Talk:Ampère's force law

A timely article
The addition of this page on Ampere's Force Law deserves warm applause. The lack of it has caused problems in other Wikipages dealing with electromagnetism. The following suggestions might improve it.

1) Change the units for the infinitessimals from amperes to metres (or 'meters', if you favour, sorry favor, the US spelling).

Rationale: The left hand side of the equation for F12 must have units of newtons. The units of mu0 are correctly stated as newtons / (ampere)2. Therefore the dimensions of the subexpression between the equals sign and the integral are also newtons. Consequently the integral part needs to be a dimensionless expression. If the infinitessimals, dI1 and dI2, both had units of length then it's easy to see how they drop out against the R2 on the denominator. If, on the other hand, the infinitessimals really had units of amperes then how is a dimensionless expression reached? All writers agree that the infinitessimals are called 'current elements', so the natural (but incorrect) assumption is that their units should be amperes.

2) Change the symbol for the infinitessimals from I to s.

Rationale: The confusion above is compounded because a lower case letter l (ell) is easy to mistake for an upper case I (eye) - a problem which still afflicts my own page on AFL, and also the BIPM (reference 6). At least one writer of repute, Brebis Bleaney, avoids the problem by using lower case s (ess) for the infinitessimals. It's then much clearer that you are dealing with a distance.

3) Change the definition of AFL from 'The force of attraction or repulsion between two current-carrying wires' to the equation for F12.

Rationale: Encyclopedic definitions require concision; and mathematics is the most concise language we have. This is the stance taken by, for example, David Cheng who states that the F12 equation is Ampere's Law.

The paradigm is surely Coulomb's law, which no one defines as 'The force of attraction or repulsion between two charge carrying spheres'. Definitions of CL always emphasise the key concepts of inverse square falloff and the product q1 q2.

4) Change the definition of AFL to that given by the BIPM: the integrand of the F12 equation.

Rationale: For one thing, the differential version looks less intimidating than the double integral one.

$$ d^2\mathbf{F}_{12} = \frac{\mu_0 I_1 I_2}{4 \pi r^2} \left \lbrace d\mathbf{s}_1 \times \left ( d\mathbf{s}_2 \times 	\hat \mathbf{r}_{12} \right ) \right \rbrace $$

The main reason is to maintain consistency with CL.

Consider that CL relates hypothetical 'point charges'. It doesn't directly correspond to any real-world electrostatics apparatus (even an electron isn't a point). This ought to be reflected in our thinking about magnetostatics. We can't verify AFL directly because an isolated current element is physically impossible. However, we don't much care about that because AFL supplies the fundamental basis on which the force between practical current carrying thick wire circuits is calculated.

Bleaney avoids the phrase "Ampere's Force Law" but he starts with the differential form as, I think, should Wikipedia. BIPM uses the phrase 'the corresponding (to CL) equation for the magnetic force'. Well, I feel that that is what AFL really is: the magnetostatics counterpart to CL. There are higher level discussions in electromagnetism which rely on linkage of CL and AFL. These will be facilitated by treating AFL in this way. True, Ampere may not have used the differential form; but then he didn't give the same double integral that the present page does.

Certainly, both the parallel wires equation and the general integral form should be retained; but as derived cases and not as starting points.

RAClarke (talk) 23:13, 15 March 2008 (UTC)


 * Well I agree with 1 and 2, and just implemented those changes. As for the other suggestions, I'd be fine with it either way, for my part. Go for it, if you want. :-) --Steve (talk) 22:26, 16 March 2008 (UTC)
 * I favor a definition in words followed by the equation because math just turns most people off immediately. Could use dℓ $$d \ell$$ for elementary length to avoid use of dl $$dl$$. Brews ohare (talk) 00:56, 17 March 2008 (UTC)

Hi Brews,

I favor a definition in words followed by the equation because math just turns most people off immediately.

Speaking as an engineer whose grasp of vector calculus is shakier than Galloping Gertie, I have much sympathy with your viewpoint. The CL page indeed uses words first. The challenge, though, is to devise words that mean something more specific than the present vagueness yet are less mind numbing than an end user licence agreement:

Ampere's Force Law states that the electromechanical force exerted upon a conductor of negligible thickness, length and curvature is inversely proportional to the square of the distance separating that conductor from a second, separate, such conductor and is directly proportional to both the product of the current magnitudes and the sine of the angle made by a line joining the conductors.

I admit this is neither watertight nor snappy. Have you anything better?

Usually, a picture is worth a thousand words but the vectors, being necessarily three dimensional in nature, when drawn in 2D (by Bleaney, for example) just confuse. In the very long term I'd like to try something with Java 3D. Does Wikipedia support that?

RAClarke (talk) 13:05, 20 March 2008 (UTC)

Image mistakes
I1 and I2 symbols should switch places. Arrows for B1 and B2 fields can not really be drawn like than, there should be two circles going in the same direction, rather than two arrows pointing in opposite directions.


 * I believe the labels are correct. I rewrote the caption to be more explicit. Do you understand now or is it still unclear? --Steve (talk) 02:25, 3 April 2010 (UTC)

That's ridiculous and incorrect. You should draw magnetic field lines of B1 *around* the wire I1, and magnetic field lines of B2 *around* the wire I2. Understand this:

1. magnetic field lines are circular, you should draw them around each wire as is depicted in Ampere's circuital law, Biot-Savart law or 'Magnetic field around current carrying wire' article.

2. since the current in both wires goes in the same direction so will the field lines of B1 and B2 go in the same direction - not radial, but circular - anticlockwise in that particular case. You should not be solving some superposition, calculating interaction, induction or applying vector calculus in that diagram - just simply represent each entity as it is.

3. there are indeed two vectors pointing up and down, you get those "arrows" automatically in between the wires as soon as you draw first circles around those wires - and you will see there is a point where circles meet and arrows of one field point up and of the other down. The mistake is to draw those arrows going RADIALLY from the wires when they are actually *in-between* the wires.


