Potential difference of a ring rolling in magnetic field

In summary: I think I see it.The force is not the same on both sides, it is towards the center for the right hand side and away from the center for the left hand side. So the net force is in the upward direction.I think the mistake I made was that I thought the forces on both sides are the same, but they are not. I understand it now.In summary, The potential difference between points A and O in a rotating ring with a stationary center is caused by the lateral motion of each half circle through the magnetic field. This creates a net force in the upward direction, resulting in a non-zero motional emf for the portion of the ring from A to O. This
  • #1
songoku
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Homework Statement
A conducting ring of radius 2R rolls on a smooth horizontal conducting surface as shown in figure. A uniform horizontal magnetic field B is perpendicular to the plane of the ring. Find the potential of A with respect to O
Relevant Equations
Faraday and Lenz Law
1646443720640.png

I don't understand why there is potential difference between point A and O. Is there any change in magnetic flux experienced by the ring? I think the magnetic field passing through the ring's cross sectional area is constant

Thanks
 
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  • #2
songoku said:
I think the magnetic field passing through the ring's cross sectional area is constant
Yes, but doesn’t that only mean there is no current around the ring? It does not prevent a p.d. between A and O.
Maybe consider just one path from A to O. What is happening to that?
 
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  • #3
haruspex said:
Yes, but doesn’t that only mean there is no current around the ring? It does not prevent a p.d. between A and O.
If there is p.d between A and O and there is closed path, shouldn't current flow around the ring?

What causes the p.d. between A and O?

haruspex said:
Maybe consider just one path from A to O. What is happening to that?
Sorry I am not sure. A will rotate to position O and O will rotate to A in half of rotation?

Thanks
 
  • #4
songoku said:
I don't understand why there is potential difference between point A and O. Is there any change in magnetic flux experienced by the ring? I think the magnetic field passing through the ring's cross sectional area is constant
Is the radius really ‘2R’? Seems a silly naming convention, but assume it is correct.

Can you find the emfs (the ##V_{OA}##’s) for:
1) a non-rotating ring with its centre moving at velocity ##v##?
2) a rotating ring with a stationary centre?

Can you use the above results to give ##V_{OA}## for the rolling ring?
 
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  • #5
songoku said:
If there is p.d between A and O and there is closed path, shouldn't current flow around the ring?
No. If I have two wires in parallel joining two points at different potentials I do not expect a current. There is no circuit.
songoku said:
What causes the p.d. between A and O?
The lateral motion of each half circle through the field, as @Steve4Physics indicates.
 
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  • #6
To the OP: All of this assumes the magnetic field in question fills all of the space (that you are interested in). As soon as there is a magnetic edge or if the ring were to wobble out of the page then current would flow in the ring. You are correct to feel a little worried I believe, but the answer is as described.
 
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  • #7
Steve4Physics said:
Is the radius really ‘2R’? Seems a silly naming convention, but assume it is correct.

Can you find the emfs (the ##V_{OA}##’s) for:
1) a non-rotating ring with its centre moving at velocity ##v##?
2) a rotating ring with a stationary centre?

Can you use the above results to give ##V_{OA}## for the rolling ring?
I am sorry I can't answer the questions since I still don't understand why there is emf between A and O.

I know there are 2 ways to produce emf:
(1) from the change of magnetic flux
(2) from the force acting on charge particle

I understand about motional emf:
1646536904959.png


Using Fleming left hand rule, the force acting on electrons in the rod will be downwards so point B will be at higher potential to point A and there is emf between A and B. The formula of the emf is Blv and I also understand how to derive the formula

I also understand about rotating rod in magnetic field:
1646537028749.png


The force acting on electrons in the rod will be towards point Q so point O will be at higher potential and there is emf between O and Q. The formula is ##\frac{1}{2}Bl^2 \omega## and I also understand how to derive that.

hutchphd said:
To the OP: All of this assumes the magnetic field in question fills all of the space (that you are interested in). As soon as there is a magnetic edge or if the ring were to wobble out of the page then current would flow in the ring. You are correct to feel a little worried I believe, but the answer is as described.
I also understand this. If there is part of the ring going out of magnetic field or if the ring rotates with vertical axis of rotation through its center, there will be induced current since there will be rate of change of magnetic flux based on faraday's law.

