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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
Zero
B
$\frac{Bv\pi r^2}{2}$ and P is at higher potential
C
$\pi rBv$ and R is at higher potential
D
2rBv and R is at higher potential
Detailed Solution
The semicircular ring is equivalent to a straight conductor PR of length 2r moving with speed v.
Induced emf $= Bv(2r) = 2rBv$, with R at the higher potential.