Question

The aurora is caused when electrons and protons, moving in the earth's magnetic field of ≈5×10−5 T, collide with molecules of the atmosphere and cause them to glow. What is the radius of the circular orbit for (a) an electron with speed 1.0×106m/s and (b) a proton with speed 7.0×104m/s. What is re in meters and what is rp in meters?

Answer 1

a) 0.114 meters is the radius of the circular orbit for an electron with given speed.

b) 14.6 meters is the radius of the circular orbit for a proton with given speed.

Step-by-step explanation:

Force on the charge particle :

`F = qvB`

Since, the particle is revolving around in a circular path ;

`F'=(mv^2)/(r)`

particle in constantly moving in circular path under action of magnetic field:

F = F'

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

`r=(mv)/(qB)`

Where :

r = radius of the circular path

m = mass of the particle

v = velocity of the particle

B = magnetic field

q = magnitude of charge on the particle

a) Velocity of the electron =
`v=1.0* 10^6 m/s`

Mass of electron = m =
`9.11* 10^(-31) kg`

Charge on an electron , q=
`1.602* 10^(-19) C`

B =
`5* 10^(-5) T`

Radius of the circular orbit for an electron with given speed =
`r_e`

`r_e=(9.11* 10^(-31) kg* 1.0* 10^6 m/s)/(1.602* 10^(-19) C* 5* 10^(-5) T)`

`r_e = 0.114 m`

0.114 meters is the radius of the circular orbit for an electron with given speed.

b) Velocity of the proton =
`v=7.0* 10^4 m/s`

Mass of electron = m =
`1.67* 10^(-27) kg`

Charge on an electron , q=
`1.602* 10^(-19) C`

B =
`5* 10^(-5) T`

Radius of the circular orbit for a proton with given speed =
`r_p`

`r_e=(1.67* 10^(-27) kg* 7.0* 10^4 m/s)/(1.602* 10^(-19) C* 5* 10^(-5) T)`

`r_p = 14.6 m`

14.6 meters is the radius of the circular orbit for a proton with given speed.