Answer :

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.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

A. Obtain the expPhysics - Board Papers

A charged particlPhysics - Exemplar

Describe the motiPhysics - Exemplar

State the underlyPhysics - Board Papers

In a cyclotron, aPhysics - Exemplar

A magnetic field NCERT - Physics Part-I