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In a chamber, a uniform magnetic field of $6.5 \;G \left(1 \;G =10^{-4} \;T \right)$ 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.
$\left(e=1.5 \times 10^{-19} \;C , m_{e}=9.1 \times 10^{-31}\; kg \right)$
Solution
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, $r=4.2 \,cm =0.042\, m$
Frequency of revolution of the electron $=v$ Angular frequency of the electron $\omega=2 \pi \theta$
Velocity of the electron is related to the angular frequency as:
$v=r \omega$
In the circular orbit, the magnetic force on the electron provides the centripetal force. Hence, we can write:
$e v B=\frac{m v^{2}}{r}$
$e B=\frac{m}{r}(r \omega)=\frac{m}{r}(r 2 \pi v)$
$v=\frac{B e}{2 \pi m}$
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:
$v=\frac{6.5 \times 10^{-4} \times 1.6 \times 10^{-19}}{2 \times 3.14 \times 9.1 \times 10^{-31}}$
$=18.2 \times 10^{6}\, Hz$
$\approx 18 \,M\,Hz$
Hence, the frequency of the electron is around $18 \,M\,Hz$ and is independent of the speed of the electron.
