Equivalence of Inertial Reference Frames 2 [duplicate]

this question is a modified version of my previous question:https:[//physics.stackexchange.com/q/875717/589276] Consider the coordinate system S'. In the S' frame, there is a uniform magnetic field of magnitude B' in the x'y'-plane. The magnetic field lines are perpendicular to the x'y'-plane and point outward from the plane. There is also a uniform electric field E' in the x'y'-plane, directed downward.In the x'y'-plane, there is a conducting structure as shown in Figure. side $a_3$ is free to move in the y′-direction(just side $a_3$ can move) and it is connected to $a_2$ with two springs. If rod $a_3$ moves with a constant velocity u' in the +y'-direction, the magnetic flux through the upper rectangle changes. Consequently, a measurable electric current is produced in the structure. the current in the upper rectangle is clockwise . Now consider the same situation from the perspective of observer S. From the perspective of observer S, the coordinate system S' is moving with a constant velocity v in the positive x-direction. If the condition E'>cB' is satisfied, the relative velocity v between the two frames can be chosen such that, from the perspective of observer S, the magnetic field within the moving S' frame is zero: B=0. Now, from the perspective of observer S, only the electric field E exists within the S'.and E=Ey=γ(-E'+vB') The electric field E is directed along the y'-axis and points downward. If the motion of the rod $a_3$ with velocity u(u is velocity of rod $a_3$ respective to s) in the y'-direction is considered, there is no reason, from the perspective of observer S, for an electric current to be produced in the upper loop. Because magnetic field is zero and electric field is constant. In this configuration, observer S cannot consider the polarized structure to be moving in such a way as to produce a magnetic field in the vicinity of side $a_3$ and thereby create a clockwise current in the upper loop.(consider $L\gg l$ )