Ask Question
16 April, 17:26

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?

+5
Answers (1)
  1. 16 April, 21:17
    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 you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers