ambros

You need to start with this, as you did: [math] \oint_{C_1} \frac{ds_1 \sin \theta}{r^2_{12}}[/math] ...and explain how you got this: [math]=> \frac{\cos \theta_1 - \cos \theta_2}{R}[/math] cos(0) - cos(3.142) = 1 - 0.998 = 0.002 // How did you get cos(3.142) = -1? Your diagram describes 'supplementary angles', but then your statement would be incorrect, as if one of these angles goes to zero the other will go to towards 180, I also do not see how the value of Pi fit into any of that - that's certainly not a part of the equation or given variables, where did those values and conclusions come from? -- Can you make it clear where did original theta angle go and where is each of these three angles on that diagram exactly, describe them by saying where the both of their two "hands" are. Depends how you integrate, it can represent different certain and/or uncertain distances or lengths, but most importantly it represents the direction component of the electron drift velocity vector. I always use it as a unit vector, as I said. Does that answer? Tell me why are you asking so I can address exactly whatever is the point you want to make.

[math] u = \frac{\cos \theta_1 - \cos \theta_2}{R} [/math] There is nothing in post #72 that would explain that step. That does not compute at all. Why is your result now in Newtons and previously was in N/m? How come BIPM force equation gives result in Newtons, and Ampere's force law in N/m?

Hahaa. "Which it does"? You're mumbling again, wrong answer. [math]\Rightarrow F_1 = \frac{\mu_0 I_1 I_2}{2\pi s} \mathbf{\hat{r}} \ \ N/m^2?[/math] What are the units of this, write it down. [math]\mathbf{F}_1 = I_1 \int{d\mathbf{l} \times \mathbf{B}_2} \ \ N[/math] This equation is in Newtons actually, so it obviously does not need to be force per unit length. Can you decide? Which it does?

Why the middle equation has units in Newtons and the next one in N/m?

Yes, I saw it, I'll get to that, but I can say now it very unclear. There is no left term on most of those equations, so we can not analyze units as we do not know whether it is being solved for field or force, or something else. I would like darkenlighten and you show your work too. So, again, can you, yourself, solve that problem and make it more consistent in notation and more clear what is being evaluated so we can do dimensional analysis, can you solve it please? Merged post follows: Consecutive posts merged You did not solve this equation given by the BIPM: http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf [math] \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} [/math] Can you show your work? Merged post follows: Consecutive posts merged Is that force F(12)? The force with which one of these two wires will move towards the other, or in other words - the force exerted on only ONE wire due to magnetic field of the other wire? Where are the left terms in your derivation/integration? You start with F12, then you take out some part and is not known what is all that supposed to represent - field or force, or what units it has. Can you show where are those angles theta1=0 and theta2=3.14, where do you see those, do you have any diagram to explain where did you get all those values and what do they represent? It's very unclear and all over the place, I'll get back to that post and be more specific.

Ahhm, by "everything" I meant everything except that. I'm not a computer, nor Chuck Norris... I can make a slip or two by writing this BOOK that this thread has become, so just instead of asking, please show the numbers right away, that is the most convincing, and highly appreciated.

Good' date=' can you show you work please: r=1m; L1=L2=1m; I1=I2=1A; µ0= 4π*10^-7 -------------------------------------------- |----- C2 -----| | | ========|-----ds2----->|=============infinite wire===>> I2 | | |r ^ | | r_hat | | ========|-----ds1----->|=============infinite wire===>> I1 | | |<---- L ----->| |----- C1 -----| [math']\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} [/math] * The double line integration sums the force upon each element of circuit 2 due to each element of circuit 1 * ds1 and ds2 (dl1 and dl2 alternatively) are infinitesimal vector elements of the paths C1 and C2, measured in metres * The vector [math]\hat{\mathbf{r}}_{12}[/math] is a unit vector along the line connecting the element pair [from s1 to s2] * r is the distance separating these elements http://en.wikipedia.org/wiki/Ampere%27s_force_law If there is any objection to the diagram or definition of terms then feel free to correct it. -- We should at least get our terms straight, and please, less of prose and words, bring on some math and numbers.

