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AIIMS AIIMS Solved Paper-2014

  • question_answer
    A closely wound solenoid of 2000 turns and area of cross-section \[1.5\times {{10}^{-4}}{{m}^{2}}\] carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field \[5\times {{10}^{-2}}{{m}^{2}}\] T, making an angle of\[30{}^\circ \] with the axis of the solenoid. The torque on the solenoid will he

    A) \[\text{3}\times \text{1}{{0}^{-\text{3}}}\text{ N}-\text{m}\]                   

    B) \[1.5\times {{10}^{-3}}N-m\]

    C) \[1.5\times {{10}^{-2}}N-m\]                     

    D) \[3\times {{10}^{-2}}N-m\]

    Correct Answer: C

    Solution :

    \[\tau =NiBA\sin \theta \] \[=\text{2}000\times \text{2}\times \text{5}\times \text{1}{{0}^{-\text{2}}}\text{ }\times \text{1}.\text{5}\times \text{1}{{0}^{-\text{4}}}\text{ }\times \text{sin3}0{}^\circ \] \[=\text{2}000\times \text{5}0\times \text{1}{{0}^{-\text{6}}}\times \frac{1}{2}=\text{ 1}.\text{5 }\times \text{1}{{0}^{-\text{2}}}\text{N}-\text{m}\]


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