First of all, i am sorry if i overlooked this question although i browsed the list of questions on the topic. I have difficulty understanding that when a changing magnetic flux in a coil induces an emf and current in the second coil and if this emf is also time varying ( e.g. suppose an ac is applied to the first coil or an exponentially increasing current with small initial value is made to flow in the first coil) then the induced emf in the second coil will also vary with time, will not this induced current induce again a current in the first coil and the process goes on and on?? We only take time variation of flux once in the second coil while understanding the concepts of induction without taking into account such a thing ? What am i missing !
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$\begingroup$ Have a look at mutual inductance. $\endgroup$– FarcherCommented Nov 6, 2018 at 12:14
1 Answer
That's pretty much what happens.
The changing current in the first coil induces an EMF in the second coil. Depending on the attached circuit, a changing current will flow in the second coil, inducing an EMF in the first coil.
If the attached circuit is purely resistive, the current in the 2nd coil will be proportional to the EMF. For an sinusoidal current in the 1st coil, the current in the 2nd coil will have $\pi/2$ phase difference to the current in the 1st coil, and the corresponding EMF in the 1st coil will have $\pi$ phase difference to the current in the 1st coil (it opposes to the current in the 1st coil).
This EMF will, depending on the circuit in the 1st coil, induce a current.
However, if your 1st coil is attached to an AC current source, the current source will produce the prescribed current in the 1st coil whatever is attached to it, within its technical possibilities of course. This means that the source that is attached to the 1st coil will adapt to the EMF induced from the 2nd coil into the 1st coil.
Practically, the emf produced by the source will need to be slightly larger if there is a 2nd coil, to compensate for this double induction, than if the 2nd coil is not present. The corresponding power difference will mostly match the resistive losses (Joule heating) in the 2nd coil (ignoring the difference in radiated power, and other non-ideal effects such as increased losses in the 1st circuit).
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$\begingroup$ I got the thing, Thanks Nicolas. I was missing the fact that second coil isn't yet connected to any element (e.g. resistor inductor etc.). $\endgroup$ Commented Nov 9, 2018 at 16:11