ambros

Can you print out the complete code for the complete equation? -- There is no 't', no time, as I tried to explain at the beginning. I'm the only one here who actually got any result that makes sense from that equation, and I did it in three different ways, here are two: TWO PARALLEL WIRES, ampere unit setup: I1=I2= 1A; r= 1m *DO NOT MODIFY EQUATIONS, USE THEM & CALCULATE: No.2 ================================================ [math]F_{12} = \frac {\mu_0}{4 \pi} \oint_{C_1} \oint_{C_2} \frac {I_1*dl_1 \times (I_2*dl_2 \times \hat{r}_{21} )}{|r|^2}[/math] [math]=> \frac {\mu_0*Q_1*v_1 \times (Q_2*v_2 \times \hat{r}_{21} )}{4 \pi*|r|^2}[/math] [math]=> \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}}[/math] [math]=> \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})[/math] [math]=> 10^-7_{N/A^-2}*1_A \times 1_A \ = \ 10^-7 \ N[/math] TWO PARALLEL WIRES, ampere unit setup: I1=I2= 1A; r= 1m *DO NOT MODIFY EQUATIONS, USE THEM & CALCULATE: No.3 ================================================ [math]B= \frac{\mu_0*Q*v \times \hat{r}}{4\pi*|r|^2}[/math] [math]=> \frac{4\pi*10^-7_{N/A^-2}*1_C*1_{m/s} \times 1}{4\pi*1_{m^2}}[/math] [math]=> \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}[/math] [math]=> \frac{10^-7_{N/A}}{1_m} \ = \ 10^-7 \ N/A*m \ (1Tesla)[/math] [math]F = Q*v \times B[/math] [math]=> 1_C*1_{m/s} \times 10^-7_{N/A*m}[/math] [math]=>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[/math] That one would be zero, but it does not qualify as distance as long as there are not at least TWO entities, BETWEEN which the distance can be measured. You are dreaming. Go ahead, show us that post where we can see how you put REAL VARIABLES of: I1= 1A, I2= 1A and r=1m, in THIS EQUATION: [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] What post, what number? What say you? Merged post follows: Consecutive posts merged This is ridiculous. post #23: http://www.scienceforums.net/forum/showpost.php?p=556655&postcount=23 [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] Cap'n Refsmmat: ...gives us the force on a piece of wire -- its units are Newtons, not Newtons per meter. * Ampere's force law IS in N/m. Fm =MEANS= F/1meter * That equation given by BIPM is in Newtons (N) and that is NOT Ampere's force law but Biot-Savart-law and Loerntz force as demonstrated above.

Ok, can you then tell us how to input this equation with different parameters of integrals paths: [math] 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} [/math] There is no such thing as zero distance, just like there is no edge to the universe. I explained this in previous thread. -- It is 'line integral', why is that under question? The trick is that if you do "point integral" it still works. Merged post follows: Consecutive posts merged I laugh at your hands waving through the air. -- Your equation is wrong because it does not have and reference to LENGTH which you insist is so important. -- If it is "mathematical disaster" than you should be able to demonstrate it by using MATHEMATICS, stop waving hands. Can you show you work? [math]B = \frac{\mu_0*I}{2\pi r} [/math] Where is the length, and where did 2 Pies go? Where are the freaking variable angles in your equation???

So, your variables do not have initial values and yet you managed to produce that result of 2*10^-7 all by itself, out of symbols? Those are either completely unnecessary or your program does not compute at all. -- Do you realize those two paths share the same coordinates? [math] \mathbf{F}_{12} = \frac{\mu_0 I_1 I_2}{4 \pi} \int_0^1 \int_{-\infty}^{\infty} \frac{\mathbf{dC2}(s2) \times \left( \mathbf{dC1}(s1) \times \frac{\mathbf{C1}(s1) - \mathbf{C2}(s2)}{||\mathbf{C1}(s1) - \mathbf{C2}(s2)||}\right) }{||\mathbf{C1}(s1) - \mathbf{C2}(s2)||^2} [/math] You are not making any angles, this term equals +/- ONE: [math]\frac{\mathbf{C1}(s1) - \mathbf{C2}(s2)}{||\mathbf{C1}(s1) - \mathbf{C2}(s2)||} [/math] You have to tell me exactly how do you initialize those values to obtain that result of 2*10^-7 N, numbers can not become out of symbols.. t = 0? 1? 0 to 1? 0 to infinity? - to + infinity? something else? s1 = 0? 1? 0 to 1? 0 to infinity? - to + infinity? something else? s2 = 0? 1? 0 to 1? 0 to infinity? - to + infinity? something else?

