An electron moves at 2.50*106 m/s through a region inwhich there is a magnetic field of unspecified direction andmagnitude 7.40*10-2 t. (a) what are the largest andsmallest possible magnitudes of the acceleration of the electrondue to the magnetic field? (b) if the actual acceleration of theelectron is one-fourth of the largest magnitude in part (a), whatis the angle between the electron velocity and the magneticfield?

To solve this problem it is necessary to apply the concepts related to the magnetic field and the magnetic force.

By definition we know that the magnetic force is given by

Where,

v = velocity

B= magnetic field

q= charge of a Proton

We can write this equation as function of the angle between the velocity and magnetic field, i.e,

PART A) Using Newton's second law, we can write that

Equation both equation:

The planes between velocity and magnetic field are perpendicular, then

Replacing

Therefore the magnitude of maximum and minimum acceleration is

PART B) We can now calculate the angle between the electron velocity and the magnetic field through:

According the statment the electrone is one-fourth of the largest magnitude of acceleration in a. Then:

Therefore the angle between the electron velocity and the magnetic field is 14.47°

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge