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7 August, 04:33

A proton enters a uniform magnetic field of strength 0.5 T at 200 m/s. The magnetic field is oriented at an angle of 30° to the proton's velocity. What is the magnitude of the force that the proton experiences while it moves through the magnetic field?

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  1. 7 August, 05:29
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    The force experienced is 8 x 10⁻¹⁸ N

    Explanation:

    When any charge moves into the magnetic field, it experiences the force

    The force can be calculated by the relation F = q v B sinθ

    here q is the charge and v is its velocity.

    B is magnetic field strength and θ is the angle between field and velocity vectors.

    Thus F = 1.6 x 10⁻¹⁹ x 200 x 0.5 x sin 30

    = 8 x 10⁻¹⁸ N
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