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    The Motion of a Charged Particle in Electric and Magnetic Field

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    Please review my solution to the problem and explain in detail what I may be doing wrong and what concepts I may not be applying correctly. I am not sure if I am apply the crossed fields concept correct in presuming that B is perpendicular.

    The problem states:

    An electron has an initial velocity of (12.0j + 15.0k) km/s and a constant acceleration of (2.00 x 1012 m/s2)i in a region in which uniform electric and magnetic fields are present. If B = (400μT)i, find the electric field E.

    We know:
    This is a crossed fields problem with the two fields perpendicular to each other.
    B is perpendicular to v, out of the page.
    Since B is perpendicular, it does no work on the electron but instead deflects it in a circular path.

    q = 1.6 x 10-19 C
    v = (12.0j + 15.0k) km/s = square root(12.02 + 15.02) = 19.21 km/s = 1.921 x 104 m/s
    B = 400μT = 400 x 10-6 T = 400 x 10-6 N/Cm/s
    Φ = 90º

    E = qvBsinΦ
    E = (1.6 x 10-19 C)(1.921 x 104 m/s)(400 x 10-6 N/Cm/s)(sin90º)
    E = 1.229 x 10-18 N

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    https://brainmass.com/physics/electricity-magnetism/motion-charged-particle-electric-magnetic-field-140935

    Solution Summary

    An electron moves with constant acceleration in presence of electric and magnetic field both. The electric field is calculated.

    $2.49

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