I'm wondering about the solution (Electrodynamics)

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The discussion centers on the application of total magnetic flux in electrodynamics, specifically referencing problem 7.24 and problem 7.15. The total flux is defined as Φ = NΦs, where N = nl and Φs represents the flux from a single loop. The user proposes an alternative solution using the equation Φ = (nl)Bπs² = μ₀n²lπs²I(t) and discusses the path integral of the electric field, concluding that E = - (1/2πs)(dΦs/dt). This highlights the importance of understanding the relationship between total flux and electric fields in solving electrodynamics problems.

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Total flux is Φ = NΦs where ## N = nl## and Φs is flux due to single loop.In problem 7.24 He solved the problem by using defined of the total flux.
Why problem 7.15 He does not solve the problem with using by using defined of the total flux.

such as (my solution)

Φ = (nl)Φs = (nl)Bπs2 = μ0n2lπs2I(t)

and use ∫E⋅dl = -## (dΦ)/(dt) ##
 

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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
$$ \int E \cdot dl=-\frac{d\Phi}{dt} \Rightarrow E\cdot nl 2\pi s=-(nl)\frac{d\Phi_s}{dt}\Rightarrow E=-\frac{1}{2\pi s}\frac{d\Phi_s}{dt}$$
 

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