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JIPMER Jipmer Medical Solved Paper-2004

  • question_answer
    A proton enters a magnetic field of intensity \[1.5\,\,Wb\text{/}{{m}^{2}}\]with a velocity \[2\times {{10}^{7}}\text{ }m\text{/}s\] in a direction at an angle \[30{}^\circ \] with the field. The force on the proton will be (charge on proton   is \[1.6\times {{10}^{-19}}C)\]:

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

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

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

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

    Correct Answer: A

    Solution :

    Here: \[q=1.6\times {{10}^{-19}}C,\]\[B=1.5\,Wb\text{/}{{m}^{2}},\] \[\upsilon =2\times {{10}^{7}}m\text{/}s,\]         \[\theta =30{}^\circ \]  or \[\sin 30{}^\circ =\frac{1}{2}\] 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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