A proton moving in the positive x direction with a speed of 9.9 105 m/s experiences zero magnetic force. When it moves in the positive y dir

Question

A proton moving in the positive x direction with a speed of 9.9 105 m/s experiences zero magnetic force. When it moves in the positive y direction it experiences a force of 1.6 10-13 N that points in the positive z direction. Determine the magnitude and direction of the magnetic field.

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Thu Thảo 3 years 2021-08-18T13:40:04+00:00 1 Answers 13 views 0

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  1. Cherry
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    2021-08-18T13:41:26+00:00
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    Answer:

    The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

    Explanation:

    In order to calculate the magnitude and direction of the magnetic field, you take into account the following equation for the magnetic force on the proton:

    \vec{F_B}=q\vec{v}\ X\ \vec{B}       (1)

    v: speed of the proton = 9.9*10^5 m/s

    q: charge of the proton = 1.6*10^-19C

    B: magnetic field = ?

    FB: magnetic force on the proton = 1.6*10^-13N

    When the proton travels in the positive y direction (^j), you have that the proton experiences a force in the positive z direction (+^k). To obtain this direction of the magnetic force on the proton, it is necessary that the magnetic field points in the negative x direction, in fact, you have:

    ^j X (-^i) = -(-^k)=^k

    To obtain the magnitude of the magnetic field you use:

    F_B=qvBsin90\°=qvB\\\\B=\frac{F_B}{qv}=\frac{1.6*10^{-13}N}{(1.6*10^{-19}C)(9.9*10^5m/s)}\\\\B=1.01T

    The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

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