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JEE Main & Advanced AIEEE Solved Paper-2006

  • question_answer
    The flux linked with a coil at any instant t is given by \[\phi =10{{t}^{2}}-50t+250\] The induced emf at\[t=3\text{ }s\]is     AIEEE  Solved  Paper-2006

    A) \[-190V\]       

    B)        \[-10V\]              

    C)        \[10V\]                

    D)        \[190V\]

    Correct Answer: B

    Solution :

    Given that the variation of flux with time as \[\phi =10{{t}^{2}}-50t+250\] From Faraday's law of electromagnetic induction, \[e=-\frac{d\phi }{dt}\] \[=\frac{-d(10{{t}^{2}}-50t+250)}{dt}\] \[\Rightarrow \]\[e=-[10\times 2t-50]\] \[\therefore \]\[e{{|}_{t=3s}}=-[10\times 6-50]\] \[=-10V\]


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