Current Density and Current

[Sanjay101]

Near Earth, the density of protons in the solar wind (a stream of particles from the Sun) is 8.90 cm-3 and their speed is about 490 km/s.

a. Find the current density of these protons.



b. If Earth's magnetic field did not deflect the protons, what total current would Earth receive?

 

[vanesch]

You should:

1) post homework questions in our homework section

2) follow the rules there, of which the main rule is: show some work first.

I move this to hw, but you shouldn't expect any answer before you show some work.

 

[m2003]

a. To find the current density of these protons, we can use the equation J = nqv, where J is the current density, n is the number density of protons, q is the charge of a proton, and v is the speed of the protons. Plugging in the given values, we get:

J = (8.90 cm^-3)(1.602 x 10^-19 C)(490 km/s)
= 6.89 x 10^-14 A/cm^2

Therefore, the current density of these protons near Earth is 6.89 x 10^-14 A/cm^2.

b. If Earth's magnetic field did not deflect the protons, the total current received by Earth would be the product of the current density and the cross-sectional area of Earth's magnetic field. Assuming Earth's magnetic field has a radius of 6371 km, the cross-sectional area would be π(6371 km)^2 = 1.28 x 10^14 cm^2. Therefore, the total current would be:

6.89 x 10^-14 A/cm^2 x 1.28 x 10^14 cm^2 = 8.81 A

This is a significant amount of current that Earth would receive from the solar wind if not for the protection of its magnetic field. This current could potentially cause disruptions to electronic systems and communication networks on Earth.