I Magnetic vector potential of infinite straight wire

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1. Aug 30, 2016

DavideGenoa

The magnetic field generated by an infinitely long straight wire represented by the straight line $\gamma$ having direction $\mathbf{k}$ and passing through the point $\boldsymbol{x}_0$, carrying a current having intensity $I$, if am not wrong is, for any point $\boldsymbol{x}\notin \gamma$ $$\boldsymbol{B}(\boldsymbol{x})=\frac{\mu_0 I}{2\pi}\frac{ \mathbf{k}\times (\boldsymbol{x}-\boldsymbol{x}_0) }{\| \mathbf{k}\times (\boldsymbol{x}-\boldsymbol{x}_0) \|^2}.$$
I wondered whether the vector potential $\boldsymbol{A}$ such that $\nabla\times \boldsymbol{A}=\boldsymbol{B}$ exists and what function it is...
I $\infty$-ly thank you for any answer!

2. Sep 2, 2016

DrDu

Try to convert your problem to cylindrical coordinates.

3. Sep 2, 2016

DavideGenoa

After some trials and a bit of intuition I have found a vector field whose curl is $\boldsymbol{B}$:$$-\frac{\mu_0 I}{2\pi}\ln\|\mathbf{k}\times(\boldsymbol{x}-\boldsymbol{x}_0)\|\mathbf{k}$$