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Suppose we have a nonmagnetic conducting cylinder of radius $ \alpha $ directed in the z direction, with a current density $ J_{0} \hat{a_{z}} $. What is the magnetic vector potential for $\rho < \alpha$?

I know that we should start with the magnetic field, because the direct volume integral of $ \int_{-\infty}^{\infty}\int_{0}^{2\pi}\int_{0}^{a}\frac{\rho\prime d\rho\prime d\phi\prime dz\prime}{\sqrt{\rho^2 + \rho^2\prime -2cos(\phi - \phi\prime)}}$ to find the potential directly is absolutely disgusting.

Using Ampere's Circuit Law, we can state that $H2\pi\rho = J_{0}\pi\rho^2$ and therefore that $\vec{H} = \frac{J_{0}\rho}{2}\hat{a_{\phi}}$. I'm then supposed to make use of $\mu_{0}\vec{H} = \nabla \times \vec{A}$. My textbook suggests that the solution is as trivial as integrating $\mu_{0}\frac{J_{0}\rho}{2}$, however I am unsure as to what the integration variable would be. Am I correct in asserting that I am to integrate with respect to $\rho$, as the $\hat{a_{\phi}}$ component of the curl operation in cylindrical coordinates is given by $\frac{\partial{A_{\rho}}}{\partial{z}} - \frac{\partial{A_{z}}}{\partial{\rho}}$ and that the vector potential will be given by $-A_{z}\hat{a_{z}}$ because all of the current is z directed and the potential should therefore be in that direction?

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  • $\begingroup$ Is the current uniformly distributed on the surface of the cylinder, or is the cylinder actually a rod and current flows also inside? How long is the cylinder? $\endgroup$ – Ján Lalinský Mar 27 '18 at 0:49
  • $\begingroup$ @JánLalinský The cylinder is infinitely long, and the it’s a rod with current flowing inside (else the direct integral would be a surface integral and the current density would be K) $\endgroup$ – ImInfinite313 Mar 27 '18 at 19:16
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My assumption was correct. Because the current is $\hat{a_{z}}$ directed, the $\vec{H}$ field will be $\hat{a_{\phi}}$ directed, and therefore, we can equate $\frac{J_{0}\rho}{2}\hat{a_{\phi}}$ to $(\frac{\partial{A_{\rho}}}{\partial{z}} - \frac{\partial{A_{z}}}{\partial{\rho}})\hat{a_{\phi}}$. Because the current density is only in a single direction, we should expect the vector potential to be along the same axis, albeit in the opposite direction. Therefore we would have that $\vec{A} = -\frac{J_{0}}{2}\int_{0}^{\rho}\rho\prime d\rho\prime \hat{a_{z}}$. Solving this integral gives us the magnetic vector potential of $-\frac{J_{0}\rho^2}{4} \hat{a_{z}}$.

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