Let the electromagnetic field has $u$ as its energy density (amount of energy per unit volume in the field) and let $\bf S$ represents the energy flux- the amount of energy per unit time flowing across a unit area perpendicular to the flow).
Now electromagnetic field can interact with matter and do work on them; so this energy interaction must be considered when discussing energy conservation.
The field energy inside a volume $V$ is $\displaystyle \int_V u\,\mathrm dV\;.$ Amount of energy flowing out of the volume $V$ is given by $\displaystyle \int_\Sigma \mathbf S\cdot n\,\mathrm da \;.$
Now, the work done per unit time by the field on the matter inside the volume $V$ is given by $\displaystyle \int _V Nq(\mathbf E + \mathbf v\times \mathbf B)\cdot \mathbf v\,\mathrm dV$ where $N$ is the number of particles per unit volume; this can be written as \begin{align}\int _V Nq(\mathbf E + v\times \mathbf B)\,\mathrm dV& = \int_V Nq\mathbf E\cdot v\,\mathrm dV\\ &= \int_V \mathbf E\cdot \underbrace{(Nq\mathbf v)}_\textrm{current-density}\,\mathrm dV\\ &= \int_V E\cdot \mathbf J\,\mathrm dV \end{align}
So, the continuity equation is written thus: $$\underbrace{-\frac{\partial}{\partial t}\int_V u\,\mathrm dV}_\textrm{rate of change of energy inside volume $V$}=\underbrace{\int_\Sigma \mathbf S\cdot \mathbf n\,\mathrm da}_\textrm{amount of $\mathbf{field\, energy}$ flowing out of volume $V$ per unit time} + \underbrace{\int_V \mathbf E\cdot \mathbf J\,\mathrm dV}_\textrm{work done per unit time by the field on the matter inside volume $V$} \;. $$ This definitely implies the conservation of energy and moreover, the equation which OP wrote in his query is the offshoot of this same continuity equation.
Energy density and Poynting Vector:
We can write the differential form of the continuity equation above as:
$$-\frac{\partial u}{\partial t}= \textrm{div}\,\mathbf S + \mathbf E\cdot \mathbf J$$ (since, all the derivation is done in Cartesian coordinates, then $\textrm{div}\equiv \mathbf \nabla$).
Or, we can write $$\mathbf E\cdot \mathbf J= -\frac{\partial u}{\partial t}-\mathbf E\cdot \mathbf J\tag 1$$
In order to find out what $u$ and $\bf S$ actually are, it is assumed that they solely depend on the fields $\bf E$ and $\bf B\;.$
Now, using $$\mathbf J = \epsilon_oc^2\,\mathbf \nabla\times\mathbf B- \epsilon_0\frac{\partial\mathbf E}{\partial t}\,,$$ we can write $\mathbf E\cdot \mathbf J$ as $$\mathbf E\cdot \mathbf J= \epsilon_oc^2\,\mathbf E\cdot (\mathbf \nabla\times\mathbf B)- \epsilon_0\mathbf E\cdot\frac{\partial\mathbf E}{\partial t}\;.$$
Now, \begin{align}\mathbf E\cdot (\mathbf \nabla\times\mathbf B)&= (\mathbf \nabla\times\mathbf B)\cdot \mathbf E\\ &= \mathbf \nabla\cdot (\mathbf B\times\mathbf E)+ \mathbf B\cdot (\mathbf \nabla \times \mathbf E)\;.\end{align}
Therefore, \begin{align}\mathbf E\cdot \mathbf J&=\epsilon_oc^2\,\mathbf \nabla\cdot (\mathbf B\times\mathbf E)+\epsilon_oc^2\,\mathbf B\cdot (\mathbf \nabla \times \mathbf E)-\epsilon_0\mathbf E\cdot\frac{\partial\mathbf E}{\partial t}\\ &=\epsilon_oc^2\,\mathbf \nabla\cdot (\mathbf B\times\mathbf E)+\epsilon_oc^2\,\mathbf B\cdot (\mathbf \nabla \times \mathbf E)-\frac{\partial}{\partial t}\,\left(\frac12 \epsilon_0\,\mathbf E\cdot \mathbf E\right)\\ &=\mathbf \nabla\cdot (\epsilon_oc^2\,\mathbf B\times\mathbf E)+ \epsilon_oc^2 \mathbf B\cdot \left(-\frac{\partial\mathbf B}{\partial t}\right)-\frac{\partial}{\partial t}\,\left(\frac12 \epsilon_0\,\mathbf E\cdot \mathbf E\right)\\ &=\mathbf \nabla\cdot (\epsilon_oc^2\,\mathbf B\times\mathbf E) - \frac{\partial}{\partial t}\left(\frac{\epsilon_oc^2}{2}\,\mathbf B\cdot \mathbf B+ \frac{\epsilon_0}2 \,\mathbf E\cdot \mathbf E\right) \tag 2\end{align}
Comparing $(1)$ and $(2)\,,$ we get $$u = \frac{\epsilon_0}2 \,\mathbf E\cdot \mathbf E+\frac{\epsilon_oc^2}{2}\,\mathbf B\cdot \mathbf B\\ \mathbf S=\epsilon_oc^2\,\mathbf E\times\mathbf B \;.$$
This vector $\bf S\,,$ that is the energy flux vector we considered at the beginning of the making up of the continuity equation, is called the Poynting Vector.
