2002 II 6

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The radial electric field of a cylinder of constant charge is:

\frac{\partial E_{r}}{\partial r}=4\pi q\bar{n}
E_{r}\left(a\right)=2\pi aq\bar{n}

The magnetic field is given by:

\nabla\times\mathbf{B}=\frac{4\pi}{c}\mathbf{J}
B_{\theta}=2\pi a\bar{n}\frac{v}{c}

To be in force balance:

q\left(\mathbf{E}+\frac{\mathbf{v}}{c}\times\mathbf{B}\right)=0

Plugging in for the radial direction:

q\left(2\pi aq\bar{n}-\frac{v}{c}2\pi a\bar{n}\frac{v}{c}\right)=0

Which implies:

\frac{v^{2}}{c^{2}}=1

This page was recovered in October 2009 from the Plasmagicians page on Generals_2002_II_6 dated 06:09, 14 May 2007.

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