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JCECE Medical JCECE Medical 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 0.5T. 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, \[F=Bil=B\left( \frac{Bvl}{R} \right)l\] \[\left[ \because \,\text{induced}\,\text{current,}\,i=\frac{Bvl}{R} \right]\] \[\therefore \] \[P=B\left( \frac{Bvl}{R} \right)lv\frac{{{B}^{2}}{{v}^{2}}{{l}^{2}}}{R}\] \[=\frac{{{(0.5)}^{2}}\times {{(2)}^{2}}\times {{(1)}^{2}}}{6}=\frac{1}{6}W\]


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