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3 April, 01:16

An electron traveling toward the north with speed 4.0 * 105 m/s enters a region where the Earth's magnetic field has the magnitude 5.0 * 10-5 T and is directed downward at 45° below horizontal. What is the magnitude of the force that the Earth's magnetic field exerts on the electron? (e = 1.60 * 10-19 C)

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  1. 3 April, 04:48
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    Speed of electron

    v = 4 * 10^5 •j m/s

    Magnetic field

    B = 5 * 10^-5 T at angle of 45° to horizontal

    Charge of electron

    q = 1.6 * 10^-19C

    Magnitude of force F?

    The Force exerted in an electric field is given as

    F = q (v*B)

    Now, x component of the magnetic field

    Bx = BCos45 = 5*10^-5 Cos45

    Bx = 3.54 * 10^-5 •i T

    Also, y component

    By = BSin45 = 5 * 10^-5Sin45

    By = 3.54 * 10^-5 •j T

    B = 3.54 * 10^-5 •i + 3.54 * 10^-5 •j T

    Now, F = q (v*B)

    Note that,

    i*i=j*j=k*k=0

    i*j = k, j*k = i, k*i = j

    j*i = - k, k*j = - i and i*k = - j

    Therefore

    F = q (v*B)

    F = 1.6*10^-19 (4*10^5•j * (3.54 * 10^-5 •i + 3.54 * 10^-5 •j T))

    F = 1.6*10^-19 (4*10^5 * 3.54 * 10^-5 (j*i) + 4*10^5 * 3.54 * 10^-5 (j*j))

    F = 1.6*10^-19 (14.14 (-k) + 0)

    F = - 2.26 * 10^-18 •k N

    It is in the negative direction of z axis

    The magnitude of the force the field experience is 2.26 * 10^-18 N
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