The whole derivation is based on the continuity equation, which is the mathematical expression of conservation of energy.
OP can conclude the equation he wrote in the question by converting the differential form of equation $(2)$ into integral form; using the definition of $u$ and $\bf S$ derived above and lastly using Gauss's theorem.
References:
$\bullet$ Lectures on Physics by Feynman, Leighton, Sands.
you're saying that energy conservation is needed as an independent equation, in order to deduce that the Poynting vector gives energy flux. I don't think this is true. I think the energy continuity equation can be shown as a consequence of Maxwell's equations.
No! Never, ever I've made such statement. I've answered to your query I know the books say it represents energy flux, but how do you prove it represents energy flux?; that's it. What I've done is to clearly present the continuity equation before OP and have tried to clearly interpret each term and how they together imply conservation of energy. The continuity equation can indeed be derived from Maxwell's equation and this is what I've done to define $u$ and $\bf S\;.$ I'm really pretty upset on the allegation OP made.
But this is not a proof of conservation of energy.
It's indeed a proof of conservation of energy.
Contrast this with conservation of charge within electromagnetism. Conservation of charge is a consequence of Maxwell's equations, it does not have to be assumed independently.
Okay, let's fix this thing. Conservation of charge implies this continuity equation(this can be derivable from Maxwell's equations):
$$-\frac{\partial \rho_V}{\partial t}= \mathbf \nabla \mathbf J\;.$$
This suggests conservation of energy would look like:
$$-\frac{\partial u}{\partial t}= \mathbf \nabla \mathbf S\;.$$
But that is incomplete as it is the total energy and not just field energy which is conserved; the interaction of field with matter has to be taken into consideration.
is energy conservation a consequence of Maxwell's equations or something assumed independently from it?
Yes. The continuity equation meant for conservation of energy is derived from Maxwell's equation; that's how the definition of $u$ and $\bf S$ have been computed. OP must be confused to see in my answer first the continuity equation as if I'm saying continuity equation should be considered independently in order to consider conservation of energy. No! That was intentionally done in order to conceive that the Poynting Theorem indeed implies conservation of energy.
If the increase of current is fast enough, the induced electric field will be substantial and then it will contribute to net field energy. However, these experiments (increasing electric current through a coil) are usually meant to be done in a quasistatic way, in which case the induced electric field is very small, so its field energy is negligible. If we changed the current very fast (by increasing voltage on the coil very fast), there would be jump in induced electric field, a radiation-like disturbance and then we are in the realm of EM radiation theory, which is complicated.
The induced electric field is proportional to $dI/dt$, so it is also proportional to acceleration of the mobile charges in the wires. If the field is due to single accelerated charge, it is also called "acceleration field", so we can think of induced field as a sum of many acceleration fields.
This induced field is very different from the Coulomb field - it is solenoidal (non-conservative, curly). Its changes (due to change in $dI/dt$) do propagate out of the system to infinity with speed of light and decay as $1/r$, so in that sense it is always a radiation field. But in usual (low frequency) AC circuits, $dI/dt$ is low enough or geometry is radiation-inhibiting enough that this radiation field carries negligible energy away from the system and we can pretend there is no radiation, only exchange of energy between system components. But if the geometry is radiative or frequency of the current oscillations is high enough, the energy loss may be substantial and then the radiation character must be taken into account, then it is necessary to take the more correct view and talk about the acceleration field being radiative.
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So, as noted, we use Poynting's theorem to get: $$\frac{1}{\mu_0} \oint_{\partial V} (\vec{E} \times \vec{B}) \cdot d\vec{S} = -\int_V \vec{E} \cdot \vec{J} dV$$ [The static version of Poynting's theorem is just: divergence theorem, $\nabla \cdot (\vec{E} \times \vec{B}) = \vec{B} \cdot (\nabla \times \vec{E}) - \vec{E} \cdot (\nabla \times \vec{B})$, then $\nabla \times \vec{E} = 0$ and $\nabla \times\vec{B} = \mu_0\vec{J}$]
The electric field is just the negative gradient of electric potential. We can use integration by parts in higher dimensions:
$$-\int_V \vec{E}\cdot\vec{J} dV = \int_V (\nabla \phi) \cdot \vec{J} dV = \oint_{\partial V} \phi \vec{J} \cdot d\vec{S} - \int_V \phi(\nabla \cdot \vec{J}) dV$$
The current density doesn't diverge. So it's just the first term, basically, the surface integral of voltage times current.
Of course your boundary condition says there's no current out of the surface, still, the previous statement holds in static situations even without that boundary condition.