 * The magnetic field is a vector field: At any given point, the magnetic field is a vector. Vectors are traditionally depicted with a small straight arrow. That's why B1 is written as a straight arrow, not a circle.


 * What you're thinking of is the field line depiction of the magnetic field, which is indeed a circle that wraps around a straight wire. If I were making the diagram I would use field lines to depict the magnetic field because it's more familiar to most people (including you). Yes, the diagram isn't as clear as it could be, but it's not incorrect. Maybe sometime I will edit it to make it clearer and simpler.


 * You should think of the arrow B1 as a tiny piece of a circle that wraps around wire 1. This is the basic relation between vectors and their field lines. The circles you're talking about are field lines that depict the magnetic field everywhere at once, the B1 arrow is a vector that correctly depicts the magnetic field at just one point.


 * The total (superposition) magnetic field due to the two wires together has nothing to do with the figure. B1 is part of the magnetic field of wire 1 alone, and B2 is part of the magnetic field of wire 2 alone. (If I were making the diagram I would have left out B2, to make it a bit clearer and simpler.) If you're interested, here is the total magnetic field due to two parallel wires carrying current in the same direction -- -- and in opposite directions -- . (From this webpage). The wires are shown "head-on", sticking straight out of the computer screen. --Steve (talk) 07:31, 3 April 2010 (UTC)

You only need to show those two arrows (B1 and B2) are actually part of a circle going (originating) around the other wire, so it is obvious what wire is producing what field and so people can associate: I1->B1 and I2->B2. Something like this:.


 * I added my new image. Does that help? --Steve (talk) 21:33, 3 April 2010 (UTC)

I'm happy with that. Very good. Though I imagined it all to be a part of the same picture where this additional drawings would be "half-visible" or kind of "transparent" so to not clutter the original image. But this is surely explains it and is enough to make me stop complain. Thank you. —Preceding unsigned comment added by 203.211.93.86 (talk) 05:18, 4 April 2010 (UTC)

F(12), F(21), F(m), ds1, ds2, and "ds" vs "dl"
F(12) is supposed to act on wire I1 due to field B2, not on wire I2, so I think both diagrams and equation need to exchange these parameters around.

$$ \mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2} \ $$

...it should go like this:

$$ \mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \int_{L_1} \int_{L_2} \frac {d \mathbf{l_1}\ \mathbf{ \times} \ (d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2} \ $$

This formula is not really Ampere's force law but some combination of Biot-Savart law and Lorentz force, so I make my conclusions based on that. And in regards to F(12) and F(21) switching places, I look at electric fields and vector calculus so the force F(12)=q1q2/r^2 is the force you use to calculate acceleration of charge q1, not q2. There should also be some explanation about the meaning of F(m) - is that combined or total force of F(12) and F(21)? -- Btw, would not the original equation of Amperes law be more accurate if it had distance *squared* too, like Biot-Savart law and this other equation?


 * It says in the article "F12 is the total force on circuit 2 exerted by circuit 1". I guess you would prefer the opposite notation. Who cares? F12 is clearly defined right here; no one will be confused because 3 months earlier they learned a different equation where F12 was defined differently. I don't think either F12 or F21 is universally preferred in textbooks or papers, and I don't think either is more intuitive.


 * Ampere's force law is some combination of Biot-Savart law and Lorentz force. Nothing more, nothing less. How do you interpret it?


 * The article says Fm is "the force per unit length one wire exerts upon the other". To me, that's perfectly 100% clear. Could you explain again how that might be ambiguous? What do you mean by "combined or total force"? Do you mean F21+F12? Because by Newton's third law, F21+F12=0 (the forces are equal and opposite). Anyway F21+F12 is not described by that quote, as I read it.


 * The Fm equation is required be r not r^2, you can derive this equation by evaluating the double integral expression. With r^2 the equation would be incorrect. You could also check this experimentally if you wanted to. :-) --Steve (talk) 18:44, 4 April 2010 (UTC)

Who cares?

I see you make a lot of decisions around here, are you regular user or do you have higher authority than me to make changes in Wikipedia articles? How much higher does your opinion rate than mine, why and according to what rules?

F12 is clearly defined right here...

I care, because it is wrong. $$\ \overrightarrow{AB} \ $$ It is a vector, remember? LEFT = start, RIGHT = end, ok? -- By your logic $$F_{\overrightarrow{12}}$$ means 'vector acting on, or starting from wire 2 and pointing towards wire 1', while it clearly should be the other way around. Note the little arrow.

http://en.wikipedia.org/wiki/Vector_(geometry)

How do you interpret it?

1.) $$F_m \ \ \ != \ \ F_{12} $$

2.) $$2 k_A \frac {I_1 I_2 } {r} \ \ \ != \ \ \ \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2} $$

Ampere's force law != Biot-Savart & Lorentz force. Notice that Biot-Savart law has distance squared just like this equation given by the BIPM SI Units brochure, 8th Edition, PAGE 105. But I see you, for some reason, believe that somehow the distance SQUARED can magically become just 'distance' when this equation is solved, applied or "integrated", right?

The article says Fm is "the force per unit length one wire exerts upon the other"

And that is incorrect, see BIPM definition, Fm must be the force *BETWEEN* the TWO wires.


 * $$F_{12} =$$(force acting on, or starting at wire 1, points towards wire 2) = 10^-7 N/m
 * $$F_{21} =$$(force acting on, or starting at wire 2, points towards wire 1) = 10^-7 N/m


 * $$F_m = F_{12} + F_{21} = 2 * 10^-7 N/m $$

Could you explain again how that might be ambiguous?

What does the force BETWEEN TWO wires mean? That certainly does not sound like "total force on circuit 2 exerted by circuit 1", does it? -- You think just by solving the force for one wire will give you the correct result for the force BETWEEN the TWO parallel wires as defined by the definition of ampere unit, don't you? It will not, it will give you only half and you need both F12 and F21 to correctly express the force BETWEEN these TWO wires.

