Protons in cosmic-rays strike the earth's over the earth's surface of `10` protons `m^(-2)s^(-1)`. What dioes the total current the earth receive from beyond its atmosphere in the form of incident consmic ray proton ? The earth's radius is `6.4 xx 10^(6)m`

Answer

`0.82 xx 10^(-7) A`

Explanation

Change on proton `= + e = 1.6 xx 10^(-19)C`
Surface area of earth`= 4 pi R^(2)`
`= 4 xx 3.14 xx (6.4 xx 10^(6))^(2)`
`= 5.14 xx 10^(14)m^(2)`
Charge receved by earth per second per `m^(2)`
`= n e = 10 xx 1.6xx 10^(-19)C`
So current = rate flow of change
`= n e = 4 pi R^(2)`
`= (10 xx 1.6 xx 10^(-19)) xx 5.14 xx 10^(14)`
`= 0.82 xx 10^(-3) A`