ElectroMagnetic InductionHard
Question
A thin semicircular ring of radius R is falling with its plane vertical in a horizontal magnetic induction 
. At the position MNQ the speed of the ring is v and the potential difference across the ring is
. At the position MNQ the speed of the ring is v and the potential difference across the ring is Options
A.zero
B.BvπR2 / 2 and M is at higher potential
C.πBRv and Q is at higher potential
D.2BRv and Q is at higher potential
Solution
Induced motional emf MNQ is equivalent to the motional emf in an imaginary wire MQ i.e.,
eMNQ = eMQ = Bvl = Bv(2R) [l = MQ = 2R]
Therefore, potential difference developed across the ring is 2BRv and Q is at higher potential.
eMNQ = eMQ = Bvl = Bv(2R) [l = MQ = 2R]
Therefore, potential difference developed across the ring is 2BRv and Q is at higher potential.
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