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Today, 10:01

An electron experiences a magnetic force of magnitude 4.60 x 10^-15 N when moving at an angle of 60.0° with respect to a magnetic field of magnitude 3.50 x 10^-3 T. Find the speed of the electron.

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  1. Today, 10:24
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    9.49 * 10⁶ m/s

    Explanation:

    Data provided in the question:

    Magnitude of Magnetic force, F = 4.60 * 10⁻¹⁵ N

    Angle, θ = 60°

    Magnitude of magnetic field, B = 3.50 * 10⁻³ T

    Now,

    we know

    F = qVBsin (θ)

    here,

    q is the charge of electron = 1.6 * 10⁻¹⁹ V

    V is the speed of electron

    Therefore,

    4.60 * 10⁻¹⁵ = (1.6 * 10⁻¹⁹) * V * (3.50 * 10⁻³) * sin (60°)

    or

    V = 9.49 * 10⁶ m/s
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