Two long concentric cylindrical conductors of radii a and b `(b lt a)` are maintained at a potential difference V and carry equal opposite current I. Show that an electron with a particular velocity u parallel to the axis may travel undeviated in the evacuated region between the conductors.

Explanation

The electric field E in the region between the conductors varies with the diameter r from the axis as `E=(lambda)/(2pi 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=(lambda)/(2pi epsilon_(0))ln ((a)/(b))`
Combining these expressions to eliminate `lambda` gives
`E=(V)/(r ln ((a)/(b)))`
The magnetic field B varies with r as
`B=(mu_(0)I)/(2pir)`
If the inner cylinder is at a positive potential w.r.t the outer cylinder, the electric field 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 travelling with velocity u into the page, as a radial distance r from the axis of the cylinder. It experience an electric force Ee raidally outwards. In order for the electron to be undeviated, these forces must balance so that `u=E//B`. Sucstituting out expressions for E and B gives.
This expression in independent of r, so the electrons's position does not matter.