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KVPY Sample Paper KVPY Stream-SX Model Paper-5

  • question_answer
    A particle of charge Q and of negligible initial speed is accelerated through a potential difference of U. The particle reaches a region of uniform magnetic field of induction B, where it undergoes circular motion. If potential difference is doubled and B is also doubled then magnetic moment of the circular current due to circular motion of charge Q will become -

    A) Double             

    B) half

    C) four times                     

    D) remain same

    Correct Answer: D

    Solution :

    [D]
    K.E. of charge Q accelerated with potential difference of U is given by  K.E. = QU
    Magnetic moment \[=i\,\,\times \,\,Area\]\[=\frac{Q}{T}\,\,\times \,\,\pi {{R}^{2}}\]
    Where T is time period of circular motion of charge Q and R is radius of circular motion
    \[\therefore T=\frac{2\pi m}{QB}\]
    \[R=\sqrt{\frac{2mKE}{QB}}=\sqrt{\frac{2mU}{QB}}\]
    Magnetic moment \[=\frac{{{Q}^{2}}\,\,\times \,\,B}{2\pi m}\times \frac{\pi \,\,\times \,\,2m\,\,\times \,\,U}{{{Q}^{2}}{{B}^{2}}}\]
    Magnetic Moment \[\propto \frac{U}{B},\] so if U and B both are double magnetic moment remain unchanged.


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