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MGIMS WARDHA MGIMS WARDHA Solved Paper-2015

  • question_answer
    A current of 6.4 A is passed through the coil of galvanometer having 500 turns and each turn has an average area of\[3\times {{10}^{-4}}{{m}^{2}}\]. If a torque of 2 Nm is required for this coil carrying some current to set it parallel to a magnetic field then the strength of magnetic field will be

    A)  31T                       

    B)  33.34 T

    C)  32T                       

    D)  40T

    Correct Answer: B

    Solution :

                    The magnetic moment of the current loop will be \[M=NiA=500\times 0.4\times 3\times {{10}^{-4}}=6\times {{10}^{-2}}A-{{m}^{2}}\] Also torque     \[\tau =M\times B\] \[\Rightarrow \]               \[|\tau |=MBsin\theta \] Here     \[\theta =90{}^\circ ,\tau =2N-m\] So          \[B=\frac{\tau }{M}=\frac{2}{6\times {{10}^{-2}}}\] \[B=33.34T\]


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