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Instead of the differential form of Maxwell's equations, I find the
integral form a little easier to visualize here.
If I take your question as-is and don't turn it into a question that
may be easier to answer, I have to conclude that the question is not
answerable. If I form a circular loop in the x-y plane and integrate
around that circle, I can easily find a value for Integral(E dot ds)
= - d(B_flux)/dt. The result is dB/dt times the area of the circle.
The EMF around this loop has this same value. If a circular loop of
wire is placed in the same position as the loops we integrated over,
the current through the wire loop will be the EMF/R (where R is the
resistance of the loop). Since no boundaries are specified, I can
move the loop to any position that I want in the x-y plane and I will
get the same current in the same direction (clockwise or
counterclockwise). If I now take two loops of wire and place then so
their edges are almost just touching, the current in one is flowing
in the opposite direction as in the other at the point where they
almost touch. So if we are thinking in terms of electric fields
pushing an charge, the field obviously has two opposite directions at
the same point - depending on which loop I am looking at.
I take this
absurdity to indicate that the question cannot be answered exactly as
it is posed.