Vector potential in a coaxial cable

[koroljov]

Homework Statement

I have to find the magnetic vector potential in a round coaxial cable. The internal conductor has a finite (known) conductivity. The external conductor is a perfect electrical conductor. Both the radius of the internal and the external conductors are known.

I have to assume that everything happens in sinusoidal regime, hence the use of phasors.

Furthermore, the current and the vector potential have only a component along the z-axis. I have to use the Coulomb gauge, and magneto-quasi-static approximations.

First of all, I had to show that the z-component of the vector potential, Az, obeys a certain differential equation in the inner conductor. (see "relevant equations" below). That was no problem. I had to solve this equation which was no problem either. To find the actual solution, I needed two boundary conditions. This too was no problem.

The actual problem is that I have to find a third boundary condition to find E0.

Homework Equations

The differential equation:

Laplacian(Az) - j*omega*mu0*sigma*Az = -mu0*sigma*E0

with E0 a constant, and j the imaginary unit. E0=dV/dz,the derivative of the scalar potential (this can be shown to be constant easily using the restrictions on the components of the E and A vectors, and the law of Faraday).

The solution of this equation:
Az(r) = BesselJ(0,(-mu0*sigma*omega*j)^(1/2)*r)*c-1/omega*E0*j

where c is a constant that can be determined from the boundary conditions. Another bessel function was thrown away because it has a singularity at r=0.

The boundary conditions:
-The B-field must be 0 for r=0 (no extra infrmation follows from this)
-the B-field must be equal to mu0*I_totaal/(2*Pi*a) at r=a
where I_totaal is the total current, and a is the radius of the inner conductor.

The Attempt at a Solution

Knowing all this, I can solve the differential equation completely. It is surprising that nor the E-field, nor the B-field depend on E0. Look:
Ez = -dAz/dt + E0
thus
Ez = -j*omega*Az + E0
Ez = -j*omega*Az + E0
Ez = -j*omega*(something -1/omega*E0*j) + E0
Ez = -j*omega*something -E0 + E0
Ez = -j*omega*something
where "something" does not depend on E0.

For the B-field, I have to take the curl of A, which implies taking spatial derivatives. E0 will disappear, since it is independent of position.

Yet I still have to find a "third condition" to determine E0. I think that, since there are no free charges anywhere in the problem, and in the magneto-quasi-static approximation one ignores the slight charge buildups that could be associated with electrical waves in a conductor, the scalar potential V has to be constant (independent of position), and hence, that E0 has to be zero. Unfortunately, this sounds slightly too easy to be true. Am I overlooking something?

Thanks in advance.

 

[koroljov]

I found the answer. Thanks anyway.

 

[wpoely]

Can you share your solution to this problem with us?
I need to solve a very simular problem.

Thanks in advance.

 

[marlon]

Homework Statement

I have to find the magnetic vector potential in a round coaxial cable. The internal conductor has a finite (known) conductivity. The external conductor is a perfect electrical conductor. Both the radius of the internal and the external conductors are known.

Do you study Burgerlijk Ingenieur in the Ugent ?

"Indien ja, is dit het project van DeVisschere"



marlon

 

[wpoely]

Do you study Burgerlijk Ingenieur in the Ugent ?

"Indien ja, is dit het project van DeVisschere"



marlon

Yep, heeft hij dit mss ooit al eens gevraagd? Ik vond het nogal raar om exact onze opgave terug te vinden op internet.

 

[koroljov]

Hmmm. The solution, which is well-known by now, so it doesn't matter anymore anyway, is to demand that the vector potential becomes zero at r=infinity. (hence it becomes zero at r=b). There are other possibilites that will yield the correct solution too.