Question

A proton moves in a circular path of the same radius as a cosmic ray electron moving at 6.5 × 10^6 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 of the circular path the electron follows?

Answer 1

3.7 m

Step-by-step explanation:

Velocity, v = 6.5 x 10^6 m/s

magnetic field, B = 1 x 10^-5 T

charge, q = 1.6 x 10^-19 C

mass, m = 9.1 x 10^-31 kg

Let r be the radius of the path.

`r=(mv)/(Bq)`

`r=(9.1* 10^(-31)* 6.5 * 10^(6))/(1* 10^(-6)* 1.6* 10^(-19))`

r = 3.7 m

Answer 2

3.7 m

Step-by-step explanation:

We are given that

Mass of electron,m=
`9.1* 10^(-31) kg`

`Speed of electron,v=6.5\time 10^6 m/s`

Magnetic filed,B=
`1.0* 10^(-5) T`

Charge on electron,q=
`1.6* 10^(-19) C`

We have to find the radius of circular path the electron follows.

We know that

Radius of circular path,r=
`(mv)/(qB)`

Using the formula

`r=(9.1* 10^(-31)* 6.5* 10^6)/(1.6* 10^(-19)* 1* 10^(-5))`

`r=3.7 m`

Hence, the radius of circular path followed by electron=3.7 m