Do you mean F21+F12? Because by Newton's third law, F21+F12=0

Yes, I mean F12+F21. No, you are wrong to combine those vectors like that - they do not act on the same object. Think about how would you measure the force BETWEEN the TWO objects.... SPRING, put a spring between them, that's how, and then you will see those vectors add.

With r^2 the equation would be incorrect. Ay, caramba! Ampere's force law has 'r', but THE ACTUAL DEFINITION given by the BIPM has r^2. Are you saying the old 'Ampere's force law' is more accurate than this GENERAL equation given by BIPM, that has "r^2"?

You could also check this experimentally if you wanted to. :-)

I did, which means that you did not, or your instruments need re-calibration. Are you referring to some empirical look-up table where I can see some experimentally established values for B field around different conductors in relation to distance, or something like that? -- Have you ever performed two parallel wires experiment and measured the force BETWEEN the TWO wires?


 * When I said "who cares" I didn't mean to insult you. I just meant the obvious fact that, as we all know philosophically, using a letter or symbol to represent a physical quantity is inherently arbitrary, so it doesn't matter unless one symbol is much clearer or more common than another, or if they're not defined and explained. Of course if you really find 21 much easier to understand than 12 feel free to change it. In my experience, when I see F12 I don't know what it means and look for the definition, because I know that different physicists define this symbol in both ways. What would really be best is to look at a few different textbooks and lecture notes. If 80% of them define F12 the same way, then wikipedia absolutely should use the same definition. I haven't done this "survey" and I don't know what the result would be. My guess is it would be 50-50, but it's just a guess.


 * The force between two long parallel wires goes as 1/r, where r is the distance between them. Although you're right that I haven't measured this, many many people have over the past 100 years and they all have found this conclusion, and this is the conclusion required by the laws of electromagnetism as they are understood today. :-) If you actually tried this experiment and found 1/r^2 then I suspect your parallel wires were not long enough...remember they must be much longer than their separation for that equation to hold. I hope you're capable of evaluating the double integral and I encourage you to try. You'll see that r^2 becomes r. If you don't want to try, you can find the math spelled out here or in most electromagnetism textbooks.


 * OK, now I understand where you're coming from regarding "force between" and "total force". For example, a spring pulls together two objects, and has a tension T. Then the force on one object is T, and the force on the other object is T. The "force applied by the spring" is T, and the "attractive force between the two objects" is T. The quantity 2T never ever comes up in any context. But everyone thinks that 2T is relevant, including myself when I first took high school physics, including students that I've taught this to...everyone. It's very intuitive to want to talk about 2T, but still wrong. Therefore let me ask the question: How do you suggest we rephrase this? If the current wording is not clear to you than it is probably not clear to many other readers, and we should change it to something clearer. I hope this paragraph makes clear what I think we should be describing, but an important question is how to describe it clearly and unambiguously and intuitively.


 * To be clear, what I'm talking about includes the BIPM definition of Ampere: For two wires of length L, one meter apart, one wire gets pulled by a force of L*2E-7 newtons, and the other wire gets pulled by a force of L*2E-7 newtons. This is what BIPM means when they say "attractive force", because this is the universal definition of "attractive force" that all physicists always use.


 * It's clear that you're putting great effort to ensure that this article is clear and accurate, so thank you!! I hope you find my comments as educating as I find yours. :-) Best wishes, --Steve (talk) 20:58, 5 April 2010 (UTC)

I have a lot to tell you, but it just happens that I'm already in the middle of the discussion about all this right now, on line, so please feel free to join and we can discuss it here if you like: http://www.scienceforums.net/forum/showthread.php?t=50010&page=6

There I'm refuting Maxwell's equations and just got to this point of two parallel wires, I think you'll enjoy it and I welcome you there, and if that is not to your liking, then I'll come back here. -- Btw, I like to see you cheerful, I thought we are supposed to hate each other, but if not, then soon we will, as, in the name of science, people must fight for their ideas, views and opinions, right? Cheers, Ambros. —Preceding unsigned comment added by 203.211.104.137 (talk) 12:40, 6 April 2010 (UTC)

My guess is it would be 50-50, but it's just a guess.

THAT IS MATHEMATICAL STANDARD, it's how VECTORS *are* denoted. It is not up for discussion or matter of presumption, just like writing any other words from right to left would be unnaceptable in English language, and so vector notation is exact standard in the language of MATHEMATICS, where correctness is paramount.

http://en.wikipedia.org/wiki/Vector_(geometry)

- "A Euclidean vector is frequently represented by a line segment with a definite direction, or graphically as an arrow, connecting an initial point A with a terminal point B, and denoted by:

$$\overrightarrow{AB}$$

- "The magnitude of the vector is the distance between the two points and the direction refers to the direction of displacement from A to B."; FROM A - TOWARDS B, form left to right.

All the pages denoting vectors should be updated and use the notation with the little arrow.

I hope you're capable of evaluating the double integral and I encourage you to try.

Hope you need not, as you are in luck. Already you are talking to the master of integration, numerical modeling, dynamical systems, of parallel information and temporal animation processing... discoverer of many discoveries. -- So, first of all, F(m) - 'm' stands for "per meter", or alternatively "per unit length". Correct notation would be: F/1m = 2*Ka*i1*i2/r ..and even more accurate if: F/1m = 2*Ka*i1*i2/r^2

