Earth's atmosphere is constantly bombarded by cosmic ray protons that originate somewhere in space. If the protons all passed through the atmosphere, each square meter of Earth's surface would intercept protons at the average rate of 1800 protons per second. What would be the electric current in amperes intercepted by a 14 x 107 km2 area on the planet?
1 km^2 = 10^6 m^2
Current = charge/unit time = charge of proton * number of
protons/second
Current = charge of proton * protons/m^2/s^* area
= 1.6x10^-19 C * 1800 protons/m^/s * 14x10^7x10^6 m^2 =
0.04032A
Earth's atmosphere is constantly bombarded by cosmic ray protons that originate somewhere in space. If the...
Question 9 Earth's atmosphere is constantly bombarded by cosmic ray protons that originate somewhere in space. If the protons all passed through the atmosphere, each square meter of Earth's surface would intercept protons at the average rate of 1900 protons per second. ha would be the electric current nam ere ntercepte b a 2· km2 area on the planet? ' Number the tolerance is +/-590 Click if you would like to Show Work for this question: Units Open Show Work
5. Protons in cosmic rays strike the Earth's upper atmosphere at an average rate of 1200 protons per square meter per second. What total current (in Coulombs per second, or Amperes) does the Earth receive from these cosmic rays? (For the radius of the Earth, use Re- 6.38 x 10s meters.)
Protons in cosmic rays strike the Earth’s upper atmosphere at an average rate of 1200 protons per square meter per second. What total is the current (in Coulombs per second, or Amperes) does the Earth receive from these cosmic rays? (For the radius of the Earth, use R^E = 6.38 x 10^6 meters.)
A large number of energetic cosmic-ray particles reach Earth's atmosphere continuously and knock electrons out of the molecules in the air. Once an electron is released, it responds to an electrostatic force that is due to an electric field E produced in the atmosphere by other point charges. Near the surface of Earth, this electric field has a magnitude of |E| = 150 N/C and is directed downward. Calculate the change (in J) in electric potential energy of a released...