Small question about maxwell's equation curl of H

[Tom 4 billion tom]

The first page of this short pdf from MIT sums the starting point to formulate my question:
https://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture29.pdf

We can see that
∇xH = J_free
and
∇xB =μ_o (J_free +∇xM)
∇xB =μ_o (J_free +J_B)

And now my question is, if K_B ≠ 0 , how I can't see its contribution to B in the last equation?
I have been solving problems in class and in them it appears a contribution to B , after using the curl theorem, like this
∫K_B·dl

Thanks for your time

 

[jtbell]

I assume that by $\vec K_B$ you mean the bound surface currents on a magnetized material. These are a subset of the bound current density $\vec J_B$, namely the bound currents that lie on the surface of the material (idealized as an infinitesimally thin sheet) rather than being distributed through the volume of its interior.

Mathematically I think you might incorporate $\vec K_B$ into the last equation by multiplying it by a Dirac delta function that represents the surface, thereby converting it to a volume current density.

 

[Charles Link]

The evaluation of $\nabla \times \vec{M}$ includes surface currents, and, as by Stokes theorem, the discontinuity in $\vec{M}$ at the boundary results in a surface current per unit length $\vec{K}_m=\vec{M} \times \hat{n}$, that comes from $\nabla \times \vec{M}=\vec{J}_m$. $\\$ Basically, this derivation, (of $\nabla \times \vec{H}=\vec{J}_{free}$), starts with $\vec{B}=\mu_o (\vec{H}+\vec{M})$, and you take the curl of both sides. The derivation of the equation $\vec{B}=\mu_o (\vec{H}+\vec{M})$ can be somewhat lengthy, and the derivation is often omitted in many E&M textbooks. They like to use the analogous electrostatic equation $\vec{D}=\epsilon_o \vec{E} +\vec{P}$, to justify it.

 

[Tom 4 billion tom]

Thank's both, and sorry for replying late. I thing I understood the general idea of your replies but for the moment I am going to look up in the bibliography the said discontinuity at the boundary.

 

[Charles Link]

You will see a similar thing in electrostatics with $-\nabla \cdot \vec{P}=\rho_p$, and $\nabla \cdot {E}=\frac{\rho_{total}}{\epsilon_o}$, where $\rho_{total}=\rho_{free}+\rho_p$ The equation $\nabla \cdot \vec{D}=\rho_{free}$ is a result of using the definition $\vec{D}=\epsilon_o \vec{E}+\vec{P}$. Upon taking the divergence of both sides of the equation, the result $\nabla \cdot \vec{D}=\rho_{free}$ follows. There is a surface polarization charge density $\sigma_p=\vec{P} \cdot \hat{n}$, but this is all part of $-\nabla \cdot {P}=\rho_p$ by applying Gauss' law to the discontinuity in $\vec{P}$. $\\$ And it may also interest you that there is a "pole" model of magnetostatics, analogous to the electrostatic "pole" method, that works with magnetic "pole" density $\rho_m=-\nabla \cdot \vec{M}$, where the magnetic "poles" are sources of $\vec{H}$ using the inverse square law, and magnetic surface currents $\vec{K}_m$ are ignored. The "pole" method also uses the equation $\vec{B}=\mu_o (\vec{H}+\vec{M} )$. Very surprisingly, both methods get the exact same answer for the magnetic field $\vec{B}$. In this "pole" model, free currents in conductors are also considered to be sources of $\vec{H}$, with the $\vec{H}$ determined by a Biot-Savart type equation, but with $\vec{H}=\frac{\vec{B}}{\mu_o}$.