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EAMCET Medical EAMCET Medical Solved Paper-1996

  • question_answer
    A charged particle is moving in a uniform magnetic field in a circular path. The radius of the circular path is R. When energy of the particle is doubled, then new radius will be:

    A)  \[\sqrt{2}\]R                                    

    B)  R/\[\sqrt{2}\]

    C)  R                                           

    D)  R/2

    Correct Answer: A

    Solution :

     From the relation, \[\frac{m{{v}^{2}}}{r}=qvB\] \[r=\frac{mv}{qB}\] So,          \[r\propto v\]                 \[KE=\frac{1}{2}m{{v}^{2}}\] \[\Rightarrow \]               \[KE\propto {{v}^{2}}\] or            \[v\propto \sqrt{KE}\] Therefore,     \[r\propto \sqrt{KE}\] When energy of particle is doubled, then  radius will become \[(\sqrt{2})\] times.


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