From: PointedEars@web.de   
      
   Thomas 'PointedEars' Lahn wrote:   
   > Kuan Peng wrote:   
   >> 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.   
   >   
   > For the current to *change*, work has to be done, thus energy is transferred   
   > to or away from the coil.  Thus the energy stored in the coil is not   
   > conserved, but nobody claimed that it would be then; the law is, and this is   
   > what happens here, too, that the *total* energy stored in the coil *and* of   
   > its environment is conserved.   
   > >> The duration of each phase is Δt. According to Faraday's law, voltages   
   are   
   >> induced in coils A and B,   
   >   
   > No, that is NOT what Faraday's law of induction is about.  It states that an   
   > electric _current_ is induced by a changing _magnetic field_:   
   >   
   >   ∇ × E = -(1/c) ∂B/∂t,   
   >   
   > where E is the electric field vector, B is the magnetic flux density field   
   > vector, and t is time.   
   >   
   > But you already presume a changing *current*.   
      
   To elaborate:   
      
   When an electric current is flowing through coil A, that produces a non-zero   
   magnetic field around the wire of the coil:   
      
     ∇ × B = (4π/c) J + (1/c) ∂_t E.   [Ampère--Maxwell Circuital Law]   
      
   Assuming a constant external electric field, this reduces to   
      
     ∇ × B = (4π/c) J.   
      
   The magnetic field can be calculated using either the Biot--Savart Law in   
   general, or (as here) the Kelvin--Stokes theorem considering the cylindrical   
   symmetry of the problem (here: for a cylindrical conducting wire):   
      
          ∬_A dA ⋅ (∇ × B)  = (4π/c) ∬_A dA ⋅ J   
     <==> ∮_C dL ⋅ B         = (4π/c) I(r)   
     <==> ∫_0^{2π} dφ r B(r) = (4π/c) I(r)   
     <==>         2π r B(r)  = (4π/c) I(r)   
     <==>              B(r)  = 2 I(r)/(c r),   
      
   where C is a circular path with radius r centered on the axis of symmetry of   
   the wire, and dL is an infinitesimal piece of it:   
      
        R = r [cos(φ), sin(φ)]^T,   
       dL = dR/dφ = r [-sin(φ), cos(φ)]^T = r E_φ,   
     B(r) = B E_φ,   
      
   where by B the magnetic (flux density) field strength is meant, and E_φ is a   
   unit vector in the φ-direction of cylindrical coordinates.  [The actual   
   magnetic field is H = B/μ₀.  The direction of the magnetic field lines,   
   parallel to E_φ, i.e. counter-clockwise when looked at towards -E_z when the   
   current is flowing parallel to E_z, is represented by Ampère's right-hand   
   grip/(cork)screw rule.]   
      
   Thus you can see that the strength of the magnetic field *around* a wire is   
   proportional to the electric current flowing in it.   
      
   In order to change the current in coil A, somebody or something has to do   
   work: Energy is added to or removed from the system.  [Typically the current   
   changes by changing direction: it is an alternating current. Thus the   
   polarity of the magnetic field changes as well.]   
      
   When the current changes, so does the magnetic field.   
      
   A changing magnetic field B induces a non-zero electric field E:   
      
     ∇ × E = -(1/c) ∂_t B,             [Faraday's Law of Induction]   
      
   where by E and B we mean the vector fields again.   
      
   This electric field produces a force F = q E, thus an electric current in   
   the nearby coil B (which will in turn produce a non-zero magnetic field   
   around it that will to some extent influence coil A [self-induction].)   
      
   So you can see that *in total*, no energy is lost and none is gained.   
      
   --   
   PointedEars   
      
   Twitter: @PointedEars2   
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