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−

Question

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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Thu Cúc 3 years 2021-07-22T15:27:54+00:00 1 Answers 126 views 0

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  1. Vodka
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    2021-07-22T15:29:14+00:00
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    Answer:

    Explanation:

    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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