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WB JEE Medical WB JEE Medical Solved Paper-2007

  • 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)  16 cm                                  

    D)  2 cm

    Correct Answer: C

    Solution :

                     \[r=\frac{mv}{Bq}\] \[\Rightarrow \]               \[r=\frac{v}{B\frac{q}{m}}=\frac{2\times {{10}^{5}}}{0.05\times 2.5\times {{10}^{7}}}\] \[=\frac{2\times {{10}^{7}}}{12.5\times {{10}^{7}}}=\frac{200}{12.5}\text{cm}\,\text{=}\,\text{16}\,\text{cm}\]


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