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@partzman
The induction in B is classic, linked to the variation of I in A. It is only the form of the variation, a constant ramp, that causes the voltage and current in B to be constant. And I agree about the constant voltage required on A, but in practice only in a transient phase, because then the current becomes constant at V/Ra, so the voltage must be increased further to ensure that di/dt=k.
As you say, the flux through B is not constant. In fact, in addition to the constant Φb created by the DC current in B, there is a variable component linked to A, since its current increases.
For energy, we have W = 1/2.La.Ia² + 1/2.Lb.Ib² + M.Ia.Ib
For power, since dIb/dt=0, we have dW/dt = La.Ia.k + M.Ib.k and since Ib=-M.k/R, dW/dt = La.Ia.k - M².k²/R
Since Pa = La.Ia.k is the power supplied by A, Pa = dW/dt + M².k².R. We understand that the generator of current Ia has to supply surplus power to compensate for that dissipated by R. No free energy.
However, this tells us nothing about how the current in B could oppose the current in A. We can only guess that part of the flux variation from A to B would tend to increase the current in B, and that the flux variation produced by B, caused by this increase in current, is immediately compensated by its dissipation in R. In other words, the direct current has no role, but there is a hidden flux variation that self-compensates between A and B. B takes magnetic energy from the field generated by A, so A must produce more to maintain di/dt=k when B supplies the resistance.
This part of Maxwell's electromagnetism is completely unclear. The magnetic field is a misleading intermediary, as is the associated Maxwell-Faraday law. We can do without it.
It is much simpler with the electric field. The movement of charges in circuit A changes the topology of their Coulomb field, and therefore the electric field E by relativistic contraction of lengths. The non-constancy of Ia causes a non-zero tangential component of the electric field to appear around circuit B. The charges are therefore driven into circuit B, slowed down by resistance R, with the reaction effect of slowing down the charges in circuit A as they are reciprocally influenced by the Coulomb field of the charges in circuit B. The generator of Ia will therefore have to use more power.
This is conceptually and qualitatively clear, and here perfectly symmetrical between A and B. The calculation, however, is more complicated.
The Maxwellian model using the magnetic field breaks this symmetry and gives rise to false paradoxes.
« Last Edit: 2026-02-16, 11:12:15 by F6FLT »
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Topic: Free energy is easy, ask an intelligent question... (Read 16491 times)
