A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic induction B fig. (12). At the position MNQthe speed of ring is V and potential difference developed across the ring is
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a
BVπr22 and M is at higher potential
b
π R B V and Q is at higher potential
c
π R B V and Q is at higher potential
d
2 R B V and Q is at higher potential
answer is D.
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Detailed Solution
The induced e.m.f. in a conductor moving in a magnetic field is given bye= I(V x B)In the given equationsl=MQ=2R. V is ⊥ to B∴e=2RVBThe direction of induced e.m.f. will be from Q to M. So, Q is at higher potential.