I said already those are TWO DIFFERENT OBJECTS. Do we cancel force vectors when they do NOT act on the same object? No! -- Why don't you try to put your finger BETWEEN TWO permanent magnets and tell me if the force you feel would be combined (magnitude sum), or that of only one magnet? Ay, caramba! I was solving the problem numerically and I FORGOT that term as it made no difference then, but we discussed it later and I corrected it, so it was suppose to be concluded - post #106: http://www.scienceforums.net/forum/showpost.php?p=555790&postcount=106 [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] This is standard derivation and what can be converted to point charge with the term: I*int(dl) = q*v = 1C*1m/s ...while "your" equation has only 2Pi and can not be equated like that. Which is the point I forgot about earlier on. My equation does have speed, but your one is missing LENGTH. YOU: 1C = 1A*1s ME: 1C*1m/s = 1A*1m Hehe, it's a double irony. It is my ultimate point to show how ampere unit is the most ridiculous unit and circularly defined mess of self referencing nonsense, which makes all but three equations rather dubious. -- However, those three equations are accurate and applicable to the real world without infinite wires, the errors are not related to integration at all, as these equation can be very successfully applied to free charges, like electron beams, Z-pinch applications and such.

No, I can not see and you obviously do not care to be understood. If you wish me to look at it, and also make it easier for everyone else, then you will write it down. Uh, think as you wish, line or not has nothing to do with whether it should be solved with antiderivative or not. Enough has been said in a form of an opinion, show your work and let us see the math. I can not make any meaningful comment on any of that unless the equation is on the table, sorry. It does not work with ether of these two: http://integrals.wolfram.com http://www.wolframalpha.com/input/ Can you make it work online? -- What are the s1 and s2, what numerical value(s) do they have? What value did you say 't' has in your equation? How come online version is able to print out those integrals in normal notation and you can not do that? Or, can you?

Infinity is undefined, so how can we define the ampere unit with that? Length units get canceled in a way to show that distance does not matter, but I don't think I wish to pursue that argument, it's kind of boring and I really forgot what was my punch-line. Merged post follows: Consecutive posts merged You made me repeat so many times. Do you have a memory loss problem? [math]B® = \frac{\mu_0 I * dl}{4\pi r^2} \ N/A*m[/math] ---- These equations are what defines all of the electronics and electrics, that people DO USE in the real world. These equation are EXACT and not approximation, we would not used them to define units if they were not, would we? -- If they were not exact or inapplicable to regular wires in real world, than no one would be able to properly measure anything that has to do with electric currents. -- How do you imagine instruments that measure amperes work and based on what formulas? Are you saying there are some better formulas that one given by BIPM?

And when you come back to real life where both wires are ACTUALLY 1 meter long, then this equation will not work correctly? And since one wire is much shorter than it is SUPPOSED to be, that then means the result in reality would be much less than predicted by your result, right? Those are not really definite integrals, it's more of Antiderivative actually. Your choice to make one wire 1m long and the other infinite is completely arbitrary, you should solve that with indefinite integrals or antiderivatives, alternatively, make them both 1m, both infinite, of make them both infinitesimal. (mu * 1 * 1)/(4 \[Pi]) \!\( \*SubsuperscriptBox[\(\[integral]\), \(0\), \(1\)]\( \*SubsuperscriptBox[\(\[integral]\), \(-\[infinity]\), \(\[infinity]\)] FractionBox[\(CrossProduct[dC2[s2], \ CrossProduct[dC1[s1], \*FractionBox[\(C1[s1] - C2[s2]\), \(Norm[C1[s1] - C2[s2]]\)]]]\), SuperscriptBox[\(Norm[ C1[s1] - C2[s2]]\), \(2\)]] \[DifferentialD]s1 \[DifferentialD]s2\)\) \ // N {0., 2.*10^-7, 0.} You are not hiding anything, eh? Write the equation properly so it is readable.

