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MGIMS WARDHA MGIMS WARDHA Solved Paper-2003

  • question_answer
    A particle of unit mass and specific charge s is thrown from the wall perpendicularly to a wall at a distance d from the wall with speed \['\upsilon '.\] The minimum magnetic field produced so that the particle does not touch the wall, is:

    A) \[\frac{\upsilon }{sd}\]                                 

    B) \[\frac{2\upsilon }{sd}\]

    C) \[\frac{\upsilon }{2sd}\]                                               

    D) \[\frac{\upsilon }{4sd}\]

    Correct Answer: A

    Solution :

    For the particle not to be collided with the wall we have centripetal force = force in magnetic field \[\frac{m{{\upsilon }^{2}}}{r}=q\upsilon B\] \[\frac{m{{\upsilon }^{2}}}{d}=s\upsilon B\]        \[(\because r=d\,and\,q=s)\] \[\frac{m\upsilon }{d}=sB\]         \[(\because m=1)\] \[B=\frac{\upsilon }{sd}\]


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