Response to inductance of a straight wire question

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: Jared E Dwarshuis <jdwarshui@xxxxxxxxx>

Subject  Re: Re Secondary Resonance LC and Harmonics


Original poster: FIFTYGUY@xxxxxxx

In a message dated 6/29/05 5:49:15 PM Eastern Daylight Time,
tesla@xxxxxxxxxx writes:

We can examine the classic equation for an air cored inductor: L = u
Nsqrd A / l
Multiply the numerator and denominator by 4pi  we get:

L = u Nsqrd 4pi sqrd  r sqrd / 4 pi l  =  u (2pi r N)sqrd / 4 pi l

Since 2pi rN equals wire length we can write:

L = u (wire length)sqrd / 4 pi l

Letting the solenoid height ( l ) equal the wire length,  we find
that the classic equation predicts that the inductance of a straight
wire is simply:

Lstwire = u wire length / 4 pi

     But using the above "classic equation" also collapses the
inductance
to zero as the radius "r" and the turns "N" go to zero.

-Phil LaBudde

 >True if R and N are both zero then we have no wire and hence no
 >inductance.

     My intent was to show quite the opposite: If R and N are both
zero
then we have a STRAIGHT wire with minimum inductance. Actually,
minimum
inductance is obtained by doubling the wire back on itself so one half
cancels the other's inductance.

 > I think what you meant to ask is what happens as R
 > approaches zero for a constant length of wire.

     Sure...

 >As R approaches zero the solenoid length (l) approaches the wire
 >length. The substitution we used where (l)= wirelength does not lead
 >to a contradiction, it should be a legitimate step.
 >To help wrap your mind around this geometry consider what happens
when
 >you take wire from a fat inductor and wind it around progessivly
 >skinnier forms. Sure enough as the radius gets smaller the length (l)
 >aproaches that of the straight wire.

     With you so far.


 >Of course the really interesting implication is that at R=0, N is
 >infinite.

    I disagree. Because we have not changed the thickness of the wire,
as
R=0, at some point we must GIVE UP turns to lay the wire on top of
itself.
At some very small R, the wire must physically "corkscrew" and have
spaced
turns in order to complete the turns. In this way, we have used the
same
wire to form less turns.
    Which brings me back to my original point: as both R and N go to
zero,
so does the inductance. The only way you get a straight wire (R=0) to
have
any turns ("N") is via eddy currents.


 >Jared and Larry
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Hi again:

Real wires must have thickness, abstract wires have no such
restriction. It is a matter of convergence. N goes to infinite as r
goes to zero as (l) goes to wire length.

Jared and Larry
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