To solve this problem, we need to determine the potential difference across the semicircular ring.$\\$Given:$\\$• Radius of the semicircular ring $r. \\$• Magnetic field $B$ is horizontal.$\\$• Speed of the ring $v. \\$When a conductor moves in a magnetic field, an emf is induced due to the motion.$\\$The formula for the motional emf ($\varepsilon$) induced in a conductor of length $L$ moving with velocity $v \\$perpendicular to a magnetic field $B$ is:$\\ \varepsilon = B \cdot v \cdot L \\$For a semicircular ring, the effective length $L$ is the diameter of the semicircle, which is $2r. \\$Thus, the induced emf is:$\\ \varepsilon = B \cdot v \cdot 2r \\$This emf causes a potential difference across the ends of the semicircular ring.$\\$Therefore, the potential difference developed across the ring is $2Bvr. \\$Since the ring is moving downwards, the direction of the induced current will be such that the right side (R) is at a higher potential.$\\$Thus, the correct option is:$\\ 2Bvr,$ and R is at higher potential.$\\$This corresponds to Option 4.