Re: Current in a wire

["Prof. John P. Ertel" <JPE@NADN.NAVY.MIL>]

I've tried to stay out of this thread as I wasn't sure just what I could
off that would actually contribute to understanding (as opposed to my
enhancing the confusion by supper posing my own).  But it seems like
it's time for my 2 cents (see threaded comments below).

THOMAS SANDIN wrote:
>         In a normal metal, the current is uniformly distributed across the
> cross section of the wire for dc and low frequency ac.

I apologize that we (or at least I) cannot yet adequately do math over
e-mail (except by attachment of a binary document).  Suffice it to say
that all of what follows is well documented in all Jr-level E&M texts
and graduate E&M texts.  The essentials are fairly well recounted in
most calculus based introductory E&M texts. But here is, I believe, an
accurate, if NOT brief, summary.

Maxwell's Equations (MaxEqs) in the differential form (not the integral
form which just shows average effects) guarantees only that (absent any
external influences) the current density, j(vect), in a long straight
wire is "sensibly uniform".  While it cannot guarantee that it is
exactly uniform (dj/dR = 0), symmetry and MaxEqs do demand that there is
NO azimuthal dependence.

But let's suppose that j(vect) is uniform enough that it is "OK" to pass
over the differential or point form of MaxEqs and look at the integral
or average form.

For any alternating current, we are guaranteed (requiring self
consistent solutions) that the current density is of a form that allows
for an exponential drop-off of j(R) from the surface over a radial
distance (the skin depth) that depends (to first order) reciprocally
upon the frequency of the current.  As the frequency increases, the 1/e
drop-off thickness decreases so that in the high frequency limit, the
current resides only on the outside surface of the conductor.
Conversely, the low frequency asymptotic solutions are NOT independent
or the radius --- jar) must increase with R.  Only in the "low frequency
limit" does the current density again become sensibly uniform ---
[dj(R)/dR](limit as f->0) = 0.

>         (In contrast, all the current is on the surface if the wire is
> superconducting (Type 1). This distribution occurs because of the
> Meissner effect--there is no magnetic flux within the bulk of such a
> superconductor.)

>         Excess charge appears on the surface at bends of a normal
> wire--required to bend the electric field along the axis of the wire.
>                                                 Tom Sandin

Not only is this last statement correct, but it is the only
self-consistant model/explanation for the "bending of the electric
field" to follow j(vect) around any curvature of a conductor.  Remember
that the scalar macroscopic V = I*R (Ohm's Law) must be replaced in the
microscopic sense with the vector expression E(vect) = Rho*j(vect).
Therefor E(vect) must be everywhere parallel to j(vect) --- but j(vect)
must of course follow the wire while being guided by the acceleration
provided point-for-point by E(vect).  Which is the cause and which is
the effect.  Well, you get the idea!

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