When you put it that way, I agree. Yes, but I forgot now what point I was trying to make. On a brighter side I just e-mailed BIPM request to explain the definition, equation and their relation in regards to what is actual force on one wire and what is the actual result predicted by that equation. Merged post follows: Consecutive posts merged What is the length of that path in meters? [math]1 \mathrm{C} = 1 \mathrm{A} \cdot 1 \mathrm{s}[/math] Where did that meter disappear from here? Those two do not equate, or do they?

Please copy/paste that here in text format so I can easier refer to it. * int[54] is wrong, move y to 1 * why is the x coordinate marked as "t", some relation to time? * where did you get idea that 2nd integral is from - to + infinity? * "t" is not initialized and it does not appear in the final equation? * write down final equation, so I can see better and point to rest. Will you please plug in the equation EXACTLY as it is written, which means that both of those two integrals integrate over the same distance - IN PARALLEL... and see what Mathematica will tell you. Close, but you are still mixing two very different equations. Magnetic force IS NOT DEFINED with Ampere's force law, it is defined with Biot-Savart law and Lorentz force equations. This is Ampere's force law, it is some OLD equation that specifically relates only to parallel wires and force per unit length: [math]F_m = F/1_m = 2 k_A \frac {I_1 I_2 } {r}[/math] This below is NOT Ampere's force law, and these two equations ARE NOT EQUAL: [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] ...this is Biot-Savart and Lorentz force combined, and the difference should be obvious, beside that the units are different, as you finally have realized now. We are getting closer to truth. Very good, I'm proud of you. Yes, and if that doesn't work just use a bigger hammer.

You closed my last thread by saying it is going nowhere and now instead of talking about what this thread is about you are trying to get off topic?! CAN YOU RESPOND TO THIS: [math] I\int dl \ => \ 1 A = 1\tfrac C s \ => \ \mathrm{C} = 1 \mathrm{A} \cdot 1 \mathrm{s} [/math] What speed is that of 1C/s? How many meters per second is that? AMPERE: -"6.242 × 10^18 electrons passing a given point each second constitutes one ampere." Passing a given POINT? What speed is that?

No, we are integrating over distance. That is extremely different from your example. No other examples are like this, we are talking about the real world here and there are very specific things that I want to point out, which I can not do on anything else but on this very particular unit and the definition of ampere. This is the integral we are talking about here: [math]I\int dl \ => \ 1 A = 1\tfrac C s \ => \ \mathrm{C} = 1 \mathrm{A} \cdot 1 \mathrm{s}[/math] You think there is some time integral or spatial derivation involved here? Tell me then, what speed is that of 1C/s? How many meters per second is that? AMPERE: -"6.242 × 10^18 electrons passing a given point each second constitutes one ampere." -- Passing a given POINT? What speed is that? How many meters per second is that? What is the LENGTH of this "point"?

At least? Do you not see that is THE EQUATION, the one that defines the ampere unit, and with that all the rest of equations that have anything to do with any el. currents in classical electromagnetism. You do not need to split equation to parts, you did the same thing as I did only messier and without showing all the steps properly, but at least you arrived at correct conclusion, finally. Look again, all you did is to write those two "dl" as "1m". BIPM: -"The ampere is that constant current which, 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 2 x 10^-7 newton per metre of length." What you just proved is that the force on ONE wire, as defined by the BIPM equation for magnetic force that you just dimensionally solved, gives the result that is equal to 10^7 N, (the constant). And so to get the correct result using this equation both F12 and F21 need to be taken into account. F(between) = F12(magnitude)+F21(magnitude) = 2*10^-7 N F(m) = F/1m = 2*10^-7 N/m

Numbers tell you what is the constant' date=' and in what units that constant ends up. How about you first learn this: unit for time = "s" - second unit for length = "m" - meter unit for mass = "kg" - kilogram http://en.wikipedia.org/wiki/Newton_(unit) Is that supposed to be a news to someone here? I said equation is in Newtons, didn't I? [math] => 10^-7_{N/A^-2}*1_A \times 1_A \ = \ 10^-7 \ N \ = \ 10^-7 \ kg*m/s^2 [/math] Is this better? And your conclusion is then, what? Are you saying that equation given by the International Bureau of Weights and Measures (BIPM), that defines the ampere unit, and therefore all the electronics and electrics on this planet, is not correct, or that I made some mistake in my dimensional analysis?

