Taking a solenoid as an example, when constant current $i$ is passed through the solenoid constant flux is seen. But when the current starts to vary, the magnetic field created by a changing current in the solenoid itself induces a voltage in the same circuit. And Self-induced voltage creates a self-induced current in the solenoid that has opposite direction with respect to the original current, and this should be opposing the changing current, so when it does, the changing current and self-induced current have to get balanced and cancel each other. And there should never be a change in current because as soon as we try to change the current, self-induced current is produced in the solenoid and it opposes the change. But this doesn't seem to happen, the current still changes. Why?

  • $\begingroup$ "And Self-induced voltage creates a self-induced current in the solenoid" - this is false. For example, one can place an ideal voltage source across an ideal inductor and the voltage across the inductor is fixed by the voltage source. The only non-contradictory solution is that there is steadily changing current through the inductor such that the emf and applied voltage are equal in magnitude. I've downvoted your question because it is a word salad without a single equation to justify your reasoning or conclusion. $\endgroup$ – Alfred Centauri Jul 17 '19 at 0:56

The problem with your reasoning is that you assume the induced EMF always completely cancels the changing magnetic flux due to a time varying current, preventing any change in the current in the first place, which is not correct. If there were no change in the current, there wouldn't be an induced EMF.

To analyze how the current will be changing, it's not very useful to think of it as being made up of two components: the applied current, and the current induced by the changing magnetic flux. Instead, we consider the total current (which is what is measurable) that includes both of these components.

Suppose that changes in the magnetic flux through the inductor is only caused by the applied current, i.e. there is no external time-varying magnetic flux through the inductor. A current $i$ through the inductor induces a magnetic flux of $\Phi = Li$, where $L$ is the inductance. According to Faraday's Law, the induced EMF is $$\mathcal{E}=-\frac{d \Phi}{dt}=-L\frac{d i}{dt}.$$ This EMF opposes sudden changes in the current through the inductor. In practice, we usually say that the "voltage" across the inductor is $v=L\frac{di}{dt}$, and demand that Kirchhoff's Voltage Law hold (sum of voltage drops along a loop is zero). This gives the correct voltage drops across the rest of the circuit.

What does this mean? How does the current change? It depends on the circuit the inductor is connected to. If you connect an ideal inductor to an ideal voltage supply, and suddenly increase the supply voltage from $0$ to $V_0$, you must solve $$V_0=L \frac{di}{dt}$$ with initial condition $i(0)=0$. The solution is $$i(t)=\frac{V_0}{L}t,$$ i.e. the current will linearly increase in time.

If on the other hand you have a current source with a parallel output resistance $R$ connected to an inductor, and turn on the current to $I_0$ at $t=0$, the solution is $$ i(t) = I_0 \left[1 - \exp\left(\frac{-t}{L/R}\right) \right], $$ i.e. the current will gradually increase to the supply current.

  • $\begingroup$ So it means, the induced EMF only cancels a part of the changing magnetic flux due to a time-varying current, and that part depends upon 'L'. Is it? $\endgroup$ – Mohammad Vajid Jul 17 '19 at 4:50
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    $\begingroup$ I wouldn't put it that way exactly, because the time-varying current through the conductor and the induced EMF are consequences of each other in terms of the ultimate time variation of the current. What I would say is that the induced EMF opposes changes in the current (due to a change in a supply voltage/current, change in a resistance etc.). How "hard" it tries to keep the current smooth, and how fast the current can vary, is quantified by the self-inductance $L$. For a very large $L$ the current will either change very slowly, or induce very large voltages to keep it changing fast enough. $\endgroup$ – Puk Jul 17 '19 at 5:16
  • $\begingroup$ FWIW, (induced) emf and magnetic flux are so different that I wonder how the first sentence is to be understood. For an ideal inductor, the emf always equals in magnitude (and opposite in sign) the voltage across the inductor (period). Also, with regards to the 2nd paragraph, what is "the applied current"? If, for example, an (ideal) inductor were connected to an (ideal) current source, the inductor current is fixed by the 'applied current', i.e., there is no degree of freedom left for a "current induced". $\endgroup$ – Alfred Centauri Jul 18 '19 at 1:13
  • $\begingroup$ What I meant by the first sentence is that it's not true that the self induced EMF always dictates the current in such a way that the magnetic flux, and therefore the current itself, remains unchanging. This was to point out what was wrong with the line of thought in the original question. $\endgroup$ – Puk Jul 18 '19 at 1:33
  • $\begingroup$ As for your second point, I don't disagree with what you have said. I can't precisely define for you what "applied current" means. As I argue in my answer there is really one and only one current through the inductor, and it is not necessarily meaningful to speak of the "original" current and the "self-induced" current as being two distinct quantities, as the question seemed to suggest. $\endgroup$ – Puk Jul 18 '19 at 1:34

Well as much of your question that I could perceive is the induced current should cancel given current and net current should be zero If this is what you asked then yes it is true but just for an ideal case in which, when applied current is changed the flux through coil changes and induces an equal but opposite emf to the applied emf and hence cancels the current but in practical inductors the induced emf never equals the applied emf and net current flows through the inductor!

  • $\begingroup$ Why is that so? what's stopping the induced current from not becoming equal to the changing current? $\endgroup$ – Mohammad Vajid Jul 16 '19 at 19:01
  • $\begingroup$ Basically, it's not really induced current but induced emf that opposes the applied emf (changing) and cancels part of it , and the applied current is because of applied emf so when a part of applied emf is cancelled the the current that would flow because of it won't flow! $\endgroup$ – Yawar Naseem Jul 16 '19 at 19:06
  • $\begingroup$ So it's similar to saying induced current opposes a part of changing current, because induced emf, however, generates induced current. But just a "part". what's really stopping it to cancel the whole changing emf in a practical sense. $\endgroup$ – Mohammad Vajid Jul 16 '19 at 19:21

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