JEE Main & Advanced Sample Paper JEE Main Sample Paper-35

  • question_answer
    A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic induction\[\vec{B}\]. At the position MNQ the speed of the ring is v and the potential difference developed across the ring is:             

    A)  zero

    B)  \[B\pi v{{R}^{2}}/2\] and M is at higher potential

    C)  \[\pi BRv\] and Q is at higher potential

    D)  \[2RBv\] and Q is at higher potential

    Correct Answer: D

    Solution :

    \[{{e}_{MNQ}}={{e}_{MQ}}\,=BvL\,=Bv\,(2R)\]


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