Interesting' date=' I would have never realized that, thank you. [math']\mathbf{B} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} = \frac{\mu_0 I*d\mathbf{l}*sin(alpha)}{4\pi r^2} = \frac{\mu_0 I*d\mathbf{l}*sin(90)}{4\pi r^2} = \frac{\mu_0 I*d\mathbf{l}}{4\pi r^2} \ N/A*m [/math] ...which makes me wonder where did that term go from "your" equation? I never noticed you were insisting on anything, especially since I have trouble getting answers to my own questions. -- Anyway, can you now solve the force between two parallel wires and see if the dimensional analysis still hold for your equation? I wonder how will you enter those angles and wire segments length without breaking up your units, can you do that?

I know your opinion, but there is no need to assume - just calculate it. I cited them several times and will do it again. -- But, it is you who instead of expressing opinion how these integrals might work out, should use the math and ACTUALLY CALCULATE IT. We have perfect example to apply this and it will definitively prove or disprove validity of either equation, so please, just solve those integrals with given variables as defined by the two parallel wires and ampere unit scenario. 1.) I say this is correct formula to calculate magnetic field of a straight current carrying wire in relation to distance, and I quote my source: http://en.wikipedia.org/wiki/Biot%E2%80%93Savart_law - "The Biot–Savart law is used to compute the magnetic field generated by a steady current, for example through a wire... The equation in SI units is: [math]\mathbf{B} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} [/math] I also say this "other" equation: [imath]\frac{\mu_0 I}{2\pi R}[/imath] will not produce same results as above equation, and which one is generally correct can only be determined by the reality, and/or our definition of unit system, that is by solving the example of two parallel wires as defined by the ampere unit. Since you do not believe me, you then absolutely must solve that equation given below yourself, especially if you mean to keep saying my integration of it was wrong. --- CONFIRMATION EXPERIMENT --- 2.) In the case of two parallel wires, I say this is correct formula to calculate magnetic force exerted on ONE wire due to the B field of the other wire, and I quote my source (in integral form): http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf [math]F_{12} = \frac {\mu_0} {4 \pi} \oint_{C_1} \oint_{C_2} \frac {I_1*d \mathbf{l_1}\ \mathbf{ \times} \ (I_2*d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] I also say that to calculate the force BETWEEN the two parallel wires, as defined by the unit ampere, we need to take into account BOTH forces, that means: F(m) = F(12)+F(21) = 2 * 10^-7 N/m. -- Both of these claims can be refuted or confirmed with one thing, and all you need to do is to solve that one equation for that particular case scenario given by the BIPM and ampere unit, and then see what result will you actually get. Can you do that please? ...alternatively, you may use "double differential" form as given by the BIPM, or this one with slightly different notation: http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law [math] \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} [/math]

Ok, but there is nothing to integrate really, especially in this example as all those are unit vectors and magnitudes of value one so everything on the right becomes "one", and then all what's left is magnetic constant. Wikipedia is actually very clear as in the very first paragraph it states that this force is defined by Lorentz force and Biot-Savart law. It is really a gentle way of saying that Ampere's force law itself is not good enough. Then, further down the page they give this equation we are now talking about as is defined by BIPM, but if you look at it you will see it is not really Ampere's law, but fusion of proper Biot-Savart law and Lorentz force: [math]\mathbf{B} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} \ \ \&\ \ \ F = I*L \times B\ => \ F_{12} = \frac {\mu_0} {4 \pi} \frac {I_1*L_1 \mathbf{ \times} \ (I_2*L_2 \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] Merged post follows: Consecutive posts merged Yes, and do you realize now we are actually solving the same problem from the very beginning with SINGLE ELECTRON. Yes, we are solving for INFINITESIMAL segments "dl", and that means we are solving for point charges again, only these magnetic fields in wires theoretically do NOT EXIST in front or behind their perpendicular plane, not in any of our equations anyway. In two parallel wires, these "magnetic slices" do not interact with any other slices but only the ones of the other wire whose slices are in that same plane. There are no varying angles and no superposition here.

