Question

A non-relativistic charged particle flies through the electric field of a cylindrical capacitor and gets into a uniform transverse magnetic field with induction $B$ (Fig.). In the capacitor the particle moves along the arc of a circle, in the magnetic field, along a semi-circle of radius $r$. The potential difference applied to the capacitor is equal to $V$, the radii of the electrodes are equal to $a$ and $b$, with $a<b$. Find the velocity of the particle and its specific charge $q/m$.

Answer 1

Inside the capacitor, the electric field follows $a\left(\frac{1}{r}\right)$ law, and so the potential can be written as

$\phi =\frac{V\ln \frac{r}{a}}{\ln \frac{b}{a}},$

$E=\frac{-V}{r\ln \frac{b}{a}}$

Here $r$ is the distance from the axis of the capacitor.

Also, $\frac{m{v}^2}{r}=\frac{qV}{r\ln \frac{b}{a}}$

$\Rightarrow m{v}^2=\frac{qV}{\ln \frac{b}{a}}$

On the other hand,

$mv=qBr$ in the magnetic field.

Thus,$\nu =\frac{V}{Br\ln \frac{b}{a}}$

$m=\frac{\nu }{Br}=\frac{V}{B^2{r}^2\ln \frac{b}{a}}$