Can the Newton's third law be violated in wires at right angles?

[hikaru1221]

Your analysis requires that you go to the edge of one of the straight wires,the shape and orientation of the field at this place being central to the analysis you are making.

I don't get it. I consider the whole straight wire, not only a small part at the edge. So even if at the edge, the B field is a bit different (not circular), but at most points on the straight wire, it is still circular, so this makes little difference.

Hikaru when Biot and Savarts law is used to find the value of B around a straight wire the integration is carried out from one end of the wire to the other.In this analysis we are located exactly at one end of a wire where there is a sudden change of direction and any integration needs to be carried out over the whole circuit.

I don't get it neither. Do you mean that when calculating B, we MUST ALWAYS compute the integral over the whole complete circuit? And what do you mean by "we are located exactly at one end of a wire"?

First, look at my second picture. The B due to connecting wires is zero because of their shape, and the power supplies are located very far from the considered points. So even if I do the integration over the whole circuit, the result will turn out to be the same as when I only consider the straight section.

Second, I calculate B in order to calculate force, this is my main purpose. In the example about 2 parallel long straight wires, to calculate force between them, we calculate B due to ONE wire and then use the formula F=BIL to calculate force on the other wire. In other words, calculating B due to everything in my problem leads to nowhere, because I cannot calculate the force between the straight sections from that result!

How would you calculate the force that one straight section exerts on the other straight section? I think I will comprehend your point if you answer this question. Thank you very much.

 

[Dadface]

Hello hikaru,
in your opening post you marked in the force direction on the "straight wires".Before we take this any further can I ask you to look at the circuit you drew(post 25) and mark in the force direction on all four sides of each circuit.Having done this do you still think the third law is violated?

 

[hikaru1221]

Hello hikaru,
in your opening post you marked in the force direction on the "straight wires".Before we take this any further can I ask you to look at the circuit you drew(post 25) and mark in the force direction on all four sides of each circuit.Having done this do you still think the third law is violated?

Consider the left circuit.
The upper side consists of many parallel sections, or many couples of parallel sections which carries currents in opposite direction. Because in each couple, the sections are close to each other and so the B fields at the sections are approximately the same. Thus, the total force on the side is zero. And the same thing for the opposite side.
For the connecting wire connected to the power supply, I'm stumped, because the B field due to the power supply is unpredictable. I can't conclude anything here.

Okay, back to a point that I want to make clear. Again, I cite some lines I wrote before:
"In a set of 3 things A, B and C:
_ Force on A = force of B on A + force of C on A
_ Force on B = force of A on B + force of C on B
_ Newton's law: (force of B on A) = - (force of A on B)
The forces due to the whole things are (force on A) and (force on B), but the forces between A and B are (force of B on A) and (force of A on B)."
I don't consider the forces that the 2 circuits exert on each other. I only care about the forces that the straight sections exert on each other.

 

[Dadface]

Hello,I will take your second point first.
I can understand your reasoning if,for example, A,B and C were masses and we were calculating gravitational attractions.In this case there would be a force between A and B even if C weren't there.I think your reasoning works for other types of interaction as well but I can't yet see how it works for the system we are discussing.Suppose A and B are defined to be the straight sections of wire on the left hand and right hand circuits respectively and let C be the rest of the right hand circuit.We can describe the force on A or on a section of A but we cannot describe how much of this is due to B only.Unlike the gravitational or other cases we cannot ignore C because without C, B would not have a current through it and would not set up a magnetic field.The force in A is due to the field from the whole of the opposite circuit.
As for your first point just apply the left hand rule to all four sides of each circuit just as you did in your opening post where you applied it to one side only.The fields set up by each circuit will bear some similarities to the field shape set up by a circular coil the most important feature being that the field lines cut the plane containing the circuits at 90 degrees.
hikaru,I hope this is getting you thinking as much as it is me

 

[hikaru1221]

Hello,I will take your second point first.
I can understand your reasoning if,for example, A,B and C were masses and we were calculating gravitational attractions.In this case there would be a force between A and B even if C weren't there.I think your reasoning works for other types of interaction as well but I can't yet see how it works for the system we are discussing.Suppose A and B are defined to be the straight sections of wire on the left hand and right hand circuits respectively and let C be the rest of the right hand circuit.We can describe the force on A or on a section of A but we cannot describe how much of this is due to B only.Unlike the gravitational or other cases we cannot ignore C because without C, B would not have a current through it and would not set up a magnetic field.The force in A is due to the field from the whole of the opposite circuit.

