NEET Sample Paper NEET Sample Test Paper-15

  • question_answer An electron of mass m is accelerated through a potential difference of V and then it enters a magnetic field of induction B normal to the lines. Then, the radius of the circular path is:

    A) \[\sqrt{(2\,eV/m)}\]                 

    B) \[\sqrt{(2Vm/e{{B}^{2}})}\]

    C) \[\sqrt{(2Vm/eB)}\]                 

    D) \[\sqrt{(2Vm/{{e}^{2}}B)}\]

    Correct Answer: B

    Solution :

    \[Bev=\frac{m{{v}^{2}}}{r}\]or \[r=\frac{mv}{Be}\] As,\[mv=\sqrt{2mT},\]where \[T=K.E.\] So, \[r=\frac{\sqrt{2mT}}{Be}\] As the electron has been accelerated from rest through a potential difference of V volt, then\[T=eV\] \[r=\frac{\sqrt{2mVe}}{Be}=\sqrt{\frac{2mV}{{{B}^{2}}e}}\] Hence, the correction option is [b].

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