I checked everything. Now, you have to check it yourself. -- There is no need to be wondering about units, both equations have the same units but there is only one valid relation to correctly describe the magnitude of magnetic field B, of straight current carrying wire, in relation to distance: [math]B® = \frac{\mu_0 I}{2\pi r}[/math] -OR- [math]B® = \frac{\mu_0 I}{4\pi r^2}[/math] ...and to find out which one is correct you have to use THE DEFINITION given by the -BIPM SI Units brochure, 8th Edition, PAGE 105- solve it for the given values that define the unit of ampere, and see for yourself which one can produce that exact result. Merged post follows: Consecutive posts merged

Wrong, you picked up completely different equation. Amere's force law does not define the unit of ampere, but LORENTZ FORCE and BIOT-SAVART LAW, as is clear from the Wikipedia article and from THE DEFINITION given by BIPM SI Units brochure, 8th Edition, PAGE 105": http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf [math]F_{12} = \frac {\mu_0} {4 \pi} \oint_{C_1} \oint_{C_2} \frac {I_1*d \mathbf{l_1}\ \mathbf{ \times} \ (I_2*d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] You have to use the correct definition, try again. -- All the other comments are completely unnecessary, and though amusing I've responded several times to each of those, the truth will be clear to anyone who actually starts using these equations on practical case scenarios, based on physics and real world, not hallucinations and false mathematics. B(r) | | |r | | angle=90 ======================|---dl1--->|====================>> I1 | | |<---L1--->| ...hence' date=' the correct "shortcut" is this: [math']\mathbf{B} = \frac{\mu_0 q \mathbf{v}}{4\pi} \times \frac{\mathbf{\hat r}}{r^2} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} =wires=> \frac{\mu_0 I}{4\pi r^2}[/math] ...why? We must always remember what Cap'n Refsmmat told us: - "magnetic field must be perpendicular to the wire, so sin(alpha) is 1". [math]B®= \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} = \frac{\mu_0 I sin(alpha)}{4\pi r^2} = \frac{\mu_0 I sin(90)}{4\pi r^2} = \frac{\mu_0 I * 1}{4\pi r^2} = \frac{\mu_0 I}{4\pi r^2}[/math] Do not agree - fine, just solve the unit of ampere scenario from above and you should realize what equation is more correct, just please stop making any more assertions before you actually solve those two parallel wires as that is the only thing that is 100% unambiguous here, unfortunately. Merged post follows: Consecutive posts merged What angle? Cap'n Refsmmat: - "magnetic field must be perpendicular to the wire, so sin(alpha) is 1". You are being ignored because your arrogance wasted my time. Do you see how is the force defined by BIPM SI Units brochure, 8th Edition, PAGE 105": http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf ?? Stop talking and start learning. Do you think you are able to use THE definition and solve for two parallel wires and unit of ampere? I'm only interested in physics, not crazy theories, silly insults and hand-waving.

