What Does the Voltmeter Read When a Current Increases in an Inductor Loop?

[ehild]

Yes, but when I do that, I seem to get the electric field pointing in the wrong direction. How can the current always flow agains the eletric field? There must be something I'm missing here.

Yes, it looks difficult to find out what happens inside the wire. I do not know, and I also wondered what happens with the electrons there. Therefore one excludes the wire, and choose an integration path outside the wire but close to it. There is no current outside the wire...

By the way, the current flowing in the loop and providing the magnetic field is an "external current" coming from a current source, and it can not be influenced theoretically.

I don't understand what you mean by this step.
Since I want the voltage across the inductor, how can I calculate it with a loop not including the inductor?

You want the voltage across the voltmeter. And then you say that the terminals of the voltmeter are at the same potential as the terminals of the coil. But the voltage can be determined from E only where the field is conservative.
Just remember the equation with the vector potential: E=-grad V -dA/dt (sorry, "d" means partial now) So the integral of E is not the potential difference but it includes the term containing A. You would need A. But I do not know really, I never used this vector-potential formalism, and I hate the magnetic fields...

The integration path I suggested includes a segment that surrounds the loop, but avoids the magnetic field inside it.
The tangential component of the electric field is continuous at the supposed boundary between the conservative and non-conservative domains, so it can be taken the same as inside the loop.

You see how useful the integral forms are: you can avoid the problematic places...


ehild

 

[JustinLevy]

The current is flowing against the induced emf. There is nothing wrong with that. The induced emf is not driving the current (I = at). Something else is. The induced emf is simply opposing the increase in current. You appear unwilling to accept that. I suggest you end this thread and study what has been posted.

AM

I am very frustrated that you continue to insist the answer has been given to me, when in actuality the answer you have given is wrong.

You insist that the voltmeter reading is negative, despite texbooks (and simple experiments) showing that it is positive. You also previously claimed we don't need another force since the resistance is zero. Now you claim of course there is something else (yet you still don't seem to think that ignoring that force would affect the calculation). I have also explained that if the voltmeter reading is actually - L dI/dt, that this could actually increase the change in current through a circuit instead of opposing it. Your claims are contradictory.

What is this "something else"? There is no chemical, etc. forces to worry about like in a battery. It must be an electric field. There is nothing else that can be in the wire to give the charges a force.

It is most likely this missing field that is causing the problem. How do we calculate this field?

Look, we can all learn from this discussion. It sounds like none of us has a very good way of solving this yet, but we all need to learn from our mistakes to move on to solving this thing.

Yes I agree that what appears to be simple logic leads us to the conclusion that the voltmeter reading would be negative. Do you agree now though that this must be incorrect?

 

[JustinLevy]

But the voltage can be determined from E only where the field is conservative.

Based on your other comments, I'm still not sure what you mean by this. However a friend showed me a possible way to get the answer that used this in part of the process. I'll get to that in a bit, and you can tell me if that is what you were thinking.

The integration path I suggested includes a segment that surrounds the loop, but avoids the magnetic field inside it.
The tangential component of the electric field is continuous at the supposed boundary between the conservative and non-conservative domains, so it can be taken the same as inside the loop.

\nabla \times \mathbf{E} = - \frac{\partial}{\partial t} \mathbf{B}
So the field is "non-conservative" where ever the magnetic field is changing. This isn't constrained to just the region inside the loop. The magnetic field is changing outside the loop as well.

Yes, it looks difficult to find out what happens inside the wire.

Yes it looked a bit too difficult to me.
But a friend has showed me a "trick" for how to work this out. His idea is to consider the inductor to be made of a material that has some resistance (resistivity \sigma).

Electric field being conservative or not, Ohm's law is a simple linear response theory and is still correct in resistive media. So j = E \sigma. The current HAS to be in the direction of the total electric field.

Let's make this into a 1-dimensional problem and just integrate over the cross section. So E = I R (where it will be understood in this problem that R is units of resistance per length). Therefore, if the current in the inductor is I = at, then the total eletric field in the wire of the inductor is E = a t R.

We already know from Maxwell's equations, that there is an induced eletric field from the inductor of - L (dI/dt) / length of wire in inductor. Therefore, the eletric field that must be supplied by the "current source" is therefore E = a t R + L (dI/dt) / length of wire in inductor.