Now for the question in OP itself, there is no change in magnetic flux so if there is emf in the ring between O and A, I think the possible explanation is due to force acting on electrons in it. But since the ring rotates clockwise, when using fleming left hand rule, I get the force acting on the electrons is always to the center of the ring so the electrons won't be collected at a certain point hence no emf. From all the replies, my reasoning is wrong and I don't know why there would be emf between OA

haruspex said:
The lateral motion of each half circle through the field, as @Steve4Physics indicates.
Sorry I still don't understand this part.

Thanks
 
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  • #8
Yes the EMF due to changing magnetic field (the so called "transformer EMF ")is zero for this problem,

the motional EMF for the whole ring is also zero,

however the motional EMF for the portion of ring from A to O(as we go from to A to O from the left or from the right) is not zero.

I reckon we must use the general formula for motional EMF:
In short you must do the integral $$\int_0^{\pi} (\vec{B}\times\vec{v})\cdot \vec{dl}$$ to find the requested EMF.
  • ##\vec{B}## is everywhere constant around the circumference of the ring
  • ##\vec{v}## is not constant, it varies for the various ##\vec{dl}## of the ring
  • ##dl=Rd\theta## where ##\theta## the polar coordinate, it varies from 0 to ##\pi## for the semi portion of the ring
 
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  • #9
songoku said:
I get the force acting on the electrons is always to the center of the ring
This is true from the ring rotating.
But it is also translating to the right, giving a net vertical force on each half of the ring in the same direction (up) . As one rounds the ring this averages to zero but going up either side gives the same nonzero result.
 
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  • #10
songoku said:
I know there are 2 ways to produce emf:
(1) from the change of magnetic flux
(2) from the force acting on charge particle

The change of magnetic flux is the only way, which breaks down to two ways if we do the math :

  1. change of magnetic flux due to changing magnetic field (transformer EMF)
  2. change of magnetic flux due to motion of the boundary (motional EMF)
Check the following link in wikipedia to see how mathematically the two types of EMF arise from the flux integral.
https://en.wikipedia.org/wiki/Faraday's_law_of_induction#Proof
 
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  • #11
@songoku, see if this helps.

In Post #7 you have a rod moving to the right with velocity v. The upper end will become positive and the lower end negative.

This is true for any shape conductor moving to the right. Don’t let the shape/topology of the ring distract/confuse you. It's just a moving conductor.

The top of the ring (point A) becomes positive and the bottom of the ring (point O) becomes negative, just like the rod.

If it helps, consider the ring as consisting of two semi-circles (left and right halves of the ring). Each semi-circle is just a bent rod.

Question: Suppose the top of the ring is +5V and the bottom is -5V. Would current flow clockwise or anticlockwise round the ring?

Answer: The ring is acting as a cell (positive terminal A, negative terminal O) which just happens to have an internal circular construction. There is no current around the ring because the ring is not acting as a circuit here.
 
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  • #12
Delta2 said:
Yes the EMF due to changing magnetic field (the so called "transformer EMF ")is zero for this problem,

the motional EMF for the whole ring is also zero,

however the motional EMF for the portion of ring from A to O(as we go from to A to O from the left or from the right) is not zero.
hutchphd said:
This is true from the ring rotating.
But it is also translating to the right, giving a net vertical force on each half of the ring in the same direction (up) . As one rounds the ring this averages to zero but going up either side gives the same nonzero result.
Steve4Physics said:
@songoku, see if this helps.

In Post #7 you have a rod moving to the right with velocity v. The upper end will become positive and the lower end negative.

