darkenlighten Posted April 5, 2010 Posted April 5, 2010 Why the middle equation has units in Newtons and the next one in N/m? Because you left out the rest of it: [math]\Rightarrow F_1 = \frac{\mu_0 I_1 I_2}{2\pi s} \mathbf{\hat{r}}[/math] per unit length, which it does. Since we are looking at a infinite wire you can't just say the force, since [math] L = \int d\mathbf{l} = \infty [/math]. So it needs to be force per unit length.
ambros Posted April 5, 2010 Author Posted April 5, 2010 [math]\Rightarrow F_1 = \frac{\mu_0 I_1 I_2}{2\pi s} \mathbf{\hat{r}}[/math] per unit length' date=' which it does. Since we are looking at a infinite wire you can't just say the force, since [math'] L = \int d\mathbf{l} = \infty [/math]. So it needs to be force per unit length. 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?
Cap'n Refsmmat Posted April 5, 2010 Posted April 5, 2010 (edited) "dl" in my final equation represents a vector of 1 meter in length, which is why I do not have integral sign, I do not integrate - I can solve this in one step, with only one derivative. -- Anyway, as all that will be automatically sorted out when we resolve the ultimate equation, so I'd rather concentrate on this: What physical quantity does dl represent? [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? Indeed. But if you set the length to infinite (for infinitely long wires), you will get an infinite force, which is clearly not very useful. So we divide by length to get force per unit length. Merged post follows: Consecutive posts mergedYes, but with the different equation for B than this one. The whole point of this new equation is for you to demonstrate whether that 1st solution was indeed correct, or not. [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \int \int \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}[/math] Will you solve this equation now, which means - to plug in the numbers given by the two parallel wires and ampere unit scenario and obtain the value for F(12)? I did this already. Edited April 5, 2010 by Cap'n Refsmmat Consecutive posts merged.
darkenlighten Posted April 5, 2010 Posted April 5, 2010 ambros stop being so hasty. You are reading it wrong. I was not saying it was my answer divided by a length, but the force per unit length so: [math]\Rightarrow F_1 = \frac{\mu_0 I_1 I_2}{2\pi s} \mathbf{\hat{r}}[/math] So [math] F_1 [/math] per unit length [math] \Rightarrow F_1 \ \ (N/m) = \frac{\mu_0 I_1 I_2}{2\pi s} \mathbf{\hat{r}} \ \ (N/m) [/math]
ambros Posted April 6, 2010 Author Posted April 6, 2010 [math] u = \oint_{C_1} \frac{ds_1 \sin \theta}{r^2_{12}} [/math] Using the same steps as in post #72 in this thread' date=' we can turn this integral into: [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] [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. [math]F_{12} = \frac{\mu_0 I_1 I_2 L}{2\pi R} \ N[/math] [math] F_m = \frac{\mu_0}{2\pi} \frac{1 \cdot 1}{1} = 2 \times 10^{-7} \ N/m [/math] 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?
Cap'n Refsmmat Posted April 6, 2010 Posted April 6, 2010 (edited) There is nothing in post #72 that would explain that step. That does not compute at all. How come one theta becomes two angles? By the magic of definite integration! Ooh, basic calculus makes me all tingly inside. [math]\frac{\mu_0 I}{4\pi R} \int_{\theta_1}^{\theta_2}\sin \theta \, d\theta = \frac{\mu_0 I}{4\pi R} \left[- \cos \theta\right]_{\theta_1}^{\theta_2} = \frac{\mu_0 I}{4\pi R} \left[ -\cos \theta_2 + \cos \theta_1 \right][/math] Similarly, we can do the same thing to u, integrating with respect to theta. [math]ds_1 = r \, d\theta[/math] and so on. (Basic angles there.) The rest is explained in post #72, with the trig substitutions. 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? Because I divided by length, as you can see by the conspicuous absence of L in the second equation. Incidentally, BIPM gives results in Newtons per meter: 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 × 10−7 newton per metre of length. Page 113 of that PDF you're so fond of. Edited April 6, 2010 by Cap'n Refsmmat
ambros Posted April 6, 2010 Author Posted April 6, 2010 (edited) --1--------------------------------------------------------2-- theta \ ----------------------------A--------------------------------- Imagine here that theta is the angle between the wire' date=' A, and point 1. If each wire is infinitely long, theta will approach 0. Now look at the angle between the wire, A, and point 2. As the wire becomes infinitely long, it approaches [imath']\pi[/imath]. ... By the magic of definite integration! Ooh, basic calculus makes me all tingly inside. [math]\frac{\mu_0 I}{4\pi R} \int_{\theta_1}^{\theta_2}\sin \theta \, d\theta = \frac{\mu_0 I}{4\pi R} \left[- \cos \theta\right]_{\theta_1}^{\theta_2} = \frac{\mu_0 I}{4\pi R} \left[ -\cos \theta_2 + \cos \theta_1 \right][/math] Note also that in the case of an infinite wire, we have [math]\theta_1 = 0 \ \ and \ \ \theta_2 = \pi, \ so \ \cos \theta_1 - \cos \theta_2 = 1 - (-1) = 2[/math] 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. What physical quantity does dl represent? 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. Edited April 6, 2010 by ambros
swansont Posted April 6, 2010 Posted April 6, 2010 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] It's described in post #72. Which step is the problem? cos(0) - cos(3.142) = 1 - 0.998 = 0.002 // How did you get cos(3.142) = -1? It's pi radians, not degrees. cos(pi) = cos(180º) = -1
