JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    A coil having n turns and resistance\[R\,\Omega \]. is connected with a galvanometer of resistance\[4R\,\Omega \]. This combination is moved in time t seconds from a magnetic field\[{{W}_{1}}\]weber to\[{{W}_{2}}\] weber. The induced current in the circuit is

    A) \[\frac{{{W}_{2}}-{{W}_{1}}}{5Rnt}\]      

    B)        \[-\frac{n({{W}_{2}}-{{W}_{1}})}{5Rt}\]

    C)                        \[-\frac{({{W}_{2}}-{{W}_{1}})}{Rnt}\]   

    D)        \[-\frac{n({{W}_{2}}-{{W}_{1}})}{Rt}\]

    Correct Answer: B

    Solution :

    The rate of change of flux or emf induced in the coil is, \[e=-n\frac{d\phi }{dt}\] \[\therefore \]Induced current\[i=\frac{e}{R'}=-\frac{n}{R'}\frac{d\phi }{dt}\] Given,\[R'=R+4R=5R,d\phi ={{W}_{2}}-{{W}_{1}},dt=t.\] (here,\[{{W}_{1}}\]and\[{{W}_{2}}\]are flux associated with one turn) Putting the given values in Eq. (i), we get \[i=-\frac{n}{5R}\frac{({{W}_{2}}-{{W}_{1}})}{t}\]


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