When the boundary condition is taken into account, why doesn't the perpendicular component of the magnetic field interact with the material?

Question

When the boundary conditions in Maxwell's equations are examined, I cannot understand why the vertical component of the magnetic field vector does not have the permeability factor as in the horizontal component.

We know that

$$H_{\parallel}^{1}-H_{\parallel}^{2}=j_{s} \tag{1}$$

As for the perpendicular parts,

$$B_{\perp}^{1}-B_{\perp}^{2}=0 \tag{2}$$

This is where I came to the conclusion. And this is where I am confused. So I thought that the perpendicular component of the magnetic field never interacts with the material whatsoever. But I'm not sure about that because the magnetic field oscillates sinusoidally just like the electric field. In this case, it is analogous to the electric field. So, I could not understand why the material does not have a magnetic response for the vertical component.

Yes, mathematically speaking there is no magnetic monopole, so the divergence of B is zero. We can easily understand this from the surface integral because the magnetic field always appears in the form of a close path.

$$\int_V dV \ \nabla \cdot \mathbf{B} = \int_{\partial V} \mathbf{B} \cdot d\mathbf{S}$$

where $\partial V$ is the surface area surrounding the volume $V$. The magnetic field has zero divergences, which means that

$$\int_{\partial V} \mathbf{B} \cdot d\mathbf{S}= 0$$

At this point, I have some information, but I still can't think of a deep understanding.

Answer 1

Apply your last equation to a cylinder whose endcaps are parallel to the interface and at a distance of $\epsilon$ from the plane of the interface. The flux through the sides will be negligible (in the limit as $\epsilon \to 0$) and the flux through the endcaps will be $+ B^1_\perp A$ for one endcap (where $A$ is the area of an endcap) and $-B^2_\perp A$ for the other. Your last equation then implies $B^1_\perp = B^2_\perp$.