Question

a uniform magnetic field of 6.5multiplied by 10^{- 4} T is maintained in a chamber. An electron enters into the field with a speed of 4.8 multiplied by 10⁶ m/s normal to the field. Determine its frequency of revolution in the circular orbit.

Answer 1

Dear Student,

Angle between the shot electron and magnetic field, θ = 90°

Magnetic force exerted on the electron in the magnetic field is given as :
F = evB sinθ

This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.
Hence, centripetal force exerted on the electron,
${\mathrm{F}}_{\mathrm{c}}=\frac{{\mathrm{mv}}^2}{\mathrm{r}}$
In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force, i.e.,
$$\begin{gathered} {\mathrm{F}}_{\mathrm{c}}=\mathrm{F} \\ \frac{{\mathrm{mv}}^2}{\mathrm{r}}=\mathrm{evB}\mathrm{sin\theta } \\ \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{eB}\mathrm{sin\theta }} \\ \mathrm{where},\mathrm{m}=\mathrm{mass}\mathrm{of}\mathrm{electron} \\ \mathrm{e}=\mathrm{charge}\mathrm{of}\mathrm{electron} \\ \mathrm{r}=\frac{9.1\times {10}^{-31}\times 4.8\times {10}^6}{1.6\times {10}^{-19}\times 6.5\times {10}^{-4}\times \sin 90°} \\ \mathrm{r}=4.2\times {10}^{-2}\mathrm{m} \\ \mathrm{time}\mathrm{period}\mathrm{of}\mathrm{revolution},\mathrm{T}=\frac{2\mathrm{\pi r}}{\mathrm{v}} \\ \mathrm{frequency}\mathrm{of}\mathrm{revolution},\mathrm{f}=\frac{1}{\mathrm{T}} \\ \mathrm{f}=\frac{\mathrm{v}}{2\mathrm{\pi r}} \\ \mathrm{f}=\frac{4.8\times {10}^6}{2\times 3.14\times 4.2\times {10}^{-2}} \\ \mathrm{f}=1.82\times {10}^7\mathrm{Hz} \end{gathered}$$