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# eddy curr in a finite tube

- To: femm@yahoogroups.com
- Subject: eddy curr in a finite tube
- From: Roberto Brambilla <rlbrambilla@cesi.it>
- Date: Tue, 10 Apr 2001 11:36:59 +0200

    Deare David,
    thank you for your good solution.
    I also obtained the same results using
    the vector potential A=(0,0,Az(r)) instead
    of your H=(0,Hphi(r),0).
    All that is good for infinite tube!
    The major difficuty is considering a FINITE tube
    since you have a problem in two cylindrical variables (r,z).
    Eddy currents have r and z components.
    So A=(Ar(r,z),0,Az(r,z)) and the potential equations are:

    CylindricalVectorLaplacian(A)-(k^2)A=0 in metal, k^2=jw(mu)(sigma)
    CylindricalVectorLaplacian(A)=0 in air.

    The z-dependence of potentials makes the problem much more
    difficult for the analytical solution.
    I have found a reference, IEE Proc. vol. 129 A (1982)p.54-61,
    by Kriezis et al. but I have some doubts (eq. 63?).
    So other references are well accepted.
    Yours, Rob
    Roberto Brambilla
    CESI
    Via Rubattino 54
    20134 Milano
    tel +39.2.2125.5875
    fax +39.2.2125.610
    rlbrambilla@cesi.it

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