CET Karnataka Medical CET - Karnataka Medical Solved Paper-2013

  • question_answer
    A rectangular coil of 100 turns and size\[0.1m\times 0.05m\] is placed perpendicular to a magnetic field of 0.1 T. If the field drops to 0.05 T in 0.05 second, the magnitude of the e.m.f. induced in the coil is

    A)  \[\sqrt{2}\]   

    B)  \[\sqrt{3}\]

    C)  \[\sqrt{0.6}\] 

    D)  \[\sqrt{6}\]

    E)  None of the Above

    Correct Answer: E

    Solution :

     : Here, Area of coil, \[A=0.1m\times 0.05m=5\times {{10}^{-3}}{{m}^{2}}\] Number of turns, N = 100 Initial flux linked with the coil \[{{\phi }_{1}}=BA\cos \theta =0.1\times 5\times {{10}^{-3}}\cos {{0}^{o}}\] \[=5\times {{10}^{-4}}Wb\] Final flux linked with the coil \[{{\phi }_{2}}=0.05\times 5\times {{10}^{-3}}\cos {{0}^{o}}\] \[=25\times {{10}^{-5}}Wb=2.5\times {{10}^{-4}}wb\] The magnitude of induced emf in the coil is \[\varepsilon =\frac{N|\Delta \phi |}{\Delta t}=\frac{N|{{\phi }_{2}}-{{\phi }_{1}}|}{t}\] \[\therefore \] \[=\frac{100|2.5\times {{10}^{-4}}-5\times {{10}^{-4}}|}{0.05}\] \[=\frac{100\times 2.5\times {{10}^{-4}}}{0.05}V=0.5V\] * None of the given options is correct


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