# Vector potential of a linear current

First of all, I apologize in advance for my English, as I know it is not very good. So thank you for making the effort of understanding me.

Well, here is my problem. I am trying to prove that if I have a stationary current $$\mathbf{J}=J\hat e$$, where $$\hat e$$ is a constant unit vector, then there is a vector potential $$\mathbf{A}=\alpha \hat e$$, where $$\alpha$$ does not depend on the coordinate along the $$\hat e$$ axis.

So, I did the following. I rotated my problem so that $$\hat z=\hat e$$, and tried to directly integrate the vector potential:

$$\mathbf{A}=\frac \mu {4\pi} \int {\frac {\mathbf{J}(\mathbf x')} {|\mathbf {x} - \mathbf {x'}|} }$$

But the integral diverges. This integral is equal to the electrostatic potential for a line charge, and I noticed now that it diverges too. So, how can I define potential in both cases? I'm really confused. Besides, I would like some help with my original problem...

This equation is supposed to be used when the boundary integral goes to $$0$$ at infinity. See Helmholtz decomposition. In your case, you are implicitly considering that your $$\vec{B}$$ field goes faster to $$0$$ than $$\frac{1}{r}$$, but you get divergence, thus basically showing that this is not the case. In conclusion, you cannot use this formula.
• You need to do it directly. Put $\vec{B} = \nabla \times \vec{A}$ in Ampere's law and solve for $\vec{A}$. I don't see any other way... – Rol Jun 21 '15 at 19:48