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16 September, 04:24

What is the magnetic force (in newtons) on a particle traveling in a 1.5 T magnetic field if q = 7.5 microcoulombs and v = 1.75 * 106 m/s at a 45° angle to the magnetic field? Answer with three signficant digits.

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  1. 16 September, 04:53
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    Given:

    B (Magnetic field) : 1.5 T

    q = 7.5 microcoulombs

    v = 1.75 x 10 ∧6 m/s

    The angle ∅ between B and v is 45 °.

    Now we know that F = qvB sin ∅

    Substituting these values we get:

    F = 7.5 x 10∧-6 x 1.75 x 10∧6 x 1.5 x sin 45

    F = 16.752 N
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