Q12 of 31 Page 169

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.


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