એક ઓરડામાં, $6.5 \;G \left(1 \;G =10^{-4} \;T \right)$ જેટલું નિયમિત ચુંબકીય ક્ષેત્ર રાખેલું છે. આ ક્ષેત્રમાં લંબ રૂપે એક ઇલેક્ટ્રૉન $4.8 \times 10^{6} \;m s ^{-1}$  ઝડપે છોડવામાં આવે છે. વર્તુળાકાર કક્ષામાં ઈલેક્ટ્રૉનના ભ્રમણની આવૃત્તિ શોધો. શું આ જવાબ ઈલેક્ટ્રૉનની ઝડપ પર આધાર રાખે છે? સમજાવો.

$\left(e=1.5 \times 10^{-19} \;C , m_{e}=9.1 \times 10^{-31}\; kg \right)$

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.