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NEET AIPMT SOLVED PAPER MAINS 2010

  • 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 \[5\times {{10}^{-2}}T\] making an angle of \[30{}^\circ \] with the axis of the solenoid. The torque on the solenoid will be                                                                                  

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

    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 :

    Given, \[N=2000,A=1.5\times {{10}^{-4}}{{m}^{2}}\] \[i=2.0AB=5\times {{10}^{-2}}T,\]and\[\theta ={{30}^{o}}\]                 Torque\[\tau =NiBA\sin \theta \]                 \[=2000\times 2\times 5\times {{10}^{-2}}\times 1.5\times {{10}^{-4}}\times \sin {{30}^{o}}\]                 \[=2000\times 50\times {{10}^{-6}}\times \frac{1}{2}\]                 \[=1.5\times {{10}^{-2}}Nm\]


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