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JEE Main & Advanced Sample Paper JEE Main Sample Paper-25

  • question_answer
    In magnetic field of \[0.05\] T are of coil changes from 101 \[c{{m}^{2}}\] and 100 \[c{{m}^{2}}\] without changing the resistance which is \[2\,\Omega \]. The amount of charge that flow during this period is:

    A)  \[25\times {{10}^{-6}}C\]                   

    B)  \[2\times {{10}^{-6}}C\]

    C)  \[{{10}^{-6}}C\]                                

    D)  \[8\times {{10}^{-6}}C\]

    Correct Answer: A

    Solution :

    \[B=0.5\,T,\,\,{{A}_{1}}=101\,c{{m}^{2}}\,=101\,\times {{10}^{-4}}\,{{m}^{2}}\] \[R=2\Omega \] Amount of charge \[=q=\frac{B\Delta A}{R}\] \[=\frac{0.5\times \,(101\,\times {{10}^{-4}}-100\,\times {{10}^{-4}})}{2}\] \[=\frac{0.5\,\times 1\times {{10}^{-4}}}{2}=2.5\,\times {{10}^{-6}}C\]


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