Then answer the question or go away and stop trolling. This is not about solving integrals, it is simply about writing units down and performing the dimensional analysis. I'm not interested in your hallucinations, show your work and point exactly what term in what step do you imagine is incorrect, if you can - if you can not, then you are not qualified nor competent to be giving any advice. You only keep demonstrating how much you do not know, as no integrals need to be solved, for example this is how you do dimensional analysis of Biot-Savart law: [math]B = \int \frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} = \frac{\mu_0*I*\int dl \times \hat r}{4\pi*|r|^2}[/math] [math] => \frac{4\pi*10^-7_{N/A^-2}*1_A* 1m \times 1}{4\pi*1_{m^2}} = \frac{10^-7_{N/A}}{1_m} = \ \ 10^-7 \ N/A*m \ = \ 10^-7 \ Tesla[/math] That scares me as much as if I was threatening you that I will never log in, or post on this forum again. Keep it up, and when all is left is three of you with bunch of teenagers parroting from the text books and unable to think, then you will surely enjoy every discussion, because then all will be saying exactly the same thing... without ever realizing or even being bothered if what they say is actually true or has any connection with reality. That is the only kind of place where you can fool yourself to know anything, but considering how my previous thread got closed and how many people was involved in that discussion and what they have said, I think you're doing pretty good job at it already, and once you get rid of me, you will again rule these barren lands, and 4+2 can go back to equal 71, or 55, as you see fit.

Sadly, no one managed to realize what are the units of this equation and point if there are any mistakes in this dimensional analysis: [math]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}[/math] [math]=> \frac {\mu_0*Q_1*v_1 \times (Q_2*v_2 \times \hat{r}_{21} )}{4 \pi*|r|^2}[/math] [math]=> \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}}[/math] [math]=> \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})[/math] [math]=> 10^-7_{N/A^-2}*1_A \times 1_A \ = \ 10^-7 \ N[/math] This is not some kind of "taboo", is it now? This is *science* forum, right?

[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] Did you just say that this equation given by the BIPM that DEFINES all the electronics and electrics needs some corrections and can not be applied to the real world as it is?

I already told you that your "solution" is not how equations are used, you made a mess of mathematics and physics there and you are explaining by referencing nonexistent and presumed information... ...you only needed to plug in the numbers GIVEN and CALCULATE the result. That equation is EXACT, it MUST NOT BE MODIFIED IN ANY WAY, it is what DEFINES 99% of all the electronics and electrics. Only for infinite wires?!?! Sheesh! Yes, thanks for asking. Do you know you did not explain that step you wanted me to tell you about? [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] Q: Is that "theta" - really theta or theta_1, or theta_2? You got from theta to having theta_1 and theta_2, explain that please. Q: Are you saying it is Ampere's force law or BIPM magnetic force equation, or both, that do not apply to real real world situations? Q: What would then be the actual result if wires were indeed 1 meter long, less or more? Q: Previously you said in post #80: ------------------- [math]\frac{F}{L} = IB \, \sin \alpha[/math] and consider that the magnetic field must be perpendicular to the wire, so sin(alpha) is 1. -------------------- ...so, why your magnetic fields must not be perpendicular anymore? Where is that "alpha" angle now, is that not supposed to be "theta" and why don't you just cancel it like you did then? That was supposed to be the same situation, right? Q: What units do you say this equation has and can you do dimensional analysis of it, please? [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] Merged post follows: Consecutive posts merged No' date=' that is not the same equation. [math']\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi R}[/math] The whole point of this discussion is for you to realize that equation is WRONG according to this equation: [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] ...you will never realize they are different if you do not perform dimensional analysis and by refusing to do so you are only making it more interesting for me. So, what units that equation has, what say you?

