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

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- Subject: 回复: [femm] Re: Calculation of self and mutual inductances
- From: "si hang" <stsc_sh@sina.com>
- Date: Fri, 29 Dec 2000 16:42:36 +0800

    Before you caculate inductance or mutul inductance, you have to formulate
    your problem in such a way that all the currents (space, surface or linearly
    distributed) but one are set to zero. There must be no permanent magnets in
    your model. In that case equation above becomes xtremely simple and you can
    get inductance value as :
    E
    L = 2 * -----
    I^2
    where E is stored magnetic energy and I is the only current(amp-turns).
    This question is used for selfinductance caculation. It's more exact than
    flux linkage caculation.

    Also, you can get mutul inductance use flux linkage method by the below
    equation:

    φ
    L = ----
    I
    where φ is the flux linkage with the coil excited by current I(amp-turns).

    for plannar problem:

    ∫A ·dS ∫A ·dS
    φ = RightSide(--------- ) - LeftSide(---------)
    ∫dS ∫dS

    where A is the vector magnetic potential; RightSide and LeftSide denote the
    right and the left side of the coil accordingly.

    The difficult to use this equation is that we can't get dS from femm. But if
    we choose a uniform mesh density in coil cross section and the mesh is dense
    enough,
    we can assume that each dS is a constant in the cross section. So the
    equation become easy to caculate by femm:

    φ = RightSide( ∫A ) - LeftSide( ∫A )

    Hope this can help you.

    si hang
    -----Original Message-----
    ·¢¼þÈË: Rob <rtickle@visi.com>
    ÊÕ¼þÈË: femm@egroups.com <femm@egroups.com>
    ÈÕÆÚ: 2000Äê12ÔÂ29ÈÕ 16:58
    Ö÷Ìâ: [femm] Re: Calculation of self and mutual inductances

    --- In femm@egroups.com, P.Neumaier@z... wrote:
    > Hello,
    >
    > 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 * A?quot;. But this integral yields incorrect
    > results, which can be prooved 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*m?quot; which is obviously not the unit for an inductance.
    >
    > Thanks for any answer in advance.
    >
    > Peter

    Are you dividing the result FEMM gives you by the current in the coil
    squared? If so are you using the correct value for the current? If
    you have multiple turns the current scales with N... I did some
    calculations like this when I started using FEMM and got some
    incorrect answers due to faulty current values. Be sure to add up all
    cross sectional areas of conductor multiplied by their respective
    current densities. This solved the problems I was having...

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