# Vector potential in a coaxial cable

1. Dec 8, 2006

### koroljov

1. The problem statement, all variables and given/known data
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 wich 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.
2. Relevant 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.

3. 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, wich 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?

Last edited: Dec 8, 2006
2. Dec 8, 2006

### koroljov

I found the answer. Thanks anyway.

3. Dec 18, 2006

### wpoely

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

4. Dec 18, 2006

### marlon

Do you study Burgerlijk Ingenieur in the Ugent ?

"Indien ja, is dit het project van DeVisschere"

marlon

5. Dec 18, 2006

### wpoely

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

6. Dec 19, 2006

### koroljov

Hmmm. The solution, wich 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.