If the current source is far enough away that we can ignore the magnetic field, and the wires used to connect it to the inductor are of negligible resistance, then we have (now letting R be the total resistance) the voltage across the current source is:
V = IR + L (dI/dt)

In the limit the resistance goes to zero and we have:
V = L (dI/dt)

If we put a voltmeter across this, similarly far enough away that we can ignore the magnetic field and the wires used are of negligible resistance, it would of course read the voltage across the current source as:
V = L (dI/dt)

Tada! We did it!
There WAS an electric field we were ignoring. This of course was suggested by some previously, but now we finally have a way to explicitly include this! And it yields the correct answer that the voltage across the inductor is of course positive!


So, what do you think?
It seems to make sense. But maybe I'm just won over by the fact that it gets the correct answer, and I'm ignoring some flaws in doing so.

This seems to say that if we removed the current source, and had an external device make a changing magnetic field such that the flux in the loop changed exactly the same... In that situation, we WOULD get a negative voltage. But since current would flow the opposite way, I guess we'd still have V = + L dI/dt ?

 

[ehild]

Justin,

The method your friend suggested to find the voltage across the inductor by applying a counter emf is really exclent. It can be applied also when there is no current source, only the time varying magnetic field. In this case, zero current should be maintained by the counter-emf. This method is the working principle of the potentiometer :).

Accept it now, because it is a good working principle. And I am sure and I do hope you will find arguments against it after a while, which will make you think and understand electricity better and better.

ehild

 

[jambaugh]

The problem has been in the phrasing of the definition of inductance. The phrase "opposing change in current" is causal. Whereas the problem is to deduce a voltage which is "affecting" not "effected by" the change in current.

Again it is a treatment of the induced electric potential as a reactive force in the same sense of an inertial force in the sense of D'Alembert's principle. Again I say it is clear when you invoke a mechanical analogue. Replace the flow of current through a wire with the wire being a string. Imagine an inductive coil or loop as being a massive pulley. The mass of the pulley "opposes" the acceleration of the string (increasing current) in the sense of the "equal and opposite" force on the implied wench pulling the string through. The fact that the string is accelerating implies an applied force "promoting" the acceleration.

If a mass is being accelerated then the applied force is F = MdV/dt. The inertia of the mass itself is defined as "opposing" its acceleration in the sense that whatever object is applying the force F is then experiencing an opposing force -F.

If the current through an inductive circuit element is increasing then there is an applied voltage V = L dI/dt. Here we do not have as direct an equation since we are dealing with scalar quantities but the understanding is clear. The direction of increasing current is in the direction promoted by the voltage i.e. from higher potential to lower.

In effect the energy of the B field of the inductive current is analogous to a mechanical kinetic energy term. Throw a capacitance in there an you have the analogue of a mechanical spring with 1/C the spring constant.

I've had to deal with confusion over the "inductance opposes change of current" phrasing with many students. I have found that invoking this mechanical analogue has been the best way in my experience for them to keep signs straight and instill in them intuition about which they can feel confident (and be correct).

 

[cabraham]

I know this is an old thread, but I just found it. Interesting, it is. Anyway, I just thought I'd chime in.

An inductor, in the pure sense, cannot be current-driven, but it must be voltage-driven. Otherwise, an infinite voltage would be required at time t = 0. When a constant current source, dc, is connected to an open switch, & the other side of the switch connected to the 1-turn inductor loop, what happens when the switch is closed at t=0?

If the ccs (constant current source) current value is Ics, & the inductor initial current is 0, then we have a problem. We would get an infinite voltage at t=0 when the switch closes, since changing the inductor current from 0 to Ics in zero time span results in infinite voltage per v(t) = L*di(t)/dt.

But as we know, an ideal pure inductor cannot exist. There is always a little capacitance presented at the input. The surfaces of the input terminals present a very small capacitance which shunts the input. The ccs charges up this capacitance at t=0. The voltage across the input terminals increases very rapidly, & the current energizes the inductor. As the inductor energizes, the voltage falls. Eventually, the voltage at the input terminals is zero.

The difficulty w/ this problem lies in the fact that the inductor current cannot instantly change from 0 to Ics w/o infinite voltage & infinite power. But, the tiny capacitance at the input terminals alleviates the infinite voltage/power conditions.

The steady state voltage is zero since the loop has no resistance. The ccs is outputting a current of Ics, a voltage of 0, & a power of 0, in the steady state. Did I help at all?

Claude