A cosmic-ray proton in interstellar space has an energy of 10.0 MeV and executes a circular orbit having a radius equal to that of Mercury’s

Question

A cosmic-ray proton in interstellar space has an energy of 10.0 MeV and executes a circular orbit having a radius equal to that of Mercury’s orbit around the Sun (5.80 × 1010m). What is the magnetic field in that region of space?

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Jezebel 4 years 2021-08-03T23:10:50+00:00 1 Answers 527 views 1

Answers ( )

  1. thnahdat
    0
    2021-08-03T23:12:29+00:00
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    Answer:

    The value is B = 7.7106 *10^{-12 } \ T

    Explanation:

    From the question we are told that

    The energy is E = 10.0 MeV = 10.0 *10^{6} \ eV = 10.0 *10^{6} * 1.60*10^{-19} = 1.6 *10^{-12} \ J

    The  radius  is  R = 5.80 *10^{10} \ m

    Generally the magnetic field is mathematically represented as

           B = \frac{m * v }{q * R }

    Where m is the mass of  proton with value  m = 1.6*10^{-27} \ kg

     v is the velocity of the proton which is mathematically deduced  from the formula for kinetic energy as

                       v = \sqrt{ \frac{E }{ 0.5 * m } }

    Here E is also equivalent to kinetic energy of the proton so

                       v = \sqrt{ \frac{1.6 *10^{-12} }{ 0.5 * 1.6*10^{-27} } }

                     v = 4.47214 *10^{7} \ m/s

    So  

         B = \frac{1.60 *10^{-27} * 4.47214 *10^{7}}{ 1.60 *10^{-19} * 5.80*10^{10}}

          B = 7.7106 *10^{-12 } \ T

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