I'm wondering about the solution (Electrodynamics)

Another · Sep 1, 2018

Textbook problem 7.24 on a toroidal coil carrying an alternating current, with parts (a) and (b) asking for the induced emf, the resistor current, and the back emf ratio.

Handwritten-style printed solution to Problem 7.24 showing flux, EMF and current derivations with boxed results of 2.61 x 10^-4 sin(wt) volts, 5.22 x 10^-7 sin(wt) amperes and -2.74 x 10^-7 cos(wt) volts, plus a ratio of amplitudes line on page 131.

Text of Problem 7.15 about a long solenoid with radius a and n turns per unit length carrying current I(t) in the phi direction, asking for the electric field inside and outside the solenoid in the quasistatic approximation.

Solution to Problem 7.15 in the quasistatic approximation: B is mu0 n I z-hat for s less than a and zero for s greater than a. Inside, flux, line integral and E field are given, with E = -(mu0 n s/2)(dI/dt) phi-hat; outside, E = -(mu0 n a^2/(2s))(dI/dt) phi-hat.

Total flux is Φ = NΦ_s where ## N = nl## and Φ_s 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)Φ_s = (nl)Bπs² = μ₀n²lπs²I(t)

and use ∫E⋅dl = -## (dΦ)/(dt) ##

Delta2 · Sep 3, 2018

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

I'm wondering about the solution (Electrodynamics)

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