content(warnings): Add W17-W28, X8-X12, N7 — E&M misconceptions, cross-refs, notes

concepts/em/u13/e13-5-inductance.tex

@@ -12,6 +12,8 @@ Equivalently, from Faraday's law of induction, a changing current induces an EMF
 \]
 where the negative sign reflects Lenz's law: the induced EMF opposes the change in current. The SI unit of inductance is the henry (H), where $1\;\mathrm{H} = 1\;\mathrm{V\!\cdot\!s/A} = 1\;\mathrm{kg\!\cdot\!m^2/(s^2\!\cdot\!A^2)}$.}
 
+\wc{An inductor opposes \emph{change} in current, not current itself}{An ideal inductor with steady (DC) current has zero voltage drop across it ($V=0$ when $dI/dt=0$). It only produces a back-EMF $\mathcal{E}=-L(dI/dt)$ when the current is \emph{changing}. A steady current passes through an ideal inductor as freely as through a wire.}
+
 \nt{Inductance is purely a geometric property. For a fixed geometry (and no ferromagnetic material near the coil), $L$ is constant and independent of the current. The larger the coil, the more turns, and the greater the flux linkage per unit current, the larger the inductance. A coil with $L = 1\;\mathrm{H}$ and $dI/dt = 1\;\mathrm{A/s}$ develops a $1\;\mathrm{V}$ back EMF.}
 
 \thm{Solenoid self-inductance}{For an ideal long solenoid of length $\ell \gg R$, total turns $N$, cross-sectional area $A = \pi R^2$, and vacuum permeability $\mu_0 = 4\pi\times 10^{-7}\,\mathrm{T\!\cdot\!m/A}$, the magnetic field inside is $B = \mu_0\,n\,I$ where $n = N/\ell$. The flux through each turn is $\Phi_B = B\,A = \mu_0 N I A/\ell$. Therefore