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J & K CET Engineering J and K - CET Engineering Solved Paper-2013

  • question_answer
    A conductor of length \[5\text{ }cm\]is moved  parallel to itself with a speed of \[2\text{ }m/s,\] perpendicular to a uniform magnetic field of\[{{10}^{-3}}\text{ }Wb/{{m}^{3}}\]. The induced e.m.f. generated is

    A)  \[2\times {{10}^{-3}}V\]        

    B)  \[1\times {{10}^{-3}}V\]

    C)  \[1\times {{10}^{-4}}V\]        

    D)  \[2\times {{10}^{-4}}V\]

    Correct Answer: C

    Solution :

    \[e=-Bvl\] Given,  \[l=5cm=5\times {{10}^{-2}}m,\,v=2m/s\] \[B={{10}^{-3}}Wb/{{m}^{2}}\] \[e={{10}^{-3}}\times 2\times 5\times {{10}^{-2}}\] \[e=10\times {{10}^{-5}}V\] \[e=1\times {{10}^{-4}}V\]


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