An electric field of 1.32 kV/m and a magnetic field of 0.516 T act on a moving electron to produce no net force. If the fields are perpendic

Question

An electric field of 1.32 kV/m and a magnetic field of 0.516 T act on a moving electron to produce no net force. If the fields are perpendicular to each other, what is the electron’s speed?

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Xavia 2 weeks 2021-08-28T18:08:48+00:00 1 Answers 0 views 0

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    2021-08-28T18:09:49+00:00

    Answer:

    The speed of the electron is 2.55\times 10^3\ m/s.

    Explanation:

    Given that,

    The magnitude of electric field, E=1.32\ kV/m=1.32\times 10^3\ V/m

    The magnitude of magnetic field, B = 0.516 T

    Both the magnetic and electric fields are acting on the moving electron. Then,  the magnitude of electric field and magnetic field is balanced such that :

    evB=eE\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.32\times 10^3}{0.516}\\\\v=2558.13\ m/s

    or

    v=2.55\times 10^3\ m/s

    So, the speed of the electron is 2.55\times 10^3\ m/s. Hence, this is the required solution.

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