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JEE Main & Advanced JEE Main Paper (Held On 11 April 2014)

  • question_answer
    A coil of circular cross-section having 1000 turns and \[4c{{m}^{2}}\]face area is placed with its axis parallel to a magnetic field which decreases by \[{{10}^{-2}}\] Wb \[{{m}^{-2}}\] in 0.01 s. The e.m.f. induced in the coil is:   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

    A) 400 mV                                

    B) 200 mV

    C) 4 mV                     

    D) 0.4 mV

    Correct Answer: A

    Solution :

    Given: No. of turns N = 1000 Face area, \[A=4c{{m}^{2}}=4\times {{10}^{-4}}{{m}^{2}}\] Change in magnetic field, \[\Delta B={{10}^{-2}}wb{{m}^{-2}}\] Time taken, t\[t=0.01s={{10}^{-2}}\sec \] Emf induced in the coil e = ? Applying formula, Induced emf, \[e=\frac{-d\phi }{dt}\] \[=N\left( \frac{\Delta B}{\Delta t} \right)A\cos \theta \] \[=\frac{1000\times {{10}^{-2}}\times 4\times {{10}^{-4}}}{{{10}^{-2}}}=400mV\]


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