As for your first point just apply the left hand rule to all four sides of each circuit just as you did in your opening post where you applied it to one side only.The fields set up by each circuit will bear some similarities to the field shape set up by a circular coil the most important feature being that the field lines cut the plane containing the circuits at 90 degrees.
hikaru,I hope this is getting you thinking as much as it is me

What you mean here is we are unable to find a force on A due to B only; only the total force on A can be found, right? But how about the rest of the circuit containing A? I think we should include it in C.

Okay, before digging deeper, I want to ask you some more:

1. You say we cannot find the force that A exerts on B only. Is this due to:
- the limitation of our ability to perceive (?!)
- or because A and B cannot exist without C,
- or the force doesn't exist?

No comment for the 1st one. For the 3rd one, if the force due to each part doesn't exist, then the force due to everything won't exist. And for the 2nd one, it is obvious that no C, no A and B, no force; but the situation here is that A, B and C exist at the same time, so A and C should exert forces on B together. If we can find the total force A and C exert on B, that means we can find the force A exerts on B, because A is current and C is also current, they share the same nature (the magnetic field is due to current, not circuit, because even if there is one moving electron, magnetic field exists; so I think only the characteristics of current matters here. Besides A can no way perceive that it is permitted to exert force on B because there is C; A should act independently from C).

The only reasonable explanation I found is that the force on B is due to neither A nor C; it is due to the magnetic field - another "thing" besides A, B and C, which leads me to the questions in post #19.

2. This idea is quite silly, but just to expand the problem. Imagine that I can make a machine which can shoot "FAST electron beam", and the machine has a structure which prevents the EM field generated by the machine to the outside. Now back to the 1st picture. Instead of 2 wires, I place 2 machines shooting beams which are perpendicular to each other and don't intersect. No circuit, but there are currents. The force's nature changes, but we can find the forces between the beams.

3. I don't get your point about the forces in the 2nd picture in post #25. The circuits in this picture are special, as I have already pointed out in some posts. The most important thing in this picture is that at the points near the red straight sections, total B field = B field due to the sections only.

Arguing with me is quite tiring, huh?

 

[Dadface]

hikaru,I think all your questions will be most easily answered if you carry out the little exercise I described in post 32.You did it for your straight wires in the opening post so you should be able to manage it for the rest of the circuit.Consider your left hand circuit and the resultant field this sets up.The field lines make closed loops and the most relevant feature of the pattern is that field lines cut the plane of both circuits by 90 degrees.If you can't see it use any suitable text to look at the field set up by a narrow circular coil.For a square,rectangular or other shaped circuit the field pattern will be different but the 90 degree angle still applies.Now look at this field passing through your right hand circuit and work out the direction of the force on each side in turn.You will see that if the force direction on the top wire is upwards(as in your op) then the force direction on the bottom wire is downwards,the force direction on the left vertical wire is to the left and the force direction on the right vertical wire is to the right(sorry to be long winded but I don't know how to send diagrams).The force on each side weakens with distance from the closest edge and from this you will see that the resultant force on the right hand circuit will be pointing up at an angle to the horizontal and towards the left hand circuit.By extending your analysis you will see that the resultant force on the left hand circuit points towards the right hand circuit.
Initially I found the analysis here to be quite difficult but when I got a mental picture of the resultant field pattern due to each circuit it became much easier.

If course this doesn't prove that the forces are exactly opposite in direction and nor does it prove that the forces are equal in magnitude.

 

[hikaru1221]

Have you looked at my circuit at post #25?

