ambros

Thanks for the laughs, sweetheart.

Please, print out the complete equation in Mthematica code so everyone can see it and so I can test your one just like you tested mine, wouldn't that be fair? Merged post follows: Consecutive posts merged Really? Can you name one or two problems, or practical scenarios that you believe can not be solved by that Mathematica code I gave? What examples did you use to test it?

Reality is that you can not express "your theory" so that Mathematica can integrate it properly. There is nothing to deny even, I'm simply being a skeptic, so just as if someone wanted you to start believing in ghosts you would surely need to see one first, or you would not even consider the idea seriously, isn't that right, Mr Skeptic? This Mathematica syntax correctly represents that equation given by the BIPM: integral[0,1] integral[0,1] 10^-7Newtons/1^2 {1,0,0} cross {1,0,0} cross {0,-1,0} ...and anyone can see that for themselves by simply applying it in whatever case scenario involving magnetic forces they can come up with, just by plugging the given values and execute the code here: http://www.wolframalpha.com/input/ [math]F_{12} = \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4 \pi*|r|^2} [/math] VARIABLE: c1 c2 r I1 I2 r_hat integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} x {1,0,0} x {0,-1,0} Now, you have a chance to test this for yourself easily in Mathematica online, SEE the "ghost" with your own eyes and find out whether it indeed applies to reality, or not. However, if you refuse to even test that code against the reality, than it is you who is in denial. Skepticism is one thing, but when you refuse to "look through my telescope" it's called dogmatism.

It can not be tried because it does not compute. A.) YOUR CODE DOES NOT EVEN COMPUTE - can you refute this? B.) MY CODE CAN SOLVE ANY MAGNETIC PROBLEM - can you refute this? Yes, I'm afraid, your understanding of this subject is horrifying. You are even so lazy to realize that my code can solve any example involving any magnetic forces. Why? Because it is correct implementation of this equation given by the BIPM that defines ampere unit and all the other equations in that have anything to do with el. currents: [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] Once you are ready to stop waving hands, then you can either find some example that I can not compute with my code, or you can finally make "your" code actually produce something else than ERROR. -- Do you accept the challenge, what say you? Merged post follows: Consecutive posts merged Wolfram Alpha can surely CALCULATE ANY INTEGRAL, finite and indefinite, just as full version can, as long as you can write them correctly. -- You have to have NUMERICAL VALUES and let Mathematica integrate the result, you can not have 15 undefined variables when only 5 are given by the equation, you can not produce numerical result with symbols, so we must be able to enter numerical input which is strictly defined by the BIPM equation and your code must be mathematically identical to that equation, only rewritten in different syntax. I have tried player and it crashed, I can not run you code, but I see the same algorithm in the source code as you gave previously. I simply then converted your CrossProduct functions to mathematical symbols and denoted integral limits at the beginning so it is more readable, but that is that, from there you need to convert it to vector format and get rid of all the variable but 5: I1, I2, dl1, dl2 and r.

That is the same code you gave at the beginning which we discussed and I explained many things that are wrong with it, the most important one is that it does not compute in Mathematica, that is simply not correct syntax to produce working integral functions with variable parameters. You and I know that does not work and why or who you are trying to fool is beyond me. [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] I converted your code with syntax used in Mathematica online: *** YOUR CODE: integral[0,1] integral[-infinity, +infinity] dC2[s2] cross dC1[s1] cross (C1[s1]-C2[s2])/Norm[C1[s1]-C2[s2]] /(Norm[C1[s1]-C2[s2]]^2) (does not compute, incorrect syntax & undefined variables) *** MY CODE: integral[0,1] integral[0,1] 10^-7Newtons/1^2 {1,0,0} cross {1,0,0} cross {0,-1,0} (computes, can solve any situation involving magnetic forces) Your code can not be computed as that is not how variables are defined and used in integrals. You need to have very certain input precisely defined with the function you are trying to rewrite, nothing less nothing more. [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] INPUT IS DEFINE BY THE FUNCTION: I1, I2, r, int(dl1), int(dl2) 1.) You can not have C1, dC1 and s1 all at once. You specify all three of these in the beginning with "integral[start,end]", and you must let Mathematica handle the integration and it will inegrate those values for you, if you let it, and instruct it properly that is. 2.) You must have a way to input different magnitudes of I1 and I2, but your vectors are defined outside of the equation, so you first must rewrite the whole thing in vector form and that will help you get rid of duplicate variables. 3.) Make it COMPUTE if you want me to look at it any further, so far the result predicted by your code for any scenario is - ERROR.

http://www.wolframalpha.com/input/ VARIABLE: c1 c2 r I1 I2 r_hat integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} x {1,0,0} x {0,-1,0} Now let me see your calculation, ok? No one else even managed to get any results. No one else, including you, does not have any calculation to show, and especially the one that will compute in Mathematica. Let me see ONE, JUST ONE of those calculations, show me only one of those that I can input in Mathematica and get correct answers for parallel and perpendicular wires, or any result for that matter. I guarantee you will not be able to provide any of those calculations, can you prove me wrong? What say you? Merged post follows: Consecutive posts merged I'm actually downloading that sheeshbab, and curse you if I'm wasting my time again.

