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The $\mathbf{H}$ field can be derived from the potential $\psi$:

$$\mathbf{H}=\dfrac{\mu_0}{4 \pi} \int_{V'} \rho \dfrac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dV' + \dfrac{\mu_0}{4 \pi} \oint_{S'} \sigma \dfrac{ \mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dS'=\mathbf{H}^{V}+\mathbf{H}^{S}$$

where $\rho=-\nabla \cdot \mathbf{M}$ and $\sigma=\mathbf{M} \cdot \hat{n}$

$\mathbf{H}^V$ is defined and continuous throughout space. $\mathbf{H}^S$ is defined and continuous everywhere except at boundary $S'$. At the boundary, $\mathbf{H}^S$ is undefined and has a discontinuity of $\mu_0\ \sigma (\hat{n})$. Consequently at the boundary, $\mathbf{H}$ is undefined and has a discontinuity of $\mu_0\ \sigma (\hat{n})$. Consequently at the boundary, $\mathbf{B}=\mathbf{H} + \mu_0 \mathbf{M}$ is undefined and has a discontinuity of $\mu_0\ \sigma (\hat{n})+\mu_0\mathbf{M}$.

How does it make sense to talk about the divergence at a point where the field is undefined? I know that $\nabla \cdot \mathbf{B}=0$ everywhere except boundary $S'$.

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  • $\begingroup$ There are no discontinuities in B-field. div B=0 everywhere. H-field can be discontinuous and have non-zero divergence. $\endgroup$ – Rob Jeffries Apr 27 at 9:59
  • $\begingroup$ Then can you please show how $\mu_0\ \sigma (\hat{n})+\mu_0\mathbf{M}=0$ $\endgroup$ – N.G.Tyson Apr 27 at 10:20
  • $\begingroup$ Because $\nabla \cdot \vec{H} = -\mu_0 \nabla \cdot \vec{M}$ $\endgroup$ – Rob Jeffries Apr 27 at 10:27
  • $\begingroup$ But can you please show how $\nabla \cdot \vec{H}=\mu_0 \nabla \cdot \sigma (\hat{n})$? $\endgroup$ – N.G.Tyson Apr 27 at 10:33
  • $\begingroup$ I do not see how to prove $\mu_0\ \sigma (\hat{n})+\mu_0\mathbf{M}=0$ $\endgroup$ – N.G.Tyson Apr 27 at 10:52

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