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Angular velocity acquired by a non-conducting ring

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zorro
#1
Dec11-10, 01:49 PM
P: 1,394
1. The problem statement, all variables and given/known data
A thin, non-conducting ring of mass m, carrying a charge q, can rotate freely about the axis. At the instant t=0 the ring was at rest and no magnetic field was present. Then suddely a magnetic field B was set perpendicular to the plane. Find the angular velocity acquired by the ring.


3. The attempt at a solution

Before proceeding towards attempt at the solution, I would like to find out why will the ring start rotating?
The ring is non-conducting, so there should not be any induced current. Which essentially means that there is no current flow even if there is an induced electric field. I don't get this part.
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Tusike
#2
Dec11-10, 04:00 PM
P: 130
The magnetic field changed from 0 to B in a small amount of dt time, so the magnetic flux changed. This induced an eletric field where the E-lines are circles. And so these E-lines exerted a force on the ring.

Visual representation of the induced E-lines:
http://maxwell.ucdavis.edu/~electro/faraday/ans4.html

-Tusike
zorro
#3
Dec11-10, 04:08 PM
P: 1,394
3 questions-
1)Will there be any difference in the answer if the question was about a conducting ring. If yes then how and why?
2)Does the induced electric field create another induced magnetic field?
3)If you see the direction of force due to magnetic field, it will be inward towards the centre.Will this force count in providing the centripetal acceleration and hence circular motion? Or is it only the force due to the electric field which causes a sudden rotation.

Tusike
#4
Dec12-10, 02:11 AM
P: 130
Angular velocity acquired by a non-conducting ring

1) If the question was about a conducting ring, then I think instead of the ring starting to rotate, the electrons in it would and current would flow through the ring (supposing it's resistance is 0)

2)Yes, it does, and then that induces another electric field etc... However these aren't really significant so you shouldn't be bothered by them, I'm not even sure even can calculate taking all that into account... At least I can't.

3) Yes, the magnetic field will create a force that points towards the center, and it will help provide the centripetal force.
zorro
#5
Dec12-10, 03:16 AM
P: 1,394
OK.
Here is how I proceed-
The increasing magnetic field produced an induced electric field which causes a tangential force qE on each electron.
I got E = -r/2 dB/dt
So force experienced = =qr/2 dB/dt
Torque experienced = qr2/2 dB/dt

How do I find out the angular velocity now?
Where should I take into consideration the centripetal force provided by the magnetic field ( I mean what is that equation) ?
Tusike
#6
Dec12-10, 04:01 AM
P: 130
You almost got it right.
Be careful, in you equations it's not dB/dt, it's dFlux / dt, and dFlux = dB*A = B*A
So you should get the following:
E = Br / 2 / dt
F = Brq / 2 / dt
and the torque M = Bqr^2 / 2 / dt

After that, you need to write the angular momentum theory (or whatever it's called), which says that dN = M*dt. You should be able to figure out the change in angular momentum, and so the angular velocity. Post what you got and I'll see if it's correct.

-Tusike
Tusike
#7
Dec12-10, 04:05 AM
P: 130
Oh, sorry, you got it right:) You don't need to be careful about anything. Only notice that B increased from 0 to B, sorry in all your equations you can substitute in B for dB, that's all:)
zorro
#8
Dec12-10, 04:14 AM
P: 1,394
The magnetic field increases at a constant rate (sudden magnetic field appears) so dB/dt

E = Br / 2 / dt
F = Brq / 2 / dt
This seems meaningless to me. There is no differential in the numerator. Only dt in the denominator.

EDIT : I did not see post 7
zorro
#9
Dec12-10, 04:16 AM
P: 1,394
"After that, you need to write the angular momentum theory (or whatever it's called), which says that dN = M*dt. You should be able to figure out the change in angular momentum, and so the angular velocity. Post what you got and I'll see if it's correct."

So the magnetic force towards the centre does not cause any change in angular momentum?
Tusike
#10
Dec12-10, 04:16 AM
P: 130
So? why is that a problem? After writing dN = N = M*dt, even dt disappears. You don't have to think of dB and dt being linked in any way. All you is that that dB = B(2) - B(1) = B - 0 = B. Similarly to how dN = N(2) - N(1) = N - 0 = N.
Tusike
#11
Dec12-10, 04:19 AM
P: 130
The magnetic force towards the center is a radial force, and so it doesn't apply any torque tangentially, and has no effect on the angular momentum or velocity. It only has the effect that the inside forces which keep the ring together, need to exact a smaller force:)
hikaru1221
#12
Dec12-10, 04:29 AM
P: 799
Quote Quote by Tusike View Post
1) If the question was about a conducting ring, then I think instead of the ring starting to rotate, the electrons in it would and current would flow through the ring (supposing it's resistance is 0)
Both will actually.

