Small question about maxwell's equation curl of H

  • #1
The first page of this short pdf from MIT sums the starting point to formulate my question:
https://ocw.mit.edu/courses/physics...gnetism-fall-2006/lecture-notes/lecture29.pdf

We can see that
∇xH = Jfree
and
∇xB o (Jfree +∇xM)
∇xB o (Jfree +JB)

And now my question is, if KB ≠ 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
∫KB·dl

Thanks for your time
 

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  • #2
jtbell
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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.
 
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  • #3
Charles Link
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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.
 
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  • #4
Thank's both, and sorry for replying late. I thing I understood the general idea of your replies but for the moment Im going to look up in the bibliography the said discontinuity at the boundary.
 
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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} ##.
 
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