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JIPMER Jipmer Medical Solved Paper-2005

  • question_answer
    An electron moves at right angle to a magnetic field of \[1.5\times {{10}^{-2}}\] tesla with a speed of\[6\times {{10}^{7}}m\text{/}s\]. If the specific charge of the electron is\[1.7\times {{10}^{11}}coul.\text{/}kg.\] 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 :

    The formula for radius of circular path is \[r=\frac{m\upsilon }{eB}=\frac{\upsilon }{\left( \frac{e}{m} \right)B}\]                 ?(1) Given: e/m of electron \[=1.7\times {{10}^{11}}C\text{/}kg\]and \[\upsilon =6\times {{10}^{7}}m\text{/}s\] \[B=1.5\times {{10}^{-2}}T\] So,      \[r=\frac{6\times {{10}^{7}}}{1.7\times {{10}^{11}}\times 1.5\times {{10}^{-2}}}\] \[=2.35\times {{10}^{-2}}m\] \[=2.35\,cm\]


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