From the Ohm's microscopic law, the current density: [tex]\vec{J}=\sigma\vec{E}[/tex]. We have: [tex]\vec{J}=nq\vec{v}[/tex] where [tex]\vec{v}[/tex] is the velocity of the electrons. So [tex]\vec{v}\propto\vec{E}[/tex]. For instance, E=const then v=const. That means, the electrons undergo not only the force from [tex]\vec{E}[/tex] but also the interaction between them and the conductor. Hence the ring must move, provided that friction is not large enough.

By the way, the fact that [tex]\vec{v}\propto\vec{E}[/tex] is quite a curious fact: while [tex]\vec{E}[/tex] represents force, we have another relation by Newton's 2nd law: [tex]\vec{F}=md\vec{v}/dt[/tex] (not [tex]\vec{v}\propto\vec{F}[/tex] !!!). To know why, search Google for the Drude model

3) Yes, the magnetic field will create a force that points towards the center, and it will help provide the centripetal force.
Oh, then how about the situation in the picture?
Attached Thumbnails
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Tusike
#13
Dec12-10, 04:36 AM
P: 130
@hikaru1221: The interaction between the electrons and the conductor is the cause of it's resistance, no? So if it's a perfect conductor, with zero resistance, should only the current flow?

And yes of course if the magnetic field create a force that didn't point towards the center, it would work against the direction of the centripetal force, and so the inside bonding forces would have to be greater.
hikaru1221
#14
Dec12-10, 06:27 AM
P: 799
@Tusike: I wouldn't dare saying much about super conductor and perfect conductor (the former deals with quantum effect, while the latter is just a theoretical concept). Anyway, say, a model for such perfect conductor is a frictionless tube containing negatively charged balls (electrons). The tube is positively charged, and so, E-field exerts forces on it. However, the tube has very large mass, so the effect is not as recognizable as that on the balls.
For the 2nd point, I would like to point out for Abdul Quadeer that magnetic force is not essentially equal to centripetal force, which might be misunderstood from your statement.
zorro
#15
Dec12-10, 06:54 AM
P: 1,394
Quote Quote by Tusike View Post
So? why is that a problem? After writing dN = N = M*dt, even dt disappears. You don't have to think of dB and dt being linked in any way. All you is that that dB = B(2) - B(1) = B - 0 = B. Similarly to how dN = N(2) - N(1) = N - 0 = N.
You wrote
E = Br / 2 / dt
F = Brq / 2 / dt

The e.m.f. depends on the rate of change of magnetic field in this case. So for a small change in B -> dB within a time interval dt, e.m.f. induced is E. E doesnot depend only on time - there has to be either a change in magnetic field / area enclosed. So writing dB as B is meaningless


Quote Quote by hikaru1221 View Post
By the way, the fact that [tex]\vec{v}\propto\vec{E}[/tex] is quite a curious fact: while [tex]\vec{E}[/tex] represents force, we have another relation by Newton's 2nd law: [tex]\vec{F}=md\vec{v}/dt[/tex] (not [tex]\vec{v}\propto\vec{F}[/tex] !!!). To know why, search Google for the Drude model
Looks interesting

Quote Quote by hikaru1221 View Post
For the 2nd point, I would like to point out for Abdul Quadeer that magnetic force is not essentially equal to centripetal force, which might be misunderstood from your statement.
Why is it not equal?
If you consider a very small part of the conductor, you can see that due to the flow of electrons, there is a magnetic force which acts exactly towards the centre.

Btw is it true that an induced electric field gives rise to another induced magnetic field and the process goes on as Tusike said?
If yes then why do we neglect it?
zorro
#16
Dec12-10, 06:58 AM
P: 1,394
Quote Quote by hikaru1221 View Post
Both will actually.

