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AMU Medical AMU Solved Paper-2012

  • question_answer
    The magnetic field in a certain region of space is given by \[B=8.35\times {{10}^{-2}}i\,T\]. A proton is shot into the field with velocity\[v=(2\times {{10}^{5}}i+4\times {{10}^{5}}j)\,\,m/s\]. The proton follows a helical path in the field. The distance moved by proton in the x-direction during the period of one revolution in the yz- plane will be

    A)  0.053 m              

    B)  0.136 m

    C)  0.157m               

    D)  0.236m

    Correct Answer: C

    Solution :

                    Given, \[B=8.35\times {{10}^{-2}}iT\]                 \[v=(2\times {{10}^{5}}i+4\times {{10}^{5}}j)\] The distance covered by proton                 \[P=T\,(v)\]                 \[=2\pi \frac{m}{qB}(v)\]                 \[=2\times 3.14\times \frac{1.67\times {{10}^{-27}}}{1.6\times {{10}^{-19}}\times 8.35\times {{10}^{-2}}i}\]                                                 \[\times (2\times {{10}^{5}}i+4\times {{10}^{5}}j)\] P = 0.157 m                 (Mass of proton \[=1.67\times {{10}^{-27}}kg\])


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