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JCECE Engineering JCECE Engineering Solved Paper-2014

  • question_answer
    The couple acting on a magnet of length \[10\,\,cm\] and pole strength\[125\,\,A-m\], kept in a field of \[B=2\times {{10}^{-5}}T\], at an angle of \[{{30}^{o}}\] is

    A) \[1.5\times {{10}^{-5}}N-m\]

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

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

    D) \[1.5\times {{10}^{-6}}N-m\]

    Correct Answer: A

    Solution :

    \[C=MB\sin \theta \]                 \[=(m\times 2l)\times 2\times {{10}^{-5}}\sin {{30}^{o}}\]     \[=150\times {{10}^{-2}}\times 2\times {{10}^{-5}}\times \frac{1}{2}=1.5\times {{10}^{-5}}N-m\]


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