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An electron accelerates from rest to the right, in a horizontally directed electric field. The electron then leaves the electric field at a speed of 4.0 × 106, entering a magnetic field of magnitude 0.20 T (Tesla) directed into the screen. Calculate the magnitude and direction of the magnetic force on the electron.

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  1. x to right
    y into screen
    z up
    F = q ( V cross B)
    q = -1.6 * 10^-19 C scalar
    V = 4 * 10^6 m/s in x direction
    B = 0.20 T in y direction
    so if q were POSITIVE force would be in UP (z direction)
    however q is NEGATIVE, so force is in -z direction, down
    since V and B are at 90 degrees, just multiply, sin of 90 = 1
    F = 1.6*10^-19 * 4*10^6 * 0.20 Newtons downward

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    Damon

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