How Is Current Density Calculated for Protons in the Solar Wind?

[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?For part a I used Current density (J) = nqv, we know q and v. I tried finding n(number of protons) by finding the volume by m/density. Then using Avogadro's number/volume to get n.

For part b I said that the Current would just me Part a answer multiplied by the cross sectional area, but i did not get part a right.

 

[vanesch]

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.

For part a I used Current density (J) = nqv, we know q and v. I tried finding n(number of protons) by finding the volume by m/density. Then using Avogadro's number/volume to get n.

In n.q.v, n represents the NUMBER density (that means, the number of charges with charge q per unit of volume). In this case, this would mean the number of protons per m^3, or per cm^3 (depending on the unit system in which you work). But that's given! The solar wind consists of 8.9 PROTONS per cubic cm (and not, as you seem to think, 8.9 GRAM of protons per cubic cm - which would be a terribly dense charge density!)

 

[Sanjay101]

o, i cannot believe i mistook that info. Thanks.

 

[mussgo]

im having trouoble with part a myself i am using vne=J
i have to convert km/s to m/s cm^-3 to m^-3 and e is 1.6e-19

i plug
500000*840*1.6x10^-19 but its wrong can i get help U_U ?