Does the induced Electric field develop instantly or lags by $\frac rc$? Consider a loop of radius $r_0=3 \times 10^8$ cm. A thin bar magnet is passed through its center. This implies that the magnetic flux through the loop will change. Now according to Faraday's law a non conservative electric field must develop inside the wire of that loop.
Question: 

Let the magnet be cross though the centre at time $t_0$. Then would the electric field develop in the wire at $t_0$ or at $t_0+\dfrac {r_0}c$?

Commentary
According to the Faraday's Law of Electromagnetic induction the electric field $E(t)$ developed inside the wire is given by the following mathematical relationship,  
$$\oint \mathbf{E}(t_0) \cdot d\boldsymbol{\ell} = - \dfrac{d \Phi(t)}{dt}|_{t=t_0}$$  
According to this, the electric field developed at time $t_0$, $E(t_0)$ has a magnitude which directly depends upon the rate of change of magnetic flux at time $t_0$. This implies that the information of the motion of the magnet can be transmitted at any distance instantly. I might be wrong. If I am not wrong then this fact can be used to create paradox by applying special theory of relativity.   
The second case can be that the electric field doesn't develop instantly and lages its cause by $\dfrac {r_0}{c}$. If this is the case then the current in the loop will develop after a time interval of $\dfrac {r_0}{c}$, then this current will create its own magnetic field which will cause a change in magnetric flux, and a net change in magnetic flux will be observed after $\dfrac {r_0}{c}$, this net change in the flux will cause its effect on $E$ at time $\dfrac {2r_0}{c}$. But we know that in the analysis of a pure conductor, V-I characteristic is solved by differential equations which gives a solution free of $c$. So either this second case is incorrect or the V-I characteristic of the pure inductor.  
Thank You.
 A: This should be considered a provisional answer.
First, let's make the setup concrete.


*

*There is a thin conducting circular loop in the $xy$ plane with
radius $r_0$.

*There is an ideal magnetic dipole aligned with and located on the $z$
axis and initially at rest at $z = z_0$.

*At time $t = t_0$, the dipole begins accelerating along the $z$ axis.


Now, let the elapsed time since $t = t_0$ be
$$\Delta t = t - t_0 $$
and the (static) magnetic field at time $t = t_0$ 
$$\vec B_0 = \vec B(t_0)$$
Then, according to special relativity, the changing magnetic field and associated electric field cannot reach the loop until an elapsed time of
$$\Delta t_l = \frac{\sqrt{r^2_0 + z^2_0}}{c}$$
For $\Delta t < \Delta t_l$, the magnetic field outside of a sphere of radius $c\Delta t$ and centered on $(0,0,z_0)$ is just $\vec B_\text{outside} = \vec B_0$ while inside, the magnetic field is $\vec B_\text{inside} = \Delta \vec B(t) + \vec B_0$.
This means that, while $\Delta t < \Delta t_l$, no magnetic field lines of $\Delta \vec B$ thread the conducting loop. $^1$
So, here's the provisional answer:  the flux of $\Delta \vec B$ through the surface bounded by the conducting loop is proportional to the number of lines of $\Delta \vec B$ threading the loop. 
Thus, for the surface bounded by the conducting loop,
$$-\frac{d\Phi(t)}{dt} = 0\; ,t < (t_0 + \Delta t_l) $$

$^1$ To be clear, I mean that no field line of $\Delta B$ can cut the loop before $t = t_0 + \Delta t_l$
A: An experimentalist's answer
It is experimentally established that the underlying framework of nature  is quantum mechanical. Changes in electric and magnetic fields create electromagnetic waves. Thus photons propagate the changes/information in your experiment, they are the quantum of energy in electromagnetism. 
It is also experimentally established that special relativity holds. This means that the information of changes in magnetic and electric fields cannot be propagated  faster than the velocity of c that controls the behavior of photons. 
There is no instantaneous propagation of energy as far as we have established by experiments and observations.