Superposition again? These equations do not do that - this is not numerical integration or time integral, these are 'line integrals'. "MY EQUATION": [math]\mathbf{B} = \frac{\mu_0 I}{4\pi r^2}[/math] "YOUR EQUATION": [math]\mathbf{B} = \frac{\mu_0 I}{2\pi R}[/math] Where do you see anything like what you describe? Did you not say magnetic angle in wire must always be perpendicular, so how can distance change if wires are supposed to be parallel and magnetic fields perpendicular, what is there to change, and why would it change in "your" but not in "my" equation? Please, I showed all the steps and we are talking about mathematics here, so if something does not compute then it does not compute, and you should be able to point exactly what and where. All the variables are visible from the diagram and given by the definition of ampere unit, alternatively explained in that Wikipedia article. -- You have not solved this one yet. We are now talking about the very specific equation, which is THE DEFINITION, this is it, to be or not to be: - "the official documentation regarding definition of the ampere BIPM SI Units brochure, 8th Edition, PAGE 105": http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf ...alternatively: If you mean to claim my integration is wrong, then again, you need to first solve it yourself so we all can see what is the difference, what is wrong and what is right. This is the moment of truth, you really can not get but one result if you do it correctly here, and if you do, then the result will be the same as mine. =====================|----dl1--->|====================>> I1 | | |r | | =====================|----dl2--->|====================>> I2 AMPERE UNIT: r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 ------------------------------------------------------------- [math]F_{12} = \frac {\mu_0} {4 \pi} \oint_{C_1} \oint_{C_2} \frac {I_1*d \mathbf{l_1}\ \mathbf{ \times} \ (I_2*d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] Pick your favorite form of this equation and solve it please.

Yes I did, did you? Do you not see the L is "LINE INTEGRAL", 1m in length? I did the same thing as you and exactly what Swansont wrote: L = integral(dl) 1.) can you please say what exactly is wrong? 2.) please show how do you integrate this equation:

I solved it couple of times by now, there is nothing to integrate as Cap'n Refsmmat demonstrated with his "F/L" term - both of these: "dl x r" are unit vectors, there are no any angles but 90 degree, and dB(r, i=const)=0, so it is linear function. -- It's as if that crazy equation on the right is product of some "proper integration", it is not. It is derivation according to some fictional scenario that has no bearing in reality: Our scenario looks like this: =====================|----dl1--->|====================>> i1 | | |r | | alpha= 90 =====================|----dl2--->|====================>> i2 ...hence, the correct "shortcut" is this: [math] \mathbf{B} = \frac{\mu_0 q \mathbf{v}}{4\pi} \times \frac{\mathbf{\hat r}}{r^2} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} =wires=> \frac{\mu_0 I}{4\pi r^2} [/math] ...why? We must always remember what Cap'n Refsmmat told us: - "magnetic field must be perpendicular to the wire, so sin(alpha) is 1". [math] B®= \frac{\mu_0 I d\mathbf{l} \times \mathbf{r}}{4\pi r^2} = \frac{\mu_0 I sin(alpha)}{4\pi r^2} = \frac{\mu_0 I sin(90)}{4\pi r^2} = \frac{\mu_0 I * 1}{4\pi r^2} = \frac{\mu_0 I}{4\pi r^2} [/math] Well, it's a little bit wrong, so I will use THE EQUATION: - "the official documentation regarding definition of the ampere BIPM SI Units brochure, 8th Edition, PAGE 105": http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf STEP 1: small modifications... [math]F_{12} = \frac {\mu_0} {4 \pi} \oint_{C_1} \oint_{C_2} \frac {I_1*d \mathbf{l_1}\ \mathbf{ \times} \ (I_2*d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] STEP 2: pif, paf, puf... [math]F_{12} = \frac {\mu_0} {4 \pi} \frac {I_1*L_1 \mathbf{ \times} \ (I_2*L_2 \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] -OR- [math]\frac {F_{12}}{L} = \frac {\mu_0} {4 \pi} \frac {I_1 \mathbf{ \times} \ (I_2 \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] STEP 3: solve for r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 [math]F_{12} = \frac {\mu_0} {4 \pi} = 10^{-7} N[/math] -OR- [math]\frac {F_{12}}{L} = \frac {\mu_0} {4 \pi} = 10^{-7} N/m[/math] Conclusion: - to get the correct result for the combined force between the two wires according to the definition of unit ampere, we need to use the multiplier '2', i.e. we need to include this force too: [math]F_{21}[/math]

