In 1897, J. J. Thomson "discovered" the electron by measuring the

charge-to-mass ratio of "cathode rays" (actually, streams of electrons, with charge qand mass m)as follows:

(a) First he passed the beam through uniform crossed electric and magnetic fields $\overrightarrow{\mathbf{E}}$and $\overrightarrow{\mathbf{B}}$(mutually perpendicular, and both of them perpendicular to the beam), and adjusted the electric field until he got zero deflection. What, then, was the speed of the particles in terms of $\overrightarrow{\mathbf{E}}$and $\overrightarrow{\mathbf{B}}$)?

(b) Then he turned off the electric field, and measured the radius of curvature, R,

of the beam, as deflected by the magnetic field alone. In terms of E, B,and R,

what is the charge-to-mass ratio (qlm)of the particles?

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

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

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

$\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{q}\overrightarrow{\mathbf{E}}\mathbf{+}\mathbf{q}\overrightarrow{\mathbf{v}}\mathbf{\times }\overrightarrow{\mathbf{B}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

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 $\frac{E}{B^{2}R}$.