Electrons in Earth’s upper atmosphere have typical speeds near 5.87 ✕ 105 m/s. HINT (a) Calculate the magnitude of Earth’s magnetic field if

Question

Electrons in Earth’s upper atmosphere have typical speeds near 5.87 ✕ 105 m/s. HINT (a) Calculate the magnitude of Earth’s magnetic field if an electron’s velocity is perpendicular to the magnetic field and its circular path has a radius of 6.81 ✕ 10−2 m. T (b) Calculate the number of times per second that an electron circles around a magnetic field line.

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Mít Mít 3 years 2021-08-30T16:53:42+00:00 1 Answers 45 views 0

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  1. kimcuc
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    2021-08-30T16:54:51+00:00
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    Answer:

    a) 4.8×10^-5T

    b) 7.38×10^-7seconds

    Explanation:

    The centripetal force in a magnetic field is given by:

    qvB= mv2/r

    Where q= charge of the ion

    m= mass of the ion

    r= radius of the circular path

    B= magnetic field

    V= velocity

    Rearranging to make B subject of formular gives:

    B=mv/r

    B= [(9.11×10^-31)×(5.87×10^5)]/ [(1.6×10^-19)×(6.81×10^-2)]

    B = (5 34×10^-25)/(1.099×10^-20)

    B= 4.86×10^-5T

    b) T =( 2 pi × r)/v

    T = 2 × 3.142 x (6.81×10^-2) / (5.87×10^5

    T = 0.428/ (5.87×10^5)

    T = 7.38 × 10^-7seconds

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