Under a uniform time dependent magenetic field, which Amperian loop shall we choose to determine the induced electric field?

Question

I was reading the Griffin's Electrodynamics textbook and I came up with this question:

Suppose we have a uniform time-dependent magnetic field, I know that by Faraday's Law, we can calculate the induced electric field at the dashed Amperian loop to be $E=-\frac{s}{2}\frac{dB}{dt}\hat{\phi}$. However, if we choose another Amperian loop that intersects with the previous loop, then at the two intersections (the two blue dots), Faraday's Law will give a different direction of the induced electric field. May I ask how should I understand this difference?

My current thought is the induced electric field won't make physical sense if we don't have an actual "receiptor". That is to say the actual direction and magnitude of the electric field would be dependent on the shape of the actual circuit I put into the magnetic field. Therefore, the two imaginary loops in my question actually correspond to two different physical questions. Therefore the induced electric field will be different. Is my thinking correct?

Answer 1

Sorry this is not the answer but comment field is very limited then I write here.

If there is no wire circle in the uniformly varying magnetic B-fields, there are B-fields and electric E-fields (emf) in space. Although the B-fields are uniform over the space positions, the emf E-fields become some non-uniform distributions.

If a wire circle is placed at the "B-field center" so as to maintain axis symmetry, a uniform electric current $\vec{J}$ is induced inside the wire circle in accordance with the uniform E-field distribution along the wire.

If the wire circle is placed at some deviated point of the "B-field center", then the axis symmetry is broken and the tangent component of the emf E-field along the wire circle differs from place to place. In this case, $-\text{grad}\phi$ distribution is generated both inside wire and space, and electric current distribution becomes uniform along wire. Note that electric scalar potential $\phi$ is single valued function; no contributions to emf.