Never did you or anyone, except me did dimensional analysis of this equation: [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, I'm asking you again to show your work and make it clear what are the units of this equation, or point exactly to what post are you referring to. Can you do that please?

What would be the actual result if wires were indeed 1 meter long, less or more? No' date=' you are keep repeating your opinion, without making any specific objection as to where the error you imagined might have occurred, in what step exactly, and most importantly you are repeatedly failing to demonstrate that you can actually solve that problem and use the equation yourself, or deduce the units of it. Solve the problem with two parallel wires, given: I1=I2 = 1A; r=1m ---------------------------------------------------------------- [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] Show your work and make it clear what are the units of this equation. Merged post follows: Consecutive posts merged You do not understand what BETWEEN TWO wires mean. If the wanted to say "the force acting on ONE wire", they would have said so. If I hit you, and you hit me back with equal force from opposite direction, then BETWEEN TWO of us we exchanged two hits, and if someone told you the actual force felt "BETWEEN US" only "count" when I hit you, then you would object, wouldn't you? Solve the problem with two parallel wires, given: I1=I2 = 1A; r=1m ---------------------------------------------------------------- [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] It is silly to be waving hands around if you can not solve this, so will you?

'Unit vector' is a "fudge"?! It is even *impossible* for "dl" to be a fudge factor, because, it always gets divided by itself and it is always ONE, it is because in this particular case with two parallel wires the LENGTH absolutely have no impact, which is what I demonstrated with equations for POINT CHARGES, when instead of 1m, I let them be infinitesimally small. http://en.wikipedia.org/wiki/Biot_savart http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law * ds1 and ds2 are infinitesimal vector elements * 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 DIFFERENTIAL ELEMENT = Infinitesimal http://en.wikipedia.org/wiki/Infinitesimal It means, that LENGTH of the wire DOES NOT MATTER, and if the force per meter is 100 N/m, then the force per 2 meters will be 200 N/m, simple as that, it is linear function, ok? Really? What universe is the one where they make infinite wires? And, if the wire was 1m long for real, what would be the ACTUAL result then, less or more? -- Isn't that kind of thing I'm supposed to say - that Ampere's force law can NOT be used in the REAL WORLD. Is that what you saying now too? Merged post follows: Consecutive posts merged What would be the actual result if wires were indeed 1 meter long, less or more? -- What exactly are you saying? What equation do you say we can not use in the REAL WORLD? What equation do you say works ONLY for infinite wire: a.) Ampere's force law does not work correctly with finite (normal) wires? b.) BIPM's MAGNETIC FORCE equation does not work correctly with finite (normal) wires? Merged post follows: Consecutive posts merged 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] 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. That is not the example of how to do INTEGRATION, but is specific DERIVATION of the Biot-Savart law, that apperantly is far away from the equation given by the BIPM. -- You need to PLUG THE NUMBERS in the BIPM's magnetic force equation, the GIVEN NUMERICAL VALUES which all happen to be "1", so what you will be doing really is dimensional and unit analysis. That is all you have to do and you do not need any sources for that but CALCULATOR, or knowledge of mathematics. Besides, all those sources use different notation and say different thing, there is no clear diagram of TWO wires and they do not solve this problem for TWO PARALLEL wires, so please decide on one source which indeed talks about TWO wires and has a diagram so we can see all the angles, and what is what. I gave more reference to support "your" equation than you yourself, and I'll do it again as you are being very mysterious. Do you want to use this source given by the Wikipedia: http://info.ee.surrey.ac.uk/Workshop/advice/coils/unit_systems/ampereForce.html Is this ok? Is this what you want to be talking about? [math] B = \frac{\mu_0 I}{2\pi r} [/math] What is the physical meaning of 'ds' (dl) in your equation? Merged post follows: Consecutive posts merged No, BIPM *DEFINITION OF UNIT AMPERE* wants force per unit length. The magnetic force - Lorent'z force itself is always in Newtons, just like electric force - Coulomb's force is always in Newtons too. -- Did you forget that "your" equation is in Newtons as well? [math]F_{12} = \frac{\mu_0 I_1 I_2 L}{2\pi R}[/math] Show your work, what are the units of this equation: [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]

