VIT Engineering VIT Engineering Solved Paper-2007

  • question_answer
    The proton of energy \[1MeV\] describes a circular path in plane at right angles to a uniform magnetic field of \[6.28\times {{10}^{-4}}T\]. The mass of the proton is \[1.7\times {{10}^{-27}}kg\]. The cyclotron frequency of the proton is very nearly equal to:

    A)  \[{{10}^{7}}Hz\]       

    B)  \[{{10}^{5}}Hz\]

    C)  \[{{10}^{6}}Hz\]

    D)  \[{{10}^{4}}Hz\]

    Correct Answer: D

    Solution :

    Cyclotron frequency is given by \[v=\frac{qB}{2\pi m}\] \[\therefore \] \[v=\frac{1.6\times {{10}^{-19}}\times 6.28\times {{10}^{-4}}}{2\times 3.14\times 1.7\times {{10}^{-27}}}\] \[=0.94\times {{10}^{4}}\] \[\approx {{10}^{4}}Hz\]


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