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Manipal Medical Manipal Medical Solved Paper-2002

  • question_answer
    A straight conductor of length 4m moves at a speed of 10 m/s. When the conductor makes an angle of\[30{}^\circ \]with the direction of magnetic field of induction of 0.1 wb. per\[{{m}^{2}}\]then induced emf. is:

    A)  8V             

    B)  4V

    C)  1V             

    D)  2V

    Correct Answer: D

    Solution :

     Induced emf. is given by \[I=B\upsilon l\text{ }sin\text{ }\theta =0.1\times 10\times 4\text{ }sin\text{ }30{}^\circ \] \[e=2\text{ }volt\]


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