# Why are the electric and magnetic waves produced by an antenna in phase?

From my understanding, an antenna produces an electric field when charges are separated, and produces a magnetic field when they are flowing. However, when the magnetic field is at a maximum, the electric field is at a minimum. If this is the case, why does the resulting electromagnetic wave have E and B components which are in phase?

The electric field is generated by separating the charges on both ends of the antenna (as you say). However, if you look to the Maxwell equations, there are two different sources of the magnetic field $$\nabla \times \mathbf{H} = \mathbf{J} +\frac{\partial \mathbf{D}}{\partial t} \,.$$

The first term describes the generation of the magnetic field via the current density $$\mathbf{J}$$ through the antenna. The second term describes the generation of the magnetic field from the change in the electric field. Typically, the first term can be neglected for antennas (in the near field) because the current density is too small. The resulting wave equations then become
$$\nabla \times \mathbf{E} =-\frac{\partial \mathbf{B}}{\partial t}$$ $$\nabla \times \mathbf{H} =\frac{\partial \mathbf{D}}{\partial t}$$ which results in wave-like solutions with the electric and magnetic field in phase with each other.

• This makes sense, thank you! Mar 24 '20 at 19:17

The fields produced by a given antenna can actually vary in phase quite significantly from one point near the antenna to another. This is known as its near field and typically extends a wavelength or so from it. Farther away, the characteristic impedance of free space dominates the phase relationship of the radiated energy, and this is resistive. Hence the far field components are always in phase.

Any energy not radiated to the far field or otherwise absorbed locally is reflected back down the antenna feed to its source. This energy is traditionally measured by its VSWR (pronounced "viswah") or voltage standing wave ratio.