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 Message 178,619 of 178,646 
 Stefan Ram to Kuan Peng 
 Re: A Derivation of Faraday's law from C 
 20 Jan 26 20:58:44 
 
From: ram@zedat.fu-berlin.de

Kuan Peng  wrote or quoted:
>Le 20/01/2026 à 14:13, ram@zedat.fu-berlin.de (Stefan Ram) a écrit :
>The second coil has an emf acting on it by Faraday's law. This EMF is
>constant. So, the current in the second coil is constant.

  Ok.

>>a field arises from this current
>The field from this current is constant.

  Ok.

>>A portion of this field creates a magnetic flux through the first coil,
>This magnetic flux is constant because the current in the second coil is
>constant.

  Ok.

>>which leads to an EMF in the first coil by Faraday's law,
>Constant magnetic flux does not change, so " which leads to an EMF " which
>is zero in the first coil.

  I see that I have not addressed this point before. Let me give
  it a try!

  The (increasing) current I1(t) in the first coil creates a flux

u11(t) = L1 I1(t)

  through the first loop, where L1 is the self-inductance of
  the first loop (by the definition of inductance). The "(t)"
  is intended to indicate the time dependency.

  The (constant) current I2 in the second coil creates a flux

u12 = M I2

  through the first loop where M is the mutual inductance of the
  loops (by the definition of the mutual inductance).

  The total flux through the first loop is

u1(t) = u11(t) + u12
      = L1 I1(t) + M I2.

  The sign of I2 is opposite that of I1 by Lenz's law.

  So one can write: u1(t) = L1 |I1(t)| - M |I2|.

  The flux u1(t) is reduced by M |I2| even if I2 is constant.

  Thus, to get the same flux as without the other coil, |I1(t)|
  must be greater, which requires more energy than without the
  other coil.

--- SoupGate-Win32 v1.05
 * Origin: you cannot sedate... all the things you hate (1:229/2)

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