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Punjab Medical Punjab - MET Solved Paper-2003

  • question_answer
    A proton enters a magnetic field of intensity \[1.5\,\,wb/{{m}^{2}}\] with a velocity \[2\times 10\,\,m/s\] in a direction at an angle \[{{30}^{o}}\] with the field. The force on the proton will be (charge on proton   is\[1.6\times {{10}^{-19}}C\]):

    A) \[2.4\times {{10}^{-12}}N\]                        

    B) \[4.8\times {{10}^{-12}}N\]

    C) \[1.2\times {{10}^{-12}}N\]                        

    D) \[7.2\times {{10}^{-12}}N\]

    Correct Answer: A

    Solution :

    Here:\[q=1.6\times {{10}^{-19}}C,\,\,B=1.5\,\,wb/{{m}^{2}}\], \[v=2\times {{10}^{7}}m/s,\,\,\theta ={{30}^{o}}\]or\[\sin {{30}^{o}}=\frac{1}{2}\] Force on proton is given by                 \[F=qv\,\,B\sin \theta \]                 \[=1.6\times {{10}^{-19}}\times 2\times {{10}^{7}}\times 1.5\times \frac{1}{2}\]                 \[=2.4\times {{10}^{-12}}N\]      


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