JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    A metal conductor of length 1m rotates vertically about one of its ends at angular velocity 5 rad/s. If the horizontal component of the earth's magnetic field is\[0.2\times {{10}^{-4}}T\]then the emf developed between the two ends of the conductor is

    A) \[5\mu V\]   

    B)                        \[50\mu V\]      

    C)        \[5mV\]              

    D)        \[50\,mV\]

    Correct Answer: B

    Solution :

    The emf induced between ends of conductor, \[e=\frac{1}{2}B\omega {{L}^{2}}\] \[=\frac{1}{2}\times 0.2\times {{10}^{-4}}\times 5\times {{(1)}^{2}}\] \[=0.5\times {{10}^{-4}}V\] \[=5\times {{10}^{-5}}V\] \[=50\mu V\]


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