PROBLEM: two parallel wires, ampere unit setup, given: I1=I2= 1A; r= 1m


 * Solution No.1 - BIPM's equation & 'unit length': $$L = \int dl = 1m$$

$$F_{12} = \frac {\mu_0} {4 \pi} \int \int \frac {I_1*d \mathbf{l_1}\ \mathbf{ \times} \ (I_2*d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2} \ = \ \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4\pi*|r|^2} $$

$$=> \frac {\mu_0} {4 \pi} \frac {I_1*L_1 \mathbf{ \times} \ (I_2*L_2 \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2} $$

$$=> \frac {4 \pi *10^-7_{N/A-^2}* 1_A*1_m \times \ (1_A*1_m \ \times 1 )}{4 \pi *1_{m^2}} = \ 10^-7 \ N $$

$$Q = 6.242 *10E18 \ electrons; \ 1C = 1A*1s$$ $$=> 1A = 1C/1s \ (1e*6.242*10E18 \ m/s)$$
 * Solution No.2 - BIPM's equation & 'point charges': $$L = \int dl = infinitesimal$$

$$F_{12} = \frac {\mu_0}{4 \pi} \int \int \frac {I_1*dl_1 \times (I_2*dl_2 \times \hat{r}_{21} )}{|r|^2} = \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4\pi*|r|^2} $$

$$=> \frac {\mu_0*Q_1*v_1 \times (Q_2*v_2 \times \hat{r}_{21} )}{4 \pi*|r|^2} $$

$$=> \frac {4 \pi*10^-7_{N/A^-2}*1_C*1_{m/s} \times (1_C*1_{m/s} \times 1)}{4 \pi*1_{m^2}} $$

$$=> \frac {4 \pi*10^-7_{N/A^-2}*1_A*1_m \times (1_A*1_m)}{4 \pi*1_{m^2}} \ =OR= \ 10^-7_{N/A^-2}*1_{C/s} \times (1_{C/s}) $$

$$=> 10^-7_{N/A^-2}*1_A \times 1_A \ = \ 10^-7 \ N$$

$$Q = 6.242 *10E18 \ electrons; \ 1C = 1A*1s$$ $$=> 1A = 1C/1s \ (1e*6.242*10E18 \ m/s)$$
 * Solution No.3 - Biot-Savart/Lorentz force & 'point charges': $$L = \int dl = infinitesimal$$

$$B= \frac{\mu_0*Q*v \times \hat{r}}{4\pi*|r|^2} $$

$$=> \frac{4\pi*10^-7_{N/A^-2}*1_C*1_{m/s} \times 1}{4\pi*1_{m^2}} $$

$$=> \frac{10^-7_{N/A^-2}*1_A*1_m }{1_{m^2}} \ =OR= \ \frac{10^-7_{N/A^-2}*1_{C/s}}{1_m} $$

$$=> \frac{10^-7_{N/A}}{1_m} \ = \ 10^-7 \ N/A*m \ (Tesla) $$

$$F = Q*v \times B $$

$$=> 1_C*1_{m/s} \times 10^-7_{N/A*m} $$

$$=>1_{C/s}*1_m \times 10^-7_{N/A*m} \ =OR= \ 1_A*1_m \times 10^-7_{N/A*m} \ = \ 10^-7 \ N $$

Let us now test the equations on some REAL WORLD, and some imaginary, examples...

CASE A:                  CASE B:          CASE C:  +                   -     +          - | I2= 1A           |     | I2= 1A   |        I2= 1A |===================|    |==========|     + ==================infinite >> - |                    |                    |           |1m                   |1m                  |1m |                    |                    | |===================|     |==========|     + ==================infinite >> - | I1= 1A           |     | I1= 1A   |        I1= 1A |<--- 9m -->|    |<-- 1m -->| +                  -     +          -


 * Ampere's force law

CASE A: $$F_m = 2 k_A \frac {I_1 I_2 }{r} = 2* \frac{4\pi*10^-7_{N/A^2} *1_{A} *1_{A}}{4\pi*1_m} = 2*10^-7 \ N/m $$

CASE B: $$F_m = 2 k_A \frac {I_1 I_2 }{r} = 2* \frac{4\pi*10^-7_{N/A^2} *1_{A} *1_{A}}{4\pi*1_m} = 2*10^-7 \ N/m $$

CASE C: $$F_m = 2 k_A \frac {I_1 I_2 }{r} = 2* \frac{4\pi*10^-7_{N/A^2} *1_{A} *1_{A}}{4\pi*1_m} = 2*10^-7 \ N/m$$


 * BIPM's magnetic (Lorentz force) equation (ver-1A)

CASE A: $$F_{12} = \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4\pi*|r|^2} = 10^-7_{N/A^2}*1_A* 9 \times (1_A* 9) = 81* 10^-7 \ N $$

CASE B: $$F_{12} = \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4\pi*|r|^2} = 10^-7_{N/A^2}*1_A* 1 \times (1_A* 1) = 10^-7 \ N $$

CASE C: $$F_{12} = \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4\pi*|r|^2} = 10^-7_{N/A^2}*1_A* \infty \times (1_A* \infty) = \infty \ N $$


 * Biot-Savart law and Lorentz force (ver-2B)

CASE A: $$B= \frac{\mu_0*Q*v \times \hat{r}}{4\pi*|r|^2}= \frac{10^-7_{N/A^-2}*1_{C/s}}{1_m} = 10^-7 \ N/A*m$$

$$F = Q*v \times B = 1_{C/s} \times 10^-7_{N/A} \ = \ 10^-7 \ N $$

CASE B: $$B= \frac{\mu_0*Q*v \times \hat{r}}{4\pi*|r|^2}= \frac{10^-7_{N/A^-2}*1_{C/s}}{1_m} = 10^-7 \ N/A*m$$

$$F = Q*v \times B = 1_{C/s} \times 10^-7_{N/A} \ = \ 10^-7 \ N $$

CASE C: $$B= \frac{\mu_0*Q*v \times \hat{r}}{4\pi*|r|^2}= \frac{10^-7_{N/A^-2}*1_{C/s}}{1_m} = 10^-7 \ N/A*m$$

$$F = Q*v \times B = 1_{C/s} \times 10^-7_{N/A} \ = \ 10^-7 \ N $$

If you don't want to try, you can find the math spelled out here..

That article refutes itself in the very beginning with its postulates. QUOTE: Parallel long wires: -"Field lines produced by wire 1 which intersect wire 2 do so at right angles both to wire 2 and --??--> a line joining 1 to 2 normally, and vice versa." -- That is not geometrically, logically or physically possible, it's disappointing. Can you get BIPM to provide explanation so we do not need to argue?

