2

The magnetic field strength, H looks very different inside a bar magnet as shown below enter image description here

I have two main questions regarding this,

  1. why can we regard ‘magnetisation poles’ as the sources of H. When I look at the equations ($H=\frac{B_0}{\mu_0} - M$) I just can’t see where it’s coming from?

  2. why do the field lines in the centre become more spaced (I.e. the field gets weaker) and thus isn’t uniform like B?

Qmechanic
  • 201,751
Jake
  • 503

2 Answers2

1

The two equations of Maxwell (Ampere) when specialized for static magnetic fields are $$\textrm{div}\textbf{B}=0 \\ \textbf{curl} \textbf{H}=0$$ The relationship between the two kinds of magnetic vector fields is $$\textbf{B}=\mu_0(\textbf{H}+\textbf{M}).$$ Here $\textbf{M}$ is the magnetic dipole moment density of the ponderable matter in which the two fields exist simultaneously. Of course, $\mu_0$ is just a dimensional conversion scaling factor without physical significance. In empty space (vacuum) $\textbf{M}=0$ and the two fields B and H are essentially the same just being measured in different physical units although some physicists would also argue over this.

Now consider Helmholtz's decomposition theorem, see 1, and write $\textbf{M}$ as the sum of a lamellar field and of a solenoidal field, and assume that the integration is carried over all space and that the dipole density is zero at infinity, then:

$$\textbf{M}= -\textrm{grad} \Phi + \textbf{curl}\textbf{A}$$ where the $\Phi$ is the scalar and $\textbf{A}$ is the vector potential: $$ \begin{align} \Phi(\mathbf{r}) & =\frac 1 {4\pi} \int_V \frac{\textrm{div} '\cdot\mathbf{M} (\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} \, \mathrm{d}V' \\[8pt] \mathbf{A}(\mathbf{r}) & =\frac 1 {4\pi} \int_V \frac{\textbf{curl}'\mathbf{M}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} \, \mathrm{d}V' \end{align} $$ This is an identity for any "wellbehaving" smooth vector field.

Now we identify the terms as $$\textbf{H}= \textrm{grad} \Phi \\ \textbf{B}= \mu_0\textbf{curl}\textbf{A}$$ therefore upon substitution

$$ \begin{align} \textbf{M}= -\frac 1 {4\pi} \textrm{grad} \int_V \frac{\textrm{div} '\mathbf{M} (\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} \, \mathrm{d}V' + \frac 1 {4\pi} \textbf{curl} \int_V \frac{\textbf{curl}'\mathbf{M}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} \, \mathrm{d}V' \end{align} $$ And here we can see explicitly that in the presence of polarized matter the source of the H-field is the divergence of M, while the source of the B-field is the curl of M.

When looking at a finite piece magnet where M discontinuously changes to zero outside the magnet neither the div nor the curl exist and the integrands get replaced by terms $$\textrm{div}'\mathbf{M}\mathrm{d}V' \rightarrow -\textbf{n}\cdot \textbf{M}' \mathrm{d}S'\\ \textbf{curl}'\mathbf{M}\mathrm{d}V' \rightarrow \textbf{n}\times \textbf{M}' \mathrm{d}S'\\ $$

The interesting term here is $\mathrm{d} q_m'=-\textbf{n}\cdot \textbf{M}' \mathrm{d}S'$ representing an infinitesimal surface charge $\mathrm{d}q_m'$ over an infinitesimal area $\mathrm{d}S'$, where $\textbf{n}$ is the outward normal of the surface of the magnetic material, and thus we have a magnetic surface charge density $\sigma'=-\textbf{n} \cdot \textbf{M}'$ caused by the discontinuity in M. Its field (the gradient term in the decomposition) of this magnetic surface charge is actually in the opposite direction of the "curl" term, so it is subtracted from it.

hyportnex
  • 18,748
  • Ahh I see, this is a little beyond what my course was doing but I think I understand the main points – Jake Jan 06 '19 at 18:37
  • if I may give an advice for your future studies of magnetism in ponderable matter then do not think of H as being secondary and B being primary, which is the 20th century fashion (in the 19th century the opposite was prevalent) but rather think of them as being equally important. One may even go so far as to claim that inside a magnet while their mathematical meaning is clear there is no direct physical meaning to be assigned to either of them unless there is a clear experimental demonstration is also associated. – hyportnex Jan 06 '19 at 19:34
  • For example in a homogeneous toroid with a tight winding around it we know what H is the same as the flux density without the core, but if we introduce a gap this is not true anymore. The difference between the two cases we can explain with surface charges as above but that is only an interpretation of the mathematical development. There is no ambiguity of interpretation outside the magnet, of course. – hyportnex Jan 06 '19 at 19:37
1

the things were explained in great detail in the previous answer. To be able to put it simply, one can note that the Maxwell equation in the absence of free currents is written rot (H) = 0 and div (B) = 0 which gives div (H) = -div ( M) We see that they are the same equations as those of electrostatics with a density of "magnetic charge" fictive -div (M). For uniform magnetization, -div (M) gives a surface density -M on the right and + M on the left in the figure. As a result, the field H in the figure is the same as the electrostatic field between the plates of two uniformely charged plates (wit the presence of edge effects if th plates are not infinite). If the plates are "small", the field is dipolar. Maybe it can help to understand its form without calculation. It is also understood that it is "demagnetizing" because it is directed in the opposite direction to M. (Sorry, I am new to this forum and I do not master the equation editor and English is not my native language).

Vincent Fraticelli
  • 4,841