AIPMT | 2014Magnetic field due to currentHard

Question

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 u, is
       

Options

A.Zero
B.Bvπr2/2 and P is at higher potential
C.πrBv and R is at higher potential
D.2rBv and R is at higher potential

Solution

ε = BLeffu(Leff = Diameter)
= B2Ru

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