Will a changing $E$ field induce a current in a loop similar to a changing $B$ field? An induced current in a wire loop that is caused by a changing B field is a common EM question. However, I couldn't find examples online where the B field was substituted for a changing E Field.
The following question was given on a test and the goal was to find the current flow caused by a varying B Field first, then a varying E Field.  My answer is illustrated below.

While it was simple to deduce the direction of the current with a changing B field (clockwise), when the E field was subbed in below, my answer was completely different. Instead, I ended up with an induced B field that was counterclockwise on the outside of the loop and clockwise on the inside of the loop.
 A: Keep in mind three facts:

*

*If you look at the Lorentz force, a static magnetic field never imparts kinetic energy onto a charged particle; it only curves its trajectory. You need electric field to speed up or down a charge.

*If you look at Faraday's law, you will see the curl of the electric field is zero when the time derivative of magnetic field  $B$ is zero everywhere.

*Maxwell's equations are coupled. A changing electric field also produces a magnetic field.

If the electric field is curlless, then there is no closed loop that accelerates charges around the whole way through: so, when  exposed to a new, curlless electric field, charges just rearrange themselves, without developing a net current around the loop.
A changing electric field can produce a magnetic field, but this magnetic field cannot directly speed up charges around the loop. You still need the line integral of $\mathbf{E}\cdot \mathbf{n}$ around the the loop to be nonzero, which requires $\nabla\times\mathbf{E}=-\partial\mathbf{B}/\partial t$ to be nonzero somewhere. So unless a changing magnetic field that produces an EMF is produced/present somehow, just changing the electric field is not enough.
A: $\nabla\times E = -\frac{\partial B}{\partial t}$ says there is "circulation" of E around the loop. E will push charges along the loop.
This follows from Stoke's Theorem: $\int_{loop} E \cdot \space dl= \int_{surface} \nabla \times E \space\text{ds}$
$\nabla \times B = \frac{\partial E}{\partial t}$ says there is "circulation" of B around the loop. That isn't going to push charges along the loop.
Both conclusions, if the charges are pushed along the loop or not, follow from $F=q(E+v\times B)=qE +q (v\times B ).$ The first term is the force due to the electric field. The second term is the force due to the magnetic field.
The charges must be moving, with velocity, $v_{\perp \space to \space B}$, perpendicular to B to feel the magnetic force $F=q(v \times B)$. The charges in the loop can't do that because they are constrained by the loop. And, if they did feel a force, that force would be perpendicular to the loop and $B$.
Feynman Lectures. Vector Integral Calculus
A: One way to see that no current can be induced in a loop by a changing E field that is normal to the plane of the loop is that if a current were induced, it would be in one direction and not the other, and that would violate a symmetry principle called "conservation of parity".
This sounds complicated but it is very simple: in all of classical physics (including E&M) there is no experiment that you can do where someone watching would be able to determine by looking at the result whether they're looking at the actual scene, or seeing it in a mirror (parity usually refers to flipping all three spatial dimensions rather than just one as happens in a mirror, but it comes out the same).
Another way that this is often said is that there is no way you could tell someone (say, a space alien, so that you have nothing environmental in common to which you could refer), using only words (no artifacts or drawings), which way is left and which is right, or which is clockwise and which counterclockwise.
But if there was a current in your experiment, it would be a way to define clockwise: "Steadily add charge to a sheet which has a conducting loop above it; looking at the loop from above, so you see the sheet behind it, watch which way the charge carriers of the same sign as those you added to the sheet move around the loop; that's clockwise."
So, the current must be zero.
You might protest that a changing magnetic field does induce a current, and seems subject to the same reasoning.  It isn't.  The math way to explain this is that B transforms as an axial vector, while E transforms as a polar vector.  As a silly example of the problem, here's how a conversation might go where you're trying to explain to space alien Bob what clockwise means by referring to a magnetically induced current:
You: Ok, so as you crank up the B field, you'll see a clockwise current flow...
Bob: Wait, you create your B field with a coil, right?
You: Yeah...
Bob: Which way do run current through the coil?
You: Counterclockwi... oh... damn
A: I think that the idea of the question is to note not only that there is a changing field, but also that the time derivative is constant.
So, your conclusion is OK in the first part. In the second part $B$ is constant around the loop, so the charges are not affected and there is no current.
The electric field acts inside the loop, not in the wire as far I understand.
