A cosmic ray electron moves at 7.65 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where field strength is 1.10 ✕ 10-5 T. What is the radius (in m) of the circular path the electron follows?

Force of a magnetic field, Fm = q * v * B

Centipetal force, Fa = (M * v^2) / r

Therefore, Fm = Fa

q * v * B = (M * v^2) / r

r = (m * v) / (q * B)

Where,

r = radius

m = mass of electron

= 9.1 * 10^-31 kg

q = electron charge

= 1.602 * 10^-19 C

B = magnetic field

= 1.10 ✕ 10^-5 T

v = velocity

= 7.65*10^6 m/s

r = (9.1 * 10^-31) * (7.65 * 10^6) / (1.6 * 10^-19) * (1.10 * 10^-5)

= 3.95 m

Radius = 3.96m

Explanation:

In a cyclotron motion, the radius of a charged particle path in a magnetic feild is given by:

r = mv/qB

Where r = radius

m = mass of particle = 9.1*10^-31kg

q = charged electron = 1.6*10^-19C

B = magnetic feid = 7.65*10^6T

V = velocity = 7.65*10^6

r = (9.1*10^-31) * (7.65*10^6) / (1.6*10^-19) (1.10 * 10^-5)

r = (6.9615*10^-24) / (1.76 * 10^-24)

r = 3.96m