In a chamber,a uniform magnetic field of $6.5 \; G$ $(1 \; G = 10^{-4} \; T)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \; m s^{-1}$ normal to the field. The radius of the circular orbit of the electron is $4.2 \; cm$. Obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain. $(e = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$

(N/A) Magnetic field strength,$B = 6.5 \times 10^{-4} \; T$.
Charge of the electron,$e = 1.6 \times 10^{-19} \; C$.
Mass of the electron,$m_{e} = 9.1 \times 10^{-31} \; kg$.
Velocity of the electron,$v = 4.8 \times 10^{6} \; m/s$.
Radius of the orbit,$r = 4.2 \; cm = 0.042 \; m$.
The frequency of revolution of the electron is given by $\nu = \frac{\omega}{2\pi}$.
Since the magnetic force provides the centripetal force,$evB = \frac{mv^2}{r}$,which simplifies to $v = \frac{r\omega}{2\pi} = \frac{r(evB/mr)}{2\pi} = \frac{eB}{2\pi m}$.
Substituting the values: $\nu = \frac{6.5 \times 10^{-4} \times 1.6 \times 10^{-19}}{2 \times 3.14 \times 9.1 \times 10^{-31}} \approx 18.2 \times 10^{6} \; Hz = 18 \; MHz$.
Since the expression $\nu = \frac{eB}{2\pi m}$ does not contain the velocity $v$,the frequency is independent of the speed of the electron.