Re: Calculation of self and mutual inductances

[jim271@xxxxxxxxxxx]

--- In femm@xxxxxxxxxxx, David Meeker <dmeeker@xxxx> wrote:
> 
> 
> P.Neumaier@xxxx wrote:
> 
> > can anybody help me? When I calculate the self inductance of e.g. 
a
> > coil encircling a copper rod using the "A.J" integral I get the
> > correct unit "Henry * A²". But this integral yields incorrect
> > results, which can be proved when defining a problem, which can 
also
> > be solved analytically. However, using the integral "A" for the
> > mutual inductance, the value is correct, but the unit is given
> > as "A*m³" which is obviously not the unit for an inductance.
> >
> > Thanks for any answer in advance.
> 
> Sorry it's taken a while for me to get back to you with this 
question.
> 
> The "A*m^3" unit on the integral of just "A" is a typo--should be 
Joule
> Meter^2 / Amp, or something consistent with this. Thanks for 
pointing this
> out.
> 
> However, I'm pretty sure that the A.J integral can be used to get 
reasonable
> values of inductance. As a giggle test, one could examine the 
approximate
> formula for the inductance of a single-turn circular loop of round 
wire:
> 
> L = mu_o r (Log[8 r / a'] - 2)
> 
> where mu_o is the permeability of free space, r is the radius of 
the loop
> measured to the center of the wire, and a' is the geometric mean 
radius of
> the wire (the geometric mean radius is a*Exp[-1/4], where a is the 
radius of
> the wire).
> 
> I have arbitrarily chosen to consider a coil with a mean radius of 
1"
> (r=0.0254 m) and a wire radius of 0.125" (a=0.003175 m), implying 
that the
> geometric mean radius of the conductor is a'=0.002473 m. I have 
imposed a
> current of 30 A in the wire.
> 
> I have attached a model of this coil. I used an asymptotic boundary
> condition (an approximation of an "open" boundary) so that the 
boundaries
> should have only a trivial effect on an inductance computation. I 
used a
> "circuit" property to impose the current in the coil.
> 
> When I do the A.J integral for this geometry, the result is 
6.865*10^(-5)
> H*A^2. The inductance of the coil is then: (6.865*10^(-5) 
H*A^2)/(30 A)^2 =
> 7.628*10^(-8) H.
> 
> Evaluating the formula for this configuration yields 7.689*10^(-8) 
H, which
> corresponds pretty well to the finite element result.
> 
> Dave Meeker
=================================================================

I've compared the FEMM inductance calculation with that of Wheeler's 
formula for several solenoidal coils and found they agree within a 
percent or two.  

Jim