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19 July, 01:38

5) An electron gun sends out a beam of electrons whose kinetic energies are all about 57 μμeV. (1 μeV=1.6*10-25 J1 μeV=1.6*10-25 J.) You need to set up a magnetic field perpendicular to the beam that causes it to turn through a 90∘∘ circular arc of length 2 mm. How strong must the magnetic field be? [Note: electrons have a mass of about 9.11*10-31 kg.]

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  1. 19 July, 02:06
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    2 x 10^-5 Tesla

    Explanation:

    Kinetic energy, K = 57 micro electron volt = 57 x 1.6 x 10^-25 J

    Length of arc = 2 mm

    Angle turn = 90° = π/2 = 1.57 radian

    Let the strength of magnetic field is B.

    mass of electron, m = 9.11 x 10^-31 kg

    K = 1/2 x mv²

    where, v be the velocity of electron

    57 x 1.6 x 10^-25 = 0.5 x 9.11 x 10^-31 x v²

    v = 4474.59 m/s

    radius = arc length / angle = 2 / 1.57 = 1.27 mm

    By the formula for the radius

    r = mv / Bq

    B = mv / qr

    B = (9.11 x 10^-31 x 4474.59) / (1.6 x 10^-19 x 1.27 x 10^-3)

    B = 2 x 10^-5 tesla

    Thus, the magnetic field is 2 x 10^-5 Tesla.
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