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RAJASTHAN PMT Rajasthan - PMT Solved Paper-1995

  • question_answer
    Atomic weight of single charge \[Li\] is \[6.06\,amu\]and its energy is\[400\,eV\]. It enter normally to a uniform magnetic field of \[0.8\] Tesla, then radius of circular path is:

    A)  \[8.35\,cm\]    

    B)                         \[9.23\,cm\]

    C)         \[10.5\,cm\]                     

    D)                         \[11.25\,cm\]

    Correct Answer: A

    Solution :

                    Radius of circular path \[r=\frac{m\upsilon }{Bq}=\frac{\sqrt{2mE}}{Bq}\] \[=\frac{\sqrt{2\times 6.06\times 1.6\times {{10}^{-27}}\times 400\times 1.6\times {{10}^{-19}}}}{1.6\times {{10}^{-19}}\times 0.8}\] \[=8.35cm\]


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