RAJASTHAN PMT Rajasthan - PMT Solved Paper-2000

  • question_answer
    An \[\alpha -\]particle moving perpendicularly to a magnetic field of \[0.2\,wb/metr{{e}^{2}},\] experiences a force of \[3.84\times {{10}^{-24}}N\]. Then speed of a particle will be:

    A)  \[6\times {{10}^{5}}\,m/s\]       

    B)         \[5\times {{10}^{-5}}\,m/s\]

    C)  \[1.2\times {{10}^{6}}\,m/s\]  

    D)         \[3.8\,\times {{10}^{6}}\,m/s\]

    Correct Answer: B

    Solution :

    Force on a moving charge in magnetic field \[F=q\upsilon B\] \[3.84\times {{10}^{10-27}}=3.2\times {{10}^{-19}}\times \upsilon \times 0.2\] \[\upsilon =5\times {{10}^{-5}}m/s\]


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