Exn, so let's say his set up is quarter wave. The quarter wave in the middle picture below cause the light to dim and the quarter wave on the right cause the light to be bright?
His setup is a quarter wave with a closed end. The voltage node is at the top. The maximum voltage is at the bottom end. The line constitutes a high impedance for the generator. The picture 2 is an open line. If it is a quarter-wave line, then the maximum voltage is at the top and the voltage node is at the bottom end. The line constitutes a low impedance for the generator. The pictures are misleading, due to the use of the same coupling coils while the line length/wave length ratio is not specified. In practice, it couldn't be the same due to a considerable impedance difference. In any case, the experimental results can be anything if the load is not negligible in comparison with the impedance between the generator and the load at the position of the load (for instance it's easy to understand that a low impedance load at mid distance on a half-wave line, transforms the first half-part of the line into a quarter-wave line with a closed end). I guess I vision two energy. One is flowing inside the wire and one flow outside. Energy outside is the field but can be represent by the Electric and magnetic field flowing in the wire. Pretty much Voltage x Current. If they are 90 degrees out of phase, then this is reactive power and represent the power flowing outside of the circuit as field.
I agree. If there is no load, the energy is the integral of the energy density 1/2*ε0*E 2+1/2*B 2/µ0 over the volume occupied by the field. If the line doesn't radiate and there is no ohmic losses (= inifinite Q), the energy is constant, you would have needed energy only to build the field at the beginning, and then you wouldn't need energy to maintain it due to the resonant conditions of the line which mean a perpetual energy exchange between voltage/currents and fields, at the rate of the frequency. And yes, energy is require to maintain the flow inside the wire.
So without further notice it's false. Energy is conserved. The need of energy implies that it goes somewhere. In the present case: heating the wire resistance and radiating EM waves (Q is not infinite). A resonant line is exactly like a LC circuit. The oscillations are damped due to the losses. You need energy only to fight the losses.
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Topic: Magnetic field from displacement currents and the TPU (Read 42495 times)
