# Doubt on why magnetic flux density is solenoidal

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 $$S'$$.

• There are no discontinuities in B-field. div B=0 everywhere. H-field can be discontinuous and have non-zero divergence. – Rob Jeffries Apr 27 '19 at 9:59
• Then can you please show how $\mu_0\ \sigma (\hat{n})+\mu_0\mathbf{M}=0$ – N.G.Tyson Apr 27 '19 at 10:20
• Because $\nabla \cdot \vec{H} = -\mu_0 \nabla \cdot \vec{M}$ – Rob Jeffries Apr 27 '19 at 10:27
• But can you please show how $\nabla \cdot \vec{H}=\mu_0 \nabla \cdot \sigma (\hat{n})$? – N.G.Tyson Apr 27 '19 at 10:33
• I do not see how to prove $\mu_0\ \sigma (\hat{n})+\mu_0\mathbf{M}=0$ – N.G.Tyson Apr 27 '19 at 10:52