Question

In a chamber , a uniform magnetic field of $6.5G(1G={10}^{-4}T)$ 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.Explain why the path of the electron is a circle.Determine the radius of the circular orbit.($e=1.6\times {10}^{-19}C,m_e=9.1\times {10}^{-31}kg$)

Answer 1

Force acting on an electron is always perpendicular to direction of motion and towards center of the circular path and hence its path will be a circle.

$\frac{m{v}^2}{r}=qvB$

$\therefore r=\frac{mv}{qB}=\frac{9.1\times {10}^{-31}\times 4.8\times {10}^6}{1.6\times {10}^{-19}\times 6.5\times {10}^{-4}}$

$=4.2\times {10}^{-2}m$

$=4.2cm$