The electric field in terms of the electric scalar potential, and the magnetic vector potential is:

$E = -\nabla\phi - \frac{\partial A}{\partial t}$, where $A$ is such that $B = \nabla \times A$.

In this field configuration:

a change in the poloidal current sheet $J$ will cause a change to propagate through $A$ and a corresponding electric field pulse, coextensive with the changing $A$. Is this $E$ field accompanied by a non-0 $B$ field other than the one already inside of $J$? Does the change propagate at the speed of light?

  • $\begingroup$ Can you write down a functional form for the "propagating electric field pulse"? $\endgroup$ – ProfRob Nov 2 '14 at 18:10
  • $\begingroup$ Rob, That would be another StackExchange question. A related question has already been asked, but not answered: "Measurement of speed of static electric field propagation?" physics.stackexchange.com/questions/101413/… $\endgroup$ – James Bowery Nov 2 '14 at 22:10
  • $\begingroup$ OK I don't understand. An electric pulse that propagates must have the form $E = f(r-ct)$? To satisfy Gauss's law, E must be perpendicular to the pulse direction and therefore has a curl and therefore from Faraday's law generates a magnetic field. How is this avoided? $\endgroup$ – ProfRob Nov 2 '14 at 22:22
  • $\begingroup$ Rob it is not avoided. The magnetic field is segregated from the propagating electric field pulse by the surface of the torus, which is the geometry of the Rogowski coil: A poloidally wound toroidal coil with $B$=0 and $A$≠0 everywhere outside the minor radius of the torus. $\endgroup$ – James Bowery Nov 2 '14 at 23:14
  • $\begingroup$ I just discovered this is more problematic than I thought. Two different experiments come up with two different empirical values for the speed of propagation of an electric field. One is consistent with instantaneous propagation: Measuring Propagation Speed of Coulomb Fields arxiv.org/pdf/1211.2913v1.pdf and one is finite speed: Coulomb interaction does not spread instantaneously cds.cern.ch/record/468803/files/0010036.pdf $\endgroup$ – James Bowery Nov 2 '14 at 23:46

If one has a coil, which in the steady state has a finite interior B-field, a zero B-field outside the coil, but a finite, curl-free A-field outside the coil; then you change the current in the coil rapidly: I think the following should happen.

As you say, a pulse of E-field will propagate outwards associated with the changing A-field. To satisfy Maxwell's equations: if the pulse E-field has the form $f(kr-\omega t)$ - e.g. something like $E_0 \exp[-(kr -\omega t)^2/2\sigma]$ - then Gauss's law should ensure that the E-field is perpendicular to $r$. Furthermore, the E-field will have a curl and then through Faraday's law will have an associated, time-dependent B-field and a non-zero Poynting vector. Finally, Amp`ere's law in vacuum will yield that $\omega/k =c$, the pulse speed.

I deduce from this that the curl of the A-field does not remain exactly zero outside the coil at all $r,t$.

  • $\begingroup$ I've clarified my question. $\endgroup$ – James Bowery Nov 3 '14 at 14:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.