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BCECE Medical BCECE Medical Solved Papers-2011

  • question_answer
    An electron moves at right angle to a magnetic field of \[1.5\times {{10}^{-2}}T\] with a speed of \[6\times {{10}^{-7}}\]m/s. If the specific charge on the electron is \[1.7\times {{10}^{11}}\] C/kg, the radius of the circular path will be

    A)  2.9 cm          

    B)  3.9 cm

    C)  2.35cm         

    D)  2cm

    Correct Answer: C

    Solution :

    Radius of circular path,                 \[r=\frac{mv}{Bq}\]                 \[=\frac{v}{\left( \frac{e}{m} \right)B}=\frac{6\times {{10}^{7}}}{1.7\times {{10}^{11}}1.5\times {{10}^{-2}}}\] = 2.35 cm


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