# Use Kirchoff's loops rule in the circuit?

1. Jul 7, 2008

### atavistic

The emf given $$-\frac {d \Phi}{dt}$$ produces a non conservative field right? Then is it right to use Kirchoff's loops rule in the circuit?

Also if a coil with self induction L, is kept in a constant magnetic field perpendicular to the plane of the coil, and the flux associated through the coil due the magnetic field is $$\Phi$$ = LI. Then I = $$\frac{ \Phi}{L}$$. What is this I?Where did it come from?

2. Jul 7, 2008

### dynamicsolo

Re: Electrodynamics

As for the first question, you are correct in that induced emf does not have an associated potential function. (This is because the induced current is the result of magnetic force acting locally on each charge in a loop, unlike the impressed electric field and associated force acting on charges from a voltage source, which does have an electric potential.) So you can't really use Kirchhoff on the induced current in the conductive loop.

When you have an inductor in a circuit, though, you are looking at that coil as part of a larger system that will follow Kirchhoff's laws, since from the outside, it is sensible to talk about the voltage drop across the inductor as $$V_L = -L\frac{dI}{dt}$$ . I think that's right, as long as mutual inductance in the circuit is negligible.

I'm finding your other question a bit obscure. Are you asking about a coil placed in an external magnetic field? The expression $$\phi = LI$$, I believe, is referring to the axial field in the inductor, with I being the current provided by a circuit.

3. Jul 7, 2008

### atavistic

Re: Electrodynamics

For your reply to first question, I dont see how we can apply kirchoff's rule to the rest of the elements of the circuit because the induced E in the circuit is itself non-conservative.

For the second reply,yes I mean an external constant magnetic field.

4. Jul 7, 2008

### dynamicsolo

Re: Electrodynamics

About the second question, if the external field is constant, it seems to me that it would have no effect on induced emf in the coil, yes? When it does vary, you get into the matter of mutual inductance, where the change in flux from an outside magnetic field does contribute to the induced emf in the coil. This is usually brushed over in introductory courses because it involves a coupling of the magnetic devices (the coil and the source of the external field) and is easy to deal with only in simple situations.

I'll have to think a bit about the first question. For now, I'd say this: the induced emf along the turns of the coil does not follow a potential function, but there is an "effective" voltage drop between the ends of the inductor, which, again in the simplest situation of no outside varying magnetic field, behaves like the expression for V_L I gave above. It does seem a little paradoxical, though, so I'll need to see what people say about this.

5. Jul 10, 2008

### dynamicsolo

Re: Electrodynamics

Having checked around a bit, I find that many textbooks don't really deal with this question. But here's basically what's going on. Magnetic induction drives charges in the conductor by an effective magnetic force which is the result of the flux change through the loop. This force does an infinitesimal work on a charge for each infinitesimal displacement, so there is a net work per charge along any path. This is why you can't use Kirchhoff's "loop rule" on a conductive loop: the net work per charge on the closed path is not zero. Since it isn't zero for the closed path, we can't sensibly define a potential function for induction (and why, in contemporary usage, we prefer to discuss "inductive emf", rather than "inductive voltage", which implies an electric potential, and thus a potential function.

The "voltage across an inductor", then, would be the net work done per charge performed over the length of the helical coil of the inductor. It is then sensible to talk about a "voltage drop", even though this is properly more like a "net inductive emf" between the ends of the inductor. Since the path is not closed, we dodge the issue of the lack of a potential function, but what we are really talking about is an "effective difference in voltage", rather than a difference in potential as would be the case, say, between the plates of a capacitor due to the electric field.

6. Jul 11, 2008

### atavistic

Re: Electrodynamics

Yes, this is what I also believe but you are wrong at the point that magnetic force does the work, its actually the induced electric field that does the work.

Thnx for looking into my doubt that most high school students dont understand or dont care about.

7. Jul 12, 2008

### dynamicsolo

Re: Electrodynamics

You are referring to the Maxwell equation

$$-\frac{d\phi_{B}}{dt} = \int E \cdot dl$$

and I will certainly concede that it can be said that it is the induced E which accelerates the charges.

I guess my thing is that arguing from Maxwell's equations has long seemed to me a bit like saying "the phenomenon works this way because the math says so". The equations are a description of the behavior of electromagnetic fields, but I don't feel that they necessarily give insight into what is happening physically. (Though maybe that's just me...)

In "motional emf", you can obtain the magnitude of the induced emf from Faraday/Maxwell and get the direction either from that or "Lenz' Law", but it is also possible to look at the motion of the "charge carriers" in the ambient magnetic field and obtain the same result from calculating the magnetic force through qv x B. For certain other situations, it is also possible to liken the magnetic flux change to a relative motion of charges and magnetic field to obtain a magnetic force. (This can be followed off into the interesting feature that "magnetic force" is a consequence of relative motion, and thus, to many people, not really a force at all.) I am still investigating this analogy to see how far it can actually be pushed; maybe the situation in this problem is too far for it to go...

EDIT: As I think about this a bit more, here's something else that bothers me about the "induced E field" description (though it certainly seems implied from the equation). Somehow, a magnetic flux change through a conductive loop is making an electric field that is a closed loop, without a "source" in the circuit (or maybe each point on the loop is an infinitesimal source). So it can be seen immediately from this why there's a problem with constructing a potential function for it.

... or even some lecturers and textbook authors, perhaps. The point isn't particularly discussed in most of the books I'm familiar with, but it does sort of beg the question as to why the inductor has a "voltage drop", but not electric potential. I guess I'd put it down to sloppy use of the term "voltage".

Last edited: Jul 12, 2008