[math]Q = 6.242 *10E18 \ electrons; \ 1C = 1A*1s => 1A = 1C/1s \ (1e*6.242*10E18 \ m/s)[/math] Notice that it is the same current magnitude whether there is only one electron with speed of 6.242*10^18m/s, or 6.242*10^18 electrons moving at 1m/s. So, just like I said at the very beginning this can be solved with only two "fast electrons", or alternatively with 'charge density' Q and "slow electrons", and the later one is what I will demonstrate now, using both BIPM magnetic force equation, and then by using Biot-Savart & Lorentz force equations FOR POINT CHARGES. http://en.wikipedia.org/wiki/Biot_savart http://en.wikipedia.org/wiki/Lorentz_force TWO PARALLEL WIRES, ampere unit setup: I1=I2= 1A; r= 1m *DO NOT MODIFY EQUATIONS, USE THEM & CALCULATE: No.2 ================================================ [math]F_{12} = \frac {\mu_0}{4 \pi} \oint_{C_1} \oint_{C_2} \frac {I_1*dl_1 \times (I_2*dl_2 \times \hat{r}_{21} )}{|r|^2}[/math] [math]=> \frac {\mu_0*Q_1*v_1 \times (Q_2*v_2 \times \hat{r}_{21} )}{4 \pi*|r|^2}[/math] [math]=> \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}}[/math] [math]=> \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})[/math] [math]=> 10^-7_{N/A^-2}*1_A \times 1_A \ = \ 10^-7 \ N[/math] TWO PARALLEL WIRES, ampere unit setup: I1=I2= 1A; r= 1m *DO NOT MODIFY EQUATIONS, USE THEM & CALCULATE: No.3 ================================================ [math]B= \frac{\mu_0*Q*v \times \hat{r}}{4\pi*|r|^2}[/math] [math]=> \frac{4\pi*10^-7_{N/A^-2}*1_C*1_{m/s} \times 1}{4\pi*1_{m^2}} [/math] [math]=> \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} [/math] [math]=> \frac{10^-7_{N/A}}{1_m} \ = \ 10^-7 \ N/A*m \ (1Tesla)[/math] [math]F = Q*v \times B[/math] [math]=> 1_C*1_{m/s} \times 10^-7_{N/A*m} [/math] [math]=>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[/math] Oops! ...I did it again.

Ok, let me try again, to explain. In short, only terms starting with 'd', like "dl", are to be integrated. However, to fully understand integrals and differentials it is important to understand infinity. Infinity goes both ways, as far as logic anyway, and while many understand infinity in the context of 'larger distance' and 'further away', the true limits of infinity slide along the three dimensional path better described as 'bigger - smaller', from plus infinite VOLUME to minus infinite POINT, and vice versa. It is only in mathematics where infinity is limited to 1D line or 2D plane, which is usually associated with the loss "information" or "degrees of freedom". swansont: [math]L = \int dl[/math] Cap'n Refsmmat: [math]\oint_{C_2} \frac{2}{R} \, ds_2 = \frac{2L}{R}[/math] darkenlighten: [math]q \mathbf{v} = I \int{d\mathbf{l}}[/math] And hence comes surprise, though rather obvious, it might seem very, very strange and almost unacceptable, where the "easy" part is to understand how space is infinitely large as whenever you go 'further' and 'bigger' there always 'must' be something "behind", so to go further and bigger even more and more, to infinity, and beyond. -- Well, would it be surprising if all that is true for minus infinity too, and that there is no such thing as point, in the same sense as there is no 'edge' to the universe... or, is there? -- I leave this for the curios reader to ponder about and I'm going back to the restaurant, ah, and, there are no any time-varying stuff here, there is no even 'time' in these line integrals, it's all frozen in time, and smudged through space. Why? Because that is how unit ampere is defined - "6.242 × 10^18 electrons passing a given point each second", and everyone thinks that is some kind of time integral or spatial differential... not quite! -- Therefore, here is your problem ladies and gentleman - you should not be concerning yourselves too much with the one infinity that expands to infinite lengths, but the one that shrinks to infinitely infinitesimal smallness. NUMERICAL SOLUTION OF BIPM MAGNETIC FORCE EQUATION - input variables are given by the definition of ampere unit - [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} ==> \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] [math]==> \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[/math] There are several more ways how to solve this, but it all, of course, boils down to these two below, which is also what may give some insight as to what I was talking about in the beginning and how to integrate or numerically solve that or similar equations. [math] B =\frac{\mu_0 q \mathbf{v}}{4\pi} \times \frac{\mathbf{\hat r}}{r^2} \ \ \ \ \& \ \ \ \ F = q(\mathbf{v} \times \mathbf{B})[/math]

