Re: Equivalent lumped inductance and toroidal coils

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Gerry Reynolds" <gerryreynolds-at-earthlink-dot-net> 

Hi Paul,

I think there are two facets to your response.  One deals with the R=0
problem and the other deals with multipath currents.   The second issue I
would like to defer as a separate issue for now.


 > Original poster: "Paul Nicholson" <paul-at-abelian.demon.co.uk>
 >
 > Gerry wrote:
 >  > Yes, R=0 is a problem if you assume all the current is all
 >  > concentrated at R=0.
 >
 > Worse, it is a problem if *any* current is at R=0, so splitting the
 > single filament up into several spread over the current-carrying
 > area of the wire doesn't help - you'll still come up against the
 > need for self-inductance for each filament.  (I suppose if you just
 > leave out the self inductances you'll have some error, but only
 > slight if you have many filaments per wire.)

It is certainly possible that I have missed your point about R=0, but I will
use a simple case to explain why I do not think this is a problem.  Take an
infinitely long straight wire with a non zero radius (Rw) carrying a DC
current (I).  For this case the differential current element is not
necessary and we can resort to amperes work law:

Closed integral {H.dl} = I    (I is enclosed within the path of integration)

If we pick a symetrical path where we know H is constant  (like at a
constant R) then the integral reduces to:

H * 2piR  = I    so H = I/(2piR)

Certainly this suggest there is a singularity at R = 0, but this is not the
case.  The enclosed current I is constant for R >= Rw.  But once we enter
into the wire the enclosed I gets smaller.  For this case where R < Rw:

I = J * (pi*R^2),

where the current density (J) is:

J = I/(pi*Rw^2)

If R -> 0, then I -> 0 (because the enclosed area -> 0).  H then becomes:

H = I (pi*R^2) / {(pi*Rw^2) * (2piR)}      and reduces to

H = (I * R) / (2pi*Rw^2)

At R=0:    I=0 and H=0

Next step is to go to AC where the current density is not uniform and is
smallest at R=0 and largest at R=Rw.  This case also does not have a problem
at R=0 for the same reason as with DC.

Now lets go to a single loop carrying current I.   The self inductance is
total flux thru the loop due to the current carried by the loop devided by
the current I.  If the flux contained in the copper wire is neglected (and
certainly the higher frequency the smaller this portion becomes), most of
the flux attributing to self inductance is still counted for.   My thinking
with self inductance applies to geometries where an area can be defined
(like a loop) and not to a straight wire.


 >
 > Not only that, but as soon as you represent the wire with more than
 > one filament, you have to decide how the total current is to be
 > shared across the bundle of paralleled filaments.   You can't
 > in general assume say uniform distribution over the surface.

A straight wire would have uniform distribution over the surface due to
symetry, but I don't know what happens to the distribution when the wire is
part of a coil.  If one assumed that J was only a function of (r), would
this alter the answer that much?  I suspect not.


 >
 > This problem extends to networks of separate conductors which offer
 > alternative conduction paths through the field.  For example an
 > OLTC primary might consist of several circular loops all in
 > parallel.

I'm very familiar with this problem.  I'm assuming that the alternate paths
you are describing are the multiple current filaments that represent the
wire and the varying current density contained within the wire.  Is this
complexity really necessary?  I have written a program to simulate the
multiple current path problem in the past using loop equations and iterating
numerically until boundary conditions were met.

I guess I would propose as a "next best step" to a perfectly accurate
exercise is to stop the H field sampling at the surface of the wire since
the current density will drop off once the wire is "penetrated" and not
worry about the distribution of current within the wire.  For purposes of
evaluating the differential current element, assume the current is at the
center of the wire (dimensionally speaking).

Best regards,
Gerry R.