Question

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?

Answer 1

Step-by-step explanation:

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

Answer 2

Radius = 3.96m

Step-by-step 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