Question

An electron is accelerated through 2400 V from rest and then enters a region where there is a uniform 1.70-T magnetic field. What are (a) the maximum and (b) the minimum magnitudes of the magnetic force acting on this electron.

Answer 1

The potential diffrence in which the electron is accelerating is (V) =2400V

The uniform magnetic field is given by(B) =1.70T

We know that the magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field with direction given by the right hand rule. The force is given by the charge times the vector product of velocity and magnetic field. Therefor the force acting on an electon in a uniform magnetic field is given by F=Bqvsintheta

Force becomes maximum when theta =90degrees force becomes minimum when theta =0 degrees

We know that the velocity of electon which is acccelerating in potential difference V is given by

v =Srt(2qV/me)

Now the maximum magnitude of force acting on the electron is given by

Fmax =Bqv =BqSqrt(2qV/me)

Where q is the charge of the electron q =1.6*10^-19C

me is the mass of an electron is given by me =9.1*10^-31kg

Now Fmax =(1.70)(1.6*10^-19C)Sqrt(2*1.6*10^-19*2400)/9.1*10^-31kg) =7.9018*10^-12N

b) The minimum force acting on the electron will be zero when angle is zero degrees