Questions
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 MNQ
the speed of ring is V and potential difference developed across the ring is
detailed solution
Correct option is D
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.Talk to our academic expert!
Similar Questions
A conductor rod AB moves parallel to X-axis in a uniform magnetic field, pointing in the positive X-direction. The end A of the rod gets
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