$, and the magnetic poles (fictitious), are not the sources for any magnetic field $B$. In addition, the $H$ is a mathematical construction, and not a second type of magnetic field.)

Last edited: Nov 20, 2016
8. Nov 20, 2016

### particlezoo

Now that I think about it, one thing I have wondered was what the role do microscopic inhomogeneities in the magnetic field have on the relationship between the "macroscopically-averaged microscopic magnetic fields" relative to the the energy content in the microscopic fields. In this view, alignment of magnetic dipoles would reduce the microscopic inhomogeneities and result in a real reduction of energy in terms of (1/2)B20 inside the magnet, even as it increases (1/2)B20 outside the magnet. In this way, the energy density of the magnetic field in a vacuum would be seen in this view as valid even inside materials, so as long the inhomogeneities were taken into account, but due to the obvious difficulty in treating inhomogeneous microscopic fields mathematically, we resort to ourselves to the idea H and M fields.

I've long suspected that the demarcation between H fields and M fields was somewhat arbitrary (as confirmed by the existence of different models for magnetization) and defining them such that they have a divergence was a sort of mathematical trick that allows us to ignore the uneven nature of the actual B field at the microscopic level when trying to compute the density of something like the potential energy of a permanent magnet subject to a uniform magnetic field. For example, if some small fraction of the total magnetic flux within the permanent magnet were "open" in the sense that they connected to the uniform external magnet field, the variation of the orientation of the magnet to the external field would result in a reduction of energy via a magnetic tension force. Am I understanding right that the magnetic fields are ultimately reducible to B fields if full microscopic treatment could be given and in such that its energy density would be given by (1/2)B20?

9. Nov 20, 2016

An important concept here is the energy of a magnetic moment in a magnetic field is $E=-m \cdot B$ (dot product), because the torque on the magnetic moment is $\tau=m \times B$. $(E=\int \tau \, d \theta )$ $\\$ For very refined calculations, sub-microscopic variations in the magnetic field $B$ would need to be considered, but those are details that only some very advanced applications would require. The magnetism subject is complicated by small magnetic domains in many materials, along with what is a strong nearest neighbor interaction between adjacent spins, making it "highly" energetically favorable for them to point in the same direction due to the exchange effect. To first order, I think the exchange effect simply causes the spins to cluster together, making the unit of magnetism much larger than a single electron spin. This offers an explanation for the very high Curie temperatures found in many ferromagnetic materials. $\\$ Meanwhile, the surface current approach to magnetism explains a permanent magnet in considerable detail, with a model that assumes macroscopic $B$ fields with little microscopic variation to them.

10. Nov 20, 2016

### particlezoo

What happens if B is not uniform, and yet static, such that when m is travelling some particular subpath, it "sees" a rotating magnetic field? Shouldn't there be another energy term beside the dot product then?

11. Nov 20, 2016

The $m$ here is oftentimes the electron spin (as opposed to orbital angular momentum). For reasons that I only partially understand $\mu_s=g_s \mu_B S/ \hbar$, where $g_s=2$ approximately, and where $\mu_B=e \hbar /(2mc)$ cgs units. $\$ $E=-\mu_s \cdot B$. $\\$ The atomic details of things such as the spin-orbit interaction aren't completely necessary to get a reasonably good understanding of the magnetism of a permanent magnet based on surface currents. $\\$ For a very interesting derivation of the surface currents, see also Griffith's E&M textbook, where he derives the vector potential $A$ for an arbitrary distribution of microscopic dipoles and shows/proves that the net effect is a bulk term $J_m= \nabla \times M/\mu_o$ plus a surface current per unit length $K_m=M \times \hat{n}/\mu_o$

12. Nov 21, 2016

Just an additional comment on the last post: The intrinsic spin with its $g_s=2$ arises from solutions of the Dirac equation, but the result can be easily included into a non-relativistic quantum mechanical formulation where the electron states simply pick up a perturbative term of $E=-\mu_s \cdot B$ in a magnetic field. $\\$ In a very simplistic model of ferromagnetism, the (uniform) magnetization generates surface currents which generates a magnetic field $B$ which acts on the magnetic moments to maintain the magnetization in a self-consistent manner. For the linear case, one can write $M=\chi B$, but ultimately the magnetization $M$ will saturate and the characteristic curve of $M$ vs. $B$ will level off. $\\$ In the hysteresis curves (of $M$ vs. $H$) of the material in a solenoid, the $H$ is from the solenoid current. What these curves fail to show is that the surface currents are also contributing to the internal magnetic field and the magnetization of the material. (For a ferromagnet, the surface currents are typically 100 to 1000 times larger than any solenoid current.). As shown in one of the later sections in the paper whose "link" I supplied in post #4, the peculiar shape of the $M$ vs. $H$ hysteresis curve is actually explained in its entirety if the $B$ fields from the surface currents are included. The material actually gets a magnetization $M$ that is a well-behaved function of the (internal) magnetic field $B$. THe magnetic field $B$ comes from both the solenoid currents and the surface currents. Using the equation $B=\mu_o H+M$, and overlaying this line on the well-behaved $M$ vs. $B$ curve, one can generate the $M$ vs. $H$ hysteresis curve by allowing $H$ to vary in the equation of the line $B=\mu_o H+M$. $\\$ Once the magntization and the surface currents become established in a permanent magnet, it requires a solenoid current (i.e. current per unit length) that is slightly greater and in the opposite direction to reverse the direction of magnetization, in which case you then have a permanent magnet in the opposite direction. This is basically what the hysteresis curves of $M$ vs. $H$ show in a graphical form. $\\$ For a high quality permanent magnet, the surface currents are normally so large that it requires a very hefty solenoid current with many turns per unit length to reverse its magnetization. Reversing the magnetization is much more readily performed by heating the magnet above the Curie temperature to demagnetize it and then supplying a weaker magnetic field in the other direction to magnetize it as it cools.