In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
Given:
Magnetic field strength, B = 6.5 G = 6.5 × 10-4 T
Initial velocity of electron = 4.8 × 106 ms-1
Angle between the initial velocity of electron and magnetic field, θ = 900
We can relate the velocity of the electron to its angular frequency by the relation,
V = rω …(1)
Where,
V = velocity of electron
r = radius of path
ω = angular frequency
We understand that, magnetic force on the electron is equal to the centripetal force on it, hence we can write,
Fe = Fc
e × V × B = mV2/r …(2)
From equation (2) we can write,
⇒ eB = mV/r …(3)
Now, by putting value of V from equation (1) in equation (3)
⇒ e × B = m × ω …(4)
We know that, ω = 2πν
Putting in equation (4), we have
…(5)
Now, by putting the values in equation (5) we get,
⇒
⇒ v = 18.2 × 106Hz
Hence, the frequency of rotation is 18.2 × 106Hz.
Note: The frequency of rotation is independent of the initial velocity of the electron.