Question

A cosmic ray electron moves at 6.5x 10^6 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 10x 10^-5 T. What is the radius, in meters, of the circular path the electron follows?

Answer 1

Radius, r = 0.36 meters

Step-by-step explanation:

It is given that,

Speed of cosmic ray electron,
`v=6.5* 10^6\ m/s`

Magnetic field strength,
`B=10* 10^(-5)\ T=10^(-4)\ T`

We need to find the radius of circular path the electron follows. It is given by :

`qvB=(mv^2)/(r)`

`r=(mv)/(qB)`

`r=(9.1* 10^(-31)\ kg* 6.5* 10^6\ m/s)/(1.6* 10^(-19)* 10^(-4)\ T)`

r = 0.36 meters

So, the radius of circular path is 0.36 meters. Hence, this is the required solution.