# How does the electric field pulse propagate during a change in the magnetic vector potential?

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?

• Can you write down a functional form for the "propagating electric field pulse"? Nov 2, 2014 at 18:10
• 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/… Nov 2, 2014 at 22:10
• 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? Nov 2, 2014 at 22:22
• 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. Nov 2, 2014 at 23:14
• 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 Nov 2, 2014 at 23:46

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 $r,t$.