On the definition of magnetic field intensity

Question

In magnetostatics theory, the magnetic field of any magnetic body with magnetization $\mathbf M$ is equivalent to that of a volume current density $\mathbf J_{m,v}$ and a surface current density $\mathbf J_{m,s}$ with the following relations:

$$\mathbf J_{m,v} = \nabla \times \mathbf M \\ \mathbf J_{m,s} = \mathbf M \times \mathbf n$$

In defining the magnetic field intensity (aka, auxiliary magnetic field) $\mathbf H$ it is assumed that only the volume magnetization current density $\mathbf J_{m,v}$ contributes to the free current $\mathbf J_f$: (See D.K Cheng, Field and Wave Electromagnetics)

$$\nabla \times \frac{\mathbf B}{\mu_0} = \mathbf J_f + \nabla \times \mathbf M \\ \nabla \times \left(\frac{\mathbf B}{\mu_0} - \mathbf M \right) = \mathbf J_f $$

However all the three sources $\mathbf J_{f}$, $\mathbf J_{m,v}$ and $\mathbf J_{m,s}$ should contribute to the magnetic field. I cannot see how the surface current is included.

Answer 1

The surface magnetization current is accounted for by the magnetization current term $\nabla \times \mathbf M$. On a boundary between the medium and vacuum, the term $\nabla \times \mathbf M$ is singular; when integrated over an arbitrarily small line normal to the boundary passing through the boundary, we get value $\mathbf M\times \mathbf n$, which is thus the surface current per unit length on the boundary.