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The conservation of electric charge, deduced from the divergence of the Maxwell-Ampère equation, takes the form: $$ \nabla \cdot \textbf{J} = -\frac{\partial \rho_e}{\partial t}.$$

For low frequency application, the quasi-magnetostatic approximation leads us to: $$\nabla \times \textbf{B} = \mu_0\textbf{J} \qquad\text{and}\qquad \nabla \cdot \textbf{J} = 0.$$

This relationship is typical of highly conductive media in which the phenomenon of electric charge relaxation appears. This can be illustrated by taking the divergence of Ohm's law [$\textbf{J} = \sigma \left(\textbf{E}+\textbf{u}\times\textbf{B}\right)$] associated with the Maxwell-Gauss law: $$\frac{\partial\rho_e}{\partial t}+\frac{\sigma}{\varepsilon_0}\rho_e+\sigma\nabla\cdot(\textbf{u}\times\textbf{B})=0.$$

If we first assume a static situation where $\textbf{u}= \textbf{0}$, the previous equation is simplified and the solution is: $$\rho_e(t) = \rho_e(0)\exp(-t/\tau),$$

where $\tau = \varepsilon_0/\sigma$ is a characteristic charge disappearance time and is about $10^{-18}$ s in metallic media. The charge density is therefore approximately zero in stationary conductors after a short transient regime. In the case of a moving conductor however, and by neglecting the first term of the previous equation from the previous analysis, the charge density is given by: $$\rho_e = -\varepsilon_0\nabla\cdot(\textbf{u}\times\textbf{B}).$$ It seems therefore possible to maintain a volumetric charge density in a moving conductor.

My questions are the following : If we take a moving conductive fluid and manage to produce a diverging $(\textbf{u}\times\textbf{B})$ field, can we really sustain a net volumetric charge density in the bulk (and not only a surface charge density)? If so, has it been observed experimentally? Is it measurable? If $\textbf{B}$ is oscillating, does it mean that new currents appear?

The all argument is taken from:

[1] P.A. Davidson. Introduction to magnetohydrodynamics, 2nd edition. Cambridge Texts in Applied Mathematics (2017) and

[2] J.A. Shercliff. A textbook of magnetohydrodynamics. Pergamon Press (1965).

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Yes, the simplest example is a solid conductive disk rotating in external magnetic field, with lines of force perpendicular to the disk. Charge of one sign concentrates on the rim, charge of the opposite sign is distributed in the disk (with uniform spatial density). A disk made of liquid would behave the same - as long as it is conductive and the flow in it is not too mixing (turbulent), the charge separation effect will be present.

So the effect is really due to motion of conductor in external magnetic field ($\mathbf u $ is velocity of the conductor element), whether the conductor is liquid or not does not matter.

In the inertial frame co-moving with a disk element, there is an "induced" electric field due to the magnet $\mathbf E_m'$ that has value $\mathbf E_m' = \mathbf u\times\mathbf B$, this follows from the rules of how fields transform in special relativity. This electric field pushes on the free charge in the conductor; in case of $\mathbf B$ parallel to $\boldsymbol{\omega}$, it pushes positive charge out towards the rim. When the disk is spinned up, or magnetic field is increased from zero, it thus creates a short-lived current that redistributes charge in the disk. Due to this redistribution, additional electric field $\mathbf E_{d}'$ appears in the same frame, which is due to charge in the disk: it is the sum/integral of all the Coulomb fields of the charges in the disk and its rim. Very quickly, an equilibrium charge distribution and its electric field $\mathbf E_{d}'$ is established, where in the disk, $\mathbf E_{d}' = -\mathbf E_m' = -\mathbf u\times\mathbf B$, so total force is zero (ignoring the centripetal force needed to make the charge go in circles, which is negligible for achievable angular velocities).

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  • $\begingroup$ Thank you very much. Your comment led me to the following paper that deepen your answer : Lorrain, P. (1990). Electrostatic charges in v*B fields: The Faraday disk and the rotating sphere. European Journal of Physics. 11. 94. $\endgroup$
    – Antoine
    Sep 12, 2023 at 20:27
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    $\begingroup$ @Antoine Oh my, that paper has some bizarre statements and uses the symbol $\mathbf J$ incoherently. From (1) to (5) $\mathbf J$ means conduction current only, without the advection due to rotating charge density. Then in (6) and (8), $\mathbf J$ has to be total current density, including the advection component $\rho \mathbf v$. Then in (9) to (17), he switches back to $\mathbf J$ being conduction current without $\rho \mathbf v$. What a mess. And (14) for the conduction component of current can be derived much more easily in the rotating frame of the disk. $\endgroup$ Sep 13, 2023 at 0:40
  • $\begingroup$ Thank you I will check other references. Concerning the charge density, if it is indeed present, do you know how is the phenomenon explained in the rotating frame of a conductive disk ? $\endgroup$
    – Antoine
    Sep 13, 2023 at 13:21
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    $\begingroup$ Correct evaluation of divergence of $\mathbf E'$ in any co-moving inertial frame should be like this: $\nabla'\cdot\mathbf E' = \nabla'\cdot(\mathbf E + \mathbf u_p\times\mathbf B) = \nabla'\cdot\mathbf E + \nabla'\cdot (\mathbf u_p\times\mathbf B) = \nabla'\cdot\mathbf E + \mathbf B\cdot (\nabla'\times\mathbf u_p) - \mathbf u_p \cdot (\nabla' \times \mathbf B)$. $\endgroup$ Sep 14, 2023 at 14:36
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    $\begingroup$ Since $\mathbf u_p$ does not depend on position in single co-moving inertial frame, in uniform magnetic field the magnetic term does not contribute at all, and we have $\nabla'\cdot\mathbf E' = \nabla'\cdot\mathbf E$. $\endgroup$ Sep 14, 2023 at 14:37

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