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:

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Correct answer: D. 2rBv and R is at higher potential

Here we have to calculate the emf of the conducting ring when it is falling

A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field

So, we have to calculate with the following formulae

$emf = VB{l}_{eq} = VB\left(2R\right)$

where $R$ is at higher potential and $P$ is at lower potential.

A thin semicircular conducting ring (PQR) of radius r is falling...

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