Here in this picture


you can see $I_W$ which is induced by H. But why $I_W$ is not vice versa? Because of $$rot \, \vec B = \mu_0 \, \left( \varepsilon_0 \frac{\partial \vec{E}}{\partial t} + \vec j \right)$$ ? Maybe I have to use $$\oint \vec{E}\cdot \, d\vec{r} = - \int \frac{\partial \vec{B}}{\partial t}\cdot \, d\vec{A} = U_{induced}$$

  • $\begingroup$ I'm a little unclear what you are asking. Do you want to know why the current is induced by the H field, rather than the H field being induced by the current Iw ? $\endgroup$
    – Colin K
    Feb 21 '11 at 20:20
  • $\begingroup$ @Colin K Hi! I want to know where Iw comes from? $\endgroup$
    – kame
    Feb 21 '11 at 20:32
  • 1
    $\begingroup$ The eddy currents are determined by the second equation you listed. $\endgroup$
    – Johannes
    Feb 22 '11 at 2:05
  • $\begingroup$ Okay I understand now. Because dB/dt in the Iw-Area will increase, there must be a E-field which goes anti-clockwise. (How can I mark the solution?) $\endgroup$
    – kame
    Feb 24 '11 at 16:14
  • $\begingroup$ @kame: Write your solution as an answer and then mark it as the accepted solution. You can answer your own questions. $\endgroup$
    – endolith
    Dec 5 '11 at 3:04

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