Magnetic vector potential of infinite straight wire

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SUMMARY

The discussion focuses on the magnetic vector potential of an infinitely long straight wire carrying a current intensity \(I\). The magnetic field \(\boldsymbol{B}(\boldsymbol{x})\) is defined for points not on the wire as \(\boldsymbol{B}(\boldsymbol{x})=\frac{\mu_0 I}{2\pi}\frac{ \mathbf{k}\times (\boldsymbol{x}-\boldsymbol{x}_0) }{\| \mathbf{k}\times (\boldsymbol{x}-\boldsymbol{x}_0) \|^2}\). The vector potential \(\boldsymbol{A}\) that satisfies \(\nabla\times \boldsymbol{A}=\boldsymbol{B}\) is identified as \(-\frac{\mu_0 I}{2\pi}\ln\|\mathbf{k}\times(\boldsymbol{x}-\boldsymbol{x}_0)\|\mathbf{k}\). This solution is derived through conversion to cylindrical coordinates and intuition regarding vector fields.

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DavideGenoa
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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!
 
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Try to convert your problem to cylindrical coordinates.
 
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After some trials and a bit of intuition I have found a vector field whose curl is ##\boldsymbol{B}##:$$-\frac{\mu_0 I}{2\pi}\ln\|\mathbf{k}\times(\boldsymbol{x}-\boldsymbol{x}_0)\|\mathbf{k}$$
 

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