From the Ohm's microscopic law, the current density: [tex]\vec{J}=\sigma\vec{E}[/tex]. We have: [tex]\vec{J}=nq\vec{v}[/tex] where [tex]\vec{v}[/tex] is the velocity of the electrons. So [tex]\vec{v}\propto\vec{E}[/tex]. For instance, E=const then v=const. That means, the electrons undergo not only the force from [tex]\vec{E}[/tex] but also the interaction between them and the conductor. Hence the ring must move, provided that friction is not large enough.
Do you refer here to conservation of momentum and newton's third law?
If the electrons collide with the ions of the conductor, to conserve the momentum the ring has to move?
hikaru1221
#17
Dec12-10, 07:55 AM
P: 799
Quote Quote by Abdul Quadeer
If the electrons collide with the ions of the conductor, to conserve the momentum the ring has to move?
Interaction. I don't dare saying that is a force or something, since the simple "electron flow" inside a conductor that we know so far is a model, and the real behavior of electrons is way more complicated. However, as science hasn't encountered a one-way interaction (or perhaps science has already; but that shouldn't be this case), if the lattice does something on the electrons, the electrons should do the same on the lattice.
Anyway, we won't recognize easily such movement in practice, even if there is no friction. The mass of the lattice is way too large. If the induced E-field is large enough to initialize a recognizable movement, electron flow would be large enough to make you cry with your melted ring However, that's not the case in your problem, since it's not a conducting ring.

Quote Quote by Abdul Quadeer View Post
Why is it not equal?
If you consider a very small part of the conductor, you can see that due to the flow of electrons, there is a magnetic force which acts exactly towards the centre.
Does pointing towards the center mean it is the only radial force?
Centripetal force = radial component of the TOTAL force.

Btw is it true that an induced electric field gives rise to another induced magnetic field and the process goes on as Tusike said?
If yes then why do we neglect it?
It's most likely so (except for special cases where the induced fields do not change with time), and the result is an electromagnetic wave spreading over the space.
I have been wondering about your question and haven't got an exact answer for this. Anyway, let's give it a shot. As people usually say, varying E-field induces B-field, and varying B-field induces E-field. By saying so, we might understand that changing E-field is a source of B-field, and vice versa.

We have these famous 4 Maxwell's equations (in free space, for simplicity):
[tex]div \vec{E} = \rho / \epsilon_0[/tex]
[tex]div \vec{B} = 0 [/tex]
[tex]curl \vec{E} = - \partial \vec{B} / \partial t[/tex]
[tex]curl \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \partial \vec{E} / \partial t[/tex]
Don't bother the weird math operators (curl, div). The above interpretation (i.e. varying E-field is a source of B-field) can be understood by looking at the 3rd and 4th equations. But what if I rewrite them in the following way?
[tex]div \vec{E} = \rho / \epsilon_0[/tex]
[tex]div \vec{B} = 0 [/tex]
[tex]curl \vec{E} + \partial \vec{B} / \partial t = 0[/tex]
[tex]curl \vec{B} - \mu_0 \epsilon_0 \partial \vec{E} / \partial t = \mu_0 \vec{J} [/tex]
So as you can see, I put everything related to E and B to the left side, and everything related to charge ([tex]\rho[/tex]) and current ([tex]\vec{J}[/tex]) to the right. And I'll interpret them in this way: charge and current are sources of a field, namely electromagnetic field, consisting of 2 components E and B, such that E and B behave accordingly to the 4 equations.

That means, charge and current can be attributed to creating electromagnetic field. They are the true sources of the field. Intuitively, the field should be weakening with distance from the source! And even more (or perhaps less) "intuitively", it should be weakening dramatically: by the law of energy conservation, the magnitude of the field is proportional to (distance to the source of electromagnetic wave)^2.

Back to the problem. The initial B-field must come from somewhere, and that somewhere is some mechanism that contains current. So the farther I go away from this current source, the weaker the field is. I also know that in the vicinity of the induced E-field, there is another B-field. But I already go away from the current source, so this B-field must be much weaker than my initial B-field.

That's just a naive attempt. But I don't dare to go for quantitative explanation
zorro
#18
Dec12-10, 08:17 AM
P: 1,394
Quote Quote by hikaru1221 View Post


Does pointing towards the center mean it is the only radial force?
Centripetal force = radial component of the TOTAL force.
No it does not mean.
But there is no other force which will have any component towards the center. Only two forces are operating - force due to electric field which is tangential and force due to magnetic field which is radial (ignoring forces due to further induced fields, if any)

I wonder what will be the effect of that centripetal force on the movement of electrons.

Last question - Does the ring acquire a constant angular velocity or it keeps on increasing ?

I think it will acquire a constant angular velocity due to the rotational impulse received from the electric field. B appears suddenly but it will be constant after that.


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