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My instructor put in the notes a picture of a loop without an inductor - just a battery with voltage $\epsilon$, a switch, and resistor with resistance $R$. He explains that as soon as the switch is flipped, an emf is created that will oppose the change in flux. With an inductor, I can see how this would happen. But this made me wonder what happens when there isn't an inductor. Does current immediately reach $\epsilon\ /\ R$? Or is there some back emf that opposes flux change? It would make sense thinking about it that even without an inductor, there would be a resistance to flux change inside the closed loop, so it wouldn't immediately reach $I = \epsilon\ /\ R$.

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  • $\begingroup$ A loop of wire is an inductor, it just doesn't usually have a particularly high inductance. $\endgroup$ – The Photon Apr 19 '17 at 5:27
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The answer to your question is that there is always inductive coupling between the legs of such a loop. Beyond that, transmission line theory answers your question of what happens. Basically, there's parasitic inductance and capacitance everywhere you go, and they ensure you never see instantaneous jumps in current.

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  • $\begingroup$ Thank you!! So does an inductor effectively just amplify this a ton by its solenoid shape? $\endgroup$ – rb612 Apr 19 '17 at 4:14
  • $\begingroup$ Yes. The shape of the inductor creates a stronger magnetic field than you'd get from stray wires (far stronger!). In fact, it's so much stronger that in most calculations you can ignore any magnetic effects that don't involve a bunch of loops of wire. Only at high frequencies and corner cases (like the one you mention) do we start having to pay attention to parasitic inductances. $\endgroup$ – Cort Ammon Apr 19 '17 at 4:18
  • $\begingroup$ Interestingly enough, this is one of the first lessons in the definitive book on high speed digital electronics: The Art of High Speed Digital Design: a handbook of black magic. In it, they describe how a capacitor can start to look like an inductor at high frequencies because the capacitor's two metal legs start to create a noticeable inductive effect. Trimming the capacitor legs actually changes the character of the circuit! $\endgroup$ – Cort Ammon Apr 19 '17 at 4:20
  • $\begingroup$ thank you Cort! Extremely informative! Would you be able to explain why the capacitor legs have a relatively large inductive effect? $\endgroup$ – rb612 Apr 19 '17 at 4:22
  • $\begingroup$ The legs are two parallel wires, and parallel wires couple inductively. It's not a lot of inductance, but at high frequencies (gigahertz), it doesn't take a lot of inductance to create a high impedence (resistance at a given frequency). And by "corner cases" I meant unusual circumstances like "I actually care about whether its an instantaneous rise or fall, or just a really really really fast one." $\endgroup$ – Cort Ammon Apr 19 '17 at 4:25

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