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

  • question_answer
    A conducting ring of radius r with a conducting spoke OA is in pure rolling on a horizontal surface in a region having a uniform magnetic field B as shown, v being the velocity of the centre of the ring. Then the potential difference \[{{V}_{0}}-{{V}_{A}}\]is:

    A) \[\frac{Bvr}{2}\]

    B) \[\frac{3Bvr}{2}\]

    C) \[\frac{Bvr}{2}\]

    D) \[\frac{-3Bvr}{2}\]

    Correct Answer: C

    Solution :

    [c] Considering pure rolling of OA about A: the induced emf across OA will be: \[\left| \left. {\vec{e}} \right| \right.=\frac{B\omega {{(r)}^{2}}}{2}\] From lenz law, 0 will be the negative end, while A will be the positive end. Hence   \[{{V}_{0}}-{{V}_{A}}=-\frac{B\omega {{r}^{2}}}{2}\]


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