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7 June, 02:25

An ion source is producing 6Li ions, which have charge + e and mass 9.99 * 10-27 kg. The ions are accelerated by a potential difference of 13 kV and pass horizontally into a region in which there is a uniform vertical magnetic field of magnitude B = 1.0 T. Calculate the strength of the smallest electric field, to be set up over the same region, that will allow the 6Li ions to pass through undeflected.

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  1. 7 June, 02:49
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    6.45 x 10^5 N/C

    Explanation:

    q = + e = 1.6 x 10^-19 C

    m = 9.99 x 10^-27 kg

    V = 13 kV = 13000 V

    B = 1 T

    Let v be the speed and E be the strength of electric field.

    1/2 mv^2 = eV

    v^2 = 2 e v / m

    v^2 = (2 x 1.6 x 10^-19 x 13000) / (9.99 x 10^-27)

    v = 6.45 x 10^5 m/s

    As the charge particle is undeflected, the force due to magnetic field is counter balanced by the force due to electric field.

    q E = q v B

    E = v B = 6.45 x 10^5 x 1 = 6.45 x 10^5 N/C
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