You'll see that r^2 becomes r.

Your imaginations is playing tricks on you. Above you can see what I "see", your turn.

_Ambros. —Preceding unsigned comment added by 203.211.104.137 (talk) 13:24, 8 April 2010 (UTC)


 * It usually takes people 4 to 9 months, many hours a week, to learn just the basics of classical electromagnetism, and even then, they usually end up understanding it very poorly. (I know, I've taught it myself.) And that is with the benefit of face-to-face explanations (at a chalkboard) of confusing points: A form of communication which is at least 10X more efficient than internet message-boards. I know this too: I've tried many times to communicate physics over the internet, with very smart people. It is possible but agonizingly slow, and I don't have time for that these days (unless you happen to live near me to discuss in person). Therefore, I'm sorry but I'm not going to respond to your post above. I recommend you go to a nearby college (if any) and discuss with a physics teacher there, or read the textbook Introduction to Electrodynamics by David Griffiths.


 * I did actually read through some of your previous message-board discussion, and found everyone in the conversation to be almost equally confused. No wonder it didn't go anywhere. However, the textbook Introduction to Electrodynamics is crystal-clear and fun to read too, IMO. If your read it cover-to-cover (as I have many times), you will have all your questions answered, I promise. :-) --Steve (talk) 05:01, 14 April 2010 (UTC)


 * OK OK. I put into the article a proof that the general formula and parallel-wire formula are consistent. I tried to err on the side of excruciating detail. I also did a coordinates-based proof, rather than the more typical one involving trigonometry. The math is a bit harder but the concepts and geometry are clearer. (Plus I don't need to make a diagram! :-) ) I hope you trust that my cross-products and integral over x2 are correct: I actually used Mathematica to double-check both. :-)


 * I guess I hope that seeing this proof will help you see the problems with your various attempted-derivations above. (For example, do you see now why it's wrong to pull r out of the integral?) But if you still don't see it, again I suggest you look to a local college or textbook. :-) --Steve (talk) 06:00, 14 April 2010 (UTC)

PROVIDE REFERENCE, so I can see it.

I guess I hope that seeing this proof will help you see the problems with your various attempted-derivations above.

Are you making fun of yourself?

What DERIVATION?!?! It is NUMERICAL INTEGRATION, as it should be!!

GIVEN NUMERICAL INPUT: two parallel wires: I1=I2= 1A; r= 1m


 * NUMERICAL INTEGRATION No.1 - BIPM's equation & 'unit length': $$L = \int dl = 1m$$

$$F_{12} = \frac {\mu_0} {4 \pi} \int \int \frac {I_1*d \mathbf{l_1}\ \mathbf{ \times} \ (I_2*d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2} \ = \ \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4\pi*|r|^2} $$

$$Step 1. => \frac {\mu_0} {4 \pi} \frac {I_1*L_1 \mathbf{ \times} \ (I_2*L_2 \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2} $$

$$Step 2. => \frac {4 \pi *10^-7_{N/A-^2}* 1_A*1_m \times \ (1_A*1_m \ \times 1 )}{4 \pi *1_{m^2}} = \ 10^-7 \ N $$

And I solved it in THREE DIFFERENT WAYS, hello!!! What part do you not understand? This can not be said any more clear than this, and if you can not find the flaw in my MATHEMATICS, like I did with yours, then it implies it is correct. -- There are two little steps here, so why can you not articulate and point out the "mistake"? Can you not mumble anything that is not written in your book, what say you?


 * YOUR MISTAKES:

(x1,D,0) & (x2,D,0)" < THIS IS HIDEOUSLY WRONG.

$$\hat{\mathbf{r}}_{21} = \frac{1}{\sqrt{(x_1-x_2)^2+D^2}}(x_1-x_2,D,0)$$

Where did "r" go? What is "D" equal to? Why? Explanation, reason, logic?

$$\vec{F}_{12} = \frac {\mu_0 I_1 I_2} {4 \pi} \int_{L_1} \int_{L_2} \frac {(dx_1,0,0)\ \mathbf{ \times} \ ((dx_2,0,0) \ \mathbf{ \times } \ (x_1-x_2,D,0) )} {|(x_1-x_2)^2+D^2|^{3/2}}$$

Is that approximation for LOOPS?! Did you do dimensional analysis? Units are wrong, and why are you changing the original equation at all? DO NOT MODIFY THE ORIGINAL INTEGRAL, just plug in THE NUMBERS and CALCULATE.

$$\vec{F}_{12} = \frac {\mu_0 I_1 I_2} {4 \pi} \int_{L_1} \int_{L_2} dx_1 dx_2 \frac {(0,-D,0)} {|(x_1-x_2)^2+D^2|^{3/2}}$$

Are those indefinite or definite integrals, infinity, minus infinity, zero, Pi, 43? Why did you change the original formula and the definition of the ampere unit?! Those crazy equations of yours DO NOT COMPUTE in Mathematica at all.

THIS IS THE FORMULA: d2F = mu0/4pi*r^3 * i1*dl1 x (i2*dl2 x r) http://www.bipm.org/en/si/si_brochure/chapter1/1-2.html

HERE IN MATHEMATICA SYNTAX: D[10^-7Newtons/1^2 * 1*{l1,0,0} cross 1*{l2,0,0} cross {0,-1,0}, l1,l2]

EXECUTE ONLINE HERE: http://www.wolframalpha.com/input

= 1*10^-7 N

Integrals are numerically integrated, not symbolically approximated. Sheesh!!

YOUR EQUATIONS ARE WRONG, it is not even the same equation it is supposed to be.

No I do not trust you, you do not know math, not even how vectors are denoted, so:

1.) Provide the reference for those false equations and make it available for inspection.

