Question

Electrons emitted with negligible speed from an electron gun are accelerated through a potential difference V along the x-axis. These electrons emerge from a narrow hole into a uniform magnetic field B directed along this axis. However, some of the electrons emerging from the hole make slightly divergent angles, as shown in the figure. Show that these paraxial electrons are refocussed on the x-axis at a distance $\sqrt{\frac{8{\mathrm{\pi }}^{2}mV}{e{B}^2}}$.

Answer 1

Given:
Electrons are accelerated through a potential difference = V
Let the mass of an electron be m and the charge of an electron be e.
We know:
Electric field, E = $\frac{V}{r}$
Force experienced by the electron, F = eE
Acceleration of the electron, a = $\frac{eV}{rm}$
Using the equation of motion
v² − u²= 2 × a × s,
v² = 2 × a × s (As u = 0)
Here, s = r
$$\begin{gathered} v=\sqrt{\frac{2eVr}{rm}} \\ =\sqrt{\frac{2eV}{m}} \end{gathered}$$
Time taken by electron to cover the curved path,
$T=\frac{2\pi m}{eB}$
As the acceleration of the electron is along the y axis only, it travels along the x axis with uniform velocity.

Velocity of the electron moving along the field remains v.
Therefore, the distance at which the beam is refocused, d = v × T
$$\begin{gathered} d=\sqrt{\frac{2eV}{m}}\times \frac{2\pi m}{eB} \\ d=\sqrt{\frac{8{\pi }^{2}mV}{e{B}^2}} \end{gathered}$$