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QAIPM-2014-Ph-1427: NEET 2014 Physics - Class 12 AIPM-2014-Ph-1427 Solution: A thin semicircular conducting ring (PQR) of radius r is…

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

2Bv\pi r^2

and P is at higher potential

\pi rBv

and R is at higher potential

2rBv

and R is at higher potential

Multiple Choice Options

A zero
B ${{Bv\pi {r^2}} \over 2}$ 2 B v � r 2 _ and P is at higher potential
C $\pi rBV$ � r B V and R is at higher potential
D 2rBV and R is at higher potential

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