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BCECE Engineering BCECE Engineering Solved Paper-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)  2.35 cm               

    D)         2 cm

    Correct Answer: C

    Solution :

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


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