# Induced Electric field due to magnetic field in Faraday experiment

Regarding this experiment where a magnet is moved in and out of a coil -(see the picture)

what i considered to be true is that when there is a changing magnetic flux through the coil(due to changing magnetic field coz area and orientation of coil is kept constant), it produces a circular Electric field around the coil which produces current in coil and that Electric field is given by faraday law as we all know very well. Also assume change in magnetic field to be as such that the current produced in coil is constant. Lastly this current produces a opposing magnetic field which opposes motion of bar magnet. Case Close

But now in this Answer-Click to Teleport

It says even if there is no current produced coz there is no coil there is stil some $$\frac{\partial E}{\partial t}$$ that produces the magnetic field. The problem is-(1)Why isn't this magnetic field in case we discussed above, was it missed or it wouldn't exist? (2) how different will the magnetic field produced by this $$\frac{\partial E}{\partial t}$$ be from the one produced by constant current in above case we discussed? (3) will this magnetic field be opposing the motion of magnet?

Your reasoning is basically correct that moving the magnet results in $$d{\bf B}/dt$$ and to have the matching $$\nabla \times {\bf E}$$ there must be a nonzero $${\bf E}$$, which by itself will also be time-dependent so you have $$d{\bf E}/dt$$ and there should be a matching $$\nabla \times {\bf B}$$. Since the magnetostatic field of a stationary magnet has $$\nabla \times {\bf B}=0$$, there must indeed be some change in the shape of the $${\bf B}$$-field.