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

The motional emf induced in a conductor moving in a uniform magnetic field is given by $\varepsilon = (\vec{v} \times \vec{B}) \cdot \vec{l}$, where $\vec{l}$ is the vector connecting the ends of the conductor.

For the semicircular ring PQR, the effective length vector $\vec{l}$ is the straight line from P to R, which has a magnitude equal to the diameter of the semicircle.

$l_{\text{eff}} = 2r$

Since $\vec{v}$, $\vec{B}$, and $\vec{l}$ are mutually perpendicular, the magnitude of the induced motional emf is:

$\varepsilon = B v l_{\text{eff}} = 2rBv$

The polarity of the induced emf is determined by the direction of the magnetic force on the positive charges, given by $\vec{F} = q(\vec{v} \times \vec{B})$.

The velocity $\vec{v}$ is directed downwards and the magnetic field $\vec{B}$ is directed into the plane of the paper.

By the right-hand rule, the cross product $\vec{v} \times \vec{B}$ points towards the right.

This means positive charges are pushed towards terminal R, making it at a higher potential than terminal P.

Therefore, the potential difference developed across the ring is $2rBv$ with terminal R at a higher potential.

Answer: $2rBv$ and terminal R is at higher potential