P.Neumaier@xxxxxxxx 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
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