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JEE Main & Advanced JEE Main Paper (Held On 10 April 2016)

  • question_answer
    A conducting metal circular -wire loop of radius r is placed perpendicular to a magnetic field which varies with time as \[B={{B}_{0e}}^{-t/\tau }\], where B0 and \[\tau \]are constants, at t = 0 . If the resistance of the loop is R then the heat generated in the loop after a long time \[(t-\infty )\] is:   JEE Main Online Paper (Held On 10 April 2016)

    A) \[\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}R}{\tau }\]                     

    B) \[\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}}{2\tau R}\]

    C) \[\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}}{\tau R}\]                     

    D) \[\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}}{2\tau R}\]

    Correct Answer: D

    Solution :

                 heat equated \[\int\limits_{0}^{\infty }{{{i}^{2}}Rdt}\] \[\int\limits_{0}^{\infty }{\frac{^{\varepsilon }in{{d}^{2}}}{Rt}Rdt}\] \[=\frac{1}{R}\int\limits_{0}^{\infty }{^{\varepsilon }in{{d}^{2}}dt}\] \[=\frac{1}{R}\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}}{{{\tau }^{2}}}\int\limits_{0}^{\infty }{{{e}^{-2t/\tau }}dt}\] \[\left. =\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}}{R{{\tau }^{2}}}\frac{{{e}^{-2t/\tau }}}{-2/\tau } \right|_{0}^{\infty }\] \[=\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}\tau }{2R{{\tau }^{2}}}\left\{ 0+1 \right\}\] \[\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}}{2R\tau }\]


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