Question

A cosmic ray electron moves at 5.25 × 10⁶ m/s perpendicular to the Earth’s magnetic field at an altitude where the field strength is 1.0 × 10^-5 T.What is the radius, in meters, of the circular path the electron follows?

Answer 1

The charge enters the magnetic field perpendicularly with a speed v. The magnetic force acting on the charge will provide the centripetal force to make it move in a circle.

Let R be the radius of the circular path.

Charge, e=1.6×10^-19 C

Charge's speed v=5.25×10^6 m/s

Magntic field strength, B=1×10^-5 T

Massbof the electron, M=9.11×10^-31 kg

Centripetal force = Magnetic force

R= {(9.11×10^-31 kg) × (5.25×10^6 m/s)}/{(1.6×10^-19 C)×(1.0×10-5 T)}

R= 2.99m

R= 3.0 m