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14 November, 18:27

An electron enters a region of space containing a uniform 2.71 * 10 - 5 2.71*10-5 T magnetic field. Its speed is 197 197 m/s and it enters perpendicularly to the field. Under these conditions, the electron undergoes circular motion. Find the radius r of the electron's path and the frequency f of the motion.

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  1. 14 November, 19:15
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    r = 0.0414mm

    F = 757,692.3Hertz

    Explanation:

    If the body enters space with uniform magnetic field B, the force experienced by the object is expressed as

    F = qvBsintheta ... 1

    Also, if the body undergoes a circular motion, the force experienced by the body in a circular path is given as

    Fc = mv²/r ... 2

    Equating both forces

    F = Fc

    qvBsin theta = mv²/r

    Since the body enters perpendicular to the field, theta = 90°

    The equality becomes;

    qvB sin90° = mv²/r

    qvB = mv²/r

    qB = mv/r

    r = mv/qB

    Given mass of the electron m = 9.11*10^-31kg

    Velocity of the object v = 197m/s

    Charge on the electron q = 1.6*10^-19C

    Magnetic field B = 2.71*10^-5T

    Substituting this value into the equation to get the radius r we have;

    r = 9.11*10^-31 * 197/1.6*10^-19 * 2.71*10^-5

    r = 1794.67*19^-31/4.336*10^-24

    r = 413.89*10^-7

    r = 0.0000414m

    r = 0.0414mm

    b) Frequency of the motion F = w/2π where w is the angular velocity

    Since w = v/r

    F = (v/r) / 2π

    F = v/2πr

    F = 197/2π (0.0000414)

    F = 757,692.3Hertz
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