JEE Main & Advanced Physics Magnetic Effects of Current / करंट का चुंबकीय प्रभाव Question Bank Motion of Charged Particle In Magnetic Field

  • question_answer
    One proton beam enters a magnetic field of \[{{10}^{-4}}\]T normally, Specific charge = \[{{10}^{11}}\,C/kg.\] velocity = \[{{10}^{7}}\,m/s\]. What is the radius of the circle described by it        [DCE 1999]

    A)            0.1 m                                       

    B)            1 m

    C)            10 m                                        

    D)            None of these

    Correct Answer: B

    Solution :

               \[r=\frac{mv}{qB}=\frac{{{10}^{7}}}{{{10}^{11}}\times {{10}^{-4}}}=1m\ (\because \ q/m={{10}^{11}}C/kg)\]


You need to login to perform this action.
You will be redirected in 3 sec spinner