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23 February, 13:31

An electron travels into a 0.5 T magnetic field perpendicular to its path, where it moves in a circular arc of diameter 0.028 m. What is the kinetic energy of the electron?

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  1. 23 February, 13:47
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    6.89 * 10⁻¹³ Joules

    Explanation:

    Radius of the circular arc = diameter / 2 = 0.028 m / 2 = 0.014 m

    radius of the circular arc = mev / qB

    me = 9.11 * 10⁻³¹ kg (mass of an electron)

    q of an electron = 1.6 * 10⁻¹⁹ C

    B = 0.5 T

    v velocity of the electron = r qB / me = (0.014 m * 1.6 * 10⁻¹⁹ C * 0.5 T) / (9.11 * 10⁻³¹ kg) = 0.00123 * 10 ¹² m/s = 1.23 * 10⁹ m/s

    Kinetic energy = 1/2 mv² = 0.5 * 9.11 * 10⁻³¹ kg * (1.23 * 10⁹ m/s) ² = 6.89 * 10⁻¹³ Joules
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