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A thin semicircular conducting ring (PQR...

A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in figure. The potential difference developed across the ring when its speed is v, is

A

`pirBv` and R is at higher potential

B

2rBv and R is at higher potential

C

Zero

D

`Bvpir^(2)//2` and P is at higher potential

Text Solution

Verified by Experts

The correct Answer is:
B
emf is indirect called motion emf (BLv) instead of varying the magnetic field.
`e=BLv=Bv(2r)=2Bvr`
R will be at higher potential.
By Fleming thumb rule, current flow from P to R so electrons move towards P and +ve charge increase at R .
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