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VMMC VMMC Medical Solved Paper-2012

  • 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 \[5\times {{10}^{-2}}\]making an angle of \[{{30}^{o}}\] 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, \[N=2000,A=1.5\times {{10}^{-4}}{{m}^{2}}\] \[i=2.0A,B=5\times {{10}^{-2}}T,\] 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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