Potential difference of a ring rolling in magnetic field

[songoku]

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

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

 

[haruspex]

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?

 

[songoku]

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?

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

 

[Steve4Physics]

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?

 

[haruspex]

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.

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

The lateral motion of each half circle through the field, as @Steve4Physics indicates.

 

[hutchphd]

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.

 

[songoku]

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:

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:

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.

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

The lateral motion of each half circle through the field, as @Steve4Physics indicates.

Sorry I still don't understand this part.

Thanks

 

[Delta2]

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

 

[hutchphd]

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.

 

[Delta2]

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

 

[Steve4Physics]

@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.

 

[songoku]

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.

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.

@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.

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?

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

 

[Delta2]

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.

 

[haruspex]

1) a non-rotating ring with its centre moving at velocity v?

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.

 

[Steve4Physics]

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.

 

[songoku]

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

Also the problem states the radius as $2R$ so the result should be $4BvR$, without the minus sign.

I understand

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?

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?

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

 

[haruspex]

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.

 

[Steve4Physics]

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.

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!)

 

[songoku]

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

 

[rude man]

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.

 

[songoku]

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

 

[rude man]

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

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?

 

[songoku]

Think two batteries of identical emf connected + to + and - to -.
Is there an emf between the + and the - contacts?

Yes

Is there current?

No, because the total emf of the circuit is zero

Thank you very much rude man

 

[rude man]

Yes, just as Mr. Faraday said: $\oint \bf E \cdot \bf {ds} = emf = -d\phi/dt$ !

 

[hutchphd]

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.

 

[Herman Trivilino]

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 $v$.

 

[ergospherical]

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*}

 

[TSny]

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 $d\mathbf{r} = 2Rd\theta \mathbf{e}_{\theta}$?

 

[ergospherical]

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.

 

[TSny]

The radius of the ring seems to be quoted as $2R$, so for internal consistency's sake I overlooked the potential typo.

Ah. Thanks. It's all good.

 

[berkeman]

Thread closed for Moderation...

 

[berkeman]

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.

 

[rude man]

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".

 

[berkeman]

Thread closed again for Moderation...

 

[berkeman]

Sigh, after multiple thread derailments, this thread will remain closed.