# Homework Help: Induced EMF (Faraday's law)

1. Apr 14, 2013

### NaOH

1. The problem statement, all variables and given/known data
It is often discussed that a changing magnetic field will create a non-conservative electric field, which would then induces an EMF and hence an induced current, but what happens when we connect voltmeters?

Consider putting a circular loop of wire with some resistance in a changing magnetic field. The magnetic field can be said to only affect the loop. Now I attach an ideal voltmeter to 2 points on the loop, what would be observed on the voltmeter?

2. Relevant equations

Faraday's law, E = change in flux wrt time

3. The attempt at a solution

This is qualitative, so not much. There could be a few cases.

One: I just get the full EMF value.

Two: I get a fraction of the full EMF value proportional to arc length tapped. Runs into problem when I consider what the positive sense of the potential is when it is tapped at the diameter of the loop.

Three: 0. Either because EMF is the same at any point, or that whatever EMF the arc length subtended is consumed by the current and resistance of the subtended arc.

Four: not useful to define potential difference because electric field is not conserved. it does not explain how the voltmeter will respond.

2. Apr 15, 2013

### domenico

If ρ is the resistance of the wire loop per unit length, and r is the radius of the loop, then the total resistance would be: R = ρ*(2πr). Applying Ohm's law, the induced current is: I = E/R. If you place the voltmeter, the measured value would be: V = I*ρ*l, where l is the arc length from the (+) terminal of the voltmeter to the (-) terminal, in the sense of current flow. So, I think the right answer would be the second one.

3. Apr 15, 2013

### NaOH

But by using only I = E/R = E/[ρ*(2πr)]; V= Ir = E/[ρ*(2πr)]*ρ*l = E*l/[2πr] it feels like only the resistance matters, and it does not include the EMF contribution of the arc the voltmeter measured...

My TA said: We go 1 round and the EMF is E. We go a fraction round and it is E*l/[2πr]
So, V = EMF gain + loss in potential due to resistance
V = E*l//[2πr] - E*l/[2πr] = 0

But my lecturer said something like: it's like a battery. when we tap the end using a voltmeter, the internal resistance is not measured, we get the EMF, like in this case... which suggests [2]
But the problem is when i put the 2 voltmeters across the diameter (which i refer to as a and b) of the circle of a uniform resistance loop, one would measure a>b, and the other would measure potential at b>a (refer to attached)

if I only use 1 voltmeter, it would seem like what potential i measure depends on how i place the voltmeter...say only V1 is used and without changing anything I just move V1 to V2's position, then I would measure V2! I don't think it explains the correct physics... Neither would it suggest what I measure if I place V1 at a position above the loop (with magnetic shielding).

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4. Apr 16, 2013

### collinsmark

I encourage you to think about this. emf is kind of a difficult concept for everyone. The more you think about it on your own, perhaps the better you will gain an intuitive understanding of it. So I'll be brief.

Lumped parameter models don't directly apply to the situation that you describe. Not only is the emf distributed around the closed wire loop, but the resistance is distributed too. Lumped parameter models are not so easy at modeling distributed parameters. But that's not to say that we can't use lumped parameter models at all. Just for a related exercise, try out the following problems and ask yourself how they relate to your problem.

Consider the situation that you mentioned where the voltmeter's leads are attached at opposite sides of the loop. Consider this lumped parameter model. What's V0?

Okay, suppose that the voltmeter's leads weren't at opposite ends of the closed loop, but separated by only a small part of the arc. What's V0 in this case?

[Edit: revised graphics for clarity.]

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5. Apr 16, 2013

### NaOH

I need more guidance.

For the symmetric case, it would be similar to how the ring is imagined to be tapped with a voltmeter on opposite ends: whatever EMF gained is promptly consumed by the resistance:

Using kirchhoff's loop rule, $0.5 V_{emf} - 0.5RI + 0.5 V_{emf} - 0.5RI = 0$
So... $I = V_{emf}/R$

Applying Kirchhoff's loop rule again, $V(+) + 0.5V_{emf} - 0.5 RI = V(-)$
Thus, $V_{0} = V(+) - V(-) = -0.5 V_{emf} + 0.5 R V_{emf}/R = 0$

But I have this nagging feeling it should be $0.5V_{emf}$, but can't show why.

Similar conclusion can be drawn for the asymmetrical case. $V_{0}$

6. Apr 16, 2013

### Andrew Mason

Kirchoff's law applies only to conservative fields (path independent). Is this the case here? There are two loops involved. The first is the wire loop you are applying the leads to. The second is the loop made by the voltmeter. leads and the section of wire between the leads. If you do not know the size or orientation of the loop made by the leads can you determine the induced emf?

