I'm wondering about the solution (Electrodynamics)

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  • #1
Another
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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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  • #2
Delta2
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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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