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 2 years 2021-07-27T15:15:07+00:00 1 Answers 29 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 [tex]T = 0.262 \times 10^{-6}[/tex] sec

    Mass of proton [tex]m = 1.67 \times 10^{-27}[/tex] kg

    Charge of proton [tex]q = 1.6 \times 10^{-19}[/tex] C

    Here proton moves in circular path

        [tex]\frac{mv^{2} }{r} = qvB[/tex]

    Velocity of proton is given by,

      [tex]v = \frac{2\pi r }{T}[/tex]

    Put the value of velocity in above equation,

       [tex]\frac{m2 \pi r}{Tr} = qB[/tex]

    Now magnetic field is given by,

      [tex]B = \frac{2\pi m }{qT}[/tex]

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

      [tex]B = 0.25[/tex] T

    Therefore, the strength of magnetic field is 0.25 T

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