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# Re: Calculation of self and mutual inductances

- To: femm@egroups.com
- Subject: Re: Calculation of self and mutual inductances
- From: jim271@hotmail.com
- Date: Sun, 14 Jan 2001 03:04:49 -0000
- In-reply-to: <3A5F18E2.2ABB6F59@foster-miller.com>
- User-agent: eGroups-EW/0.82

    --- In femm@egroups.com, David Meeker <dmeeker@f...> wrote:
    >
    >
    > P.Neumaier@z... 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

- References:
  - Re: [femm] Calculation of self and mutual inductances
    - From: David Meeker <dmeeker@foster-miller.com>

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