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AMU Medical AMU Solved Paper-1999

  • question_answer
    A cyclotron is accelerating proton, where the applied magnetic field is 2 T, the potential gap is 100 kV then how much turn the proton has to move between the dees to acquire a kinetic energy of 20 MeV

    A)  100             

    B)  150

    C)  200             

    D)  300

    Correct Answer: A

    Solution :

    : Kinetic energy acquired by proton \[=(N)\times (Ve)\]in half rotation. Energy is acquired in each half turn or\[K=2NVe\]where N = Number of full rotations/turns. \[K=20\text{ }MeV=20\times {{10}^{6}}\times 1.6\times {{10}^{-19}}J\] \[\therefore \]\[N=\frac{K}{2Ve}=\frac{20\times {{10}^{6}}\times 1.6\times {{10}^{-19}}}{2\times (100\times 1000)\times 1.6\times {{10}^{-19}}}\] \[=\frac{200}{2}=100\] Number of full turns of proton between the dees = 100


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