Uh, ah, there is a whole another discussion there, about "equal" and "opposite" direction of those magnetic forces, but it is not necessary to go there in order to resolve these simple scenarios, fortunately, so I'll skip that. http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law [math]\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}[/math] So, you are saying this equation is wrong after all? EDIT: ...I just realized the internet is actually full of web-pages to support the crazy formula. That's simply insane because in the same time all the pages talking specifically about Biot-Savart law and current carrying wires will agree with me and make clear what is the actual equation for that particular case, such as all the Wikipedia articles on the subject. How can some crazy derivation make those two equal? One equation must be wrong, which one? Wiki: - "The Biot–Savart law is used to compute the magnetic field generated by a steady current, for example through a wire... The equation in SI units is: [math]\mathbf{B} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2}[/math] -not this-> [math]B = \frac{\mu_0 I}{2\pi R}[/math] No, this is not correct. The magnitude of the field at any point is the result of contributions from the length of the wire. A shorter wire will give you a weaker field. This is why the BIPM definition of current references infinite wires. Is the "wire length" to you the same thing as wire segment "dl"? If a shorter wire give weaker field, then infinite wire gives infinite field?

That is not the equation I gave. Look again, and pay attention to "4Pi" and "r^2". The Ampere's force law has nothing to do with your solution using Lorentz force. The formula I gave is Lorentz force and Biot-Savart law combined: [math]\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} [/math] http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law *** TWO PARALLEL WIRES: r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 --------------------------------------------- F(between two wires) = ? [math]\mathbf{F}_{12} = \frac{\mu_0}{4\pi} \frac{1}{r^2} = 10^{-7} N/m[/math] [math]\mathbf{F}_m = \mathbf{F}_{12} + \mathbf{F}_{21} = 2* \mathbf{F}_{12} = 2 * 10^{-7} N/m[/math] That is how it is done. -- If you were to calculate the displacement between these two wires, would you calculate that with just one force or would you include the both forces to calculate the total displacement due to attraction? The length of the wire has absolutely no impact, nor is considered in any of these equations. Length of the wire is not the same thing as segment "dl". -- In most cases these wires are supposed to be infinite anyway, but not because the length matters in any way to the magnitude of the B field in relation to distance, only to avoid dealing with the edges of the magnetic field "cylinder". I gave you completely different equation, plus we were talking about your solution where you calculate B field and force with the Lorentz force. But, in any case, that is very strange way to make that derivation. You are skewing the meanings and the value of the multiplier and constant. [math]F_m = 2* \frac{\mu_0}{4\pi} \frac{1}{1} = 2 *10^{-7}N/m[/math]

No, I'm not talking about Biot-Savart law and imaginary scenarios, I'm talking about these four monsters that you supposedly DERIVED FROM MAXWELL'S EQUATIONS: Have you derived these from Maxwell's equations and will you provide some reference finally? Merged post follows: Consecutive posts merged I like the "F/L" term, but unfortunately that's not complete, just like darkenlighten's solution. Why are you calculating only the force for one wire, do you think that is what is meant by "force between conductors"? -- I do not know what to say, where is that guy 'Bignose' - Maths Expert? You see, even if that was supposed to be the complete result for this scenario, that equation would still contradict the real one, so do you think this equation is wrong: http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law That's right.

That's mathematics. This is physics: -"The ampere is that constant current which' date=' if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to [b']2 * 10^–7 newton per metre of length[/b]." r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 --------------------------------------------- B1= ?; B2 = ?; F(total)= ? If that equation is PHYSICALLY correct you ought to be able to actually use it on this case scenario and predict the correct result of this experiment. Will you please demonstrate whether you can indeed obtain the correct result with your equation?

