Re: Formula for self-inductance?


Wimpie wrote:

First thing, before everyone starts telling me how "wrong" I am, I
have to say that what I said actually was wrong! One does not need
self-inductance of turns to calculate mutual inductance between two
coils. The simple and very accurate procedure is to replace each wire
loop in both coils with a thin filament and then use the Maxwell
relation (VOLII #701) to calculate the mutual inductance between all
pairs of loops from one coil to the next. You sum the Ms to get total
M and since you are summing over the entire geometry coils of any size
and shape can be easily (with a computer!) and accurately calculated.
If one assumes that current distribution in the wires is uniform, one
can actually go backwards from multiturn coils to one turn thick wires
by dividing by NxM where those are the number of turns in each coil.
(I think this is finally correct!)

The problem of course is finding self-inductance where the thin
filament trick fails because it drives inductance infinite.

You cannot ignore current distributions, because most of the energy
stored in the B field is close to the wire. So if you only want to go
static, you must define a current distribution. If you consider
uniform distribution inside the wire, there will also be B-field in
the wire itself that increases the inductance.

Yes, this is a worry I have. I really am not sure how accurate a
uniform current assumption is going to be. In many cases it may be OK.
My thought is if you have a coil of wire rather than a current sheet.
The coil forces the current distribution uniform over the geometry of
the coil (because all loops are in series).

You are right, for a straight wire this approach doesn't work (it
blows up to infinity for r=infinite). In that case you may use the
stored energy approach. As you know the B-field distribution, you also
know the energy in the magnetic field and may use Energy = 0.5L*I^2.

Yes, I really like the stored energy approach, only at this point I'm
not quite sure how to go about it. I'm guessing you'd have to assume
some current distribution and then calculate B everywhere, then us B
to find energy and use energy to find L. Would be better if somehow
you could FIND the current distribution rather than assume it.

Se also text above. If you assume a non-filament current distribution,
B will never go to infinity close to the wire, hence the energy
density will also not go to infinity. At large distance from the wire
segment, energy density falls of with r^4 (because H falls off with
R^2). The occupied volume as function of dr increases with R^2. So the
integral will also not go to infinity for r=infinity. The result will
be I finite stored energy in the B field, this should give you the
inductance for a straight wire segment. I hope you are better the me
in math, I should dive deeply into my books to solve the integral
exactly.

I see. Yeah, this could work. Of course the current distribution thing
could be a nightmare. As you note a uniform distribution may not be
accurate.

Are you really sure about this? As J is not infinite, B is a
continuous function that will not go to infinity. Even the thin sheet
method (so J is infinite) is used in several EM-field solvers.

As you note using J works for flux methods and energy methods but if
you try to use the idea that self-inductance is just mutual inductance
back on itself (which is what I was doing) then it doesn't seem work.
Of course my math is not so terrific either! :-)

I agree with you on the mutual inductance and self-inductance problem
in case of filaments (it is not my favorite way to see inductance). I
believe, as flux linkage does fail in some situations, the energy
approach doesn't. Off course, you have to assume a current
distribution.

If one starts thinking of the large picture, a view emerges of what
one would really like to see. It goes like this: First off,
Resistance, Capacitance, and inductance are all geometrically
determined. So if I have some conductive object in space, I ought to
have some way to start with just the geometry and then calculation
what is going on! I mean if I take this object and attach a couple of
wires to it at some point (assuming highly conductive areas of
connection) one should be able to somehow figure out the distribution
of current in the object, the amount of self-inductance the terminals
present as well as the capacitance found there. Of course it doesn't
take much geometry to make a mathematical nightmare! But this is the
age of computers. We don't have to have simple formulas and log tables
anymore! My thought is that one ought to be able to somehow, take that
geometry, divide it up into little pieces, apply formulas to the
pieces and then solve for all the parameters in question. Say start
with finding the current density for an applied voltage, then use that
to find inductance etc.

That's the theory, only I'm finding this thing is not quite so
straight-forward. Pieces of the basic theory seem missing. And worst
of all, my abilities in math are not especially terrific either! But
still, the problem seems useful!

When you take dynamics into account, and assuming that your
construction is very small with respect to wavelength, current
distribution becomes frequency depended, hence the inductance. Some
experimentally derived formulas mention two formulas for the same
geometry. One for DC (uniform distribution in wire) and one for
transient/RF (current sheet distribution around circumference).

I presume there is a transition frequency between stationary current
distributions and RF ones determined by wave propagation. Both are
interesting but for now I was hoping to stay with the low frequency
case.

By using Maxwell, the influence will be remarkable on the flux inside
the conductor (as wave propagation speed in good conductors is
extremely low). As long as your construction is very small with
respect to wavelength in air, the contribution of radiated fields is
negligible with respect to the reactive field.

Yes, even for high frequencies the assumption would have to be non-
radiation otherwise it's an antenna problem which is a much larger can
of worms!

Benj

.

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