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JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Morning)

  • question_answer
    A conducting circular loop made of a thin wire, has area \[3.5\,\,\times \,\,{{10}^{-3}}\,{{m}^{2}}\] and resistance 100. It is placed perpendicular to a time dependent magnetic field\[B\left( t \right)=\left( 0.4T \right)sin\left( 50\pi t \right)\]. The field is uniform in space. Then the net charge flowing through the loop during \[t\,\,=\,\,0\] s and \[t\,\,=\,\,0\] ms is close to: [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

    A) 6 mC

    B) 14 mC

    C) 7 mC

    D) 21 mC

    E) None of These

    Correct Answer: E

    Solution :

    [Bonus] \[\Delta \,q\,\,=\,\,\frac{\Delta \,\phi }{R}\] \[B\text{ }=\text{ }0.4\text{ }sin\text{ }50\text{ }\pi t\] \[\phi \text{ }=\text{ }B.\,A\text{ }=\text{ }\left( 0.4\text{ }sin\text{ }50\text{ }\pi t \right)\text{ }3.5\text{ }\times \text{ }{{10}^{-3}}\] \[=\,\,\,1.4\text{ }\times \text{ }{{10}^{-3}}\,sin\text{ }50\pi t\] At \[t=\text{ }0\text{ };\text{ }\phi \text{ }=\text{ }0\] \[At\text{ }t=\text{ }10\,\,\times \,\,{{10}^{-3}}\,s\] \[\phi \text{ }=\text{ }1.4\,\,\times \,\,{{10}^{-3}}\,\,sin\,(50\pi \,\,.\,{{10}^{-2}})\] \[=\text{ }1.4\,\,\times \,{{10}^{-3}}\] \[\Rightarrow \,\,\,\,\Delta \,q\,\,=\,\,\frac{1.4\,\,\times \,\,{{10}^{-3}}}{10}\,\,=\,\,0.14\,\,\times \,\,{{10}^{-3}}\] = 0.14 mC.


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