# Re: Wire length LC derivation,

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• Subject: Re: Wire length LC derivation,
• From: "Tesla list" <tesla@xxxxxxxxxx>
• Date: Sun, 13 Mar 2005 22:27:20 -0700
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Original poster: jdwarshui@xxxxxxxxx

`Part 1`

```By Ampere, the inductance for a long air cored solenoid is:
L = (u Nsqrd A) / h                       (where h is the length of
the solenoid)
(where A is Area )
(where N is the number of
turns)
Since:
2 pi r N = wire length*```

```  *actual wire length = sqrt ( (2pirN)sqrd + (h)sqrd )
but we disregard (h)sqrd as an insignificant term.```

```Then:
N = wire length / 2 pi r N```

```Substituting we get:
L = u ( wire length)sqrd / 4 pi h```

```By Maxwell:
u = 1 / Csqrd  epsilon        (Where C = 3x10 to 8 m/s)```

```Then:
L = (wire length)sqrd / (epsilon Csqrd  4 pi  h )```

`Part 2`

```For sake of simplicity in the following description of LC we will use
the derived capacitance of an isolated sphere. This would be directly
applicable to a Tesla coil secondary.```

``` From Gauss:
V = q/4 pi epsilon integral a to b for 1/ (r sqrd) dr```

```Since:
q = cap Volt```

``` Then:
cap. = 4 pi epsilon (ab / b-a)```

```Since we have a single isolated plate, b approaches infinity
we get:
cap. = 4 pi epsilon R           ( R is the radius of the sphere )```

```Now omega = 1/ sqrt (LC)
Substituting and canceling we get:```

`1/ (omega sqrd) = (wire length / 3x10 to 8 m/s)sqrd  x  ( R/h )`

`So:`

` [omega = (C/ wire length) x sqrt(h/R)`

```*R must of course be decreased to accommodate the self capacitance of
the coil.```

```Notice that the distance units cancelled for sqrt (h/R).  Wire length
and the speed of light provide the units for 1/s```

`END.`

Jared Dwarshuis, Lawrence Morris