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Building on the excellent answer of HTNW (and also thanks to the comments of Ján Lalinský), I wish to elaborate a bit about inverting the primary and the secondary.

It is true that the primary and the secondary can be inverted while keeping the same currents, so that the same magnetic field is generated in both case. But regarding the electric field, it is also dependent upon the charge distribution on the surface of the wires, according to Jefimenko's equations (see note below); this charge distribution increases (more or less linearly in most cases) in the direction of the increasing potentials for the secondary of a transformer, and decreases in the direction of the increasing potentials for the primary. So, by inverting the primary and the secondary, keeping all currents the same, the charge distribution will revert its direction in both the primary and the secondary. By symmetry, the electric field will be inverted, and
as a result, the Poynting vector will still be directed from the (new) primary to the (new) secondary. That explains the apparent paradox.

Note: Jefimenko's equations connects the EM field with the sources: $$\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \varepsilon_0} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}\rho(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2}\frac{1}{c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t} - \frac{1}{|\mathbf{r}-\mathbf{r}'|}\frac{1}{c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] dV',$$ $$\mathbf{B}(\mathbf{r}, t) = -\frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3} \times \mathbf{J}(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2} \times \frac{1}{c} \frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] dV',$$ where $\mathbf r$′ is a point in the charge distribution, $\mathbf r$ is a point in space, and $$t_r = t - \frac{|\mathbf{r}-\mathbf{r}'|}{c}$$ is the retarded time.

From these equations, it is not obvious that the E-field should be inverted if the charge distribution is inverted in the coils, but it should be possible to show the expected symmetry by a technical derivation.

Building on the excellent answer of HTNW (and also thanks to the comments of Ján Lalinský), I wish to elaborate a bit about inverting the primary and the secondary.

It is true that the primary and the secondary can be inverted while keeping the same currents, so that the same magnetic field is generated in both case. But regarding the electric field, it is also dependent upon the charge distribution on the surface of the wires, according to Jefimenko's equations (see note below); this charge distribution increases (more or less linearly in most cases) in the direction of the increasing potentials for the secondary of a transformer, and decreases in the direction of the increasing potentials for the primary. So, by inverting the primary and the secondary, keeping all currents the same, the charge distribution will invert its direction in both the primary and the secondary. By symmetry, the electric field will be inverted, and
as a result, the Poynting vector will still be directed from the (new) primary to the (new) secondary. That explains the apparent paradox.

Note: Jefimenko's equations connects the EM field with the sources: $$\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \varepsilon_0} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}\rho(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2}\frac{1}{c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t} - \frac{1}{|\mathbf{r}-\mathbf{r}'|}\frac{1}{c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] dV',$$ $$\mathbf{B}(\mathbf{r}, t) = -\frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3} \times \mathbf{J}(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2} \times \frac{1}{c} \frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] dV',$$ where $\mathbf r$′ is a point in the charge distribution, $\mathbf r$ is a point in space, and $$t_r = t - \frac{|\mathbf{r}-\mathbf{r}'|}{c}$$ is the retarded time.

From these equations, it is not obvious that the E-field should be inverted if the charge distribution is inverted in the coils, but it should be possible to show the expected symmetry by a technical derivation.

Building on the excellent answer of HTNW (and also thanks to the comments of Ján Lalinský), I wish to elaborate a bit about inverting the primary and the secondary.

It is true that the primary and the secondary can be inverted while keeping the same currents, so that the same magnetic field is generated in both case. But regarding the electric field, it is also dependent upon the charge distribution on the surface of the wires, according to Jefimenko's equations (see note below); this charge distribution increases (more or less linearly in most cases) in the direction of the increasing potentials for the secondary of a transformer, and decreases in the direction of the increasing potentials for the primary. So, by inverting the primary and the secondary, keeping all currents the same, the charge distribution will revert its direction in both the primary and the secondary. By symmetry, the electric field will be inverted, and
as a result, the Poynting vector will still be directed from the (new) primary to the (new) secondary. That explains the apparent paradox.

Note: Jefimenko's equations connects the EM field with the sources: $$\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \varepsilon_0} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}\rho(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2}\frac{1}{c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t} - \frac{1}{|\mathbf{r}-\mathbf{r}'|}\frac{1}{c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] dV',$$ $$\mathbf{B}(\mathbf{r}, t) = -\frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3} \times \mathbf{J}(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2} \times \frac{1}{c} \frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] dV',$$ where $\mathbf r$′ is a point in the charge distribution, $\mathbf r$ is a point in space, and $$t_r = t - \frac{|\mathbf{r}-\mathbf{r}'|}{c}$$ is the retarded time.

Building on the excellent answer of HTNW (and also thanks to the comments of Ján Lalinský), I wish to elaborate a bit about inverting the primary and the secondary.

