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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2006

  • 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 of 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 :

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


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