This is true for any shape conductor moving to the right. Don’t let the shape/topology of the ring distract/confuse you. It's just a moving conductor.

The top of the ring (point A) becomes positive and the bottom of the ring (point O) becomes negative, just like the rod.

If it helps, consider the ring as consisting of two semi-circles (left and right halves of the ring). Each semi-circle is just a bent rod.

Question: Suppose the top of the ring is +5V and the bottom is -5V. Would current flow clockwise or anticlockwise round the ring?

Answer: The ring is acting as a cell (positive terminal A, negative terminal O) which just happens to have an internal circular construction. There is no current around the ring because the ring is not acting as a circuit here.
I think I understand.

Steve4Physics said:
Can you find the emfs (the ##V_{OA}##’s) for:
1) a non-rotating ring with its centre moving at velocity ##v##?
2) a rotating ring with a stationary centre?
But sorry I am still not able to answer this question. Is integral needed to answer this or maybe there is another way?

Delta2 said:
I reckon we must use the general formula for motional EMF:
In short you must do the integral $$\int_0^{\pi} (\vec{B}\times\vec{v})\cdot \vec{dl}$$ to find the requested EMF.
  • ##\vec{B}## is everywhere constant around the circumference of the ring
  • ##\vec{v}## is not constant, it varies for the various ##\vec{dl}## of the ring
  • ##dl=Rd\theta## where ##\theta## the polar coordinate, it varies from 0 to ##\pi## for the semi portion of the ring
Never learned about this but this is my attempt:

$$\int_0^{\pi} (\vec{B}\times\vec{v})\cdot \vec{dl}$$
$$=Bvr \int_0^{\pi} \cos \left(\frac{\pi}{2} - \theta \right) d\theta$$
$$=-2Bvr$$

Thanks
 
  • #13
This looks almost correct to me, though you have omitted some steps (in the cross product the velocity ##\vec{v}## splits into two components, the constant translational component and the varying linear rotational component ##\omega \times R##, but the whole dot product is zero for the second component-why?)

Also the problem states the radius as ##2R## so the result should be ##4BvR##, without the minus sign.
 
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  • #14
Steve4Physics said:
1) a non-rotating ring with its centre moving at velocity v?

songoku said:
Is integral needed to answer this or maybe there is another way?
Consider just one semicircular part from A to O. You can either do it with an integral or recognise that it is only the distance perpendicular to the motion that is relevant.
 
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  • #15
songoku said:
But sorry I am still not able to answer this question [about translating ring and rotating ring] Is integral needed to answer this or maybe there is another way?
The rigorous approach is to do the maths (integration). But having an intuition provides an alternative approach (at the cost of rigour). Here’s an attempt to give an intuition...

A translating rod (as in Post #7) develops an emf between its ends with ##|\mathscr E| = Blv##. Let’s assume you are OK with this.

For ##B## constant, ##lv## corresponds to the area ‘swept’ by the rod per second (e.g. by the front edge of the rod). ##Blv## is therefore the ‘rate of flux cutting’, i.e. ##\frac {d\Phi}{dt}##.

But the same argument applies for any shape of translating conductor – it doesn’t have to be a rod. It can even have a hole in the middle!

As for a loop purely rotating in a magnetic field (no translation) we can apply a symmetry argument. One diameter is indistinguishable from any other diameter. And each diameter’s orientation relative to the field is identical. From symmetry it follows that no emf exists between opposite ends of any diameter. (If it did, asymmetry would exist.)

For a rolling loop, the total emf is the sum of the translating emf and the rotating emf.
 
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  • #16
Delta2 said:
This looks almost correct to me, though you have omitted some steps (in the cross product the velocity ##\vec{v}## splits into two components, the constant translational component and the varying linear rotational component ##\omega \times R##, but the whole dot product is zero for the second component-why?)
Because they are perpendicular

Delta2 said:
Also the problem states the radius as ##2R## so the result should be ##4BvR##, without the minus sign.
I understand

haruspex said:
Consider just one semicircular part from A to O. You can either do it with an integral or recognise that it is only the distance perpendicular to the motion that is relevant.
How to recognize that only the perpendicular distance to the motion will be relevant?

