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3 November, 02:28

A proton is moving at 2.1 x 10 m/s clockwise through a constant and perpendicular magnetic field. The radius of curvature through the field is 0.6 meters. What is the strength of the magnetic field?

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  1. 3 November, 03:02
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    For this question the magnetic force provides the force required for the circular motion

    Equating

    Magnetic force = centripetal force

    Bqvsinx = mv^2/r

    as magnetic field is perpendicular x=90

    Bqvsin90 = mv^2/r

    Bq = mv/r

    B = mv/rq

    Then replace r = 0.6, v = 2.1ms and mp from formula sheet and you can obtain B which is magnetic field intensity
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