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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2011

  • question_answer
    The torque required to hold a small circular coil of 10 turns, area\[1m{{m}^{2}}\]and carrying a current of \[\left( \frac{21}{44} \right)\]A in the middle of a long solenoid of \[{{10}^{3}}\]turns/m carrying a current of 2.5 A, with its axis perpendicular to the axis of the solenoid is Solenoid

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

    B)  \[1.5\times {{10}^{-8}}N-m\]

    C)  \[1.5\times {{10}^{+6}}N-m\]

    D)  \[1.5\times {{10}^{+8}}N-m\]

    Correct Answer: B

    Solution :

    We have, \[M=NIA\] \[B={{\mu }_{0}}nI\] Torque, \[C=MB\] Here, \[C=({{n}_{1}}{{I}_{1}}A)({{\mu }_{0}}{{n}_{2}}{{I}_{2}})\] \[=\left( 10\times \frac{21}{44}\times {{10}^{-6}} \right)\left( 4\times \frac{22}{7}\times {{10}^{-7}}\times {{10}^{3}}\times 2.5 \right)\] \[=1.5\times {{10}^{-8}}N-m\]


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