I am waiting any one of you to write down this exact equation in Mathematica code so we can test it on parallel and perpendicular wires: [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] ================================== Here is mine: VARIABLE: c1 c2 r I1 I2 r_hat integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} x {1,0,0} x {0,-1,0} ...and I repeat - IT WORKS - it is as REAL AS IT GETS and it can be used in real world and applied to any situation, ANY SITUATION, I tell ya! What more do you want?

No, I got exactly the same answer. -- Can you not write your integral to be used in Mathematica online? Why not? What result do you get with two perpendicular wires or loop and a straight wire?

My little Mathematica code for Lorentz force can solve any example in your text book that has to do with any magnetic forces, it can solve it CORRECTLY, so there is no way you can call such equation "wrong". To call it wrong you should first provide what you believe is correct, or at least you should be able to find some example for which my equation will give the wrong answer, but If it always gives correct answer, then it is not wrong, right?

But how do you explain 2X coulomb can produce the same force as 2x 1meter of wire? Because the former is "per point", and the later is "per meter", that's it. You are actually close when talking about dividing distance with distance, but I explained all that long ago, and really, the only possible way to do it correctly is that little piece of code I gave above. You can seriously use it in real life and go on and solve all the problems that have to do with the magnetic force, try it.

I can demonstrate it any way you want. The question is what is your level and have you got ant idea what you talking about. 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[/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] I hope this is not above your level, so please do you see a mistake and can you point it out exactly what term in what step do you believe is not correct?

You want me to try to make a sense of your mess that does not compute? If you can not make it work, that only suggests you never actually used it and so your advices and opinions are just plain unnecessary. My equations can solve any example in your textbook, so there is no way I will waste time looking into something you yourself do not know what it is. You could just as well offered me to drink cactus, no thanks! - Make it work and only then I will look at it, sorry.

That equation does exactly that. You can change all the parameters, any number you see you can use as variable, though unit vector should stay 1. |\ | y. y. |I2= 1A*1m | . I2= 1A*1m . | --------------> |___ . z /\ . z /\ / / . /r= 1m . /r= 1m / / .--------------> x .--------------> x I= 1A*1m I= 1A*1m {1,0,0}x{1,0,0}x{0,-1,0} {1,0,0}x{0,1,0}x{0,0,-1} PARALLEL (LEFT): integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} x {1,0,0} x {0,-1,0} PERPENDICULAR (RIGHT): integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} x {0,1,0} x {0,0,-1} Say, for two parallel wires, you wanna input: C1=3.4m, C2= 2.5m, I1= 34A, I2= 12A, r = 0.3m ------------------------------------------------------------------------------------------- Write this: integrate[0,3.4] integrate[0,2.5] 10^-7Newtons/0.3^2 {34,0,0} x {12,0,0} x {0,-1,0} ------------------------------------------------------------------------------------------- Have you tested that? That does not compute here. You seem to believe to know this better than me, so please just write one of those in Mathmatica code. Can you do that please?

That code does not compute. I do not think you can use Mathematica properly if you can not make it work online, IT IS THE SAME ENGINE. -- Yes, I would like if that mess you made can actually get any result other than ERROR. Now, give us that solution for loop and wire you said you have ready, will you? Merged post follows: Consecutive posts merged Please demonstrate what are you talking about. Write down that integral/equation exactly, so I can try it in Mathematica, or you write it down in Mathematica code yourself, please.

Funny, whenever I try to integrate over infinity - Mathematica tells me that I get infinity. So, please, is Mathematica wrong or is your solution wrong, because Mathematica says - IT DOES NOT COMPUTE. Maybe I do not know how to properly do it, especially since I never saw anything even close to what you be pretending to have as a "solution. -- So, can you write down that solution of yours in Mathematica code, please? What result do you get for the loop wire perpendicularly placed around the straight wire?

Ok, but before I leave, can you please just copy/paste that solution for parallel (perpendicular) wires as calculated with BIPM magnetic force equation everyone is talking about? -- If that supposed solution can not be expressed in Mathematica code, then I do not accept it as solution, is that reasonable?

