What is the Electric Current Produced by Solar Wind Protons Near Earth?

In summary, the solar wind that streams off the sun at a speed of 400 km/s contains roughly equal numbers of electrons and protons. Near the Earth, a cube with a side length of 1 m contains approximately 8.7 * 10^6 protons with a velocity of 470 km/s. By considering the movement of these protons, we can find the electric current produced by the flow through a cross-sectional area of 1 m^2, which is equal to 6.6 * 10^(-7). While the solar wind may not directly affect the current, it is important to consider its composition when solving this problem.
  • #1
Physicsrapper
24
0

Homework Statement


Here is the Task:
The solar wind streams off of the sun in all directions at Speeds of about 400 km/s.
It contains roughly equal number of electrons and Protons.
Near the Earth we find in a cube with a side length of 1 m about 8.7 * 10^6 Protons. The velocity of These Protons is 470 km/s.

Find the electric current produced by the flow of These Protons through a cross-sectional area of 1 m^2.


Homework Equations


The solution should be: 6.6 * 10^(-7)



The Attempt at a Solution


I really don't have an idea how to integrate the solar wind into my equation.
Why is

(8.7 * 10^6 * 1m^2 * (1.6 * 10^(-19)) / 470 000 m/s

wrong?

What has the solar wind to do with that cube?
Can someone please help me and give me some tips?

Thanks in advance!
 
Physics news on Phys.org
  • #2
I don't have a clue on how to treat the solar wind. If "It contains roughly equal number of electrons and Protons" then it should be electrically neuter, so it would not affect the current in any way.

However I arrived to the solution you mention without considering the wind. Here is how I see it: if we have 8.7 · 10^6 Protons in a 1m² cube then we have [itex]8.7 · 10^6 \frac{protons}{m}[/itex] (note: longitudinal meter here).

If the protons are moving at 470 Km/s, or 470 000 m/s then we have:

[itex] 1.6 · 10 ^{-19} \frac{Coulombs}{proton} \times 8.7 · 10^6 \frac{protons}{m}\times 470 000 \frac{m}{s} = 6.5424 · 10^{-7} Coulombs[/itex]

Which agrees with your solution.
 
  • #3
Now I understood. Thank you very much for your help.
 
Back
Top