From: PointedEars@web.de   
      
   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?]   
      
   Then, if the external electric field is constant over time or the external   
   electric field strength is constantly increasing over time, too, the   
   resulting magnetic field is constant over time, too, as you can see from the   
   Ampère--Maxwell Law   
      
     ∇ × B = μ₀ J + (1/c²) ∂E∕∂t.   
      
   Take the (partial) derivative with respect to time, then   
      
     ∂/∂t (∇ × B) = μ₀ ∂J/∂t + (1/c²) ∂²E∕∂t².   
      
   If I = const., then ||J|| = dI/dA = const., and ∂J/∂t = 0, so   
      
     ∂/∂t (∇ × B) = ∇ × ∂B/∂t = (1/c²) ∂²E∕∂t².   
      
   Calculating the surface integral over some cross-sectional area A, one finds   
   by application of the Kelvin--Stokes Theorem (K/S):   
      
          ∬_A dA ⋅ (∇ × ∂B/∂t) = (1/c²) ∂²E∕∂t² ∬_A dA   
      
      K/S   
     <==> ∮_C dL ⋅ ∂B/∂t        = (1/c²) ∂²E∕∂t² ∬_A dA,   
      
   where C is the delimiting curve, so eventually   
      
                         ∂B/∂t ~ (1/c²) ∂²E∕∂t² ∬_A dA,   
      
   and unless ∂²E∕∂t² ≠ 0, then ∂B/∂t = 0.  ∂²E∕∂t² = 0 if   
   either ∂E∕∂t = 0,   
   i.e. E = const. (wrt. time), or ∂E∕∂t = const. (wrt. time) ≠ 0. ∎   
      
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   PointedEars   
      
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