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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2012

  • question_answer
        An electron moves at right angle to a magnetic field of\[1.5\times {{10}^{-2}}\]tesia 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)  2.9cm                                  

    B)  3.9cm

    C)  2.35cm                                

    D)  2cm

    Correct Answer: C

    Solution :

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


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