At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.45 ✕ 106 m/s perpendicular

Question

At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.45 ✕ 106 m/s perpendicular to the Earth’s magnetic field at an altitude where the field strength is 1.10 ✕ 10−5 T

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Delwyn 3 years 2021-08-08T01:01:01+00:00 1 Answers 25 views 0

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  1. Bo
    0
    2021-08-08T01:02:04+00:00
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    Answer:

    The speed of the proton is 4059.39 m/s

    Explanation:

    The centripetal force on the particle is given by;

    F = \frac{mv^2}{r}

    The magnetic force on the particle is given by;

    F = qvB

    The centripetal force on the particle must equal the magnetic force on the particle, for the particle to remain in the circular path.

    \frac{mv^2}{r} = qvB\\\\r = \frac{mv^2}{qvB} \\\\r = \frac{mv}{qB}

    where;

    r is the radius of the circular path moved by both electron and proton;

    ⇒For electron;

    r = \frac{(9.1*10^{-31})(7.45*10^6)}{(1.602*10^{-19})(1.1*10^{-5})}\\\\r = 3.847 \ m

    ⇒For proton

    The speed of the proton is given by;

    r = \frac{mv}{qB}\\\\mv = qBr\\\\v = \frac{qBr}{m} \\\\v = \frac{(1.602*10^{-19})(1.1*10^{-5})(3.847)}{1.67*10^{-27}} \\\\v = 4059.39 \ m/s

    Therefore, the speed of the proton is 4059.39 m/s

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