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12th Class Physics Electro Magnetic Induction Question Bank Case Based (MCQs) - Electromagnetic Induction

  • question_answer
    \[A\,0\centerdot 1m\] long conductor carrying a current of 50 A is held perpendicular to a magnetic field of \[1\centerdot 25\,mT\]. The mechanical power required to move the conductor with a speed of \[1\,m\,{{s}^{-1}}\]is

    A) \[62\centerdot 5\,mW\]

    B) \[625\,mW\]

    C) \[6\centerdot 25\,mW\]

    D) \[12\centerdot 5\,mW\]

    Correct Answer: C

    Solution :

    Here, \[l=0\,.\,1\,m,\,n=1\,\,m\,{{s}^{-1}}\] \[I=50\,\,A,\,B=\,1.\,25\,mT=1\,.25\times {{10}^{-3}}\,\,T\] The induced e.m.f. is, \[\varepsilon =B/v\] The mechanical power is \[P=\varepsilon I=B/vI=1\,.\,25\times {{10}^{-3}}\,\times \,0\,.\,1\times 1\times 50\] \[=\,6\,.\,25\times {{10}^{-3}}\,W=6\,.\,25\,mW\].


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