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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • question_answer
    A straight conductor of length 4 m 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.perm2 then induced emf. is:

    A)  8 V                        

    B)         4 V

    C)  1 V                        

    D)         2 V

    Correct Answer: D

    Solution :

    Induced emf is given by \[e=B\upsilon l\,\sin \,\text{=0}\text{.1}\times \text{10}\times \text{4}\,\text{sin 30}{}^\circ \,\text{e}\,\text{=2}\,\]volt


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