Generals 2002 II 6

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:

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