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I am working on a problem that states the following:

Imagine an infinite straight wire carrying a current $I$ and uniformly charged to a negative electrostatic potential $\phi$

I know here that the current $I$ will set up a magnetic field around the wire that abides to the right hand rule with magnitude:

$ B(r) = \frac{I\mu_0}{2\pi r}. $

However, what is the importance of there being a negative electrostatic potential $\phi$? Does this mean that the wire sets up an electrostatic $\vec{E}$ field in addition to the magnetic field?

Loonuh
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1 Answers1

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We would have to know the rest of the problem to be sure, but most likely it means what you say. A wire connected to a DC source will in fact have a current going through it and it will be at some nearly uniform potential. If we wanted to solve Laplace's equation to find the potential everywhere, the wire would work as a boundary condition.

Javier
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