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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    An\[\alpha -\]particle with a specific charge of \[2.5\times {{10}^{7}}C\,k{{g}^{-1}}\] moves with a speed of\[2\times {{10}^{5}}\] \[m{{s}^{-1}}\]in a perpendicular magnetic field of 0.05 T. Then the radius of the circular path described by it is:

    A)  8 cm                     

    B)         4 cm

    C)  16cm                   

    D)         2cm

    E)  32 cm

    Correct Answer: C

    Solution :

    \[r=\frac{mv}{Bq}\] \[\Rightarrow \]               \[r=\frac{v}{B\frac{q}{m}}=\frac{2\times {{10}^{5}}}{0.005\times 2.5\times {{10}^{7}}}\]                 \[=\frac{2\times {{10}^{7}}}{12.5\times {{10}^{7}}}=\frac{200}{12.5}cm=16\,cm\]


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