A singly charged positive ion has a mass of 2.46 ✕ 10−26 kg. After being accelerated through a potential difference of 270 V, the ion enters

Question

A singly charged positive ion has a mass of 2.46 ✕ 10−26 kg. After being accelerated through a potential difference of 270 V, the ion enters a magnetic field of 0.535 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field.

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Eirian 2 months 2021-09-01T07:30:48+00:00 1 Answers 4 views 0

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    2021-09-01T07:32:38+00:00

    Answer:

    r=0.017m

    Explanation:

    Given data

    Mass m=2.46×10⁻²⁶kg

    Voltage V=270 V

    Magnetic field B=0.5353 T

    To find

    Radius r

    Solution

    By using conservation of energy to find speed of charge

    (1/2)mv^2=qV\\v^2=2qV/m\\v=\sqrt{\frac{2qV}{m} }\\

    Substitute the given values

    v=\sqrt{\frac{2(1.6*10^{-19C})(270V)}{(2.46*10^{-26}kg)} }\\v=5.93*10^{4}m/s

    So the radius can be find as:

    r=\frac{mv}{qB}\\

    Substitute the values

    r=\frac{(2.46*10^{-26}kg)(5.93*10^{4}m/s)}{(1.6*10^{-19}C)(0.535T)} \\r=0.017m

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