Question

A cosmic ray electron moves at 7.6 x 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.02 x 10-5 T. What is the radius of the circular path the electron follows

Answer 1

r = 4.23 m

Step-by-step explanation:

To find the radius of the circular path of the electron you use the following formula:

`r=(mv)/(qB)` (1)

This formula can be used because the motion of the electron is perpendicular to the direction of the magnetic field vector.

m: mass of the electron = 9.1*10^-31 kg

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

B: magnitude of the magnetic field = 1.02*10^-5 T

v: velocity of the electron = 7.6*10^6 m/s

You replace the values of m, v, q and B in the equation (1):

`r=((9.1*10^(-31)kg)(7.6*10^6 m/s))/((1.6*10^(-19)C)(1.02*10^(-5)T))\\\\r=4.23m`

hence, the raiuds of the orbit of the electron is 4.23m