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# Re: [femm] Re: inductance - revised axisymmetric model

- To: femm@yahoogroups.com
- Subject: Re: [femm] Re: inductance - revised axisymmetric model
- From: dcm3c@aol.com
- Date: Thu, 15 Nov 2001 01:49:25 EST

In a message dated 11/14/2001 3:33:13 PM Eastern Standard Time, tborzic@yahoo.com writes:

> thanx Jim, I hope that will do with The Professor.
> I have one more question..
> As you may see, inductance is generally my problem.
> Currently i am trying to solve inductance between
> two reactangular conductors, infinite lenghth, paralell.
> they are standing like this - II.
> Numerically, inductance for such case is:
> L=(2*mi0/h)*(b+a/3)
> where h-heighth,2b distance between conductors (not
> between theirs centers), and a width.
> so if a=0.2cm, b=0.8cm (2b = distance 1.6cm),h=2.2cm
> L=9.89 e-7 H/m, while femm allways returns half of that value.
> I applied 1A current for easier computing.
> so, where is mistake?
> thanks
> tomislav

Your formula, L=(2*mu0/h)*(b+a/3), is only correct if h is really large in comparison to b and a. However, the parameters you chose for your example don't have this attribute--your formula just isn't a good approximation for this particular case.
The same thing is going on with web page that Jim cited--they are using basically the same formula. I'm pretty sure that the value that femm gives is reasonable for this problem.
Dave.

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