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15 August, 17:56

A cosmic-ray proton in interstellar space has an energy of 59 MeV and executes a circular orbit with a radius equal to that of Mercury's orbit (5.8 * 1010 m) around the Sun. What is the magnetic field in that region of space? The proton has a charge of 1.60218 * 10-19 C and a mass of 1.67262 * 10-27 kg. Answer in units of T

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  1. 15 August, 18:16
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    Let mass and velocity of proton be m and v.

    1/2 m v² = 59 x 10⁶ e V

    = 59 x 10⁶ x 1.6 x 10⁻¹⁹ J

    = 94.4 x 10⁻¹³ J

    mv² = 188.8 x 10⁻¹³ J

    v² = 188.8 x 10⁻¹³ / m

    = 188.8 x 10⁻¹³ / 1.67262 x 10⁻²⁷

    = 112.8768 x 10¹⁴

    v = 10.62 x 10⁷ m / s

    In circular path of proton, magnetic force equals centripetal force.

    m v² / r = B q v, B is magnetic field, q is charge on proton, r is radius of circular path.

    188.8 x 10⁻¹³ / 5.8 x 10¹⁰ = B x 1.6 x 10⁻¹⁹ x 10.62 x 10⁷

    B = 1.9157 x 10⁻¹¹ T.
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