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

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SUMMARY

The electric current produced by solar wind protons near Earth can be calculated using the formula: I = n * q * v, where n is the number of protons per unit area, q is the charge of a proton, and v is the velocity of the protons. In this case, with approximately 8.7 * 10^6 protons in a 1 m² area and a velocity of 470 km/s, the resulting current is 6.6 * 10^(-7) Amperes. The solar wind's composition, containing equal numbers of electrons and protons, does not negate the current produced by the protons due to their motion.

PREREQUISITES
  • Understanding of electric current and its calculation
  • Familiarity with the charge of a proton (1.6 * 10^(-19) Coulombs)
  • Knowledge of basic physics concepts such as velocity and density
  • Ability to perform unit conversions (e.g., km/s to m/s)
NEXT STEPS
  • Study the principles of electric current in plasma physics
  • Learn about the properties of solar wind and its effects on Earth's magnetosphere
  • Explore the mathematical modeling of charged particle flow
  • Investigate the relationship between particle density and electric current in different mediums
USEFUL FOR

Students in physics, particularly those studying electromagnetism and plasma physics, as well as researchers interested in solar wind interactions with planetary atmospheres.

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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!
 
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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 8.7 · 10^6 \frac{protons}{m} (note: longitudinal meter here).

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

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

Which agrees with your solution.
 
Now I understood. Thank you very much for your help.
 

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