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19 October, 22:02

A positive charge of 0.026 C moves horizontally to the right at a speed of 443.592 m/s and enters a magnetic field directed vertically downward. If it experiences a force of 22.182 N, what is the magnetic field strength?

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  1. 19 October, 23:07
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    Magnetic field, B = 1.9232 T

    Explanation:

    Given dа ta:

    Value of the charge, Q = 0.026 C

    Speed, V = 443.592 m/s

    Force experienced, F = 22.182

    Now,

    the Force (F) experienced by a charge in a magnetic field is given as:

    F = QVBsinθ

    where,

    B is the magnetic field

    Angle between the magnetic field and the velocity.

    since, the velocity is in horizontal direction and the magnetic field is downwards. Therefore, the angle θ = 90°

    thus, we have

    22.182 = 0.026 * 443.592 * B * sin90°

    or

    B = 1.9232 T
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