1
\$\begingroup\$

What happens when a potential difference is applied across an inductor? I am trying to get a picture of mechanism from start, just after potential difference is applied in detail? In resistor we can see that there are positive ions opposing and current flows at constant rate so energy dissipated in resistor is equal to that given by battery. But what is happening in conductor so that the induced EMF gets equal to E? What is the picture of the mechanism?

\$\endgroup\$
3

3 Answers 3

Reset to default
3
\$\begingroup\$

There is a well known, well studied phenomenon known as electromagnetic induction that is mathematically described by Faraday' Law of Induction.

Essentially, if the current through a loop of wire is changing, the associated magnetic field threading the loop is changing and, by Faraday's Law, there is an associated induced emf.

Now, to your question:

But what is happening in conductor so that the induced emf gets equal to E ? What is the picture of the mechanism ?

Let's assume the wire making up the loop is an ideal conductor or effectively so. Now, recall that there can be no electric field inside an ideal conductor*.

With the voltage source E connected to the loop of wire, the only way for there to be no electric field in the conductor is for there to be an induced electric field that precisely cancels the field due to the voltage source.

Since the requirement for an induced emf is a time changing magnetic field and associated time changing current, the time rate of change of current is precisely the value required to induce an emf that cancels the applied voltage.

*there are conditions in which this is not true but these are beyond the scope of this answer.

\$\endgroup\$
4
  • \$\begingroup\$ As you said the field inside any conductor is zero ... It is zero in the steady state when there is not a potential difference that is constantly maintined .... And even if we assume the field to be zero then from where are th electrons getting the energy to the increase in current di/dt ?? \$\endgroup\$
    – user28804
    Commented Oct 26, 2013 at 5:45
  • \$\begingroup\$ @user28804, for an ideal conductor, there is no electric field inside period. From Electromagnetic Theory for Microwaves and Optoelectronics (courtesy of Google books): "Then just as in the static case, there must be neither a time-varying electric field nor electric charge inside the perfect conductor...". Also, recall that the voltage source supplies energy to the inductor and that this energy is stored in the magnetic field. \$\endgroup\$
    – Alfred Centauri
    Commented Oct 26, 2013 at 11:11
  • \$\begingroup\$ You are not answering the question that i am asking " how does the value of current changes with time if the induced emf is equal but opposite to emf of cell ?? " \$\endgroup\$
    – user28804
    Commented Oct 28, 2013 at 7:44
  • \$\begingroup\$ @user28804, actually I have answered that question, in this answer and in an answer to your other, related question. But it's an answer you refuse to accept: supposing an ideal conductor, it is the only logically consistent state of affairs. This is not much different from considering massless particles in a non-relativistic context where massless particles can have non-zero, acceleration with zero net force applied. \$\endgroup\$
    – Alfred Centauri
    Commented Oct 28, 2013 at 11:29
1
\$\begingroup\$

A resistor and inductor both resist current flow, but the inductor opposes CHANGES in current while the resistor simply opposes the MAGNITUDE. The inductor opposes changes in current because changes in current cause a magnetic field (or back EMF) that pushes back against the current trying to flow.

Here is an Inductor analogy: Let's say you push a guy using a constant force. If he is a resistor, he immediately starts moving at a constant rate. And if you stop pushing him, he immediately stops. A big guy (large resistor) moves slowly and a small guy moves fast. Easy to understand, right? Now, lets say he is an inductor. The instant you push him he feels like a brick wall, but then he starts to move faster and faster and faster.... then you stop pushing him and then he continues to move but slower and slower and slower.... until he stops on his own. A big guy (large inductor) will be sluggish compared to a small guy. In this analogy, the force=Voltage and speed=Current. I suppose you can say that a resistor has no momentum properties while an inductor does. Hope this helps... now on to the math...

One of the best ways to analyze inductor circuits is to use the well known equation for an inductor:

V = L*di/dt

Rearranging, we get this: V/L = di/dt

What does this mean? It means that the rate of change of current through the inductor is constant and equal to the voltage applied divided by the inductor value. If we integrate the equation we get: i(t) = V/L * t

So, in time, the current through an inductor is a simple linear ramp (y = mx). This type of analysis is used mostly in Switch Mode Power Supplies (SMPS) where the inductor is used for DC/DC conversion (i.e. 5VDC to 3VDC).

Hope that helps. By the way, in this figure, the back EMF is equal to V because the voltage source V is an ideal source that dissipates any transient back EMF. But note that from my figure, when the switch opens up, the back EMF voltage will go to infinity because there is no ideal voltage source there to keep it from doing so. Why does it go to infinity? Because the change in current was infinite (the switch stopped the current instantly!). V = L * INFINITY!!! So, if you push an inductor guy and he is moving, you can't stop him instantaneously because he'll rip your arm off. (-: If you are superman (ideal) I guess you can do it. This is why inductive loads quite frequently produce large voltages that can blow stuff up. Fun stuff...

Ideal Inductor Characteristic Curve

\$\endgroup\$
5
  • \$\begingroup\$ To be sure, an ideal inductor does not "resist current flow". Indeed, the voltage across an ideal inductor, for any constant value of current, is zero. \$\endgroup\$
    – Alfred Centauri
    Commented Oct 25, 2013 at 23:35
  • \$\begingroup\$ as i mentioned my question was to get a picture of what is happening with the electrons and why is the induced emf equal to emf of the battery ? Because we can see that the current is increasing with time and for that to happen some energy has to be provided and if the emf induced in inductor equals emf of the battery , the electrons are getting no energy as they move along the circuit then how is the current increasing ? \$\endgroup\$
    – user28804
    Commented Oct 26, 2013 at 5:48
  • \$\begingroup\$ @user28804, if the crux of your question is regarding the energy of the electrons as they move along the circuit, then you're "barking up the wrong tree". See, for example, this: amasci.com/miscon/eleca.html#electron \$\endgroup\$
    – Alfred Centauri
    Commented Oct 26, 2013 at 11:46
  • \$\begingroup\$ To be clear, your original question did not mention anything about electrons. As far as I can tell, we've answered your initial question and follow-up questions. You'll need to restate your question more precisely to get better answers. \$\endgroup\$
    – Erik Anderson
    Commented Oct 30, 2013 at 17:49
  • \$\begingroup\$ And inductors do resist current flow if you're talking about AC. That's why they're commonly called "chokes". But Alfred's comment is true for DC - inductors do not resist DC current. \$\endgroup\$
    – Erik Anderson
    Commented Oct 30, 2013 at 17:54
0
\$\begingroup\$

enter image description here
from cnx.org
In your case R is the ESR of the battery (you may want that to be zero but then you leave the real world)

What is happening is that electric and magnetic fields are interacting. I don't think you can easily model that in terms of physical analogues.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.