Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    A proton and an \[\alpha \]-particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25\[\mu F\] to make 5 revolutions, then the periodic time for the \[\alpha \]-particle would be

    A)  50\[\mu s\]                      

    B)         25 \[\mu s\]

    C)  10 \[\mu s\]                     

    D)         5\[\mu s\]

    Correct Answer: C

    Solution :

    Time period of proton\[{{T}_{p}}=\frac{25}{5}=5\mu s\] By using\[T=\frac{2\pi m}{qB}\] \[\Rightarrow \]               \[\frac{{{T}_{a}}}{{{T}_{p}}}=\frac{{{m}_{\alpha }}}{{{m}_{p}}}\times \frac{{{q}_{p}}}{{{q}_{\alpha }}}=\frac{4{{m}_{p}}}{{{m}_{p}}}\times \frac{{{q}_{p}}}{2{{q}_{p}}}=2\] \[\Rightarrow \]               \[{{T}_{\alpha }}=2{{T}_{p}}=10\mu s\]


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