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I know Ampere's Law can be described like:

enter image description here

I am aware of how to use this law; however, as I was reading through Maxwell's equations I came across magnetization current density and polarization current density. The above law seems to be defined in terms of the electric current density vector, I was wondering how this law would be different if it were to involve magnetization current density or polarization current density vectors?

Can someone please help explain, I really want to ensure a good understanding of this equation and would appreciate the help.

Thank you and sorry for any formatting issues/missing tags, please feel free to edit as needed.

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In matter (not in vaccuum, as written in your question) the "Ampere-Maxwell" law is $$\nabla \times \mathbf H = \mathbf J + \frac {\partial \mathbf P}{\partial t} + \epsilon_0 \frac {\partial \mathbf E}{\partial t}.$$ Here $\mathbf P$ is the dipole moment density of the material (it is obvioulsy zero in vacuum) and satisfies the "Gauss-Maxwell" equation $\nabla \cdot \mathbf P + \epsilon_0 \nabla \cdot \mathbf E = \rho $ that are conventionally written with the for the field vector $\mathbf D = \mathbf P + \epsilon_0 \mathbf E$ as $$\nabla \times \mathbf H = \mathbf J + \frac {\partial \mathbf D}{\partial t}.$$ $$\nabla \cdot \mathbf D = \rho .$$ In the Ampere-Maxwell law, $ \frac {\partial \mathbf P}{\partial t}$ is the polarization current and $\epsilon_0 \nabla \cdot \mathbf E $ is the displacement current. Regarding "magnetization current", are you talking about this modification of Faraday's law? If not then what do you mean by it?

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  • $\begingroup$ Thanks for the answer, yes, with "magnetization current" I am talking about the modification of Faraday's law which uses magnetic current density I believe. Could this equation be used with magnetic current density, if so I was wondering how it would change? $\endgroup$
    – harry
    Dec 14, 2021 at 17:00
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    $\begingroup$ If you introduce magnetic currents in the Faraday law to modify $\nabla \times E$ that does not change the Ampere-Maxwell law for $\nabla \times H$ but you must modify the $\nabla \cdot B=0$ equation to \nabla \cdot B = \rho_m$ with $\nabla \cdot J _m = -\frac {\rho_m } {\partial t}$ for $B=\mu_0 (M+H)$ $\endgroup$
    – hyportnex
    Dec 14, 2021 at 17:06
  • $\begingroup$ Thank you, got it. $\endgroup$
    – harry
    Dec 14, 2021 at 20:55

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