First slide
Motional EMF
Question

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

Moderate
Solution

The induced e.m.f. in a conductor moving in a magnetic field is given by
e= I(V x B)
In the given equations
l=MQ=2R. V is  to Be=2RVB
The direction of induced e.m.f. will be from Q to M. So, Q is at higher potential.
 

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