RAJASTHAN ­ PET Rajasthan PET Solved Paper-2006

  • question_answer
    A straight conductor of length 0.4 m is moved with a speed of 7 m/s perpendicular to the magnetic field of intensity of\[0.9Wb/{{m}^{2}}\]. The induced emf across the conductor will be

    A)  3.52 V           

    B)  25.2 V

    C)  2.52 V           

    D)  0.252 V

    Correct Answer: C

    Solution :

     Induced emf of conductor \[V=BvI\,\sin \theta \] \[=0.9\times 7\times 0.4\,sin90{}^\circ \] \[V=2.52\text{ }volt\]


You need to login to perform this action.
You will be redirected in 3 sec spinner