# How does an inductor generate back emf equal to a source driving it?

I have a question about the way inductors fundamentally work. How can an inductor be modeled as accurately as possible?

I tried to approach this problem from the basics of electromagnetic fields, but I am still having some problems understanding the matter

When we connect the ends of a inductor across a battery:

1. The end at lower potential pushes the electrons in the inductor by applying electromotive force on it
2. due to this push the electrons gain acceleration
3. because of this acceleration a change in current with respect to time di/dt occurs
4. because of this di/dt an electric field is developed in the inductor to oppose it
5. now there are two forces acting on the electrons that is the force due to battery and the force due to induced electric field in the battery
6. now we see that under these two opposing forces the net acceleration of electrons has decreased and there is anew value of di/dt at that instant now since the di/dt has reduced the induced emf this time will be less compared to previous now again under the influence of these two forces a new acceleration will come and according to it will come a new di/dt
7. and so there is a jerk like increment as well as decrease in the di/dt so the problem is what is happening inside because in textbooks they write that in simple inductor and battery circuits the di/dt is constant?

Another question I have is about the induced emf to be equal to the external battery if the induced emf(opposing emf induced in loop) is equal to the external battery then how in the world is di/dt not equal to zero because if there is are two fields exactly same in magnitude and in opposite directions then how come the current increases with time?

If external emf and induced emf are same then the energy given to electrons in completing the circuit is equal to the energy taken by the inductor from the electrons then how is their kinetic energy (current) increasing?

• It is becoming apparent to me, and others I think, that you are trying to reason about this without having a firm grasp of the fundamentals. My constructive advice for you is to set this problem "on the back burner" for now, continue your studies in the fundamentals, and then come back to it. There are simply too many misconceptions evident in what you've written above for there to be any hope that one answer will clear them up in one go. – Alfred Centauri Oct 26 '13 at 12:23
• Related question from user28804: electronics.stackexchange.com/questions/86500/… – Alfred Centauri Oct 26 '13 at 12:39
• I know that there are many misconception in this and thats why i am here to gather some information and seek help from some expert because that all i have written above is the general way anyone would think ... So if you have any answer , do post it – user28804 Oct 26 '13 at 12:51
• answering questions is the main purpose of this site and sometimes the odd misconception is contained in the question and these are usually addressed in the answer however, you are building misconception on top of misconception and it's too much to unravel plus, in your almost identical questions you've shown little respect for the answerer by deleting them and, you've apparently barely taken anything onboard from those answers preferring, it seems, to jump to illogical conclusions that you fashion into a new question. Have a little thought for those who might contemplate answering. – Andy aka Oct 26 '13 at 17:20

One way to think of a inductor from the circuit's point of view is that it gives inertia to current. This gives you some intuitive feel for why current builds up slowly as voltage is applied, and why the voltage goes as high as it needs to on a attempt to shut off the current abruptly. This isn't how the physics works, but this can be a very useful mental model for understanding and designing circuits with inductors.

I think I would like to explain it this way.

A DC battery is connected across the inductor. The current will not immediately achieve its steady-state i.e constant value; not because it is facing some force against its flow but because the current is in transient phase. So, initially some amount of current flow takes place. Due to change in magnetic flux experienced within the coil , a back emf is produced. Note that , back emf only exist if current flowing across the inductor is changing i.e it resists the change in current and magnitude of back emf is depedent on how " current is changing with time" (which is independent of back emf ).

Because DC current will increase towards its steady state value , the rate with which this current changes with time will determine the change ( or no change) in the magnitude of back emf. However this time, the battery will have to do more work to pass the current (in order to establish the desired value of current in inductor) which gets stored in the magnetic field of inductor for future use. This continues till DC current has reached its steady state value and no change in current takes place ; that eventually diminishes the back-emf.

I tried to explain it from a novice point of view, hope it helped.