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JEE Main & Advanced Physics Electro Magnetic Induction Question Bank Self Evaluation Test - Electromagnetic Induction

  • question_answer
    A thin circular ring of area A is perpendicular to uniform magnetic field of induction B. A small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of circuit is R. When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer is

    A) \[\frac{BR}{A}\]

    B) \[\frac{AB}{R}\]

    C) ABR

    D) \[{{B}^{2}}\,A/{{R}^{2}}\]

    Correct Answer: B

    Solution :

    [b] \[Q=\frac{\Delta \phi }{R}=\frac{{{\phi }_{2}}-{{\phi }_{1}}}{R}=\frac{BA-0}{R}=\frac{BA}{R}\]


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