 

[ehild]

I think you will find this interesting.

http://www.df.lth.se/~snorkelf/Longitudinal/Slutdok.html

ehild

 

[Dadface]

Have you looked at my circuit at post #25?

I have looked .It was the circuits you drew in post 25 that I was referring to in post 36.Try out the exercise and hopefully you will see what I am getting at.

 

[hikaru1221]

@ehild: Thank you very much. I haven't finished reading yet but it's surprisingly interesting.
@Dadface:
This is what I see in my circuit:
_ Consider the left circuit. In the upper side, each pair of wires has 2 currents going in opposite directions, so the element dB due to the pair is zero and the external force on the pair is zero. The same for the opposite side.
_ The left side which connects the power supply is very far, so we can consider this isolated from the external magnetic field. This side experiences no external force and don't contribute to the B field at the space near the red side.
_ This leaves alone the two red sections, and so, the external force that each circuit experiences is the same as the force that each straight wire experiences and perpendicular to the straight wire.
Did I miss your point?

 

[Dadface]

This is what I see in my circuit:
_ Consider the left circuit. In the upper side, each pair of wires has 2 currents going in opposite directions, so the element dB due to the pair is zero and the external force on the pair is zero. The same for the opposite side.
_ The left side which connects the power supply is very far, so we can consider this isolated from the external magnetic field. This side experiences no external force and don't contribute to the B field at the space near the red side.
_ This leaves alone the two red sections, and so, the external force that each circuit experiences is the same as the force that each straight wire experiences and perpendicular to the straight wire.
Did I miss your point?

I think I'm missing your point.When you refer to the upper side and each pair of wires what pair are you referring to? I am guessing that you are referring to the two vertical sections of wire and if so it is the same current going in opposite direction.The field doesn't cancel out and somewhat resembles the field you would get around a very short solenoid.Its true that the more distant parts of the circuit have a smaller effect but the "connecting wires"at the near edges of the circuits are just as close as the "red wires".

 

[hikaru1221]

Short solenoid? You mean it's like "flattened solenoid"?
The field doesn't cancel out, it simply nearly equals to zero. When it's too small, why counts it?

 

[Dadface]

I think you will find this interesting.

http://www.df.lth.se/~snorkelf/Longitudinal/Slutdok.html

ehild

Thank you ehild.I scanned it and my first impression is that there is some interesting stuff but I would need to invest a lot of time to understand it properly.

 

[Dadface]

Short solenoid? You mean it's like "flattened solenoid"?
The field doesn't cancel out, it simply nearly equals to zero. When it's too small, why counts it?

I counted it because you counted it.Your circuits can be considered as flattened solenoids of square or rectangular cross section and consisting of one turn only.B is small but it's not zero.

 

[hikaru1221]

Oh so you mean the circuit is a short flattened retangular solenoid with one turn. I thought you were saying about the upper side's wire in post #41.
Yes, I agree with that, we can consider it so. However as I pointed out, the B field around the red section is mostly due to the red section only. And about the pair, yes, I referred to the 2 vertical sections carrying opposite currents (or the same current going in opposite directions). And the B field due to the pair is nearly zero. That's what I meant in post #42.

 

[Dadface]

Try the exercise hikaru.

 

[hikaru1221]

I think this is your exercise, right?

hikaru,I think all your questions will be most easily answered if you carry out the little exercise I described in post 32.You did it for your straight wires in the opening post so you should be able to manage it for the rest of the circuit.Consider your left hand circuit and the resultant field this sets up.The field lines make closed loops and the most relevant feature of the pattern is that field lines cut the plane of both circuits by 90 degrees.If you can't see it use any suitable text to look at the field set up by a narrow circular coil.For a square,rectangular or other shaped circuit the field pattern will be different but the 90 degree angle still applies.Now look at this field passing through your right hand circuit and work out the direction of the force on each side in turn.You will see that if the force direction on the top wire is upwards(as in your op) then the force direction on the bottom wire is downwards,the force direction on the left vertical wire is to the left and the force direction on the right vertical wire is to the right(sorry to be long winded but I don't know how to send diagrams).The force on each side weakens with distance from the closest edge and from this you will see that the resultant force on the right hand circuit will be pointing up at an angle to the horizontal and towards the left hand circuit.By extending your analysis you will see that the resultant force on the left hand circuit points towards the right hand circuit.
Initially I found the analysis here to be quite difficult but when I got a mental picture of the resultant field pattern due to each circuit it became much easier.

