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

  • question_answer
    A 0.1 m long conductor carrying a current of 50A is held perpendicular to a magnetic field of 1.25 mT. The mechanical power required to move the conductor with a speed of \[1\text{ }m{{s}^{-1}}\] is:

    A) 62.5 mW

    B) 625 mW

    C) 6.25 mW                      

    D) 12.5 mW

    Correct Answer: C

    Solution :

    (c) 6.25 mW Here, \[l=0.1\,m,\,\,v=1\,m{{s}^{-1}}\] \[l=50\,\,A,B=1.25\,mT=1.25\times {{10}^{-3}}T\] The induced emf is \[\varepsilon =Blv\] The mechanical power is \[P=\varepsilon l=Blvl=1.25\times {{10}^{-3}}\times 0.1\times 1\times 50\]             \[=6.25\times {{10}^{-3}}W=6.25\,\,mW\]


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