Question

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

Answer 1

Hint: Magnetic force balance by centripetal force.
Magnetic field strength, $B=6.5\times {10}^{-4}\text{T}$
Charge of the electron, $e=1.6\times {10}^{-19}\text{C}$
Mass of the electron, $m_e=9.1\times {10}^{-31}\text{kg}$
Velocity of the electron, $v=4.8\times {10}^6\text{m}/\text{s}$
Radius of the orbit, $r=4.2\text{cm}=0.042\text{m}$

Frequency of revolution of the electron $=v$
Angular frequency of the electron, $\omega =2\pi v$
Velocity of the electron is related to the angular frequency as:
$v=r\omega$

In the circular orbit, the magnetic force on the electron is balanced by the centripetal force. Hence, we can write:
$evB=\frac{m{v}^2}{r}$
$eB=\frac{m}{r}\left(r\omega \right)=\frac{m}{r}\left(r2\pi \nu \right)$
$\nu =\frac{Be}{2\pi m}$

This expression for frequency is independent of the speed of the electron.

On substituting the known values in this expression, we get the frequency as:
$\nu =\frac{6.5\times {10}^{-4}\times 1.6\times {10}^{-19}}{2\times 3.14\times 9.1\times {10}^{-31}}$
$=18.2\times {10}^6\text{Hz}$
$\approx 18\text{MHz}$

Hence, the frequency of the electron is around $18\text{MHz}$ and is independent of the speed of the electron.