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EAMCET Medical EAMCET Medical Solved Paper-2006

  • question_answer
    A coil of 1200 turns and mean area of \[500\text{ }c{{m}^{2}}\]is held perpendicular to a uniform magnetic field of induction \[4\times {{10}^{-4}}.\] The resistance of the coil is 20\[\Omega \]. When the coil is rotated through \[180{}^\circ \] in the magnetic field in 0.1 s the average electric current (in mA) induced is :

    A)  12                                         

    B)  24

    C)  36                                         

    D)  48

    Correct Answer: B

    Solution :

                     In perpendicular position in uniform magnetic field, magnetic field linked with the each turn of coil \[\text{o}{{\text{ }\!\!|\!\!\text{ }}_{1}}=BA\] On rotating coil through \[{{180}^{o}},\] flux linked \[\text{o}{{\text{ }\!\!|\!\!\text{ }}_{2}}=-BA\] \[\therefore \]  Change in magnetic flux \[\Delta \text{o }\!\!|\!\!\text{ }=\text{o}{{\text{ }\!\!|\!\!\text{ }}_{2}}-\text{o}{{\text{ }\!\!|\!\!\text{ }}_{1}}=-2BA.\] \[\therefore \]  emf induced in the coil \[e=-N\frac{\Delta \text{o }\!\!|\!\!\text{ }}{\Delta t}\] \[=\frac{2NBA}{\Delta f}\]where N is the number of turns in the coil. The current in the coil \[i=\frac{e}{R}=\frac{2NBA}{R\Delta \Tau }\] \[=\frac{2\times 1200\times 4\times {{10}^{-4}}\times 500\times {{10}^{-4}}}{20\times 0.1}\] \[=2400\times {{10}^{-5}}\] \[=24\times {{10}^{-3}}A\] \[=24\,mA\]


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