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AMU Medical AMU Solved Paper-1999

  • question_answer
    The magnetic induction in the region between the pole faces of an electromagnet is 0.7 weber/\[{{m}^{2}}\]. The induced e.m.f. in a straight conductor 10 cm long, perpendicular to B and moving perpendicular both to magnetic induction and its own length with a velocity 2 m/sec is

    A)  0.08 V            

    B)  0.14V

    C)  0.35 V            

    D)  0.07 V

    Correct Answer: B

    Solution :

    : Induced emf across conductor\[=Blv\] \[\therefore \]\[e=Blv=0.7\times \frac{10}{100}\times \frac{2}{1}=0.14V\]


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