AM

7. Apr 16, 2013

### collinsmark

Correct! Very good.

I could say, "trust your math on this one." But maybe I can help out a little more intuitively with one last set of exercises below.

Yes, that is correct too.

Following the same general idea that you just did above, consider the following figure. Assume all voltage sources shown have a potential difference of 0.125Vemf each, and each resister has a resistance of 0.125R.

• What's the electrical potential of point B relative to point A?
• What about point C relative to point A?
• D relative to A?
• And all the way up to point H relative to point A?
(Hint: This is the same sort of thing you just did with the last couple of figures. Given what you already know from the last set of exercises, you are allowed to do this one in your head if you wish.)

====================

Now let's get a little crazy with all this. Let's break the loop open. This time, as shown in the figure below, we no longer have a closed circuit.

• Okay, now what's the electrical potential of point B relative to point A?
• What about point C relative to point A?
• D relative to A?
• And all the way up to point K relative to point A?

[Edit: Oops, originally made a mistake in one of my figures and left out a voltage source and resistor. Made correction (so there's 8 of each).]

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8. Apr 16, 2013

### NaOH

The reply was in response to collinsmark's comment. So... erm... it's conservative for the circuits outlined in collinsmark's comment, but not so for the confounded loop in a changing magnetic field.

I'll give the rest of the comments further thought after a night's rest.

Zero at every point designated.

• 0.125Vemf
• 0.250Vemf
• 0.325Vemf
• 1.00Vemf

I've lost sight of how this is related to the loop in changing magnetic field... unless it's equivalent to a circuit break scenario. But it wouldn't make sense of an induced current to flow in a broken circuit, which affirms the point made that Lumped parameter models have some limitations.

I'd also like to thank everyone for their replies, especially collinsmark for taking to time to make clear diagrams.

Last edited: Apr 16, 2013
9. Apr 16, 2013

### collinsmark

Correct!

Correct again.

Well, if you have a completely* closed conducting loop in a changing magnetic field, current can and will be induced in that loop. But if you apply a voltmeter to any two points of that loop** the potential difference is always zero.

*(if all of the loop is within the changing magnetic field, and the loop has uniform resistance.)
**(let's assume the voltmeter's leads don't significantly poke inside the magnetic field too.)

We've used 8 resistors and 8 voltages sources here in our lumped parameter model. If we were to take this to its limit and model a realistic closed loop in a changing magnetic field, we would use and infinite number of infinitesimal voltage sources alternating with an infinite number of infinitesimally small resistors. The combined emf of the voltages sources is cancelled out by the combined IR drop of the resistors. Current does flow, even though all points on the loop have identical electrical potential.

This is equivalent to your hypothesis labeled "Three" in your original post.

But what if you wanted to "tap into" that emf and current, rather than just wasting it as heat in the loop? Answer: break the circuit and tap into it. (In my diagram you could tap into it via K and A.)

I agree that lumped parameter models have their limitations. But it would make sense to have induced emf in a broken circuit. Even though your toaster may not be turned on, it is plugged into the outlet/mains, although the circuit is broken since the toaster is off. But there is a generator on the other side, the emf is there and ready to go on a moment's notice whenever you want a tasty breakfast. All you have to do is put in the bread, press the lever, and you've tapped into the generator (point K to point A sort of thing). Toast is on the way.***

***(We're also assuming here that the toaster has a significantly higher resistance than the internal windings of the generator. That way the outlet/mains can still maintain a steady voltage, and you wont blow a fuse, circuit breaker, or ground fault detector. [And we're also ignoring distribution lines.])

Last edited: Apr 16, 2013
10. Apr 16, 2013

### NaOH

Well, the argument is so far good, but it doesn't seem to reconcile with what my lecturer says, which is the basic motivation for starting this topic. I swallow the potential difference is always zero at any two points better, but there seems to be something else missing...

Consider the following diagram,

Letting resistance be uniform...

"The electric field associated with a time-varying magnetic field is not conservertive:
$\oint_{C}\vec{E}(\vec{r},t).d\vec{r} = -\frac{d}{dt}\int_{S}\vec{B}(\vec{r},t) . d\vec{A}$

The magnetic field is increasing, so by faraday's law/len's law the induced EMF and thus current flows clockwise.

$V_{1} = \frac{1}{2}*\frac{d\Phi_{B}}{dt}$ (ignoring the -sign for now, 1/2 for half the resistance of loop)

V1 and V2 have the same size but:
V1 = Va - Vb > 0 (like how the field points from lower potential to higher potential in an battery)
V2 = Vb - Va > 0

Last edited: Apr 16, 2013