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

Explanation

Magnetic field strength, `B = 6.5 xx 10^(-4)T`
Charge of the electron, `e = 1.6 xx 10^(-19)C`
Mass of the electron, `m_(e) = 9.1 xx 10^(-31)kg`
Velocity of the electron, `V =4.8 xx 10^(6)m//s`
Radius of the orbit, `r = 4.2 cm = 0.042 cm`
Frequency of revolution of the electron `= v`
Angular frequency of the electron `= omega = 2nv`
Velocity of the electron is related to the angular frequency as:
`v = r omega`
In the circuit orbit the magnetic force on the electron is obtained by the centripetal force. Hence, we can write:
`evB = (mv^(2))/(r)`
`eB = (m)/(r) (romega) = (m)/(r)(r2pi v)`
`v = (Be)/(2pim)`
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 =(6.5 xx 10^(-4) xx 1.6 xx 10^(-9))/(2 xx 3.14 xx 9.1 xx 10^(-11))/(2 xx 3.14 xx 9.1 xx 10^(-11))`
`=18.2 xx 10^(6)Hz`
`~~18 MHz`
Hence, the frequency of the electron is around `18MHz` and is independent of the speed of the electron.