JEE Main & Advanced Physics Electro Magnetic Induction Question Bank Motional EMI

  • question_answer
    A metal conductor of length 1m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth's magnetic field is \[0.2\times {{10}^{-4}}T\], then the e.m.f. developed between the two ends of the conductor is    [MP PMT 1992; AIEEE 2004]

    A)            \[5\ mV\]                              

    B)            \[5\times {{10}^{-4}}V\]

    C)            \[50\ mV\]                            

    D)            \[50\ \mu V\]

    Correct Answer: D

    Solution :

               \[e=\frac{1}{2}B\omega {{r}^{2}}=\frac{1}{2}\times 0.2\times {{10}^{-4}}\times 5\times {{(1)}^{2}}=50\mu V\]


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