this question related to the electromagnetic induction

Dear Student
For the potential difference we have to reduce emf in the ring by the cell, as shown below

Here to get the direction of e1 we followed Fleming right hand rule to find out current direction.
According to which thumb is in direction of motion, fore finger is in direction of magnetic field so middle finger will direction of current from Ato B, Hence A is at high potential and B is at lower potential,
Same concept we can apply to get direction of emf for C and D,

A to B distance is 2r (diameter of first ring) and C to D distance is 4r (diameter of second ring)$$\begin{gathered} e_1 = Blv = B\left(2r\right)\left(2v\right) =4Brv \\ {e}_2 = Blv = B\left(4r\right)\left(v\right) =4Brv \\ Therefore the pd between the heighest point of the two ring will be: \\ {e}_{AC} =4Brv -(-4Brv) = 8 Brv \end{gathered}$$

Regards