Are you saying three articles in Wikipedia about Biot-Savart law, magnetic fields and Ampere's force law all gave wrong equation and all of them failed to show that equation of yours? There are no angles in my examples. To see if your equations applies to the very specific scenarios we are talking about you need to actually USE it and see if it will give you correct result. Can you do that? =====================|<---dl1--->|====================>> I1 | | |r | | =====================|<---dl2--->|====================>> I2 *** TWO PARALLEL WIRES: r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 --------------------------------------------- B1= ?; B2 = ?; F(total)= ? There is nothing to argue about if no one is able to actually demonstrate and apply that equation on practical case scenario so to confirm if it gives correct result, or not. Where are the angles, does your equation apply, can you show me? Merged post follows: Consecutive posts merged Considering where this argument brought us, that's a very good question. -- In relation to some constant distance from the wire it will be the same, that's clear from all these equations - magnitude of B field around wire depends only on current "I" and distance (1/r^2 at right angle), but in any case NOT on wire length - there is no such variable in Biot-Savart law equation. The thing is, magnetic field around wire in these equations is ONE field, it is not combination of many fields, but it's one field just like single electron has one electric field, only this magnetic field instead of being a sphere is cylindrical, and instead of originating from a point, its "origin" is a line mathematically constructed with line integral. -- So the 2m wire will have overall more magnetic potential, i.e. more magnetic energy as a whole, but measuring this potential some constant distance from the wire will be the same regardless if the wire is 1, 2 or 100 meters long. You only add up force vectors when they are acting on the same object. Imagine there is a spring between the two wires whose compression will determine the force between the wires. Both forces will add up. Say current I1= 0.5A and I2=1.5A, will their attraction be the same as when they were both 1 Amps? Yes, but you would not know that if you calculated the force from only one wire. Is there any question that I ignored or did not answer directly? It is you who still needs to provide some reference for those equations supposedly derived from Maxwell's, or you could simply say that you were mistaken.

*** TWO PARALLEL WIRES: r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 --------------------------------------------- I SAY THIS IS CORRECT: [imath]B = \frac{\mu_0 I}{4\pi r^2} [/imath] YOU SAY THIS IS CORRECT: [imath]B = \frac{\mu_0 I}{2\pi r} [/imath] F(total)= ? (1m)^2 will have the same numerical value, but 2Pi and 4Pi will not. I hope *you* can demonstrate your assertions, so can you solve this? 2 * 10^-7 N/m is correct result. Here are the numbers and equations, I'm only insisting that before insulting me and instead of waving hands you actually solve the problem yourself, and point exactly what is it you believe is incorrect in my calculation. Can you do that? =====================|<---dl1--->|====================>> I1 | | |r | | =====================|<---dl2--->|====================>> I2 ...or, in other words: r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 --------------------------------------------- B1= µ0/4π* I1/r^2 = 10^-7 * 1/1^2 = 10^-7 B2= µ0/4π* I2/r^2 = 10^-7 * 1/1^2 = 10^-7 F(1-2)= I1* dl1 x B2 = 1 * 1 x 10^-7 = 10^-7 F(2-1)= I2* dl2 x B1 = 1 * 1 x 10^-7 = 10^-7 --------------------------------------------- F(total) = 2 * 10^-7 N/m First and most importantly you need to solve the problem yourself, so everyone can see that you actually know what you talking about and so everyone can see precisely what is it you believe is the "correct" solution and what exactly do you think is wrong with my calculation.

