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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    A small piece of metal wire is dragged across the gap between the poles of a magnet in 0.4 s. If change in magnetic flux in the wire is\[8\times {{10}^{-4}}\text{Wb},\] then emf induced in the wire is:

    A)  \[8\times {{10}^{-3}}V\]                              

    B)  \[6\times {{10}^{-3}}V\]

    C)  \[4\times {{10}^{-3}}V\]                              

    D)  \[2\times {{10}^{-3}}V\]

    Correct Answer: D

    Solution :

                     \[|e|=\frac{d\text{o }\!\!|\!\!\text{ }}{dt}\] \[=\frac{8\times {{10}^{-4}}}{0.4}\] \[=2\times {{10}^{-3}}V\]


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