A proton moves through a magnetic field at 20.3 % of the speed of light. At a location where the field has a magnitude of 0.00629 T and the

Question

A proton moves through a magnetic field at 20.3 % of the speed of light. At a location where the field has a magnitude of 0.00629 T and the proton’s velocity makes an angle of 137 ∘ with the field, what is the magnitude of the magnetic force acting on the proton?

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Thu Giang 3 years 2021-07-25T15:25:29+00:00 1 Answers 4 views 0

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  1. mit
    0
    2021-07-25T15:26:38+00:00
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    Answer:

    Force on the proton will be .73\times 10^{-14}N

    Explanation:

    We have given speed of proton is 20.3% of speed of light

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

    So speed of proton v=\frac{3\times 10^8\times 23}{100}=6.9\times 10^7m/sec

    Magnetic field B = 0.00629 T

    Charge on proton q=1.6\times 10^{-16}C

    Angle between velocity and magnetic field \Theta =137^{\circ}

    Force on the proton is equal to F=qvBsin\Theta =1.6\times 10^{-19}\times 6.9\times 10^7\times 0.00629\times sin(137^{\circ})=4.73\times 10^{-14}N

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