2.) Write down that magnetic force equation in Mathematica, as given by the BIPM and DO NOT MODIFY IT, but let the Mathematica integrate the NUMBERS GIVEN: two parallel wires: I1= 1A, I2= 1A , r= 1m

THE EQUATION: d2F = mu0/4pi*r^3 * i1*dl1 x (i2*dl2 x r)

Just plug in the NUMBERS and CALCULATE, DO NOT MODIFY THE EQUATION.

3.) Write down that Mathematica code you're talking about, post it here so everyone can see it and try it out, and laugh.

Ambros-aba (talk) 11:51, 14 April 2010 (UTC)


 * OK, I'm not much surprised that we can't communicate about math or physics. But surely we can agree about something as simple as fonts. Lower case L looks identical to capital i in the font used on wikipedia. Therefore I used changed the L's to script. But you changed it back! Why?
 * Surely we can agree about something as simple as notation. You went to the effort of editing a diagram so that F12 is a force felt by wire 1. Fine with me. But the equation right next to the diagram says that F12 is a force felt by wire 2. So I changed it to be consistent. But you changed it back! Why?
 * You can tell that I actually read your edits, and decide separately which ones should stay and which I want to change. I don't just revert them all at once, even though it would be easier. Can you please do the same thing for mine?
 * You are using the wrong definitions of r and r12hat. The r in the denominator of the double-integral equation is a function of the integration variables. It is not just constant "1 meter" for two parallel straight wires 1 meter apart. Likewise, r12hat is a function of the integration variables. It is not just a fixed vector that points perpendicularly to the wire. These are functions, and it is wrong to pull a function out of an integral: $$\int_a^b f(x) dx \neq f(x)(b-a)$$. What you call "not modifying the equation" is what I call "plugging in values based on incorrect definitions which make a difficult integral into a different, trivial integral". If you want the correct definitions and how to apply them, please read a textbook or talk to a physicist in person. The physicist can show a derivation, you can point out what's wrong with it, the physicist can respond, you can respond, etc., and all this will happen 10X faster than it would over the internet. :-) --Steve (talk) 18:27, 14 April 2010 (UTC)

I'm not much surprised that we can't communicate about math or physics.

Then start talking physics and mathematics, stop mumbling.

GIVEN NUMERICAL INPUT: two parallel wires: I1=I2= 1A; r= 1m


 * INTEGRATION No.1 - BIPM's equation & 'unit length': $$L = \int dl = 1m$$

$$F_{12} = \frac {\mu_0} {4 \pi} \int \int \frac {I_1*d \mathbf{l_1}\ \mathbf{ \times} \ (I_2*d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2} \ = \ \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4\pi*|r|^2} $$

$$Step 1. => \frac {\mu_0} {4 \pi} \frac {I_1*L_1 \mathbf{ \times} \ (I_2*L_2 \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2} $$

$$Step 2. => \frac {4 \pi *10^-7_{N/A-^2}* 1_A*1_m \times \ (1_A*1_m \ \times 1 )}{4 \pi *1_{m^2}} = \ 10^-7 \ N $$

'''What part do you not understand? '''

...diagram says that F12 is a force felt by wire 2.

F(1-2): force on object 1 (due to object 2)

F(2-1): force on object 2 (due to object 1)

Three objects: A, B, C

F(A-B): force on A (due to B)

F(B-A): force on B (due to A)

F(B-C): force on B (due to C)

F(B-A,C): force on B (due to A and C)

F(C-A,B): force on C (due to A and B)

The first symbol, the left one, represents the object force is acting on. Arrow does not specify the direction of the force, notation is always "F(12)" whether the force on wire 1 is repulsive or attractive, i.e. towards or away from wire 2, it is +/- sign what defines the direction.

The physicist can show a derivation...

Your equations DO NOT COMPUTE at all. That equation given by the BIPM is EXACT, it defines the unit of ampere and with it all the rest of electrics, electronics and anything that has to do with any el. currents. NO DERIVATIONS, NO APPROXIMATIONS - JUST PLUG IN THE NUMBERS AND CALCULATE!!

THIS IS THE FORMULA: http://www.bipm.org/en/si/si_brochure/chapter1/1-2.html
 * d2F = mu0/4pi*r^3 * i1*dl1 x (i2*dl2 x r)

THIS IS GIVEN INPUT:
 * Two parallel wires, i1= I2, i2= 1A , r= 1m

HERE IN MATHEMATICA SYNTAX:
 * D[10^-7Newtons/1^2 * 1*{l1,0,0} cross 1*{l2,0,0} cross {0,-1,0}, l1,l2]

EXECUTE ONLINE HERE:
 * http://www.wolframalpha.com/input

= 1*10^-7 N

'''What part do you not understand? '''

Likewise, r12hat is a function of the integration variables.

http://www.wolframalpha.com/input

...YOUR MATHEMATICS DOES NOT COMPUTE! Who are you trying to fool?

Can you show us how do you write your nonsense in Mathematica syntax, yes/no?

$$\int_{L_1} \int_{L_2} \frac {I_1 d \vec{\ell}_1\ \mathbf{ \times} \ (I_2 d \vec{\ell}_2 \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}$$

$$=NOT \ TRUE=> \int_{L_1} \int_{L_2} \frac {(dx_1,0,0)\ \mathbf{ \times} \ ((dx_2,0,0) \ \mathbf{ \times } \ (x_1-x_2,D,0) )} {|(x_1-x_2)^2+D^2|^{3/2}}$$

$$=NOT \ TRUE=> \int_{L_1} \int_{L_2} dx_1 dx_2 \frac {(0,-D,0)} {|(x_1-x_2)^2+D^2|^{3/2}}$$

Where did "r" go?

What is "D" equal to?