Steve4Physics said:
The rigorous approach is to do the maths (integration). But having an intuition provides an alternative approach (at the cost of rigour). Here’s an attempt to give an intuition...

A translating rod (as in Post #7) develops an emf between its ends with ##|\mathscr E| = Blv##. Let’s assume you are OK with this.

For ##B## constant, ##lv## corresponds to the area ‘swept’ by the rod per second (e.g. by the front edge of the rod). ##Blv## is therefore the ‘rate of flux cutting’, i.e. ##\frac {d\Phi}{dt}##.

But the same argument applies for any shape of translating conductor – it doesn’t have to be a rod. It can even have a hole in the middle!
For a translating rod, the area 'swept' by the rod will be in the shape of rectangular. For semicircle AO, won't the area 'swept' be in the shape of segment of circle?

Steve4Physics said:
As for a loop purely rotating in a magnetic field (no translation) we can apply a symmetry argument. One diameter is indistinguishable from any other diameter. And each diameter’s orientation relative to the field is identical. From symmetry it follows that no emf exists between opposite ends of any diameter. (If it did, asymmetry would exist.)
If let say I want to find emf between the center of the ring and point A, would the emf be the same as ##Blv## where ##l## is the radius?

Thanks
 
  • #17
songoku said:
For semicircle AO, won't the area 'swept' be in the shape of segment of circle?
No. It will be a rectangle but with the vertical sides replaced by semicircles.
 
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  • #18
songoku said:
For a translating rod, the area 'swept' by the rod will be in the shape of rectangular. For semicircle AO, won't the area 'swept' be in the shape of segment of circle?
No it won't. Try the following exercise.

Consider a straight line with ends A(0,0) and B(0,2) and displace it to the right by 1 unit. The new end positions are A’(1,0) and B’(1, 2). The area ‘swept’ is a 1x2 rectangle.

Now consider a straight line with ends P(0,0) and Q(1,2) and displace it to the right by 1 unit. The new end positions are P’(1,0) and Q’(2, 2). The area ‘swept’ is a parallelogram. You should be able to show the area of the parallelogram is the same as the rectangle's. The line's 'tilt' does not affect the area swept.

When you translate a curve to find the area swept, you can consider the curve as consisting of lots of connected straight line-elements. The total area swept is the sum of the areas of the parallelograms swept by the individual elements. The shape of the curve makes no difference to the total area swept.

songoku said:
If let say I want to find emf between the center of the ring and point A, would the emf be the same as ##Blv## where ##l## is the radius?
Are you referring to pure rotation, so the top of the ring has instantaneous velocity v to the right and the bottom of the ring has instantaneous velocity v to the left? If so...

For a rotating conducting disc, radius ##l##, I think the potential difference (centre to edge) will be ##\frac 1 2 Blv##. Can you see where the ##\frac 1 2## comes from? This setup is called a homopolar generator.

However, your original question is about a rotating ring (presumably of negligible thickness) so I think there would be no potential difference between the centre and ring due to the rotation alone. (I’m sure someone will correct me if I’m wrong!)
 
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  • #19
Steve4Physics said:
For a rotating conducting disc, radius ##l##, I think the potential difference (centre to edge) will be ##\frac 1 2 Blv##. Can you see where the ##\frac 1 2## comes from? This setup is called a homopolar generator.
I think it is the same as deriving formula for rotating rod in magnetic field in post #7

Thank you very much for all the help and explanation haruspex, Steve4Physics, hutchphd, Delta2
 
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  • #20
Post 12 makes an impiortant point: the superposition of rolling and translational motion, one of which generates zero emf, enabling significant solution of the problem.

To answer the question of how there can be an emf without current: what about the right half of the ring? Doen't it generate emf too?

