A proton moves through a magnetic field at 26.7 % 26.7% of the speed of light. At a location where the field has a magnitude of 0.00687 T 0.

Question

A proton moves through a magnetic field at 26.7 % 26.7% of the speed of light. At a location where the field has a magnitude of 0.00687 T 0.00687 T and the proton’s velocity makes an angle of 101 ∘ 101∘ with the field, what is the magnitude of the magnetic force acting on the proton?

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Dulcie 4 years 2021-08-21T04:45:39+00:00 1 Answers 13 views 0

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  1. Calantha
    0
    2021-08-21T04:46:45+00:00
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    Answer:

    8.64283\times 10^{-14}\ N

    Explanation:

    q = Charge of proton = 1.6\times 10^{-19}\ C

    v = Velocity of proton = 0.267\times c

    c = Speed of light = 3\times 10^8\ m/s

    B = Magnetic field = 0.00687 T

    \theta = Angle = 101^{\circ}

    Magnetic force is given by

    F=qvBsin\theta\\\Rightarrow F=1.6\times 10^{-19}\times (0.267\times 3\times 10^8)\times 0.00687\times sin101\\\Rightarrow F=8.64283\times 10^{-14}\ N

    The magnetic force acting on the proton is 8.64283\times 10^{-14}\ N

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