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JCECE Engineering JCECE Engineering Solved Paper-2012

  • question_answer
    A wire of length \[1\,\,m\] is moving at a speed of \[2\,\,m{{s}^{-1}}\]perpendicular to its length in a homogeneous magnetic field of\[0.5\,\,T\]. If the ends of the wire are joined to a circuit of resistance\[6\Omega \], then the rate at which work is being done to keep the wire moving at constant speed is

    A) \[1\,\,W\]                                          

    B) \[\frac{1}{3}\,\,W\]

    C) \[\frac{1}{6}\,\,W\]                                        

    D)  \[\frac{1}{12}\,\,W\]

    Correct Answer: C

    Solution :

    Rate of work\[=\frac{W}{t}=P\] But         \[P=Fv\], Also,      \[F=Bil=B\left( \frac{Bvl}{R} \right)l\]                                 \[\left[ \because \,\,induced\,\,current,\,\,i=\frac{Bvl}{R} \right]\] \[\therefore \]  \[P=B\left( \frac{Bvl}{R} \right)lv=\frac{{{B}^{2}}{{v}^{2}}{{l}^{2}}}{R}\]                     \[=\frac{{{(0.5)}^{2}}\times {{(2)}^{2}}\times {{(1)}^{2}}}{6}=\frac{1}{6}W\]


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