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I found this form of Ampere's Law: $$ \sigma \frac{\partial \mathbf{A}}{\partial t} + \nabla\times\mathbf{H} = \mathbf{J}_e, $$ where $\sigma$ is the conductivity of a material, $\mathbf{A}$ is the magnetic potential and $\mathbf{J}_e$ is the externally generated current density.

I am trying to figure out how it can be obtained from Maxwell-Ampere's equation: $$ \nabla\times\mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t}. $$ As far as I understand, the current density $\mathbf{J} = \sigma\mathbf{E} + \mathbf{J}_e$ and the displacement field $\mathbf{D} = \varepsilon_0\varepsilon_r \mathbf{E}$. The electric field can be defined as $\mathbf{E} = -\nabla\varphi - \frac{\partial \mathbf{A}}{\partial t}$.

However, when I try to put it all together I am always left with some extra terms. Could someone please help me with that?

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  • $\begingroup$ Please tell us where this came from, rather than just saying "I found..." Did you find it in a book? If so, what book? Did you find it by doing our own calculations? $\endgroup$ – Ben Crowell Jan 26 at 22:48
  • $\begingroup$ @BenCrowell It is used in Comsol Multiphysics simulation software to compute magnetic field and induced current distributions. In particular, to model rotating machinery like generators. For example, it is cited in this tutorial model. $\endgroup$ – Alexander Jan 27 at 15:24
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Where did you find your "Ampere's law". I think it is just incorrect --- unless it's for a superconductor in the Coulomb gauge in which case the supercurrent is given by the London equation $$ {\bf j}= -\frac{n_se^2}{m}{\bf A}. $$ Then you get something like what you cite. In a general gauge the london equation is $$ {\bf j}= \frac{n_s e}{2m}(\nabla\theta -2e {\bf A}) $$ where $\theta$ is the phase of the superconducting order parameter.

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  • $\begingroup$ It is used in Comsol Multiphysics simulation software to compute magnetic field and induced current distributions. In particular, to model rotating machinery like generators. For example, it is cited in this tutorial model. $\endgroup$ – Alexander Jan 27 at 15:26
  • $\begingroup$ @Alexander. Ah I see. They using the $A_0=\phi=0$ gauge without telling you, and also making the slow variation aproximation to omit the $\partial {\bf D}/\partial t$ term. $\endgroup$ – mike stone Jan 27 at 16:41
  • $\begingroup$ Oh, that makes sense. Thank you very much for the clarification! $\endgroup$ – Alexander Feb 4 at 12:21

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