Let's say free electrons are contained in a small cloud because of electrostatic forces confining them. Now, if those walls vanish, the cloud will expand very fast because of the coulomb force inducing repulsion between alike charges.

I suppose there is now a current in the shape of a radially expanding flux, centered around the center of mass of the electron cloud. If one checks a small area of the sphere surrounding the cloud before its expansion, once the cloud starts expanding, a number of negative charges will move through it, so we have coulombs per second, i.e. amperes.

What does that mean for the magnetic field created by this varying electric field? Since the magnetic field is encircling the axis of the current, i.e. it is perpendicular to its direction, I would expect the ampere effect, i.e. that parallel wires with same sign current attracts each other. But that would seem to mean the lines of the electrons moving away from the center will contract and make the sphere collapse.

What is the real consequence of such a setting? A slowing down, like an inductor resists current change? Possibly a rebound or even an oscillating cloud?

EDIT: Someone explains to me how a question already assuming the correct answer is a duplicate. I understand it answers my question yes, but not as a duplicate.

  • $\begingroup$ I don't quite understand why "parallel wires with same sign current attracts each other" will lead to "the lines of the electrons moving away from the center will contract and make the sphere collapse". Would you please elaborate? $\endgroup$
    – verdelite
    Sep 30 '19 at 13:17
  • $\begingroup$ Since electric force is orders of magnitude stronger than magnetic force, I think in your case the magnetic force may not play a big role. $\endgroup$
    – verdelite
    Sep 30 '19 at 13:17
  • $\begingroup$ @verdelite: If you make all radial lines attached to a common center closer to each other, you are forced to reduce the volume because you increase density, simple as that. $\endgroup$
    – Winston
    Sep 30 '19 at 13:24
  • $\begingroup$ You are pointing to a question that already assumes the magnetic field is null. How can this be a duplicate? $\endgroup$
    – Winston
    Sep 30 '19 at 17:43
  • $\begingroup$ I am having a hard time visualizing all of this. Would a diagram help here? $\endgroup$ Sep 30 '19 at 23:15

Because you haven't given us any information about the initial distribution of the charge within the cloud, I will assume here that the distribution is uniform (specifically, $\rho(r)=\rho_0$ for $r<r_0$ and $\rho(r)=0$ otherwise). Importantly, that means that the problem has spherical symmetry.

Since the cloud is spherically-symmetric, the electric field must always be directed radially (since a non-radial electric field would look different under rotation, even though the underlying charge distribution would look exactly the same; alternatively, since every tangential contribution to the electric field from one side of the sphere is exactly canceled by an equal contribution from the other side).

Since the electric field is always radially directed, the charges in the cloud all move radially outward, and the current density is therefore also pointing entirely in the radial direction.

Since the current density and the electric field are both entirely directed radially outward, the distribution even after expanding still has spherical symmetry. This means that the magnetic field must also be directed radially, because any other field configuration would lead to the magnetic field looking different under rotation while everything that generates it looks exactly the same.

Now, we use Gauss's Law for Magnetism:

$$\iint\mathbf{B}\cdot d\mathbf{A}=0$$

where the integral is taken over a closed surface. We choose a sphere about the origin, of any arbitrary radius. Since the problem is spherically symmetric, the magnitude of $\mathbf{B}$ is constant on any sphere about the origin, and, since it's directed radially, it's always parallel/antiparallel (doesn't matter which) to the directed area vector $d\mathbf{A}$. This means that

$$\iint\mathbf{B}\cdot d\mathbf{A}=(-)^?BA=(-)^?B(4\pi r^2)=0$$

where $(-)^?$ refers to the possible presence of a negative sign. Since $r$ is not necessarily zero, it must be true that $B$ is zero, so the magnetic field is zero everywhere.

Now, if you start with an asymmetric initial distribution, then very, very different (and complicated) things will happen, since you lose all of the nice restrictions on the behavior that spherical symmetry guarantees. The specific behavior is probably best explored using a simulation, since, for an arbitrary initial distribution, you're very likely to get a set of PDEs that have no analytical solution anyway.

  • $\begingroup$ I do indeed believe the answer is not that trivial in the case of a real world setting. The charges are point like and so there is zero chance an electron gas of macroscopic size will be a "uniform charge distribution". Worse, as soon as the Coulomb explosion starts things will get extremely chaotic, every departure from uniformity will be amplified by collisions. I am pretty sure there would be EM radiation and absorption, redistributing momentum, etc. So yes, I agree a numerical simulation will be necessary and besides the effort you put in your answer, I appreciate your honesty. $\endgroup$
    – Winston
    Oct 3 '19 at 6:53
  • 1
    $\begingroup$ @Exocytosis The answer above still probably applies in reality, just on average. Basically, if you were to take a bunch of collections of point charges that were initially uniformly randomly distributed in a sphere, then if you were to simulate all of those collections, and calculate the magnetic field in all of those simulations, and average them (by vector addition), then I would expect the magnetic field vector to average out to zero in the limit of a large number of trials. $\endgroup$ Oct 3 '19 at 9:42

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