From: PointedEars@web.de   
      
   Kuan Peng wrote:   
   > Le 21/01/2026 à 02:12, Thomas 'PointedEars' Lahn a écrit :   
   >> Kuan Peng wrote:   
   >>> Le 20/01/2026 à 15:08, John Hasler a écrit :   
   >>>> Kuan Peng writes:   
   >>>>> However, Faraday’s law does not define :   
   >>>>>>   the current in coil B creates a field that pushes back against the   
   >>>>>>   change in current in coil A   
   >>>>   
   >>>> Coil B is a coil with current in it.  Faraday's law predicts that   
   >>>> it will generate a field which opposes that generated by coil A.   
   >>>   
   >>> What if the current in B is constant?   
   >>   
   >> [Those are vector fields that can depend both on time *and* position.   
   >> Therefore, it is important to state precisely with respect to what a field   
   >> is constant: Does it not vary over time, or does it perhaps vary over time,   
   >> but not in space?]   
   >>   
   > The current in A varies linearly. The current in B is constant.   
   > So, the magnetic field of A+B varies linearly.   
      
   I wonder how there is a current in (the secondary) coil B at all *before*   
   electromagnetic induction.  Usually there is not, i.e. the secondary coil   
   is NOT connected to a voltage source, but to some electric appliance:   
      
      
      
   In any case, your third statement is only superficially true because the   
   change of the current in coil A changes the magnetic field in and around   
   coil A (which is also the magnetic field in and around around coil B).  The   
   change of the magnetic field induces another current which counteracts the   
   one in coil A and perhaps even coil B (Lenz's Law).   
      
   That, in a sense, another current is induced by that change follows from   
   Faraday's law of induction (eq. 2 below) that was used to induce a current   
   in coil B in the first place; but its direction is not obvious (to me).   
      
   In vacuum:   
      
     ∇ × B  = μ₀ (J + ε₀ μ₀ ∂E/∂t) = μ₀ J + (1/c²)   
   ∂E/∂t,    (1)   
     ∇ × E' = -∂B'/∂t,                                       (2)   
      
   where (IIUC) E' is now the contribution to the electric field that is   
   induced by the change in the magnetic field B' due to the induced current,   
   that produces Lenz's opposing current ':-)   
      
   > A and B are both in this magnetic field which varies linearly.   
      
   It is not, for the reason explained above.   
      
   > According to Faraday’s law,   
      
   _of induction_ (but I guess we can drop that to simplify this discussion)   
      
   > the induced voltages in A and B are proportional to dB/dt   
      
   Yes, that follows (sort of) from the differential form above, since the   
   induced _electric field_ gives rise to a spatial difference of electric   
   potential, i.e. a voltage.   
      
   > which is constant.   
      
   It is not.   
      
   > The induced voltage and current in A with or without the presence of B is   
   > the same.   
      
   It is not.   
      
   > So, the energy dissipation is zero with or without the presence of B.   
      
   /Ex falso quodlibet./   
      
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   PointedEars   
      
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