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  1. Electric current flows through resistors. The flow of electric current leads to the formation of a magnetic field. This leads to change in magnetic flux, which causes an induced e.m.f. Can I say that the resistor behaved like an inductor for this moment?

  2. How correct I am when I say that an inductor does its job of opposing the flow of current only when the current starts flowing through it? (Assuming there's no change in flux later on and the current being D.C.)

  3. Is the e.m.f produced in an inductor equal to the applied external voltage? If yes, what drives the current?

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  1. A resistor consisting of a coil of wire would produce a magnetic field when there was a current through it. But some resistors are wound non-inductively: the wire is folded back on itself for the whole length of the coil, so there's almost no magnetic field produced. Even simple coils of high resistivity wire produce smaller magnetic fields (for a given current) than a coil of copper wire, which would have to be much longer. And many resistors are straight cylinders of high resistivity material, that would produce only small magnetic fields.

  2. The inductor opposes changes in current, by producing an opposing emf when the current changes.

  3. The emf equals the applied external voltage if the inductor has no resistance. In that case the current doesn't need driving. When (as is usually the case) the inductor does have resistance, $r$, then the emf isn't equal to the applied voltage! Instead$$\mathscr{E}_{\text{app}}-L\frac{dI}{dt}=Ir$$

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  • $\begingroup$ For the first answer, could you elaborate "folded back itself"? $\endgroup$ – Rayyan Asif Khan Mar 12 '18 at 19:19
  • $\begingroup$ Also, in your answer to my third question, the emf, though equal, will be opposite in direction to the applied voltage, right? Won't this unable the flow of current? $\endgroup$ – Rayyan Asif Khan Mar 12 '18 at 19:20
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    $\begingroup$ (1) "Folded back $on$ itself" means this. Cut a length of wire. Bend it through 180° at its midpoint. Hold the ends next to each other, so that the two halves of the wire run parallel (or antiparallel!) to each other. Then wind your coil with the two halves of the wire next to each other, so when you connect it, the current is along one half of the wire and then in the opposite direction through the other half. $\endgroup$ – Philip Wood Mar 12 '18 at 22:30
  • $\begingroup$ (2) Having emfs cancelling won't stop a current that's already flowing. Try putting $r=0$ into the equation in my answer above. Then solve this simplest of differential equations, taking $\mathscr E_\text{app}$ and $L$ as constants. Assume that $I=0$ when $t=0$. $\endgroup$ – Philip Wood Mar 12 '18 at 22:34
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Yes, real-world resistors have a bit of inductance. The explanation for how the inductance arises, as well as the description of the inductance portion of the resistor's behavior, is basically the same as the explanations for an inductor. It works to model the inductance portion of a resistor's behavior by treating the real-world resistor as being an ideal resistor in series with an ideal inductor.

However, since a resistor is usually quite short compared to the conduction path of an inductor, the resistor's inductance is generally quite small, so it's inductance can in general be neglected unless perhaps if you're dealing with very high frequencies.

A real-world inductor also has some resistance, so a real inductor can also be modeled as being an ideal resistor in series with an ideal inductor. However, a real resistor is in general going to have more resistance and less inductance than a real inductor, so the resistor is going to have a much lower "Q factor" $Q=\omega L/R$ than the inductor, which means that it's going to behave more like an ideal resistor and less like an ideal inductor than a real inductor would.

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