# Re: New Inductance Formula

```Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz <twftesla-at-qwest-dot-net>" <acmq-at-compuland-dot-com.br>

Ok, I will enter this discussion too...

It's interesting to see what is Wheeler's formula for the inductance
of a solenoid in metric units.

The inductance of a long solenoid is given, theoretically, by:

L=u0*a*N^2/l H

where u0=4*pi*10^-7, a is the area of a cross-section, and l is the
length of the coil. Using a=pi*r^2:

L=u0*pi*r^2*N^2/l H

Add a "correction factor" to the denominator, replacing l by
l+0.9*r:

L=u0*pi*r^2*N^2/(l+0.9*r) H

This is Wheeler's formula. To put it in the familiar shape for
measurements in inches, multiply r and l by 2.54/100, or simply
multiply the formula by this factor:

L=u0*pi*2.54/100*r^2*N^2/(l+0.9*r) H

Substituting u0, multiplying numerator and denominator
by 10, and putting the result in uH:

L=4*pi^2*2.54/100*r^2*N^2/(10*l+9*r) uH

L=1.0028*r^2*N^2/(10*l+9*r) uH

So, the multiplying constant ~1, in the Wheeler' formula is just:
4*pi^2*2.54/100 uH/inches.

Antonio Carlos M. de Queiroz

```