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From textbooks, you can see that the electric fields and magnetic fields are in phase for electromagnetic fields propagating from an antenna.

However, using the right hand rule, and the fact that current in an antenna is maximum when the dipoles are neutral, wouldn't that mean the electric field and magnetic field is 90 degrees out of phase?

I read somewhere that near the antenna it will be out of phase, but further out it will be in phase, but how is it even possible?

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Look at Maxwell's equations: $$\begin{align} \nabla \cdot \vec{E} & = \frac{\rho}{\epsilon_0} \\ \nabla \times \vec{E} & = - \frac{\partial \vec{B}}{\partial t} \\ \nabla \cdot \vec{B} & = \vec{0} \\ \nabla \times \vec{B} & = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}. \end{align}$$ Now, keep in mind that a derivative or integral, when applied to a pure sine wave, introduces a $90^\circ$ phase shift. In every case, $\vec{E}$ and $\vec{B}$ have the same number of derivatives applied to them, so they will have the same phase. The sources, $\vec{J}$ and $\rho$, have no derivatives so they will have a $90^\circ$ phase shift relative to the fields.

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  • $\begingroup$ I have not studied to this level. Mind if you explain pictorially in the context of antennas? $\endgroup$ – Michael Nov 4 '16 at 4:43
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In the near field the $\vec E$ and $\vec B$ are 90 degrees out of phase. In the far field they are in phase $\frac{\vec E}{\vec B}=Z_0$, $Z_0=377 \Omega$ is vacuum wave impedance, like in a plane electromagnetic wave.

See also answers in Phase-alignment in EM-waves

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