Question

In a chamber, a uniform magnetic field of $6.5$ $G$ $(1G={10}^{-4}T)$ 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. $(e=1.5\times {10}^{-19}C,m_e=9.1\times {10}^{-31}kg)$.

Also obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Answer 1

At every instant, the direction of motion of the particle is perpendicular to the direction of force. Hence the path of the electron is circular.

We know that, at equilibrium,

$evB=m{v}^2/r$

also, $v=r\omega$

$\Rightarrow f=Be/2\pi m$

$f=18MHz$

Frequency is independent of velocity.