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A thin semi-circular conducting ring of ...

A thin semi-circular conducting ring of radius R is falling with its plane verticle in 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

`Bv pi R^(2)//2` and M is higher potencial

C

`pi` RBv ans Q is at higher potential

D

2RBv ans Q is at higher potential

Text Solution

Verified by Experts

The correct Answer is:
D
Induced emf produced across MNQ will be same as the induced emf produced in straight wire MQ.
`:. E=Bvl=Bvxx2R` with Q at higher potential.
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