Two long concentric cylindrical conductors of radii $a$ and $b(b<a)$ are maintained at a potential difference and  carry equal and opposite currents $I$. Find a particular velocity $v$ of an electron parallel to the axis for which it may travel undeviated in the evacuated region between the conductors.

The electric field $E$ in the region between the conductors varies with the distance $r$ from the axis as 

$E=\frac{\lambda }{2\pi {\epsilon }_{0}r},$
where $\lambda$ is the charge per unit length on the inner cylinder, and we also know that the potential difference $V$ between the conductors is given by
$V=\frac{\lambda }{2\pi {\epsilon }_0}\ln \left(\frac{a}{b}\right)\text{. }$
Combining these expressions to eliminate $\lambda$ gives
$E=\frac{v}{r\ln \left(\frac{a}{b}\right)}.$
The magnetic field $\mathrm{B}$ varies with $r$ as
$B=\frac{{\mu }_{0}I}{2\pi r}$
If the inner cylinder is at a positive potential w.r.t. the outer cylinder, the electric field $\mathrm{E}$ is directed radially outwards. If we assume that current flows into the page along the inner cylinder and out of the page along the outer cylinder, the magnetic field will be directed clockwise as shown.

Consider an electron traveling with velocity $v$ into the page, at a radial distance $r$ from the axis of the cylinders. It experiences an electric force $E e$ radially inwards, and a magnetic force Beu radially outwards In order for the electron to be undeviated, these forces must balance so that $v=E / B$. Substituting out expressions for $\mathrm{E}$ and $\mathrm{B}$ gives
$v=\frac{\frac{V}{r\ln (a/b)}}{\frac{{\mu }_{0}I}{2\pi r}}=\frac{2\pi V}{{\mu }_{0}I\ln (a/b)}.$
This expression is independent of $r$, so the electron's position does not matter.