No, current and distance are constants, hence B is constant and so dB= 0. If you want the contribution from "dl" then of course you do that line integral. -OR- dl is a vector, whose magnitude is the length of the differential element of the wire, and whose direction is the direction of conventional current [math]\scriptstyle{\hat{\mathbf{r}}}[/math] is the displacement unit vector in the direction pointing from the wire element towards the point at which the field is being computed, and No, none of these equations have to do anything with any superposition, especially the ones for magnetic fields. These equations evaluate individual filed vectors that, if magnetic, you better do not superimpose until you get force vectors, then you can do superimposition by simple vector addition of force-vectors, don't superimpose field-lines. No, those two have nothing to do with the distance, those are directional vectors, and in practice (wires) both of them are unit vectors, so that term is not necessary, and again has no connection to the distance which is not a vector but scalar magnitude, squared. As I said term "dl x r" can not even be used with wires. We only see that perpendicular plane with those concentric circles (magnetic field-lines), and the best you can with that is to make a CYLINDER out of it, i.e. to do a line integral - no supposition here, just a smudge through the third dimension. I'm not talking about any loops, but STRAIGHT WIRES. |+ |- |+ |- | | | | | | | | |=========|---dl1--->|========| |==|---dl1--->|=| | | | | |r |r | | | | |=========|---dl2--->|========| |=========|---dl1--->|==| | | | | | | | | |+ |- |+ |- No circles or loops here, that would matter. r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 --------------------------------------------- B1= ? ; B2= ? ; F(total)= ? Other equation? There are dozen of equations on this page. -- I got correct result, so I have no idea what could you possibly be complaining about. Anyway, here is the news, integration is not the same thing as superposition, we do not integrate B fields when we integrate force. We do not need to integrate anything really, not with these equations, we can solve everything as derivatives and just combine them later in relation to whatever distance given, it's a linear relation.

Why are you saying that to me? It is the other equation that is missing integral sign... not to mention it is also missing "dl" vector, cross product, distance vector, real distance magnitude, which is not even squared, and is also missing two Pies. -- I did integrate, that is why the result is not just Force in Newtons, but Force per unit length, or Newtons per Meter "N/m". You can not derive some equation according to some imaginary scenario, get rid of all the variables and think it applies to this scenario, or any scenario. You need to start with full equation in its original form, then you will see the only term you can throw out from the general Biot-Savart law when working with wires is "dl x r", nothing else. [math]\mathbf{B} = \frac{\mu_0 q \mathbf{v}}{4\pi} \times \frac{\mathbf{\hat r}}{r^2} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} =wires=> \frac{\mu_0 I}{4\pi r^2}[/math] There are no any angles when when there is an electric CURRENT, that's why we can get rid of "dl x r". We can not look in front and behind the magnetic field in such segment because of all the other charges along the segment. We have to have a single charge or 'volume charge', only then we can use this equation in its full form in full 3D, as I demonstrated earlier, and then the term "I" becomes the term "q*v" in both Biot-Savart and Lorentz force equation. Ergo, this is the most general form, and so this equation will apply to ANY scenario, whether single charges, electron beams or current carrying wires, any shapes or sizes, including loops, pentagons and giraffes.

=====================|<---dl1--->|====================>> I1 | | |r | | =====================|<---dl2--->|====================>> I2 r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 --------------------------------------------- B1= µ0/4π* I1/r^2 = 10^-7 * 1/1^2 = 10^-7 B2= µ0/4π* I2/r^2 = 10^-7 * 1/1^2 = 10^-7 F(1-2)= I1* dl1 x B2 = 1 * 1 x 10^-7 = 10^-7 F(2-1)= I2* dl2 x B1 = 1 * 1 x 10^-7 = 10^-7 --------------------------------------------- F(total) = 2 * 10^-7 N/m There is only one correct equation to do this. How do you do it? What result? Go ahead and solve two parallel wires and ampere unit scenario. X . | . | r . |R . | . | ======|<--dl-->|================================>> I1 | | |<-- l -->| X . | r' . | . | R . | . | ======|<--dl-->|===================================>> I1 | | |<------------- l' ---------->| 1.) B field is not calculated for some infinite segment or arbitrary segment. The magnitude of B field here is evaluated specifically in relation to the segment dl, and nothing else, ok? 2.) Do you see the magnitude of B field in relation to segment dl will depend on "distance" r, not R? The distance FROM the segment dl to some point, should be measured STARTING FROM the segment, do you agree? 3.) Do you see how that final equation does not contain all the important variables from which it was derived such as: any angles, l and r? 4.) Do you realize that equation will give the same result for two completely different scenarios given above? You could move point X left-right as far away as you want from the segment dl and that equation will see no difference as long as "R" is the same, but what is R in that equation? R is not the *real distance* FROM the segment dl, hence the equation is wrong - do we agree?