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Solving for magnetic vector potential (An integral issue)

  1. Sep 11, 2011 #1
    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
    [tex]\mathbf{B}=\nabla \times \mathbf{A}[/tex]
    [tex]
    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}
    [/tex]


    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.)
    [tex]
    B_z = (\nabla \times \mathbf{A})_z
    [/tex][tex]
    = \frac{1}{\rho}\frac{\partial (\rho\, A_\phi)}{\partial \rho}
    [/tex][tex]
    \rho B_z\, d\rho = d(\rho\, A_\phi)
    [/tex][tex]
    \rho\, A_\phi(\rho) = \int \rho B_z \, d \rho
    [/tex]

    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. jcsd
  3. Sep 12, 2011 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That's what I would do :smile:
     
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