If course this doesn't prove that the forces are exactly opposite in direction and nor does it prove that the forces are equal in magnitude.

I agree that at all points in the plane of the circuits, the B field is perpendicular to the plane, there is no doubt about this.

Consider the right circuit:
_ The force on the red wire (the top one) is upward (1)
_ The force on the bottom wire is downward (2)
_ The left side: If it's straight, the force is obviously to the left. But it's more complicated when in fact, it's not. However I can predict that the force is nearly to the left, i.e. its horizontal component >> its vertical component. And the same for the right side. (3)

However:
_ For (1), it's obvious.
_ For (2), since it's too far from the rest of the circuit, plus that the B field due to the left and right sides is nearly zero as I explained, the force is nearly zero.
_ For (3), the force is nearly zero too as I explained.

 

[Dadface]

You can't see it so try this:
Take two long parallel pieces of straight wire each with connecting wires and a battery. Do you agree that the third law applies to this situation.Next take the middle section of each wire and bend it so that there is a corner with a 90 degree angle.Keep the wires in the same plane with the corners closest and facing.Do you agree that the third law applies to this situation the only difference being that the force per current element gets smaller as you proceed from the middle points to the edges.Now look at the circuits you drew but turn your diagram through 45 degrees.

 

[hikaru1221]

I still can't see your point. Your example is perfectly symmetrical. Mine is not.

 

[Dadface]

I still can't see your point. Your example is perfectly symmetrical. Mine is not.

So shift my example along so that it matches yours.It makes no difference and you can draw the circuits and how they are arranged in a whole variety of ways.It's the principle you should be looking at.

 

[hikaru1221]

So shift my example along so that it matches yours.It makes no difference and you can draw the circuits and how they are arranged in a whole variety of ways.It's the principle you should be looking at.

I still don't think they're the same. Anyway, from the file ehild gave us, I found that the 3rd law is valid in the case of 2 circuits exerting forces on each other (the Neumann force's formula is symmetrical). Though I haven't figured out how to prove the formula yet, I believe this is what you're trying to point out, am I correct? I guess I missed something in my analysis of the 2nd picture. So in short, here are your points:
_ The 2 circuits exert on each other forces that obey the 3rd law.
_ There is no way to find the element force, i.e. force between the sections on the circuits.

You were right at the first point (assume that the Neumann force's formula is right). Back to the 2nd point, which is the central issue of the problem (anyway I still care about the forces between the straight sections). From post #47:

"Take two long parallel pieces of straight wire each with connecting wires and a battery. Do you agree that the third law applies to this situation."

I found a website which contains a simulation of the experiment: http://www.magnet.fsu.edu/education/tutorials/java/parallelwires/index.html. In the simulation, they connect the straight red sections in one circuit, but I believe if we separate them into 2 circuits, there will be no problem. So if the 3rd law applies to the straight sections only, which means the effect of the rest of the circuit is not taken into account, then we still can calculate the force due to sections only, which violates your 2nd point.

 

[Dadface]

Hikaru ,lets extend this to two circuits X and Y and let them be of any shape,size and orientation.Let x be a current element(small section)of circuit X and let y be a current element of circuit Y.Now let's use Biot and Savarts law to calculate the forces on x and y.Now the value of B at x is due to the whole of circuit Y and similarly the value of B at y is due to the whole of circuit X.Because the whole of each circuit is involved we cannot pin down what the force at x would be due to y alone and what the force at y would be due to x alone.

 

[Physiana]

why would the forces be of different strength? I can't see it.

 

[hikaru1221]

why would the forces be of different strength? I can't see it.

Do you refer to the first picture? The magnitude might be the same if it's symmetrical, but the direction is not.

