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19 September, 08:39

What is the velocity of a beam of electrons that goes undeflected when passing through perpendicular electric and magnetic fields of magnitude 8100 v/m and 6.0*10?3 t, respectively?

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  1. 19 September, 09:20
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    F = qE + qV * B

    where force F, electric field E, velocity V, and magnetic field B are vectors and the * operator is the vector cross product. If the electron remains undeflected, then F = 0 and E = - V * B

    which means that |V| = |E| / |B| and the vectors must have the proper geometrical relationship. I therefore get

    |V| = 8.8e3 / 3.7e-3

    = 2.4e6 m/sec

    Acceleration a = V²/r, where r is the radius of curvature.

    a = F/m, where m is the mass of an electron,

    so qVB/m = V²/r.

    Solving for r yields

    r = mV/qB

    = 9.11e-31 kg * 2.37e6 m/sec / (1.60e-19 coul * 3.7e-3 T)

    = 3.65e-3 m
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