# Homework Help: Induced emf

1. Nov 25, 2003

### gnome

The setup is just a circular loop with a capacitor C on the right side (i.e. at 3 o'clock) and a resistor R on the left side (at 9 o'clock), completely inside a uniform magnetic field. There is no battery, etc; just the emf induced by the magnetic field.

A uniform magnetic field (directed INTO the page) decreases at a constant rate dB/dt = -K, where K is a positive constant. A circular loop of wire of radius a containing a resistance R and a capacitance C is placed with its plane normal to the field. (a) Find the charge Q on the capacitor when it is fully charged. (b) Which plate is at the higher potential. (c) Discuss the force that causes the separation of charges.

for part (a), given that
$$\frac{dB}{dt} = -K$$

The magnetic flux at any point in time is
$$\Phi_{B} = \pi r^2B$$

so the induced emf is
$$\epsilon = -\frac {d\Phi_B}{dt} = -\pi r^2\frac{dB}{dt} = \pi r^2K$$

so wouldn't the charge on the capacitor when "fully charged" be given by

$$Q = C\epsilon = \pi r^2KC$$
??

But then what role does the resistance play? (I think that the resistance affects the time it takes to charge the capacitor, but not the ultimate amount of charge. So, was the resistor placed in this question just to confuse?)

for (b):
Since the magnetic field is directed into the page and is decreasing, the induced current should be such that its magnetic field is in the same direction as the original field. Thus I'm thinking that the flow of the charging current should be clockwise, resulting in a higher potential on the upper plate. Is this correct?

for (c):
(The textbook is pretty vague on this ... at least it seems so to me, so my grasp of the concept is commensurately vague.[b(]
The force exerted on the charges in the loop is the result of a difference in potential induced by the change in the magnetic field. Here, since the direction of the field is into the page, and the field is decreasing, the effect on the charges in the loop is the same as if there was an INCREASING magnetic field directed out of the page, or if the magnetic field was steady and the loop was moving parallel to the field, INTO the page.

 OK, I've decided that that last half-sentence is completely wrong. If the field was CONSTANT, and the loop moved in a direction parallel to the field, i.e., directly into or out of the page but always with the plane of the loop oriented parallel to the page, there would be NO induced current, correct? (Because then the magnetic flux through the loop would be constant, despite the motion of the loop.)

The separation of charges progresses until the force exerted by the magnetic field is balanced by the attractive force due to the electrical field that is produced inside the conductor -- is this more or less correct?

This is very confusing. Is this completely unrelated to the magnetic force that would be exerted on a charge moving through the magnetic field, since (at least initially) the charges are not moving? BUT, what about when the charges begin to move in the loop as a result of the induced current? Isn't there then an additional force on the charges since they are moving perpendicular to the field? Do we just ignore that?

#### Attached Files:

• ###### circuit.jpeg
File size:
7.4 KB
Views:
186
Last edited: Nov 25, 2003
2. Nov 25, 2003

### Staff: Mentor

Yep.
Seems like it!
Sounds good to me.
The changing flux in the loop induces an EMF in the wire. If the flux doesn't change, the net EMF is zero.
Right!
The accumulating charge on the capacitor creates a voltage (and thus an E field in the wire) that opposes the induced emf. Once the capacitor fully charges, the net forces (along the wire) on the charges in the wire is zero.

You are correct. When the charges move, the magnetic field exerts a force on them. This force acts sideways to the wire. Ignore it.

3. Nov 25, 2003

### gnome

Thanks, Doc Al. I guess I'm not quite as lost as I thought. But I wish I could give these answers with more conviction. I'm finding applications involving Ampere's Law and Faraday's Law to be the most difficult aspects of this course.

Here's another situation that's been bothering me. The text introduces motional emf with a straight conductor of length l moving with velocity v through a magnetic field B perpendicular to v. Here, it is easy enough to see that
&Delta;V = Blv

Next we get the situation where the conductor is part of a circuit enclosing an area in which there is a uniform magnetic field, and as the conductor moves the area of the field enclosed by the circuit changes with time, and we get the nearly identical result
&epsilon; = -Blv

Then we are told that it is not necessary for the conductor to move; that we get the same induced emf in a stationary loop if there is an equivalent change in the magnetic flux passing through the loop resulting from changes in the magnetic field itself. Not quite as obvious, but I'm getting pretty comfortable with that, so OK.

But what about the case of a stationary, isolated STRAIGHT conductor (NOT part of a circuit) in a changing magnetic field? I would think that there would be a force exerted on the charges in that conductor, causing a net positive charge at one end and a net negative charge at the other. But how would one calculate that force, or the potential difference that would result?

Or, for that matter, what about a stationary, isolated point charge in a changing magnetic field? If a changing magnetic flux exerts a force on charges in a loop, shouldn't it exert a force on charges not in a loop? These considerations don't seem to be mentioned at all in the text.

Last edited: Nov 25, 2003
4. Nov 26, 2003

### Staff: Mentor

This stuff is subtle.
These are excellent, perceptive questions! Faraday's law talks about the emf induced in a loop. As you know, if the conductor moves in the magnetic field, you can understand how the magnetic field exerts a force on the charges in the conductor to cause an emf. But what if the conductor doesn't move? A magnetic field exerts no force on a non-moving charge.

What's going on here is that Faraday's law introduces new physics! Something about changing magnetic fields also causing emf. Faraday's law expresses what happens (in a special case), but it wasn't until Maxwell that we understood the mechanism of what was going on. What is going on is that a changing magnetic field is creating an electric field. This electric field is what is pushing the charges to create the emf.

So you are absolutely right: you don't need a loop---the induced emf really has nothing to do with a loop! The reason that this isn't mentioned is that the loop case is easy, but the general case is not. You'll need to understand Maxwell's theory... his infamous equations.

I don't think I've ever seen these things mentioned in an introductory text... which is a huge mistake, because it's a natural question. Not mentioning it just causes doubt and confusion and the feeling that you are missing something. (You are!)

5. Nov 26, 2003

### gnome

Thanks, Doc.

Actually, it turns out that it is mentioned in our text. (Serway & Beichner's Physics for Scientists & Engineers, 5th Ed., Section 31.4, in case you're curious.)

But when I read it I didn't really "get it", & so I didn't make the connection between that and what I was asking. Frankly, I still don't quite get it, even after my professor explained it again yesterday.

They say that the electric field generated by the changing magnetic field is nonconservative. Also, they say that (in the case of the loop in a changing magnetic field) the work done in moving a test charge around the loop is qE(2&Pi;r). I guess those two statements are related, in that if the electric field were conservative, NO work would be done in moving something all the way around and back to where it started. Is this correct?

edit: Let me add this question. Is the field tangential because it is nonconservative, or is it nonconservative because it is tangential? Or is it both nonconservative and tangential because of another factor? Or are these two things merely coincidental?

They say that the work done in moving a test charge around the loop is also equal to q&epsilon;. This puzzles me.
I remember that W = &Delta;u = q&Delta;V. But there we were dealing with a change in potential. Now, as I'm writing this it occurs to me that, in the circuits we've been analyzing (Kirchhoff's rules, etc.) &epsilon; is also a change in potential.

But in the theoretical loop there is no resistance, just a "perfect" conductor, so then the potential is the same everyplace around the loop, right? So where's the change in potential, and how does it make sense to have an "&epsilon;", and work being done?

I guess it has to do with the fact that the direction of the induced electric field is tangential and therefore different at each point around the loop, yes?

Well, even if you can't say anything to make this clearer, it's been helpful just reciting it over and over to myself.

Is this another one of those things that we have to accept, "just because"?

Last edited: Nov 26, 2003