B versus H field (and Magnetization)

  • Thread starter gahando
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  • #1
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Hi all,

Recently I learned about magnetization
I understand that you apply a magnetic field B, and that your magnetization M is proportional to the field ... H? Like so,
\mathbf{M}=\chi\mathbf{H}

What is the \mathbf{H} field, though? It is the 'auxiliary' magnetic field, but what does that mean?

Please shed some light on this (probably very trivial) matter, if possible.

Thanks,
gahando

EDIT: My apologies if this question has been asked \inf times
 
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Answers and Replies

  • #2
Meir Achuz
Science Advisor
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It really should be [itex]\mathbf{M}=\chi\mathbf{B}[/itex], but EEs prefer to use
[itex]\mathbf{M}=\chi\mathbf{H}[/itex] because H depends directly on the current, and they can
'see' H on A dial. The engineers outvoted the physicists at conclaves where units were standardized, so now all textbooks and even physicists follow the (unphysical) procedure. That is why chi varies wildly for ferromagnets,
while it would be nearly constant in terms of B.
 
  • #3
Jano L.
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I understand that you apply a magnetic field B, and that your magnetization M is proportional to the field ... H?
The law of proportionality is only approximate. It holds well for non-magnetic materials like air, wood or plastic. For magnets and ferites it is common that the magnetization is not proportional to the field, but follows magnetization curve.

What is the \mathbf{H} field, though?
$$
\mathbf H = \frac{\mathbf B}{\mu_0} - \mathbf M
$$
so that the equation
$$
\nabla \times \mathbf H = \mathbf j_f
$$
holds, where ##\mathbf j_f## is conduction current.


It is the 'auxiliary' magnetic field, but what does that mean?
There is the relation
$$
\mathbf B = \mu_0(\mathbf H + \mathbf M)
$$
which is always valid. Some people regard ##\mathbf H## as less useful than ##\mathbf B## with ##\mathbf M##, hence the name.
 
  • #4
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$$
\mathbf H = \frac{\mathbf B}{\mu_0} - \mathbf M
$$
so that the equation
$$
\nabla \times \mathbf H = \mathbf j_f
$$
holds, where ##\mathbf j_f## is conduction current.

There is the relation
$$
\mathbf B = \mu_0(\mathbf H + \mathbf M)
$$
which is always valid. Some people regard ##\mathbf H## as less useful than ##\mathbf B## with ##\mathbf M##, hence the name.
Right ... I mean, the equations make sense, but so, $$\mathbf B $$ is the applied field and then $$ \mathbf H $$ is this field that comes out of the combined effects of the applied field $$ \mathbf B $$ and the resulting magnetization $$ \mathbf M $$ ... and yet, the relation shows that

$$ \mathbf B - \mu_0 \mathbf M = \mu_0 \mathbf H $$

Is the field $$ \mathbf H $$ supposed to be understood as the field that remains after some of the $$ \mathbf B $$ field energy has gone into magnetizing the material?
 
  • #5
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Don't think about it too much, just learn the equations, there's a lot of convention and not too much content in these auxiliary fields. You have a B field and that leads to a magnetisation M. So M is a material-specific function of B. But to make maxwells macroscopic equations look pretty they invented a new vector field H, and now everyone uses B/H functions.
 
  • #6
Khashishi
Science Advisor
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The ##\mathbf{H}## field is not as fundamental as the ##\mathbf{B}## field, and it's not that important to understand. You could do a lot of research in advanced physics and never find a need for the ##\mathbf{H}## field. There is a similarly defined electric ##\mathbf{D}## field, which doesn't get much use.
 
  • #7
Jano L.
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... $$\mathbf B $$ is the applied field and then $$ \mathbf H $$ is this field that comes out of the combined effects of the applied field $$ \mathbf B $$ and the resulting magnetization $$ \mathbf M $$ ... and yet, the relation shows that

$$ \mathbf B - \mu_0 \mathbf M = \mu_0 \mathbf H $$
The field ##\mathbf B## is not easily controllable hence the adjective "applied" is not very appropriate. What can be applied is external field by some magnet or electromagnet. This field is not total magnetic field ##\mathbf B##, but only the field due to the magnet or electromagnet, and can be designed by ##\mathbf B_0##. The total field is sum of this external field plus the field due to magnetized body we consider.

On the other hand, the fields ##\mathbf B,\mathbf H## are total fields that do not carry the distinction between internal and external. Still, the field ##\mathbf H## can be sometimes directly related to the currents (which are easily controllable), hence its use.

Is the field $$ \mathbf H $$ supposed to be understood as the field that remains after some of the $$ \mathbf B $$ field energy has gone into magnetizing the material?
No, there is no such relation.
 
  • #8
DrDu
Science Advisor
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In principle we could make it without introducing H at all. This would require to work with the microscopic charge density j in the material. However it turns out that the relation between j and B (the material equation) is extremely non-local. If you consider a homogeneously magnetized isolating material, the macroscopic current density will be concentrated at the surface. Obviously, as charge is conserved in a static field div j=0, hence j must be expressible as the rotation of another vector field which we call M, i.e. rot M=j. The relation between M and B turns out to be quite local so that we can conveniently write ## M=\chi B##.
The situation is slightly different beyond magnetostatics, i.e. for finite frequencies as we also have the possibility to express j as ##\dot{P}##, i.e. the time derivative of the electric polarization. This convention is usually used in optics where only the triple P (or equivalently D), E and B is used.
 

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