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

  • question_answer
    A proton (mass \[=1.67\times {{10}^{-27}}\,kg\] and charge \[=1.6\times {{10}^{-19}}\,C)\] enters perpendicular to a magnetic field of intensity 2 \[weber/{{m}^{2}}\] with a velocity \[3.4\times {{10}^{7}}\,m/\sec \]. The acceleration of the proton should be                   [DPMT 1999]

    A)            \[6.5\times {{10}^{15}}\,m/{{\sec }^{2}}\]                  

    B)            \[6.5\times {{10}^{13}}\,m/{{\sec }^{2}}\]

    C)            \[6.5\times {{10}^{11}}\,m/{{\sec }^{2}}\]                  

    D)            \[6.5\times {{10}^{9}}\,m/{{\sec }^{2}}\]

    Correct Answer: A

    Solution :

                       F = ma = qvB Þ \[a=\frac{qvB}{m}=\frac{1.6\times {{10}^{-19}}\times 2\times 3.4\times {{10}^{7}}}{1.67\times {{10}^{-27}}}\]                = 6.5 ´ 1015m/sec2


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