It is true that the primary and the secondary can be inverted while keeping the same currents, so that the same magnetic field is generated in both case. But regarding the electric field, it is also dependent upon the charge distribution on the surface of the wires, according to Jefimenko's equations (see note below); this charge distribution increases (more or less linearly in most cases) in the direction of the increasing potentials for the secondary of a transformer, and decreases in the direction of the increasing potentials for the primary. So, by inverting the primary and the secondary, keeping all currents the same, the charge distribution will revert its direction in both the primary and the secondary. By symmetry, the electric field will be inverted, and
as a result, the Poynting vector will still be directed from the (new) primary to the (new) secondary. That explains the apparent paradox.

Note: Jefimenko's equations connects the EM field with the sources: $$\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \varepsilon_0} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}\rho(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2}\frac{1}{c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t} - \frac{1}{|\mathbf{r}-\mathbf{r}'|}\frac{1}{c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] dV',$$ $$\mathbf{B}(\mathbf{r}, t) = -\frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3} \times \mathbf{J}(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2} \times \frac{1}{c} \frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] dV',$$ where $\mathbf r$′ is a point in the charge distribution, $\mathbf r$ is a point in space, and $$t_r = t - \frac{|\mathbf{r}-\mathbf{r}'|}{c}$$ is the retarded time.

From these equations, it is not obvious that the E-field should be inverted if the charge distribution is inverted in the coils, but it should be possible to show the expected symmetry by a technical derivation.

Building on the excellent answer of HTNW, I wish to elaborate a bit about inverting the primary and the secondary.

It is true that the primary and the secondary can be inverted while keeping the same currents, so that the same magnetic field is generated in both case. But regarding the electric field, it is generated by the charge distribution on the surface of the wires; this charge distribution is responsible of the electromotive force inside the wires, and increases more or less linearly in the direction of the increasing potentials. This means that this charge distribution will change its direction of increase as the primary and the secondary are inverted (while keeping the same currents). As a result, the E-field will be inverted and the Poynting vector will still be directed from the (new) primary to the (new) secondary. That explains the apparent paradox.

EDIT (to answer to the comment of Ján Lalinský): What we need here is to connect the E field with the sources. This is the aim of Jefimenko's equations: $$\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \varepsilon_0} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}\rho(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2}\frac{1}{c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t} - \frac{1}{|\mathbf{r}-\mathbf{r}'|}\frac{1}{c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] dV',$$ $$\mathbf{B}(\mathbf{r}, t) = -\frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3} \times \mathbf{J}(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2} \times \frac{1}{c} \frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] dV',$$ where $\mathbf r$′ is a point in the charge distribution, $\mathbf r$ is a point in space, and $$t_r = t - \frac{|\mathbf{r}-\mathbf{r}'|}{c}$$ is the retarded time.

For electrical wires, the term $\frac{1}{c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t}$ is negligible, so the E-field is nearly $$\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \varepsilon_0} \int \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}\rho(\mathbf{r}', t_r)+ \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2}\frac{1}{c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t}.$$ This explain why I asserted that the E-field is mainly due to the surface charges on the wire. This is not contradictory with Faraday induction law that says that $E = \nabla \Phi - \partial_t \mathbf A$, because there are a contra-electromotive force produced by the secondary. But a rigorous solution of this apparent paradox is probably too technical for me.

Building on the excellent answer of HTNW (and also thanks to the comments of Ján Lalinský), I wish to elaborate a bit about inverting the primary and the secondary.

It is true that the primary and the secondary can be inverted while keeping the same currents, so that the same magnetic field is generated in both case. But regarding the electric field, it is also dependent upon the charge distribution on the surface of the wires, according to Jefimenko's equations (see note below); this charge distribution increases (more or less linearly in most cases) in the direction of the increasing potentials for the secondary of a transformer, and decreases in the direction of the increasing potentials for the primary. So, by inverting the primary and the secondary, keeping all currents the same, the charge distribution will revert its direction in both the primary and the secondary. By symmetry, the electric field will be inverted, and
as a result, the Poynting vector will still be directed from the (new) primary to the (new) secondary. That explains the apparent paradox.

Note: Jefimenko's equations connects the EM field with the sources: $$\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \varepsilon_0} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}\rho(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2}\frac{1}{c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t} - \frac{1}{|\mathbf{r}-\mathbf{r}'|}\frac{1}{c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] dV',$$ $$\mathbf{B}(\mathbf{r}, t) = -\frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3} \times \mathbf{J}(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2} \times \frac{1}{c} \frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] dV',$$ where $\mathbf r$′ is a point in the charge distribution, $\mathbf r$ is a point in space, and $$t_r = t - \frac{|\mathbf{r}-\mathbf{r}'|}{c}$$ is the retarded time.