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I'm wondering about the solution (Electrodynamics)
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[QUOTE="Delta2, post: 6050721, member: 189563"] You can do it that way too, however in the last equation you write the path integral of ##\int E\cdot dl## is over all the loops (the path over which we integrate will be some sort of tight helix connecting all the circular loops) so it will have approximately total length ##nl2\pi s## so the ##nl## factor will be simplified from both sides of the equation, that is it will be $$ \int E \cdot dl=-\frac{d\Phi}{dt} \Rightarrow E\cdot nl 2\pi s=-(nl)\frac{d\Phi_s}{dt}\Rightarrow E=-\frac{1}{2\pi s}\frac{d\Phi_s}{dt}$$ [/QUOTE]
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I'm wondering about the solution (Electrodynamics)
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