# Is it possible to sustain an net electric charge density in a moving conductor? (MHD application)

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).

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).
• @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. Sep 13, 2023 at 0:40
• Correct evaluation of divergence of $\mathbf E'$ in any co-moving inertial frame should be like this: $\nabla'\cdot\mathbf E' = \nabla'\cdot(\mathbf E + \mathbf u_p\times\mathbf B) = \nabla'\cdot\mathbf E + \nabla'\cdot (\mathbf u_p\times\mathbf B) = \nabla'\cdot\mathbf E + \mathbf B\cdot (\nabla'\times\mathbf u_p) - \mathbf u_p \cdot (\nabla' \times \mathbf B)$. Sep 14, 2023 at 14:36
• 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 $\nabla'\cdot\mathbf E' = \nabla'\cdot\mathbf E$. Sep 14, 2023 at 14:37