Chapter 26, Problem 012 Near Earth, the density of protons in the solar wind (a

Question

Chapter 26, Problem 012 Near Earth, the density of protons in the solar wind (a stream of particles from the Sun) is 12.1 cm-3, and their speed is 559 km/s. (a) Find the current density of these protons. (b) If Earth's Near Earth, the density of protons in the solar wind (a stream of particles from the Sun) is 12.1 cm , and their speed is 559 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? (a) Number (b) Number Units SE Units

Explanation / Answer

A.

Current density is given by:

J = q*d*V

q = 1.6*10^-19

d = density of proton = 12.1 protons/cm^3 = 12.1*10^6 protons/m^3

V = 559 km/sec = 559000 m/sec

So,

J = 1.6*10^-19*12.1*10^6*559000

J = 1.082*10^-6 A/m^2

B.

J = i/A

i = J*A

A = pi*Re^2

i = 1.082*10^-6*pi*(6.37*10^6)^2

i = 1.379*10^8 Amp.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.