Roberto Brambilla wrote:
Deare David,This is one of those situations where it's best not to use magnetic vector potential, at least for the case in which the wire goes down the center of the pipe. In this case, the solution everywhere in the air is prescribed by Ampere's loop law as:
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.
H = i/(2*Pi*r)
where i is the current in the wire and r is the distance away from the wire. The presence of the tube doesn't affect the field in the air around the tube in this case, because the net current enclosed by any loop through the air is just the current in the wire.
Then, rather than describing the field in the tube in terms of two different components of magnetic vector potential, you can just describe the field in the tube in terms of the tangential H in the tube. (This is just like using an electric vector potential formulation...) Now, you just have to tack one more term, d^2 H /(dz^2), onto the differential equation that described things as just a function of r, with H= i/(2*Pi*r) as the boundary condition on all boundaries of the tube (including the top and bottom of the tube). This PDE can be solved analytically by slogging through the usual separation of variables routine.
Of course, your Case (2) where the wire is outside of the pipe is less straightforward....
Dave.
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