Electrons and protons travel from the Sun to the Earth at a typical velocity of 3.83 ✕ 105 m/s in the positive x-direction. Thousands of mil

Question

Electrons and protons travel from the Sun to the Earth at a typical velocity of 3.83 ✕ 105 m/s in the positive x-direction. Thousands of miles from Earth, they interact with Earth’s magnetic field of magnitude 3.04 ✕ 10−8 T in the positive z-direction. Find the magnitude and direction of the magnetic force on a proton. Find the magnitude and direction of the magnetic force on an electron.

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Thành Đạt 5 years 2021-08-22T22:54:15+00:00 1 Answers 56 views 0

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  1. Philomena
    0
    2021-08-22T22:55:30+00:00
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    Answer:

    F=2.84*10^{-26}N  & -y direction

    F=2.84*10^{-26}N & +y direction

    Explanation:

    From the question we are told that:

    Speed of electron V_e=3.83 * 10^5 m/s +x direction

    Earths magnetic field B_e=3.04 * 10^-^8 +z direction

    a)

    Generally the equation for magnetic force F_m is mathematically given by

    F=q(V_e*B_e)

    where

    q=1.6*10^{-19}c\\\=i*\=z=-\=j

    F=1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)

    F=1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)

    F=-2.84*10^{-26}N \=j

    Magnitude & Direction

    F=2.84*10^{-26}N  -y direction

    b)

    Generally the equation for magnitude and direction of the magnetic force on an electron. is mathematically given by

    \=F'=-1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)

    \=F'=-2.84*10^{-26}N \=j

    Magnitude & Direction

    F=2.84*10^{-26}N & +y direction

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