Re: inductance - axisymmetric model

[tborzic@xxxxxxxxx]

--- In femm@xxxx, "Jim in_Dallas" <jim271@xxxx> wrote:

> >
> If we use the axisymmetric model for the coil, the FEMM results is 
very 
> close to the value calculated using the formula for a current loop.
> 
> Example:
> 
> current loop: L=N^2*R*mu0*Ln(R/r)
> where R=radius of loop
> r=radius of wire

Jim,in Your smallcoil femm file r is not 0.0005cm it is d,(d=2r)
so r=0.000025
so this formula works only for ....Ln(R/d)?
because you draw d=0.0005 not r (r is a half of diameter)

> 
> Using your dimensions:
> R=4cm
> r=0.0005cm
> 4 turns
> then L = 167.2nH
> 
> An alternative formula found in some references is: L=N^2*r*mu0*[Ln
(8*R/r) 
> -2]
> 
> Since the Ln(8) approx = 2, these formulae give similar results.
> 
> Using FEMM:
> Let J=1.0 then
> integral(A.J)=6.236e-9
> I = 0.1954 Amps/turn
> 
> L=163.4nH which differs by less than 3 percent. I've attached my 
fem file.
> 

when i changed gap between wires to something smaller result is 167.x 
nH. Only thing that bothers me is that radius in formula..
So, now i will bite my nails all night (it is night in Croatia, 
actually midnight) hoping that r in formula is actually d.
Or I made mistake when mesuring dimensions in smaillcoil file?
Hope You will soon read this:)))
Thanks again
tomislav