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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 30

  • question_answer
    A short wire AB carrying \[{{I}_{1}}\] current lies in the plane of long wire which carry current I upward. If wire AB is released from horizontal position, and \[{{a}_{A}}\] and \[{{a}_{B}}\]are magnitude of acceleration of points A and B respectively correct alternative. (The space is gravity free).

    A) \[{{a}_{A}}>{{a}_{B}}\]                    

    B) \[{{a}_{A}}<{{a}_{B}}\]

    C) \[{{a}_{A}}={{a}_{B}}\ne 0\] 

    D) \[{{a}_{A}}={{a}_{B}}=0\]

    Correct Answer: C

    Solution :

    [c] As force on short wire AB acts upward and the torque of magnetic force about c.m. is in clockwise sense, so acceleration of point A, \[{{a}_{A}}={{a}_{cm}}+\alpha I/2\] and for B,\[{{a}_{B}}={{a}_{cm}}-\alpha I/2.\]Thus \[{{a}_{a}}>{{a}_{B}}.\]       


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