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JEE Main & Advanced JEE Main Solved Paper-2014

  • question_answer
    A conductor lies along the z-axis at \[-1.5\le z<1.5m\]and carries a fixed current of 10.0 A in \[-{{\hat{a}}_{z}}\]direction (see figure). For a field \[\vec{B}=3.0\times {{10}^{-4}}{{e}^{-0.2x}}{{\hat{a}}_{y}}T,\]find the power required to move the conductorat constant speed to \[x=2.0m,y=0m\]in \[5\times {{10}^{-3}}s.\] Assume parallel motion along the x-axis.   JEE Main  Solved  Paper-2014

    A) 14.85 W                               

    B) 29.7 W

    C) 1.57 W

    D) 2.97 W

    Correct Answer: C

    Solution :

    Force on conductor = I L B \[B={{B}_{0}}{{e}^{-\alpha x}}\] (Due to magnetic field) [+ x−direction ] \[\left\langle P \right\rangle =\frac{1}{T}\int\limits_{0}^{T}{F.V.}dt=\frac{1}{T}\int\limits_{0}^{{{x}_{0}}}{F}.dx\]\[[{{x}_{0}}=2m]\] \[\Rightarrow \]\[\left\langle P \right\rangle =\left[ \frac{1}{T}\frac{I\,L\,{{B}_{0}}}{(-\alpha )}{{e}^{-\alpha x}} \right]_{0}^{2}\] \[\Rightarrow \]\[\left\langle P \right\rangle =\frac{I\,L\,{{B}_{0}}}{T\times 0.2}\left[ 1-{{e}^{-\alpha (2m)}} \right]\] \[=\frac{10\times 3\times 3\times {{10}^{-6}}}{5\times {{10}^{-3}}\times 0.2}\left[ 1-{{e}^{-0.4}} \right]\] \[=\frac{1.8}{0.2}\left( 1-\left( 1-(0.4)+\frac{0.16}{2}-\frac{0.064}{16}+...... \right) \right)\] \[\approx 9\times \frac{0.64}{2}=2.88\] On Exact evaluation \[\left\langle P \right\rangle =2.97W\]


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