Ask Question
6 October, 03:38

Near Earth, the density of protons in the solar wind (a stream of particles from the Sun) is 11.6 cm-3, and their speed is 563 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?

+2
Answers (1)
  1. 6 October, 03:50
    0
    a) 10462341.6*10⁻¹³ A/m²

    b) 133709238.907 A

    Explanation:

    n = Density of the protons in the solar wind = 11.6 cm⁻³ = 11.6*10⁶ m⁻³

    v = Velocity of the protons = 563 km/s = 563000 m/s

    e = Charge of a proton = 1.602*10⁻¹⁹ coulombs

    R = Radius of Earth = 6.3781*10⁶ m

    A = Area of Earth = πR² = π (6.3781*10⁶) ²=127.8*10¹² m²

    a) Current density

    J = nev

    ⇒J = 11.6*10⁶*1.602*10⁻¹⁹*563000

    ⇒J = 10462341.6*10⁻¹³ A/m²

    ∴ Current density of these protons is 10462341.6*10⁻¹³ A/m²

    b) Current

    I = JA

    ⇒I = 10462341.6*10⁻¹³*127.8*10¹²

    ⇒I = 1337092389.07*10⁻¹

    ⇒I = 133709238.907 A

    ∴ Total current Earth receives is 133709238.907 A
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Near Earth, the density of protons in the solar wind (a stream of particles from the Sun) is 11.6 cm-3, and their speed is 563 km/s. (a) ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers