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Here in this picture


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

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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 ? –  Colin K Feb 21 '11 at 20:20
@Colin K Hi! I want to know where Iw comes from? –  kame Feb 21 '11 at 20:32
The eddy currents are determined by the second equation you listed. –  Johannes Feb 22 '11 at 2:05
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?) –  kame Feb 24 '11 at 16:14
@kame: Write your solution as an answer and then mark it as the accepted solution. You can answer your own questions. –  endolith Dec 5 '11 at 3:04

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