*** SUMMARY & CONFIRMATION with the REAL WORLD... ============================================= - The length of the wire has absolutely no impact, nor is considered in any of these equations. CASE A: CASE B: CASE C: + - + - | I2= 1A | | I2= 1A | I2= 1A |===================| |==========| + ==================infinite >> - | | | |1m |1m |1m | | | |===================| |==========| + ==================infinite >> - | I1= 1A | | I1= 1A | I1= 1A |<------ 35m ------>| |<-- 1m -->| + - + - CASE A: [math]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[/math] CASE B: [math]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[/math] CASE C: [math]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[/math] CASE A: [math] F_{12} = \frac{\mu_0 I_1 I_2 L}{2\pi R} = \frac{4\pi*10^-7_{N/A^-2}*1_{A}*1_{A}*35_{m}}{2\pi*1_m} = 2*10^-7_{N}*35 = 70*10^-7 \ N[/math] CASE B: [math] F_{12} = \frac{\mu_0 I_1 I_2 L}{2\pi R} = \frac{4\pi*10^-7_{N/A^-2}*1_{A}*1_{A}*1_{m}}{2\pi*1_m} = 2 *10^-7_{N} = \ 2*10^-7 \ N[/math] CASE C: [math] F_{12} = \frac{\mu_0 I_1 I_2 L}{2\pi R} = \frac{4\pi*10^-7_{N/A^-2}*1_{A}*1_{A}*\infty_{m}}{2\pi*1_m} = 2*10^-7_{N}*\infty = \infty \ N[/math] Apparently' date=' you are mistaken - the length of the wire is not even a part of the Ampere's force law equation, and so it obviously does not contribute, nor it even makes any difference in affecting the amount of the total force, per unit length. [math]F_m = 2 k_A \frac {I_1 I_2 }{r} \ \ \& \ \ F_{12} = \frac{\mu_0 I_1 I_2 L}{2\pi R}[/math] "Your" equations do not agree with you, it would seem. There is no even any term to define wire length in Ampere's force law, so what were you talking about? Where are those angles and varying distances? -- With given steady current and two parallel wires, which means r=constant, this equation could not be more static, constant and linear, there is no any superposition here. Do we agree? [math] \oint_{C_1} \frac{ds_1 \sin \theta}{r^2_{12}}[/math] ...the step from that above to this: [math]=> \frac{\cos \theta_1 - \cos \theta_2}{R}[/math] There is nothing like it in that post' date=' this is what is in the post #72: 1.) [math']\tan \theta = R/(-s)[/math] 2.) [math]B = \frac{\mu_0 I}{4\pi} \int_{\theta_1}^{\theta_2} \frac{R}{\sin^2 \theta} \frac{\sin^2 \theta}{R^2} \sin \theta \, d\theta = \frac{\mu_0 I}{4\pi R} \int_{\theta_1}^{\theta_2}\sin \theta \, d\theta [/math] 3.) [math]\theta_1 = 0 \ \ and \ \ \theta_2 = \pi, \ so \ \cos \theta_1 - \cos \theta_2 = 1 - (-1) = 2[/math] The initial step from above is not present in that post at all. There is no any diagram so I have not idea why would anyone take a tan(theta), or where is the original theta angle and where did it go... where are those theta_1 and theta_2 angles and according to what logic one goes to zero and the other towards Pi with infinite wires? -- That example does not seem to be about two parallel wires at all, so is there a diagram that can explain all those new terms that are NOT GIVEN by the INPUT PARAMETERS?