Question

Near Earth, the density of protons in the solar wind (a stream of particles from the Sun) is 12.5 cm-3, and their speed is 564 km/s. (a) Find the current density of these protons. (b) If Earth's magnetic field did not deflect the protons, what total current Earth receive?

Answer 1

Answer:

A) 11.28 x 10^(7) A.m²

B) 2.258 x 10^(17)A

Explanation:

A) The current density is given by the formula ;

J = nqv

Where n is the density of protons in the solar wind which is 12.5 cm³ or 12.5 x 10^(-6) m³

q is the proton charge which is 1.6 x 10^(-19) C

v is velocity which is 564km or 564000m

Thus, J = 12.5 x 10^(-6) x 1.6 x 10^(-19) x 564000 = 11.28 x 10^(7) A.m²

B) the formula for the total current the earth received is given as;

I = JA

The effective area is the cross section of the earth and thus,

Area = πr² where r is the radius of the earth given as: 6.371 x 10^(6)

A = π(6.371 x 10^(6)) ²

So I = 11.28 x 10^(7) x π(6.371 x 10^(6))² = 2.258 x 10^(17)A