91Ó°ÊÓ

There is a beam of electrons with charge qand mass m.

There are uniform crossed electric and magnetic fields $\stackrel{→}{E}$and $\stackrel{→}{B}$which are mutually perpendicular and perpendicular to the beam of electrons.

The net force on a particle of charge qand velocity $\stackrel{→}{v}$moving in electric and magnetic fields $\stackrel{→}{E}$and $\stackrel{→}{B}$is

$\stackrel{\mathbf{→}}{\mathbf{F}}\mathbf{=}\mathbf{q}\stackrel{\mathbf{→}}{\mathbf{E}}\mathbf{+}\mathbf{q}\stackrel{\mathbf{→}}{\mathbf{v}}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

From equation (1), if the net force on a charged particle in an electric and magnetic field is zero,

$$\begin{gathered} qE=qvB \\ v=\frac{E}{B}.....\left(2\right) \end{gathered}$$

Thus, the speed of the particle getting zero deflection in cross electric and magnetic fields is $\frac{E}{B}$.

Linear momenta of a particle moving in a circular orbit in a magnetic field

$mv=qBR$

Here, $R$is the radius of the circular orbit.

Substitute expression for speed from equation (2)

$$\begin{gathered} m\frac{E}{B}=qBR \\ \frac{q}{m}=\frac{E}{B^{2}R} \end{gathered}$$

Thus, the charge to mass ratio of the particle is $\frac{E}{B^{2}R}$.