So the matter is I'm really confused about boundary conditions of magnetic fields. Prior to the boundary conditions, I thought that say you have an external magnetic field, $B_0$, now if I place a paramagnetic material in the field the resultant field inside is $B_0+\mu_0M$ and thus the field inside is greater than that of outside and that's why we are using soft iron cores may be.

But here lies the problem: if the field inside is greater than outside, now I reach to the boundary condition according to which field is continuous say there are no surface currents on the material then $$B_\text{inside} = B_\text{outside}.$$ But earlier we got that field inside is greater than outside.

My speculation (just thoughts):
Is it that the field of magnetisation also lies outside the material and is taken (included) as applied field and that's why we say that total field is continuous while magnetic field intensity $B_0/\mu_0$ is not continuous.

  • $\begingroup$ As far as I remember only the "H-field" $H=B/\mu_0-M$ needs to be continuous on the boundary (in the absence of currents). $\endgroup$
    – NDewolf
    Nov 14 '19 at 10:41
  • $\begingroup$ Maybe there are currents? Maybe a permanent magnet is created with permanent currents? $\endgroup$ Nov 14 '19 at 12:39
  • $\begingroup$ Maybe ${\bf J}= \nabla\times {\bf M}$? $\endgroup$
    – mike stone
    Nov 14 '19 at 12:50

The boundary condition perpendicular to the surface is indeed: $$ B_{in,\perp} = B_{out,\perp}$$

The magnetic induction is defined as $\mathbf{B} = \mu_0 (\mathbf{H} + \mathbf{M})$. Outside the material, this becomes $\mathbf{B}_{out} = \mu_0 (\mathbf{H}_{ext}+ \mathbf{H}_{s})$. Here, $\mathbf{H}_{s}$ is the stray field generated by the magnet. Inside the material, this becomes $\mathbf{B}_{in} = \mu_0 (\mathbf{H}_{ext} +\mathbf{H}_{d} + \mathbf{M})$. Here, $\mathbf{H}_{d}$ is the demagnetization field which is generated by the magnetization. Perpendicular to the surface, one finds (this is actually the derivation of the boundary condition):

$$ \nabla \cdot \mathbf{B} = 0 \Rightarrow H_{s,\perp}+ H_{d,\perp} -M_{\perp} = 0$$

Using this, and the fact that the demagnetization field always opposes the magnetization, one can show:

$$\mathbf{B}_{in,\perp} = \mu_0 (H_{ext,\perp} - H_{d,\perp} + M_{\perp}) = \mu_0 (H_{ext,\perp} + H_{s,\perp}) = \mathbf{B}_{out,\perp} $$

So, conceptually, the stray field $H_s$ outside the magnet (and also generated by the magnet) perfectly compensates the field inside the magnet $H_d$ and the magnetization $M$.


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