You 're hallucinating my young friend. What you said is a lie, amusing and interesting lie, disgusting too. -- Can you say what step exactly you imagined might be incorrect in my calculation? Why can you not print out that solution in Mathematica code so everyone can see it online? Merged post follows: Consecutive posts merged

You can not really comment on my result if you can not show how do you obtain your result. So what is your result, can you show your work here with equations and mathematics, or simply give a code for Mathematica? Let me see your solution for parallel and perpendicular wire, eh? I wonder if you get zero force with perpendicular wires or perpendicular loop around straight wire... actually, with infinity, you can hardly get anything else but infinity, which means - ERROR, DOES NOT COMPUTE. That's good, because I have been repeatedly saying it... though I can do that in many different ways, as I also demonstrated previously. -- So, you are complaining that I use "1m"? But, how else do you expect me to tell you the force PER METER, if I do not integrate OVER ONE METER? The integral itself is what give us this another dimension and with it - a new UNIT. It is absolutely necessary to ingrate over 1m, because that is how ampere unit is defined - force BETWEEN TWO wires of 2*1-^-7 Newtons per meter of length. I solved the problem in five different ways now, you have the math of it all right in front of you. I do not want to argue as I think this can not be said and demonstrated any more clear than this, but if not, then tell me - what do I need to do, or show you, or demonstrate to convince you? Merged post follows: Consecutive posts merged Having "different mathematics" can make 2+7 = 34. Why do you think you can comment on any of it if you can not obtain any result from that equation? What is the point of guessing? Either equation is wrong or I made a mistake, which it is? 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] Different results like that above? Ampere's force law (above) is not the same as that other equation for F(12) given by the BIPM, though they both can produce these same results, depending on how one integrates.

How would you know? ---TWO PARALLEL WIRES: i1=i2= 1A' date=' 1m distance --- integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} cross {1,0,0} cross {0,-1,0} [math]\int_0^1( \int_0^1((1*10^-7 N) {1,0,0} \times ({1,0,0} \times {0,-1,0}))/1^2 dx) dx = {0, \ 1*10^-7 N, \ 0}[/math] ---TWO PERPENDICULAR WIRES (loop around wire): i1=i2= 1A, 1m radius --- integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} cross {0,1,0} cross {0,0,-1} [math]\int_0^1( \int_0^1((1*10^-7 N) {1,0,0} \times {0,1,0} \times {0,0,-1}))/1^2 dx) dx = {0, \ 0N, \ 0}[/math] ...ENTER THE CODE IN MATHEMATICA, HERE: http://www.wolframalpha.com/input/

Stop repeating someone else's opinion. What do you imagine how words "is wrong" can have impact on me, and what it can possibly contribute to the discussion? So lame, pathetic... go away! Merged post follows: Consecutive posts merged Do you want to see if my result will be zero or see how I do integrals? Are you actually have experimental measurements or you're going to use Maxwell's equations, or something like that? You are not planing to use the same equation as me, do you? -- That's all fine, but it would be much better if you can find example with some experimental measurements.

y | |____ x I like examples! I'm in bed already so I will not plug in any numbers now, but tomorrow this is what I will do: 1 vector in Y direction to represent straight wire, and 1 vector in x direction representing 'tangential' vector or "dl" segment of the loop. I would then ask you about the precision you require and according to that I would split that loop into number of segments. I could even make it be square or a triangle for a start. The more accuracy you want, the more segments I will make and smaller they will be. It is like drawing a "circle" with 3 lines, 4 lines, 5 lines.... 19, 21... 50 lines, and that's where it starts to look like a real circle. The size of "dl" is a matter of resolution desired and is used to approximate geometry. It's very similar how we actually draw circles or any other curved paths on computer screen, as there are no real curves, but all is made of little line segments "dl". The smaller they are, the better the smoothness, precision and accuracy.

Hahaa. That's one of the first things I told you in that thread. AND, THIS IS HOW TO DO IT WITH REAL NUMBERS: 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[/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] Can you do that, can you point mistake? darkenlighten: - I've put in real variables/constants to show you how you get 2 x 10^-7 N per unit length. Go away liar.

darkenlighten: - I've put in real variables/constants to show you how you get 2 x 10^-7 N per unit length. In first two links there is no even mention of that equation, but this: [math]\mathbf{B}_2 = \frac{\mu_0 I_2}{2\pi s} \hat{\phi} ; \mathbf{F}_1 = I_1 \int{d\mathbf{l} \times \mathbf{B}_2} \Rightarrow F_1 = \frac{\mu_0 I_1 I_2}{2\pi s} \mathbf{\hat{r}} [/math]per unit length ...third link has this equation at the very, not used to do any calculation or plug in any values in. [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] Well done, Alice, a man with no shame.

Huh?? That is magnetic field, there is no force there. How could you be talking about N or N/m in relation to magnetic field equation? -- What is your point then, what are you saying, can you summarize what is it you believe is wrong and what is correct? It is 1Tesla = 1N/A*m.

Direction vectors defining direction of the current, hence defining the position of wires, which instead of being two are now one wire with twice the el. current. -- That explains it, you're simply calculating DOUBLE magnetic in one wire. That term next to cross product equals to one, do you not see it is the same above and below? You can not evaluate upper half of it without lower, the s1 and s2 can not be different above from below, so that term equals one (+/-).