From: relativity@paulba.no   
      
   Den 16.01.2026 12:50, skrev Kuan Peng:   
   > I have explained this violation of the law of conservation of energy in   
   > the introduction.   
      
   > https://pengkuanem.blogspot.com/2026/01/a-derivation-of-farada   
   s-law-from.html   
      
   > To illustrate this, consider the following experimental setup: suppose   
   > two coils, A and B, are positioned side by side, with coil B connected   
   > to a resistor R, as shown in Figure 1.   
   > Let the current in coil A, denoted as Ia, vary as follows: Ia increases   
   > linearly from zero to Imax, then decreases linearly back to zero. The   
   > duration of each phase is Δt. According to Faraday's law, voltages are   
   > induced in coils A and B, which we label Va and Vb, respectively. Since   
   > Ia varies linearly during each phase, Va and Vb remain constant   
   > throughout those intervals.   
      
      
   Let us first consider coil A only. Let its inductance be L.   
      
   (Just summing up, I expect you to agree)   
      
   When the current increases from zero to Imax,   
   the voltage on the current source is:   
     Va = L⋅dI/dt = L⋅(Imax/Δt)   
   Note that both Va  and Ia are positive.   
      
   The energy W stored in the magnetic field   
   at the time when the current is Imax is:   
     Wa = L⋅(Imax)²/2   
      
   When the current decreases from Imax to zero,   
   the voltage on the current source is:   
     Va = L⋅dI/dt = - L⋅(Imax/Δt)   
   Note that Va is negative while Ia is positive.   
      
   That means that energy is delivered back to   
   the current source.   
      
   The energy W released from in the magnetic field   
   at the time when the current is zero is:   
     W = - L⋅(Imax)²/2 = -Wa   
      
   While the current is increasing, the current source will   
   deliver energy to the magnetic field,   
   when the current is decreasing, the magnetic field   
   will deliver the stored energy back to the current source.   
      
   (If the resistance in the coil is different from zero,   
     some energy will be lost as heat. Some energy will also   
     be lost as em-radiation. This will be strongly dependent   
     on Δt. We ignore these losses.)   
      
   Now let us consider what will happen in coil B.   
      
   > Within resistor R, the voltage Vb generates   
   > a current Ib and dissipates electric power equal to |VbIb|, both of   
   > which are constant in each phase. Consequently, the total work performed   
   > in R after both phases is 2|VbIb|t.   
      
   This is correct.   
      
   When the current in A is increasing, the voltage over   
   the resistor R will be constant Vb and the current Ib = Vb/R.   
      
   When the current in A is decreasing, the voltage over   
   the resistor R will be constant -Vb and the current -Vb/R = -Ib.   
   Energy loss Wb = 2|VbIb|⋅Δt   
      
   Now let us consider how this will affect coil A.   
      
   >   
   > Since Ib is constant, it does not induce a voltage in coil A; therefore,   
   > the value of Va remains unchanged regardless of whether Ib is positive,   
   > negative, or zero—just as if coil B were not present. When Ia increases,   
   > the voltage in coil A (Va) is positive, and the electrical work   
   > performed in A is given by the integral of VaIa . Conversely, when Ia   
   > decreases, the voltage in A becomes -Va, and the work equals the   
   > integral of -VaIa . Consequently, the total energy consumption of coil A   
   > after both phases equals zero.   
      
   Let's take it from the beginning:   
      
   When the current in A is increasing, the magnetic field in A   
   will be increasing. Part of this field will go through B,   
   so there will be an increasing flux through B. This will induce   
   a constant voltage Vb and current Ib in B. This will give a   
   constant magnetic flux through B. Part of this flux go through A   
   and will have the opposite direction of the flux in A.   
   The result is that the flux in A will increase slower, and the   
   energy stored in the magnetic field at the time when the current   
   is Imax will be less than Wa = L⋅(Imax)²/2 . Let's call it Wa'.   
      
   When the current in A is decreasing, the magnetic field in A   
   will be decreasing. Part of this field will go through B,   
   so there will be an decreasing flux through B. This will induce   
   a constant voltage -Vb and current -Ib in B. This will give a   
   constant magnetic flux through B. Part of this flux go through A   
   and will have the same direction of the flux in A.   
   The result is that the flux in A will decrease slower, and the   
   energy stored in the magnetic field at the time when the current   
   is zero will be W = 0.   
      
   The net result is that the stored energy in A will increase from   
   zero to Wa', and then decrease back to zero.   
      
   This means that the current source has delivered the energy Wa   
   to the field, but has only got Wa' back.   
      
      Wa-Wa' = Wb   
      
      
   >   
   > Since the energy consumption in coil A is zero, A does not transfer any   
   > energy to coil B.   
      
   Strange conclusion.   
      
   The energy consumption in coil A is zero.   
   A transfers energy to B.   
   This energy is supplied by the the current source in A.   
      
   > We therefore encounter a case where B performs work   
   > equal to 2|VbIb|t while receiving no energy from A. This implies that   
   > the system consisting of coils A and B performs work without any energy   
   > input, which violates the law of conservation of energy.   
   >   
   > The cause of this violation is that Faraday's law predicts zero voltage   
   > in A when the current in coil B is constant.   
   --   
   Paul   
      
   https://paulba.no/   
      
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