Hikaru ,lets extend this to two circuits X and Y and let them be of any shape,size and orientation.Let x be a current element(small section)of circuit X and let y be a current element of circuit Y.Now let's use Biot and Savarts law to calculate the forces on x and y.Now the value of B at x is due to the whole of circuit Y and similarly the value of B at y is due to the whole of circuit X.Because the whole of each circuit is involved we cannot pin down what the force at x would be due to y alone and what the force at y would be due to x alone.

For the 1st point, I have done proving it. So I'm totally convinced that the 3rd law still applies to the forces between 2 circuits regardless of their shapes, sizes and orientations.

Back to your reasoning above. I have just read the lecture notes at MIT OpenCourseWare. They say that \vec{F}=q\vec{v}\times\vec{B} is an empirical law so far, and from this, people deduce the force on a current element: \vec{dF}=I\vec{dl}\times\vec{B}.
Therefore, for 2 circuits: \vec{F}_{1-on-2} = \int _{L2} \int _{L1} I_1I_2\frac{\mu_o}{4\pi}\vec{dl_2}\times(\frac{\vec{dl_1}\times\vec{r}}{r^3}).
From that, people usually deduce the force that element 1 exerts on element 2: d\vec{F}_{1-on-2} = I_1I_2\frac{\mu_o}{4\pi}\vec{dl_2}\times(\frac{\vec{dl_1}\times\vec{r}}{r^3})
which is mis-derived. Have I got your point?

 

[Physiana]

why would the forces be of different strength? I can't see it.

Do you refer to the first picture? The magnitude might be the same if it's symmetrical, but the direction is not.

Then I don't understand why Newton's third law would be violated.

 

[hikaru1221]

Then I don't understand why Newton's third law would be violated.

Newton's 3rd law: \vec{F_{12}}=-\vec{F_{21}}
If the forces are perpendicular to each other, the law is violated.

 

[Dadface]

Do you refer to the first picture? The magnitude might be the same if it's symmetrical, but the direction is not.



For the 1st point, I have done proving it. So I'm totally convinced that the 3rd law still applies to the forces between 2 circuits regardless of their shapes, sizes and orientations.

Back to your reasoning above. I have just read the lecture notes at MIT OpenCourseWare. They say that \vec{F}=q\vec{v}\times\vec{B} is an empirical law so far, and from this, people deduce the force on a current element: \vec{dF}=I\vec{dl}\times\vec{B}.
Therefore, for 2 circuits: \vec{F}_{1-on-2} = \int _{L2} \int _{L1} I_1I_2\frac{\mu_o}{4\pi}\vec{dl_2}\times(\frac{\vec{dl_1}\times\vec{r}}{r^3}).
From that, people usually deduce the force that element 1 exerts on element 2: d\vec{F}_{1-on-2} = I_1I_2\frac{\mu_o}{4\pi}\vec{dl_2}\times(\frac{\vec{dl_1}\times\vec{r}}{r^3})
which is mis-derived. Have I got your point?

I'm not sure about the point you are making here.Are you in agreement with me now?

 

[hikaru1221]

I'm not sure about the point you are making here.Are you in agreement with me now?

Yes I just tried to explain your point. Is that correct?

 

[Dadface]

Yes I just tried to explain your point. Is that correct?

Yes.You may know that Biot and Savart found,experimentally, the equation for the force between parallel wires and it is this that that led to their general equation.By carrying out the full integrations we can calculate the total(resultant) force between circuits and this we can measure.We cannot make experimentally verifiable predictions about separate current elements because this would necessitate that we separate them from the rest of the circuits of which they are a part,thereby making them non current elements and because,by definition,the elements are infinitessimally small.A theory is only as good as the predictions it makes.

 

[hikaru1221]

How did they find the equation for the force between parallel wires experimentally? The wires, again, are just parts of the circuits, aren't they?

And if the equation \vec{F}=q\vec{v}\times\vec{B} is fundamental, I think the force is due to the interaction between the moving charge (or current) and the magnefic field, not between currents. If so, whether the forces obey the 3rd law or not, it doesn't matter.