# Re: Ballast Inductor

```Exactly... most solenoid formulas show inductance proportional to N^2. This is
because each turn has not only the magnetic flux from itself, but also the flux
from all the other turns as well.

Consider a single turn with a changing current (formally, di/dt) flowing
through
it... It generates some changing flux, call it X.  That changing flux induces a
voltage in the turn, proportional to the flux, call it Y.  The ratio of the
induced voltage to the changing current is the inductance (V = L di/dt).

Consider a second turn, connected in series, but far away (so there isn't any
interaction between the turns.  Same changing current, but the voltage across
the pair is twice what it was across one.  So, the L is now twice what it was..
simple series combinations of inductors.

Now, consider if the flux from the first turn passes through the second.  The
changing flux through the second turn is twice what it was (the contribution
from the second turn+ the contribution from the first turn), so the voltage
across it it twice as high.  Use the same analysis on the first turn, which
also
has twice the flux passing through it.  The voltage across both turns, in
series, with the flux linked fully, is 4 times what it was across a single
turn,
but the changing current is the same. SO, the inductance, L, is 4 times what a
single turn's inductance was.

In real life, of course, the flux doesn't link fully, so you don't get
quite the
N^2 increase (although, with an iron core that isn't saturated, and the
turns on
top of each other, the coupling will be pretty high...).

Tesla list wrote:

> Original poster: "M Fabs" <the_machin_shin-at-hotmail-dot-com>
>
> Thanks for the responses to my inquiry about my ballast inductor/variac
> attempt.  Since 1.72 millihenries seems quite small (the house service would
> dislike 190 amps!) I think the easiest way for me to make use of the
> inductor is to wrap another layer (maybe 2?), another 100 feet of #10 right
> on top of the first. Putting these in series ought to help, right?  I
> imagine it won't be a simple doubling of the inductance.  Is this where the
> squared rule comes in that you mentioned Jim?
>
> I'm certain there are better ways to do this, but I'm just doing this for
> fun, so that's ok.
>
> Thanks again,
> MPF (Michael, but there are already too many of those ;-)
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