BTW it's pretty certain the radius was meant to be R, not 2R.
 
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  • #21
rude man said:
To answer the question of how there can be an emf without current: what about the right half of the ring? Doen't it generate emf too?
Yes, by the same value as the left half of the ring
 
  • #22
songoku said:
I think I understand.But sorry I am still not able to answer this question. Is integral needed to answer this or maybe there is another way?Never learned about this but this is my attempt:

$$\int_0^{\pi} (\vec{B}\times\vec{v})\cdot \vec{dl}$$
$$=Bvr \int_0^{\pi} \cos \left(\frac{\pi}{2} - \theta \right) d\theta$$
$$=-2Bvr$$

Thanks

songoku said:
Yes, by the same value as the left half of the ring
Think two batteries of identical emf connected + to + and - to -.
Is there an emf between the + and the - contacts?
Is there current?
 
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  • #23
rude man said:
Think two batteries of identical emf connected + to + and - to -.
Is there an emf between the + and the - contacts?
Yes
rude man said:
Is there current?
No, because the total emf of the circuit is zero

Thank you very much rude man
 
  • #24
Yes, just as Mr. Faraday said: ## \oint \bf E \cdot \bf {ds} = emf = -d\phi/dt ## !
 
  • #25
rude man said:
So we decided there is an E field between top & bottom of ring, right? With E=emf/2πR ?

No we did not. Why are you saying thiis? We decided correctly that there is an EMF .
You have certainly confused me.
 
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  • #26
songoku said:
Sorry I am not sure. A will rotate to position O and O will rotate to A in half of rotation?
Assume the half ring OA doesn't rotate. It just slides to the right at some speed ##v##.
 
  • #27
The assumption of a constant and uniform magnetic field means there is a well defined electric potential field ##V(\mathbf{r})##. For an arbitrarily moving conductor, ##\mathbf{E} + \mathbf{v} \times \mathbf{B} = 0## (recall: if ##\mathbf{v} = 0##, then the familiar ##\mathbf{E} = 0## for stationary conductors is recovered). Parameterise one half of the ring, ##OA##, by the angle ##\theta## from the downward vertical at some instant. Write for the velocity of each point on this part of the ring,\begin{align*}
\mathbf{v} &= V(\mathbf{e}_x + \mathbf{e}_{\theta}) \\
&= V(-\sin{\theta} \mathbf{e}_r + (1-\cos{\theta}) \mathbf{e}_{\theta})
\end{align*}where ##\mathbf{e}_r## and ##\mathbf{e}_{\theta}## are the associated polar unit vectors. Put ##\mathbf{B} =B\mathbf{e}_z##, so that\begin{align*}
\mathbf{E} = -\mathbf{v} \times \mathbf{B} = -VB(\sin{\theta} \mathbf{e}_{\theta} + (1-\cos{\theta}) \mathbf{e}_r)
\end{align*}Take the line integral along the arc ##OA## to determine the potential difference, with ##d\mathbf{r} = 2Rd\theta \mathbf{e}_{\theta}##,\begin{align*}
V(A) - V(O) = - \int_O^A \mathbf{E} \cdot d\mathbf{r} = 2RVB\int_0^{\pi} \sin{\theta} d\theta = 4RVB
\end{align*}
 