What are integral limits? What is that all about? Explanation, reason, logic? Write it down properly so we can see what will Mathematica say about it. PROVIDE THE REFERENCE SO I CAN SEE IT, MAKE IT AVAILABLE FOR INSPECTION. -- Mathematice says you are not telling the truth - THAT DOES NOT COMPUTE. Where is that Mathematica code you were talking about??! Why don't you show it? Write down your integral properly with their limits, so we know exactly what is it you are integrating here and so we can see what Mathematica says about it, ok?

$$\int_{L_1} \int_{L_2} dx_1 dx_2 \frac {(0,-D,0)} {|(x_1-x_2)^2+D^2|^{3/2}}$$

$$=NOT \ TRUE=> \frac{2}{D}(0,-1,0) \int_{L_1} dx_1$$

Wrong. That DOES NOT COMPUTE. Wrong direction, wrong mathematics, wrong logic, wrong everything. How old are you!? You simply canceled your dream-induced "(x1-x2)" terms without ever even using them, erased the exponent and equated what is not equal, changed the units and arbitrary added non-existent and undefined parameters while violating vector and basic arithmetic operators - you defeated the logic and obliterated the purpose of mathematics, bravo!

...a textbook or talk to a physicist in person.

You are talking to a physicist already, what part do you not understand? You can not grasp simple vector notation standards, you are adding cherries and watermelons, your theory and equations DO NOT COMPUTE in Mathematica, so who the hell are you and why should I be talking to you at all? You are some crazy teenager and self-proclaimed "editor in chief", aren't you? What do you think you're doing, kiddo?

Ambros-aba (talk) 04:07, 15 April 2010 (UTC)

Dimensional error in the Ampere's law.
The first equation in the article (the Ampere's law) is not correct dimensionally. The correct dimension for kA should be [N.A^(-2).m]. Regards, B.S. —Preceding unsigned comment added by 173.58.180.116 (talk) 02:50, 7 October 2010 (UTC) GUYS, the dimensional problem with the first equation in this article is CATASTROPHIC!!! Please, somebody, take a look. B.S. — Preceding unsigned comment added by 76.91.214.254 (talk) 00:26, 1 March 2013 (UTC)

Here is my proposal for changing the article:

1. The first paragraph of the article should be changed as follows, “The best-known and simplest example of Ampère's force law, underlies the definition of the ampere, the SI unit of current. The law states that if apply direct currents on two strait, parallel, infinitely long and infinitely teen conductive wires, then on any segment with length L on any of the wires a force will be generated. If the length L equals a single unit of length, then the law will state,“

2. Then the definition for the Ampere constant should be changed, as follows, KA = m0.L/4p So, the Ampere coefficient will will have correct dimension of [newton.meter/ampere^2] and the Ampere's law will be dimensionally balanced.

Regards, Boris Spasov, bspasov@yahoo.com — Preceding unsigned comment added by 76.91.214.254 (talk) 18:47, 1 March 2013 (UTC)


 * Are y'all talking about $$ F_m = 2 k_A \frac {I_1 I_2 } {r}$$? I don't see an error. Fm is force per length, which has units of Newtons per meter. The right-hand side also has units of Newtons per meter. Right? --Steve (talk) 02:34, 18 October 2013 (UTC)


 * (For reference, I came here from this)
 * Well, I certainly don't think Ampere's Law is dimensionally wrong (being the dimensional basis for the SI definition of all electrical metric units as it is XD' ), but I do agree that the notation is a bit confusing (to me too, at least). I mean of course the magnetic force constant has the right dimensions (just like in the Biot–Savart law), but I myself was confused by the equation here. I'll copy the convention used in the nice vector derivation down there to hopefully make it clearer for everyone :)
 * --RProgrammer (talk) 08:29, 5 April 2015 (UTC)


 * Okay, changed; my only concern is what the m is for. It makes Fm stand for "Force, Magnetic" right?  Not a notation for denoting force-per-unit-length? (in which case I'd say maybe the explicit F/L or F/l or F/ℓ is perhaps clearer at least for new users? ^^' ) --RProgrammer (talk) 10:14, 5 April 2015 (UTC)

Reciprocity of Ampère's force law?
Given that: $$\vec{F}_{12} = \frac {\mu_0} {4 \pi} \int_{L_1} \int_{L_2} \frac {I_1 d \vec{\ell}_1\ \mathbf{ \times} \ (I_2 d \vec{\ell}_2 \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}$$

You would think that: $$\vec{F}_{21} = \frac {\mu_0} {4 \pi} \int_{L_1} \int_{L_2} \frac {I_2 d \vec{\ell}_2\ \mathbf{ \times} \ (I_1 d \vec{\ell}_1 \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {|r|^2}$$

But, knowing that $$\hat{\mathbf{r}}_{12}=-\hat{\mathbf{r}}_{12}$$ (anti-parallel), what if $$I_2 d \vec{\ell}_2$$ is parallel with $$\hat{\mathbf{r}}_{21}$$ while $$I_1 d \vec{\ell}_1$$ is perpendicular to $$\hat{\mathbf{r}}_{21}$$ and therefore also perpendicular to $$\hat{\mathbf{r}}_{12}$$? This would mean that $$d^2\vec{F}_{12} = 0$$ while $$d^2\vec{F}_{21} \neq 0$$ (assuming non-zero magnitudes for all vectors considered). Generally speaking, it is often (but not always) the case that: $$\mathbf{A} \times ( \mathbf{B} \times \mathbf{C} ) \neq -\mathbf{B} \times ( \mathbf{A} \times \mathbf{-C} )$$ What gives?siNkarma86—Expert Sectioneer of Wikipedia undefined 06:50, 28 June 2013 (UTC)

Never mind. I found the answer to my question at (http://en.citizendium.org/wiki/Ampere%27s_equation).siNkarma86—Expert Sectioneer of Wikipedia undefined 08:47, 28 June 2013 (UTC)


 * Actually, reciprocity (Newton's third law) holds for complete circuits, but not for "current elements". For current elements (do not exists in nature) it does not. The symmetric version (can be found in linked site) applies to closed circuits only, and gives the same result as the asymmetric version. For "current elements", the asymmetric version is adopted for two reasons. It does generalize to measurable forces between moving point particles. The asymmetric version allows for introduction of the magnetic induction $B$, while the symmetric version does not. It is impossible to separate the dot product involved (there would be a $cos θ$ that can't be pulled out of an integration over the source). YohanN7 (talk) 09:27, 2 October 2015 (UTC)

