# Solving for magnetic vector potential (An integral issue)

1. Sep 11, 2011

### msimmons

1. The problem statement, all variables and given/known data
There is a cylinder of conducting ionized gas that occupies rho < a. For the given B, show that a suitable A can be found with only one non-zero component, Aphi, find Aphi which is also continuous at rho=a. (Part A was solving for a few relavant things)

2. Relevant equations
$$\mathbf{B}=\nabla \times \mathbf{A}$$
$$B(\rho) = \begin{cases} B_0\frac{\rho}{a} \hat{z} & \text{if } \rho \leq a \\ B_0 \hat{z} & \rho \gt a \end{cases}$$

3. The attempt at a solution
(Where line 2 is the curl in cylindrical coordinates, ignoring the second part because of the condition in the problem... only a phi component of the vector potential.)
$$B_z = (\nabla \times \mathbf{A})_z$$$$= \frac{1}{\rho}\frac{\partial (\rho\, A_\phi)}{\partial \rho}$$$$\rho B_z\, d\rho = d(\rho\, A_\phi)$$$$\rho\, A_\phi(\rho) = \int \rho B_z \, d \rho$$

Aaand I'm stuck. I'm not sure how to use the fact that A is continuous while dealing with this piecewise function. My first guess was to do something like breaking the integral into two parts, but I don't see that working because it would be necessary to have a definite integral.
Do I make it a definite integral, and play with it then? If so, my idea was perhaps my limits of integration would be 0 to rho, integrating over rho'. Thus, if rho is less than a we need only to look at one part of the piecewise, and if it is greater than a we have only a constant (after integrating from 0 to a) plus the integral from a to rho.

Hope that makes sense. Thanks for your time!

2. Sep 12, 2011

### Hootenanny

Staff Emeritus
That's what I would do