Question

A cosmic ray electron moves at 6.5 x 106 m/s perpendicular to the Earth’s magnetic field at an altitude where field strength is 1.24 10-5 T. What is the radius (in meters) of the circular path the electron follows? The charge of an electron is 1.6 x 10-19 C and its mass is 9.11 x 10-31m. Your answer should be a number with two decimal places, do not include the unit.

Answer 1

Answer: R=2.98 m

Explanation: In order to solve this answer we have to use the dynamic of circular movement actioned by the Lorentz force, this is given by:

Fm=qvB ( condiring that v and B are perpendicular)

By using the second Newton law fro a circular movement, we have:

Fm=m*a=m*v^2/R  where v and R are the speed and R of the circular movement electrons in the magnetic field.

then

qvB=m*v^2/R

Finally R=m*v/(qB)=9.11*10^-31*6.5*10^6/(1.6*10^-19*1.24*10^-5)=2.98