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# Re: [femm] Calculation of self and mutual inductances

- To: femm@egroups.com
- Subject: Re: [femm] Calculation of self and mutual inductances
- From: David Meeker <dmeeker@foster-miller.com>
- Date: Fri, 12 Jan 2001 09:46:58 -0500
- Organization: Foster-Miller, Inc.
- References: <92fgql+pf9f@eGroups.com>

    P.Neumaier@z.zgs.de 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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