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9 February, 17:48

What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56*104V/m and a magnetic field of 4.62*10-3T, with both fields normal to the beam and to each other, produces no deflection of the electrons?

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  1. 9 February, 20:36
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    3.38 * 10^4 m/s.

    Explanation:

    Given:

    E = 1.56*104V/m

    B = 4.62*10-3T

    F = q * V * B

    E = F/q

    E * q = q * V * B

    V = E/B

    = 1.56 * 10^4/4.62*10-3

    = 3.38 * 10^4 m/s.
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