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Uttarakhand PMT Uttarakhand PMT Solved Paper-2006

  • question_answer
    An electron moves at right angle to a magnetic field of\[5\times {{10}^{-2}}T\]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 \[r=\frac{mv}{eB}=\frac{v}{\left( \frac{e}{m} \right)B}\] ?.(1) Given: e/m of electron \[=1.7\times {{10}^{11}}C/kg\] and \[v=6\times {{10}^{-7}}m/s\] \[B=1.5\times {{10}^{-2}}T\] So,       \[r=\frac{6\times {{10}^{7}}}{1.7\times {{10}^{11}}\times 1.5\times {{10}^{-2}}}\] \[=2.35\times {{10}^{-2}}m=2.35\,m\]


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