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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2011

  • question_answer
    A particle of charge e and mass m moves with a velocity v in a magnetic field B applied perpendicular to the motion of the particle. The radius r of its path in the field is

    A)  \[\frac{mv}{Be}\]                                           

    B)  \[\frac{Be}{mv}\]

    C)  \[\frac{ev}{Bm}\]                           

    D)  \[\frac{Bv}{em}\]

    Correct Answer: A

    Solution :

    Radius of path,  \[r=\frac{mv}{qB}\] Here,                     \[q=e\] So,                          \[r=\frac{mv}{eB}\]


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