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  • #28
ergospherical said:
The assumption of a constant and uniform magnetic field means there is a well defined electric potential field ##V(\mathbf{r})##. For an arbitrarily moving conductor, ##\mathbf{E} + \mathbf{v} \times \mathbf{B} = 0## (recall: if ##\mathbf{v} = 0##, then the familiar ##\mathbf{E} = 0## for stationary conductors is recovered). Parameterise one half of the ring, ##OA##, by the angle ##\theta## from the downward vertical at some instant. Write for the velocity of each point on this part of the ring,\begin{align*}
\mathbf{v} &= V(\mathbf{e}_x + \mathbf{e}_{\theta}) \\
&= V(-\sin{\theta} \mathbf{e}_r + (1-\cos{\theta}) \mathbf{e}_{\theta})
\end{align*}where ##\mathbf{e}_r## and ##\mathbf{e}_{\theta}## are the associated polar unit vectors. Put ##\mathbf{B} =B\mathbf{e}_z##, so that\begin{align*}
\mathbf{E} = -\mathbf{v} \times \mathbf{B} = -VB(\sin{\theta} \mathbf{e}_{\theta} + (1-\cos{\theta}) \mathbf{e}_r)
\end{align*}Take the line integral along the arc ##OA## to determine the potential difference, with ##d\mathbf{r} = 2Rd\theta \mathbf{e}_{\theta}##,\begin{align*}
V(A) - V(O) = - \int_O^A \mathbf{E} \cdot d\mathbf{r} = 2RVB\int_0^{\pi} \sin{\theta} d\theta = 4RVB
\end{align*}
This looks very nice! But why the factor of 2 in ##d\mathbf{r} = 2Rd\theta \mathbf{e}_{\theta}##?
 
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  • #29
TSny said:
This looks very nice! But why the factor of 2 in ##d\mathbf{r} = 2Rd\theta \mathbf{e}_{\theta}##?
The radius of the ring seems to be quoted as ##2R##, so for internal consistency's sake I overlooked the potential typo. :oldeyes:
 
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  • #30
ergospherical said:
The radius of the ring seems to be quoted as ##2R##, so for internal consistency's sake I overlooked the potential typo. :oldeyes:
Ah. Thanks. It's all good.
 
  • #31
Thread closed for Moderation...
 
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  • #32
A couple posts were removed by the request of the poster (along with a reply to the posts), and the thread is re-opened for now. Thanks everybody for your patience.
 
  • #33
I deleted my posts (##25 ff.) since my argument was not really relevant. @ergospherical 's explanation is certainly correct but his statement "The assumption of a constant and uniform magnetic field means there is a well defined electric potential field" is IMO not obvious to an introductory physics student. Interestingly, his E field is the irrotational field formed by charge buildup while ## \bf v \times \bf B ## is the induced, and therefore non-conservative component.

Note too that if the observer were to move with the ring then the latter term would be absent, implying in its place a second E field to null the irrotational one. The physics in either case is obviously the same. If this is not accepted then we need to throw out the well-worn dictum that "the net E field in a conductor is zero".
 
  • #34
Thread closed again for Moderation...
 
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  • #35
Sigh, after multiple thread derailments, this thread will remain closed.
 
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1. What is potential difference?

Potential difference, also known as voltage, is the difference in electric potential energy between two points in an electric field. It is measured in volts (V) and is represented by the symbol ΔV.

2. How is potential difference related to a ring rolling in a magnetic field?

When a ring is rolling in a magnetic field, it experiences a force known as the Lorentz force. This force is perpendicular to both the direction of motion and the magnetic field, and it causes the ring to accelerate. The potential difference is related to the acceleration of the ring and the distance it travels within the magnetic field.

3. What factors affect the potential difference of a ring rolling in a magnetic field?

The potential difference of a ring rolling in a magnetic field is affected by the strength of the magnetic field, the speed of the ring, and the distance the ring travels within the magnetic field. It is also affected by the mass and charge of the ring.

4. How is the potential difference of a ring rolling in a magnetic field calculated?

The potential difference of a ring rolling in a magnetic field can be calculated using the formula ΔV = Bvl, where B is the magnetic field strength, v is the speed of the ring, and l is the distance the ring travels within the magnetic field.

5. What is the significance of potential difference in the study of electromagnetism?

Potential difference plays a crucial role in electromagnetism as it is a fundamental concept in understanding the behavior of electric and magnetic fields. It is used to calculate the strength of electric and magnetic fields and to determine the direction and magnitude of electric currents. Without potential difference, many of the principles and applications of electromagnetism would not be possible.

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