When an inductor is connected to a battery, it is to my knowledge that a current will want to flow. However, as the current is quickly increasing from 0 the inductor will oppose this change. It does this because the initial current, due to the potential difference (applied emf) of the battery, induces a magnetic field around the inductor (solenoid) which causes an emf to be induced in the coil. This “back” emf is of opposite polarity to the applied emf but due to the natural resistance of the inductor (the material which makes up the coil), the “back” emf will never equal the applied emf so there is still a net emf in the initial direction. This net emf induces a current which flows in the initial direction. I have looked up explanations of Inductors on the Stack Exchange site, Quora, YouTube, etc so I believe this explanation is accurate but I am not 100 percent sure so please correct me if I am wrong. Anyways… I am assuming that the induced current can only increase if the net emf increases, and because the applied emf is constant it would require the “back” emf to decrease for the net emf to increase. The “back” emf will decrease only if the rate of change of current slows down. I feel like I have the understanding of inductors down but I’m struggling to intuitively understand what causes the rate of change of current to slow down (I understand the maths behind it, I just can’t visual picture it). So simply my main question is:

What causes the rate of change of current (with respect to time) to decrease?

How and why does the rate of change of current (with respect to time) decrease?

In Conclusion, I know there is one part of my logic that is flawed because there is a paradox; the "back" emf needs to decrease in order for induced current to increase but my "back" emf can't decrease until the rate of charge of current decreases. How can this rate of change decrease without current changing (increasing)?

I am very confused.


1 Answer 1


If there is no resistance in the circuit, the rate of change of current in the circuit does not change, as explained here: inductor back EMF.

Thus, the decrease in the magnitude of the rate of change of current as time progresses must relate to there being resistance in the circuit.

In the circuit, you have $\mathcal E_{\rm Battery} - V_{\rm resistance} = \mathcal E_{\rm back} = (-)L\frac {dI}{dt}$.

As time progresses, the current in the circuit increases, so $V_{\rm resistance}$ increases. Because $\mathcal E_{\rm battery}$ stays the same, $\mathcal E_{\rm back} $ must decrease, as does the magnitude of $\frac {dI}{dt}$.

To my knowledge, if there is a back emf, it should oppose the applied emf and the sum of these "emfs" should result in a net emf. Current should only increase if the net emf increases... No?

To understand what is happening one needs to consider where the equation $\mathcal E_{\rm Battery} - V_{\rm resistance} = \mathcal E_{\rm back} = (-)L\frac {dI}{dt}$ come from and the use of the terms potential difference and emf.

From what you have written I assume that you have never queried as to what happens in a simple circuit which consists of an ideal battery and a resistor.
In such a situation the potential difference across the battery $V_{\rm battery}$ is equal in magnitude to the potential difference across the resistor $V_{\rm resistor}$ so that $V_{\rm battery}-V_{\rm resistor} =0$.
If the equation is multiplied by the current $i$ then the equation becomes $V_{\rm battery}i-V_{\rm resistor}i =0$ and it perhaps now becomes apparent that the equation is a restatement of the law of conservation of energy for electrical circuits, electrical power output from the battery is equal to electrical power dissipated in the resistor.
The only thing to add is that $V_{\rm battery}$ is also called the emf of an ideal battery, $\mathcal E_{\rm battery}$.

A slightly more complex example is an ideal battery used to power an ideal electric motor.
The steady state condition is that there is no current in the circuit because the emf of the battery, $\mathcal E_{\rm battery}$, is equal in magnitude to the back emf produced by the motor coil rotating in a magnetic field $\mathcal E_{\rm back,motor}$ thus $\mathcal E_{\rm battery} - \mathcal E_{\rm back,motor}=0$.
Now suppose that the motor is made to do some work by you pinching the axle of the motor.
The speed of revolution of the motor decreases and hence the back emf produced by the rotating coil, $\mathcal E_{\rm back,motor,load}$, decreases and what is the net input power, $\mathcal E_{\rm battery}i_{\rm load}-\mathcal E_{\rm back,motor,load}i_{\rm load}$, equal to?
In this case is is equal to the mechanical power output from the motor, the work done per second against the frictional force that you have applied to the axle.

Returning to your question one has to consider a dynamic situation.
Points to note are that connecting a battery to a length of wire will results in a current flowing in the wire.
If the circuit has inductance an emf will be induced because of the changing current (Faraday) and that induced emf will oppose the change producing it (Lenz) which in this case is the changing current.

So connect an ideal battery to and ideal (no resistance) inductor.
At the start even with no current flowing the current must change otherwise there would be no back emf from the inductor and then there is an unnatural situation with a battery being connected to a piece of wire with no resistance and no current is flowing. Now we come to the situation which you are unsure about.
Two equal and opposing emf, $\mathcal E_{\rm battery}$ and $\mathcal E_{\rm back}$, and yet the current changes and noting that if there is no change in the current $\mathcal E_{\rm back}=0$.

$\mathcal E_{\rm battery}i-\mathcal E_{\rm back}i$ relates two powers.
$\mathcal E_{\rm battery}i$ is the instantaneous power delivered by the battery.
So what is $\mathcal E_{\rm back}i$?
It is the rate of change of the magnetic energy stored in the inductor.
So in this situation the instantaneous electrical power delivered by the battery is equal to the instantaneous rate of increase in magnetic magnetic energy stored by the inductor.

In a time $\Delta t$ the battery delivers electrical energy equal to $\mathcal E_{\rm battery}i\Delta t$.
During that time the magnetic energy stored by the inductor changes by $\frac 12 L(i+\Delta i)^2 - \frac 12 Li^2 = Li\Delta i + \mathcal O \Delta i^2$ and you will note that $Li\Delta i = L \frac{\Delta i}{\Delta t} i \Delta t= \mathcal E_{\rm back} i \Delta t \,(=\mathcal E_{\rm battery}i\Delta t)$.

Perhaps do not think of the emfs as battling it out for domination rather that one emf is the source of electrical energy and the other emf is the sink (user) of electrical energy and they are equal because energy is a conserved quantity.

  • $\begingroup$ First of all... Thanks a lot for the response I appreciate it and I hope you had a good Christmas. $\endgroup$ Dec 27, 2021 at 0:14
  • $\begingroup$ Referring to your linked answer, I specifically would like to know what causes the current to "continue to flow". To my knowledge, if there is a back emf, it should oppose the applied emf and the sum of these "emfs" should result in a net emf. Current should only increase if the net emf increases... No? $\endgroup$ Dec 27, 2021 at 0:20
  • $\begingroup$ I understand your logic, in a mathematical sense, but I am still confused as to how current continues to flow after the first moment/instant. You said that as time progresses the current in the circuit increases but how? and when I say "how?" I mean physically I think I understand the equation... Not 100% sure though. $\endgroup$ Dec 27, 2021 at 0:32
  • $\begingroup$ @jessejpanda I have added to my answer but perhaps even that is not enough? $\endgroup$
    – Farcher
    Dec 27, 2021 at 11:54
  • $\begingroup$ this is much clearer now thank you for taking time out of your day (or night) to help. $\endgroup$ Dec 27, 2021 at 12:08

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