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Punjab Medical Punjab - MET Solved Paper-2010

  • question_answer
    A particle having a mass of \[0.5g\] carries a charge of\[2.5\times {{10}^{-8}}C\]. The particle is given an initial horizontal velocity of\[6\times {{10}^{4}}m{{s}^{-1}}\]. The minimum magnitude of the magnetic field that is required so that, particle will keep moving in a horizontal direction is [take\[g=10m{{s}^{-2}}\]]

    A) \[8.46T\]                             

    B) \[3.33T\]

    C) \[8.64\times {{10}^{-2}}T\]                         

    D)  None of these

    Correct Answer: B

    Solution :

    To keep the particle moving in horizontal direction\[{{F}_{B}}\]should be equal to weight or\[qvB=mg\] [For minimum value of\[B\]angle between\[\mathbf{\vec{v}}\]and\[\overrightarrow{\mathbf{B}}\]has to be\[\frac{\pi }{2}\]] \[\Rightarrow \]               \[B=\frac{0.5\times {{10}^{-3}}\times 10}{2.5\times {{10}^{-8}}\times 6\times {{10}^{4}}}=3.33T\]


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