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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole setup is moving towards right with a constant speed of \[1cm{{s}^{-1}}.\] At some instant, a part of L is in a uniform magnetic field of 1T, perpendicular to the plane of the loop. If the resistance of L is \[1.7\Omega ,\]the current in the loop at that instant will be close to:           [JEE Main Held on 12-4-2019 Morning]

    A) \[115\mu A\]               

    B) \[170\mu A\]

    C) \[60\mu A\]                 

    D) \[150\mu A\]

    Correct Answer: B

    Solution :

    Since it is a balanced wheat stone bridge, its equivalent resistance =\[=\frac{4}{3}\Omega \] \[\varepsilon =B\ell \text{v}=5\times {{10}^{-4}}V\] So total resistance \[R=\frac{4}{3}+1.7\approx 3\Omega \] \[\therefore \]\[i=\frac{\varepsilon }{R}\approx 166\mu A\approx 170\mu A\]


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