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BCECE Engineering BCECE Engineering Solved Paper-2011

  • question_answer
    A rod of length 1.0 m is rotated in a plane perpendicular to a uniform magnetic field of induction 0.25 T with a frequency of 12 rev/s. The induced emf across the ends of the rod is

    A)  18.89 V               

    B)         3 V

    C)  15 V                     

    D)         9.42 V

    Correct Answer: D

    Solution :

    Induced emf \[e=\pi {{l}^{2}}Bn=\frac{{{l}^{2}}B\omega }{2}\] \[=\frac{{{(1.0)}^{2}}\times 0.25\times 2\times 3.14\times 12}{2}=9.42\,V\]


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