Solving for magnetic vector potential (An integral issue)

[msimmons]

Homework Statement

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, A_phi, find A_phi which is also continuous at rho=a. (Part A was solving for a few relavant things)

Homework Equations

\mathbf{B}=\nabla \times \mathbf{A}
<br /> B(\rho) =<br /> \begin{cases}<br /> B_0\frac{\rho}{a} \hat{z} &amp; \text{if } \rho \leq a \\<br /> B_0 \hat{z} &amp; \rho \gt a<br /> \end{cases}<br />

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.)
<br /> B_z = (\nabla \times \mathbf{A})_z<br /><br /> = \frac{1}{\rho}\frac{\partial (\rho\, A_\phi)}{\partial \rho}<br /><br /> \rho B_z\, d\rho = d(\rho\, A_\phi)<br /><br /> \rho\, A_\phi(\rho) = \int \rho B_z \, d \rho<br />

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!

 

[Hootenanny]

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!

That's what I would do