General case of straight-line wires without infinite-wire approximation is closed-form and not too hairy! :D
By the way, if yall think it would be a good addition, I just took the unapproximated vector integral with both wires being of finite length and in arbitrary offsets--and it's closed form!! :D And also you can easily show with limits that, as one wire's endpoint goes to +∞ and then as its other endpoint goes to -∞ it becomes the normal one-wire-infinite approximated law given here! (And there's even (open-source, free) Maxima code so people can run it themselves if they wanted!) :D

(like, ahem, a certain combative person who's taken up the majority of this talk page XD )

It's up to yall if you think it would be a good/significant-enough addition to add :)

--RProgrammer (talk) 02:48, 6 April 2015 (UTC)

Symmetric Ampere's law equation is wrong
The equation under "By expanding the vector triple product and applying Stokes' theorem, the law can be rewritten in the following equivalent way:[8]" is wrong. The equation in reference [8] is correct, but it is not the same equation as given in this Wikipedia article. The version given in this Wikipedia article is wrong: it gives zero force between a current element that is perpendicular to another current element, when it should be experiencing a Lorentz-like force.

force between two particles
I deleted the following text here:


 * A second example, notable both because it relates directly to the Lorentz Force Law and because it closely parallels Coulomb's Law, is the force acting between two point charges, each moving at a time-varying (accelerating) velocity with respect to a fixed inertial reference frame. In this case, Ampere's Force Law gives the force as:
 * $$ \textbf{F}_{21} = { \mu_0 \over 4\pi } { q_1 q_2 \over r^2 } \textbf{v}_2 \times ( \textbf{v}_1 \times \hat{\textbf{r}}_{21}  ) $$
 * (etc...)

The reasons are: (1) This is not a "special case" of the normal Ampere's force law, because the normal Ampere's force law is an exact law for charge-neutral steady currents (magnetostatics) whereas this is an approximate law for free charges in complicated motion; (2) It can't be correct as written, because it predicts no force when v1 or v2 is zero, and also it seems to be inconsistent with the formula for force at Darwin Lagrangian, unless I'm misunderstanding something. Is there a reference for this expression? --Steve (talk) 03:01, 21 January 2018 (UTC)

Initial response
First, here is a reference (admittedly not academic or necessarily authoritative, but that does not mean that it is incorrect):


 * Physics Stack Exchange

Here is another reference (much more authoritative):


 * Purdue Universisty, Department of Physics

I will try to find additional references that show this form of the equation.

Second, the observation that this form of the equation predicts that the magnetic force will be zero if one or the other of the velocities is zero is fully consistent with Maxwell's Equations and the Lorentz Force Law. If one particle is moving with respect to a particular inertial frame of reference, it will create a magnetic field in the region of space near the second particle (Biot-Savart Law or Ampère's Circuit Law). But, if the second particle is stationary in the same frame of reference, then it will not experience a magnetic force even though there is a magnetic field (Lorentz Force Law). Of course, there would still be an electrical force, following Coulomb's Law, between the two particles.

Furthermore, the implication that the magnetic force depends on the frame of reference is not a concern, because that is exactly what Einstein concluded in his formulation of Special Relativity in 1905. And this result should come as no surprise, because Maxwell's Equations are, and always were, in fact Lorentz Invariant, meaning they do not change under Lorentz Transformation, whereas the electric and magnetic fields are Lorentz Covariant, meaning that they do change under Lorentz Transformation, covariant with the coordinates.

The bottom line is that the magnetic and electrical forces individually do depend on the frame of reference, according to Einstein's Special Relativity, whereas the total force, the vector sum of the electrical and magnetic forces, is Lorentz Invariant, and therefore does not depend on the frame of reference.

Third, it is very possible to derive this form of the equation using Ampere's Circuit Law and the Lorentz Force Law. If I have some time, I will add a derivation to this talk page soon.

Fourth, as for this form of the equation representing a "special case" of Ampere's Force Law, I believe that is in fact a valid assertion. An electrically charged point particle moving through space at a non-zero velocity is physically and mathematically identical to an infinitesimal element of electrical current:


 * $$ q \cdot \mathbf{v} = i \cdot \mathbf{dr} $$

Finally, perhaps this example would be better presented as a special case of the Biot-Savart Law article instead of this one.

First Harmonic (talk) 01:33, 19 August 2019 (UTC)

Video incorrect
The video clip on this page demonstrates Lorentz force (force by a magnetic field on a current carrying conductor) and NOT Amperes force law (mutual attraction/repulsion between two current carrying conductors). Please remove accordingly. — Preceding unsigned comment added by 12.168.224.50 (talk) 18:54, 19 April 2018 (UTC)


 * No, it is correct. Please read the video caption. Ampere's force law is in part derived from the Lorentz force law.--Jasper Deng (talk) 18:57, 19 April 2018 (UTC)


 * I agree with 12.168.224.50. The video should be deleted. It is not a "demonstration of Ampère's force law" as stated in the caption. Ampere's force law is the Lorentz force law in the special case that the magnetic field is produced by one current-carrying wire and the force is experienced by a second current-carrying wire, unless you think the entire article is wrong. The video shows the situation where the magnetic field is produced by a bar magnet. Now, one could change the caption to say "demonstration of the Lorentz force law (which is part of the foundation of Ampere's force law)". That would be an improvement. But it is slightly off-topic, and I think better deleted altogether. --Steve (talk) 12:18, 20 April 2018 (UTC)

Vacuum permeability no longer define as 4 pi
In the "Special case: Two straight parallel wires" section, μ0 should no longer be defined as 4π since 20th May 2019 when the SI base units were redefined. In addition, the units should be adapted to H/m (though the units are still technically equivalent). — Preceding unsigned comment added by JackHill3103 (talk • contribs) 14:29, 6 February 2020 (UTC)