A proton moves in a circular path perpendicular to a constant magnetic field so that it takes 0.262 × 10−6 s to complete the revolution. Det

Question

A proton moves in a circular path perpendicular to a constant magnetic field so that it takes 0.262 × 10−6 s to complete the revolution. Determine the strength of the constant magnetic field. The angular speed is given in radians per unit time

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Thiên Hương 6 months 2021-07-27T15:15:07+00:00 1 Answers 13 views 0

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    2021-07-27T15:17:00+00:00

    Answer:

    The strength of magnetic field is 0.25 T

    Explanation:

    Time period T = 0.262 \times 10^{-6} sec

    Mass of proton m  = 1.67 \times 10^{-27} kg

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

    Here proton moves in circular path

        \frac{mv^{2} }{r} = qvB

    Velocity of proton is given by,

      v = \frac{2\pi r }{T}

    Put the value of velocity in above equation,

       \frac{m2 \pi r}{Tr} = qB

    Now magnetic field is given by,

      B = \frac{2\pi m }{qT}

      B = \frac{ 6.28 \times 1.67 \times 10^{-27} }{1.6 \times 10^{-19} \times 0.262 \times 10^{-6}  }

      B = 0.25 T

    Therefore, the strength of magnetic field is 0.25 T

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