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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...

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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.

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  • $\begingroup$ Okey, great! I understand. But how can I prove the statement then? $\endgroup$ – Luna Sage Jun 21 '15 at 19:01
  • $\begingroup$ 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... $\endgroup$ – Rol Jun 21 '15 at 19:48

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