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From Maxwell's equations, why is the displacement current viewed as a source for a magnetic field? If the displacement current were moved to the other side of the equation, it would be like a current density gives rise to both a magnetic field and a time-varying electric field. So why is the former interpretation preferred over the latter?

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In fact, it is commonplace (especially in relativistic electrodynamics problems) to move the time-dependent terms in the Ampere-Maxwell Law (and similarly in Faraday's Law) to the left-hand side of the equation, yielding (in Gaussian units) $$\vec{\nabla}\times\vec{E}+\frac{1}{c}\frac{\partial\vec{B}}{\partial t}=0 \\ \vec{\nabla}\times\vec{B}-\frac{1}{c}\frac{\partial\vec{E}}{\partial t}=\vec{J}. $$ Writing the equations this way puts the fields entirely on the left and the sources entirely on the right. In this way, it is possible to see the electric and magnetic fields being generated solely by the charge and current sources on the right, and this was the viewpoint taken by, for example, Jefimenko in his derivation of the $\vec{E}$ and $\vec{B}$ fields entirely in terms of the (retarded) sources $\rho$ and $\vec{J}$.

So it is perfectly reasonable to treat $\vec{J}$ as a source for both the curl of $\vec{B}$ and the time derivative of $\vec{E}$, as suggested in the question. However, there are also reasons why it is often useful to treat $\partial\vec{E}/\partial t$ and $\partial\vec{B}/\partial t$ as sources on the left-hand sides of the equations in which they appear. ("A changing magnetic field produces an electric field," and vice versa.) The first reason is a purely practical one—that the time derivatives may be under direct control in an experiment. In Faraday's (and Henry's) original experimental work on induction, the magnetic field was directly under the control of the experimenter; the field could be changed either by moving a magnet or changing the current in an electromagnet. The result was that an electric field (in the form of an electromotive force) was observed. So it was natural to think of the changing magnetic field generating the electric field. The same kind of thing was true when, decades later, Hertz first observed the magnetic field inside a charging capacitor. In that case, the changing electric field (i.e. displacement current) was under direct control, and the induced magnetic field was being observed.

The second important reason for placing the time derivative terms on the right-hand side of the equations (as sources) is a mathematical one. According to the Helmholz Theorem, a (smooth) vector field $\vec{W}$ in three-dimensional space is completely determined by three things:

  • $\vec{\nabla}\cdot\vec{W}$ at all points in space,
  • $\vec{\nabla}\times\vec{W}$ at all points in space, and
  • the boundary conditions on $\vec{W}$ at $r\rightarrow\infty$.

This means that equations that give the curls and divergences of $\vec{E}$ and $\vec{B}$ are (along with suitable boundary conditions) guaranteed to determine the electric and magnetic fields uniquely. Thus it often makes sense to place Maxwell's equations in this form, with the time-dependent terms moved to the right and considered as sources.

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Displacement current can be thought of as the medium's polarization changing. The movement of the bound charges is in itself a current and thus generates a magnetic field. This works even in vacuums and can be applied to magnetic fields too.

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  • $\begingroup$ This does not answer the question: can we see current as contributing to the time derivative of E. $\endgroup$
    – my2cts
    Commented Jan 1, 2020 at 23:41
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A current density does not affect the time derivative of E. It would be misleading to rearrange the terms to suggest this, but technically it would make no difference.

Out of scope of this question, I like this site because it brings back to my attention issues that came up in first year physics but I since did not follow through since, like "what is the displacement current" ?

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The term has the dimension of the current density and may be non zero in absence of electric current. It completes the possible sources of the magnetic field.

Since the electric and magnetic time-dependent fields are coupled, we can write two equations: one of them for the second time drivative of $\mathbf{E}$ containing the current time derivative on the right-hand side ("source side").

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  • $\begingroup$ Can you elaborate a little? $\endgroup$
    – Paul
    Commented Feb 27, 2015 at 15:10
  • $\begingroup$ Well, the "displacement current" determines the same quantity - $\nabla\times\mathbf{B}$ - as the current density $\mathbf{j}$ does, so its plase is next to $\mathbf{j}$. $\endgroup$ Commented Feb 27, 2015 at 15:23
  • $\begingroup$ This does not answer the question: can we see current as contributing to the time derivative of E. $\endgroup$
    – my2cts
    Commented Jan 1, 2020 at 23:40
  • $\begingroup$ Then this should be your answer. $\endgroup$
    – my2cts
    Commented Jan 2, 2020 at 11:11

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