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

Options

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

Answer

B

Explanation

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 .