# Energy and Poynting vector in a solenoid

We have the following scenario.

A long (ideal) solenoid, with $$n$$ turns per unit lenght, is carrying a current $$I$$. At the beginning, the current is constant and we have a magnetic field

$$\mathbf{B} = \mu_0nI\mathbf{\hat{z}}$$

inside the solenoid.

At $$t=0$$, we begin increasing the current, so that the increasing $$\mathbf{B}$$ generates by induction an azimuthal electric field $$\mathbf{E}(r) = -\frac{1}{2}\mu_0nr\frac{dI}{dt}\mathbf{\hat{\phi}}$$

If we now calculate the surface integral of the Poynting vector $$\mathbf{S}$$ over an imaginary cilindrical surface with radius $$R$$ and height $$h$$, we get $$\oint_\Sigma\mathbf{S} \cdot d\mathbf{\Sigma}=-\mu_0\pi R^2hn^2I\frac{dI}{dt}$$ which we find to be equal to the opposite of the time derivative of the energy contained in the magnetic field inside this imaginary cilinder $$\partial_tU_m = \partial_t \frac{B^2}{2\mu_0}\pi R^2 h = - \oint_\Sigma\mathbf{S} \cdot d\mathbf{\Sigma}$$ as we expect from Poynting theorem since there's no work being done on charges.

My questions are the following:

• What about the energy in the electric field which is being generated by induction while we change the current? Shouldn't it be considered along $$U_m$$ in Poynting's theorem?
• Since the current is being increased, the charges in the solenoid are being accelerated. Is therefore the Poynting vector associated with an EM radiation towards the center of the solenoid? Is this radiation responsible for the increase in the magnetic field? If so, how does it happen in detail?

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.