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

  • question_answer
    The magnetic field in a region is given by \[B={{B}_{0}}\left( 1+\frac{x}{a} \right)\hat{k}\]. A square loop of edge-length d is placed with its edges along the x and y-axes. The loop is moved with a constant velocity\[v={{v}_{0}}\hat{i}.\]The emf induced in the loop is:

    A) zero      

    B) \[{{v}_{0}}{{B}_{0}}d\]

    C) \[\frac{{{v}_{0}}{{B}_{0}}{{d}^{3}}}{{{a}^{2}}}\]

    D) \[\frac{{{v}_{0}}{{B}_{0}}{{d}^{2}}}{a}\]

    Correct Answer: D

    Solution :

    [d] At \[x=0\], \[B={{B}_{0}}\], so \[{{e}_{1}}={{B}_{0}}{{v}_{0}}d\] At,\[x=d\],\[B={{B}_{0}}\left( 1+\frac{d}{a} \right)\], so \[{{e}_{2}}={{B}_{0}}\left( 1+\frac{d}{a} \right){{v}_{0}}d\] Now \[{{e}_{net}}={{e}_{2}}-{{e}_{1}}=\frac{{{B}_{0}}{{v}_{0}}{{d}^{2}}}{a}\]


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