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JEE Main & Advanced Physics Electro Magnetic Induction Question Bank Self Evaluation Test - Electromagnetic Induction

  • question_answer
    A 0.1 m long conductor carrying a current of 50 I A is perpendicular to a magnetic field of 1.21 mT. The mechanical power to move the conductor 1 with a speed of \[1\,m{{s}^{-1}}\] is                     

    A) 0.25 m W 

    B) 6.25 m W          

    C) 0.625 W 

    D) 1W

    Correct Answer: B

    Solution :

    [b] \[P=Fv=BIlV=1.25\times {{10}^{-3}}\times 50\times 0.1\times 1\,W\] \[=6.25\times {{10}^{-3}}W=6.25\,mW\]


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