Magnetization and magnet poles

[FEAnalyst]

TL;DR

Is magnetization different for each pole of a magnet?

Hi,

electromagnetism is definitely not my area of expertise. However, I sometimes get to deal a bit with simulations of electromagnetic problems. One thing that bothers me now and I can't find the answer to it is whether magnetization $M$ is different for north and south pole of a magnet. Most importantly, is there a different sign of magnetization for one of the poles ? I've seen a simulation example where one end (pole) of a horseshoe magnet had a magnetization $M_1=7500 \ A/m$ assigned while the other end had a magnetization $M_2=−7500 \ A/m$ applied. Would it be necessary to do the same for a bar magnet ? If yes, which pole should have positive magnetization and which one should have it negative ?

 

[vanhees71]

There are no magnetic monopoles in classical electrodynamics. Magnetization is by definition the magnetic-dipole density in a medium. If you want to heuristically interpret the dipole moment as if it were made of two monopoles, the north pole would by +, the south pole, -.

 

[Dale]

TL;DR Summary: Is magnetization different for each pole of a magnet?

One thing that bothers me now and I can't find the answer to it is whether magnetization M is different for north and south pole of a magnet.

Magnetization is a (pseudo) vector quantity. The vectors on the poles are roughly normal to the surface. But they are pointed outwards at the North Pole and inwards at the South Pole.

 

[FEAnalyst]

But they are pointed outwards at the North Pole and inwards at the South Pole.

Thanks for the reply. So this would mean positive magnetization for north pole and negative magnetization for south pole or does it depend on the sign convention used in a specific case/software ?

 

[Dale]

So this would mean positive magnetization for north pole and negative magnetization for south pole

Positive and negative don’t make sense for vectors. They have a magnitude and a direction. The magnitude is the same for north and south. The direction is out for north and in for south. Positive and negative simply don’t apply.

 

[FEAnalyst]

Positive and negative don’t make sense for vectors. They have a magnitude and a direction. The magnitude is the same for north and south. The direction is out for north and in for south. Positive and negative simply don’t apply.

I see but those concepts are used in simulations when defining magnetization as a boundary condition or material property. For example, on the page 99 of this document: https://www.nic.funet.fi/index/elmer/doc/ElmerTutorials.pdf

It's defined this way:

Name = IronPlus
MgDyn2d
Relative Permeability = 5000.0
Magnetization 1 = Real 750.0e3

Name = IronMinus
MgDyn2d
Relative Permeability = 5000.0
Magnetization 1 = Real -750.0e3

where IronPlus is assigned to the upper pole of a horseshoe magnet while IronMinus is assigned to the lower pole.

Magnetic field intensity vector symbols in the results of this simulation go from the lower end to the upper end:

So the lower end should be a north pole while the upper one should be a south pole, right? Then the convention is that the north pole has negative magnetization while the south pole has positive magnetization? I just wonder if it's always the case.

 

[Dale]

That doesn’t make any sense to me. It seems like that software is trying to do some sort of a workaround. That isn’t how magnetization works in physics. Maybe that is the best approximation that software can do in some sense.

Edit: it looks like the software is trying to avoid any simulation inside the material of the magnet. Magnetization is only non-zero inside the magnet material itself. That appears to be blank in the software.

 

[FEAnalyst]

Edit: it looks like the software is trying to avoid any simulation inside the material of the magnet. Magnetization is only non-zero inside the magnet material itself. That appears to be blank in the software.

It may look like this based on the image I attached but they commented it:

Note that the fields in the horseshoe itself have been masked away to demonstrate the well known field shape in the free space.

And the previous image shows the vector potential of the magnetic field also inside the magnet:

Their approach to magnetization can be somewhat strange though. I haven't heard about it in the case of other simulation software. Magnetization values are normally not defined there for magnetostatic simulations. I guess I'll have to ask on their forum.

 

[Dale]

Their approach to magnetization can be somewhat strange though

Yes, that seems accurate. I mean the very first sentence on the Wikipedia page is “In classical electromagnetism, magnetization is the vector field that expresses the density of permanent or induced magnetic dipole moments in a magnetic material”

https://en.m.wikipedia.org/wiki/Magnetization

Physically it is a vector field, so I am not sure what they are trying to capture with their approach. Hopefully it is justified in their documentation or support channels.

 

[vanhees71]

The magnetization in classical magnetostatics is equivalent to either a magnetic-dipole density or a magnetization current density. Both approaches lead, of course, to the same final result for $\vec{H}$ and $\vec{B}$. The vector potential is gauge dependent and thus isn't observable.

The two approaches both use, of course, the magnetostatic equations,
$$\vec{\nabla} \times \vec{H}=0, \quad \vec{\nabla} \cdot \vec{B}=0, \quad \vec{B}=\mu_0(\vec{H}+\vec{M}).$$
For a permanent magnet ("hard ferromagnetic material") you take $\vec{M}$ as a given vector field. The strategy for solution of the problem is then either with a scalar potential for $\vec{H}$ or a vector potential for $\vec{B}$:

(a) Scalar potential for $\vec{H}$

You make the ansatz
$$\vec{H}=-\vec{\nabla} \Phi$$.
Then you have
$$\vec{\nabla} \cdot \vec{B}=\mu_0 \vec{\nabla}(-\vec{\nabla} \Phi + \vec{M}) =0 \; \Rightarrow \; -\Delta \Phi=-\vec{\nabla} \cdot \vec{M}=\rho_{\text{m}}$$
with the solution
$$\Phi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho_{\text{m}}(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$

(b) Vector potential for $\vec{B}$

Because $\vec{\nabla} \vec{B}=0$ another approach is to use the vector potential, i.e., write
$$\vec{B}=\nabla \times \vec{A}.$$
Due to gauge invariance one can impose one gauge constraint. In magnetostatics the Coulomb-gauge condition is most conventient,
$$\vec{\nabla} \cdot \vec{A}=0.$$
Now
$$\vec{H}=\frac{1}{\mu_0} \vec{B}-\vec{M}$$
and
$$0=\vec{\nabla} \times \vec{H}=\frac{1}{\mu_0} \vec{\nabla} \times \vec{B} - \vec{\nabla} \times \vec{M},$$
from which
$$\vec{\nabla} \times \vec{B}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=-\Delta \vec{A}=\mu_0 \vec{\nabla} \times \vec{M}.$$
This means the magnetization is equivalent to a current denstiy
$$\vec{j}_{\text{m}}=\vec{\nabla} \times \vec{M}.$$
The solution is
$$\vec{A}(\vec{x})=\int_{\mathbb{R}} \mathrm{d}^3 x' \frac{\mu_0 \vec{j}_{\text{m}}(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$

 

[Dale]

Thanks @vanhees71 this software must be using the scalar potential for H approach.