Why do we calculate the curl of curl of the electric field and what does it mean physically? I was trying to derive wave equation. We know that the curl of an electric field is the rate of change of magnetic flux then why we have to calculate the curl of the curl of the electric field after that. What is the reasoning behind this
Faraday's law
$$ \nabla\ \times E = \frac{\partial B}{\partial t} $$
Taking curl of the equation
$$ \nabla \times \nabla\ \times E = \frac{\partial (\nabla \times B)}{\partial t} $$
I would like to know what is the physical significance of taking the curl of the curl of the electric field.
 A: As demonstrated here, the curl of the curl of a vector field is equivalently the difference of the gradient of the divergence of the vector field and the Laplacian of that field. This is written as,
$$\nabla\times\left(\nabla\times\textbf{E}\right)=\nabla\left(\nabla\cdot\textbf{E}\right)-\nabla^2\textbf{E}$$

I would like to know what is the physical significance of taking the curl of the curl of the electric field

To put it simply, this is a useful identity because it tells you how the curl of $\vec{E}$ changes in response to moving charges. We have from Maxwell's laws,
$$ \nabla \times \nabla \times \vec{E} = -\nabla \times\left(\frac{\partial \vec{B}}{\partial t}\right) = -\frac{\partial}{\partial t}(\nabla \times \vec{B})$$.
In free space, the current density is zero implying that $\nabla \times \vec{B} = \frac{\partial\vec{E}}{\partial t}$. Substituting this into the above equation yields,
$$ \nabla \times \nabla \times \vec{E} = - \frac{\partial^2\vec{E}}{\partial t^2}$$
Thus, ignoring boundary effects, Maxwell’s equations imply that the twice curl of the electric field is the negative of the second partial derivative of the electric field with respect to time. Substituting our mathematical identity for the twice curl of $\vec{E}$ yields the desired (source-free) wave equation, assuming the electric field is sourced by a uniform charge density (i.e. $\nabla \cdot \vec{E} = 0$),
$$ \nabla^2\vec{E} - \frac{\partial^2\vec{E}}{\partial t^2} = 0$$
A lengthy discussion can be found here.
A: The curl can be visualized as the infinitesimal rotation in a vector field. Natural way to think of a curl of curl is to think of the infinitesimal rotation in that rotation itself. Just as a second derivative describes the rate of rate of change, so the curl of curl describes the way the rotation rotates at each point in space.
The reason you are taking the curl of curl is because then the left hand side reduces to an identity involving just the Laplacian (as $\mathbf{\nabla} \cdot \mathbf{E} = 0$). On the right hand side you have $\mathbf{\nabla} \times \mathbf{B}$ which is just $\mu_0 \varepsilon_0 \partial\mathbf{E}/\partial t$.
A: Mathematical operations like this don't have physical significance, but they can reveal physical significance that would be otherwise non-obvious.  In other words, I would say that equations and relationships and mathematical properties might have physical significance, while operations allow us to massage those things into different forms which make that significance more clear.

The equation
$$\nabla \times \mathbf E = -\frac{\partial}{\partial t} \mathbf B \qquad (1)$$
expresses a relationship between the electric and magnetic fields, namely that the rate at which the magnetic field $\mathbf B$ changes is equal to the "circulation" of the electric field $\mathbf E$.  The physical significance of this equation is relatively clear - the presence of a circulating electric field necessitates - and is necessitated by - the presence of a time-varying magnetic field.
That's easy enough to see, but it has implications which are not obvious.  One such implication is found by taking the divergence of both sides.  Since the divergence of a curl is always zero, we find that
$$\frac{\partial }{\partial t} (\nabla \cdot \mathbf B) = 0\qquad (2)$$
That is, the divergence of $\mathbf B$ at every point is constant; it might vary from point to point, but it doesn't change in time.
The lesson here is that if the rate of change of one field is equal to the curl of another, then the divergence of the first field is constant in time.  Perhaps you find this obvious and perhaps you don't, but the point is that taking the divergence of (1) allows us to explore its implications in a more easily digestible form.
Taking the curl of (1) yields yet another consequence, namely that
$$\nabla(\nabla \cdot \mathbf E) - \nabla^2 \mathbf E = -\frac{\partial}{\partial t} (\nabla \times \mathbf B) \qquad (3)$$
This is not as simple as (2), obviously, but in spirit it is the same - it is a (non-obvious) consequence of (1).  We could summarize the lesson as follows:  if the rate of change of one field is equal to another, then the rate of change of the curl of the first field is equal to $\big\{$ the gradient of the divergence of the second field, minus the Laplacian of the second field $\big\}$.
The next obvious question is whether or not these consequences (2) and (3) yield any useful insights.  The answer is yes, because (1) is not the only piece of information we have about how $\mathbf E$ and $\mathbf B$ are related.  The remaining Maxwell equations tell us that $\nabla \cdot \mathbf E = \rho/\epsilon_0$, and that $\nabla \times \mathbf B = \mu_0 \mathbf J + \epsilon_0 \mu_0 \partial \mathbf E/\partial t$.  It's not clear how these pieces of information would help us if we were just staring at (1) all day, but by exploring its consequences (3), we are able to make use of the remaining Maxwell equations to decouple $\mathbf E$ from $\mathbf B$, and see how $\mathbf E$ and its evolution are determined by sources $\rho$ and $\mathbf J$:
$$\left(\nabla^2 - \epsilon_0 \mu_0\frac{\partial^2}{\partial t^2}\right)\mathbf E = \frac{1}{\epsilon_0}\nabla \rho +\mu_0 \frac{\partial}{\partial t} \mathbf J$$
