Relation between electric & magnetic fields in terms of field strength

[DaveE]

I'm getting the feeling that you're missing the impedance concept that @tech99 referred to earlier. Yes the energy is split equally. But the field strengths E, H (i.e. amplitudes) are related by the impedance of the medium they are traveling through. Free space is 377Ω, which is nearly always the practical value. But it doesn't have to be, for example ground penetrating radar, where the waves travel through "stuff". Your question, to me, is a question about the materials the wave is traveling through (primarily permittivity, permeability, and thus speed). In classical physics, it's a bit more of a definition than an explanation.

The antenna is an EM structure to launch or receive those waves and is a pretty complex situation near the antenna, but doesn't matter once the waves are created and traveling through the medium in question.

https://en.wikipedia.org/wiki/Wave_impedance

 

[Delta2]

Suppose we have two waves traveling along the z-direction, then $\vec{E_1}=E_{10}\sin(\omega_1 t-k_1z) \hat x,\vec{E_2}=E_{20}\sin(\omega_2 t-k_2z)\hat x$ and hence $\vec{B_1}=\frac{E_{10}}{c}\sin(\omega_1 t-k_1z) \hat y,\vec{B_2}=\frac{E_{20}}{c}\sin(\omega_2 t-k_2z) \hat y$, then it is easy to easy that for the total wave $|E|=|\vec{E_1}+\vec{E_2}|=c|\vec{B_1}+\vec{B_2}|$.

The key is that both waves E-fields are along $\hat x$ (and both B-fields along $\hat y$).

 

[ergospherical]

Well, of course it holds in the special case that the field vectors of both waves are aligned. There’s an infinity of other combinations for which it doesn’t work.

 

[Delta2]

Yes ok , there is indeed an infinity of other combinations which it doesn't work but I am glad I made you see my point :D.

 

[vanhees71]

Suppose we have two waves traveling along the z-direction, then $\vec{E_1}=E_{10}\sin(\omega_1 t-k_1z) \hat x,\vec{E_2}=E_{20}\sin(\omega_2 t-k_2z)\hat x$ and hence $\vec{B_1}=\frac{E_{10}}{c}\sin(\omega_1 t-k_1z) \hat y,\vec{B_2}=\frac{E_{20}}{c}\sin(\omega_2 t-k_2z) \hat y$, then it is easy to easy that for the total wave $|E|=|\vec{E_1}+\vec{E_2}|=c|\vec{B_1}+\vec{B_2}|$.

The key is that both waves E-fields are along $\hat x$ (and both B-fields along $\hat y$).

Yes, that are plane waves, for which this formula is correct!

 

[Klystron]

This excerpt from this article on ferrite antenna parameters might aid your calculations.

The ferrite rod antenna can be considered as a very small loop antenna. In view of its size, the loop is very much less than a wavelength in length and without the ferrite it would have a very low radiation resistance. Accordingly the losses due to the resistance of the wire would be exceedingly high. Placing the ferrite core in the coil has the effect of raising the radiation resistance by a factor of µ2. This brings the value into more acceptable limits and reduces the resistive losses caused by the wire.

While the introduction of the ferrite rod raises the radiation resistance of the antenna, it does introduce other losses. The ferrite itself absorbs power because energy is required to change the magnetic alignment of the magnetic domains inside the granular structure of the ferrite. The higher the frequency, the greater the number of changes and hence the higher the loss. It is for this reason that ferrite rod antennas are not normally used above frequencies of a few MHz.

(bolding added)

 

[salmanshu322]

Maxwell equation talk about the energy and the answer is yes.