eddy curr in a finite tube

[Roberto Brambilla <rlbrambilla@xxxxxxx>]

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
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rlbrambilla@xxxxxxx