ambros Posted April 6, 2010 Author Posted April 6, 2010 (edited) *** 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] The length contributes and yes there is such a variable \int dl. Since their magnetic fields depend on distance away' date=' this correlates to the length of a wire. Therefore the contribution from the whole length is not confined to the perpendicular plane. [/quote'] What is it your were talking about, again? Equations give the same result for any length, and how about conspicuous absence of LENGTH variable in Ampere's law? Am I dreaming right now, or was this all a part of _your hallucination? Distances change, angles change and fields superimpose over length, right? The magnitude of the field at any point is the result of contributions from the length of the wire. 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. -------------A-------------B --------------p-------------- To determine the force on point p, we have to know the contribution from A, B, and every other point on the opposite wire. You will notice that A and B are different distances away from p. ----1------2--------3----4------5--- -------------------A---------------- For each point on the upper wire, calculate the force caused by the current there on point A, and sum them. You will notice that r varies. As you sum the effect of each point on one wire on a point on the other wire, the distance between the points varies, so r varies. For example, for a point A on the first wire, we have to sum the effects of all points on the other wire. [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? But if you set the length to infinite (for infinitely long wires)' date=' you will get an infinite force, which is clearly not very useful. So we divide by length to get force per unit length. [/quote'] [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] Like that? The problem is you reached your limit and when you divide infinity with infinity you get 1N, and even that is not very certain. I think you misunderstood what I meant by SOLVING that equation. You were not supposed to be modifying it in any way or use any approximations or input any new variable in it. That equation is EXACT and as linear as Coulomb's law, the numerical INPUT VALUES are also EXACT and precisely defined. -- Use the "double differential" form given by the BIPM, this is no more complicated than velocity being derivative of position. Just do not do any modifications and derivations of that equation, PLUG IN THE NUMBERS given by the amperes unit setup and CALCULATE the result, with real NUMBERS. Merged post follows: Consecutive posts mergedIt's described in post #72. Which step is the problem? [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? Edited April 6, 2010 by ambros Consecutive posts merged.
swansont Posted April 6, 2010 Posted April 6, 2010 (edited) [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] I count seven steps to get to that from the original equation, not one. There is nothing like it in that post, 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? The setup is similar to that in the diagram of post 47, but as has been noted, a different angle is used in the triangle. edit: The basic setup is this: you have your wire, and a point P at a distance R perpendicular to the wire. Pick a element dl, which is at an arbitrary distance r (r≠R). The length R, r and the segment of wire form a right triangle. Choose one of the non-90º angles for your trig substitutions to relate R and r. That's your setup for finding the field at a distance R. Merged post follows: Consecutive posts merged*** SUMMARY & CONFIRMATION with the REAL WORLD...============================================= - The length of the wire has absolutely no impact, nor is considered in any of these equations. Sure it is. http://en.wikipedia.org/wiki/Ampere%27s_force_law For two thin, straight, infinitely long, stationary, parallel wires, the force per unit length one wire exerts upon the other in the vacuum of free space (emphasis added) The equation is only valid for infinite wires. Edited April 6, 2010 by swansont Consecutive posts merged.
darkenlighten Posted April 6, 2010 Posted April 6, 2010 What is it your were talking about, again? Equations give the same result for any length, and how about conspicuous absence of LENGTH variable in Ampere's law? Am I dreaming right now, or was this all a part of _your hallucination? Distances change, angles change and fields superimpose over length, right? Apparently, 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. Yea once again you are taking things out of context. As swansont said, the one we are giving is for an infinite wire, it gives you the force per unit length.
Cap'n Refsmmat Posted April 6, 2010 Posted April 6, 2010 Like that? The problem is you reached your limit and when you divide infinity with infinity you get 1N, and even that is not very certain. I think you misunderstood what I meant by SOLVING that equation. You were not supposed to be modifying it in any way or use any approximations or input any new variable in it. That equation is EXACT and as linear as Coulomb's law, the numerical INPUT VALUES are also EXACT and precisely defined. -- Use the "double differential" form given by the BIPM, this is no more complicated than velocity being derivative of position. Just do not do any modifications and derivations of that equation, PLUG IN THE NUMBERS given by the amperes unit setup and CALCULATE the result, with real NUMBERS. I did. BIPM wants force per unit length, so I gave that. Look at the equations. I never divided by infinity, for one thing. Merged post follows: Consecutive posts mergedDepends 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. There's two electron drift velocities at play here -- there are two wires, and their currents can be going in opposite directions. My point is that your "dl" in the equation is a pointless fudge introduced to make the equation seem to work, not a real physical quantity. Merged post follows: Consecutive posts mergedWhat is it your were talking about, again? Equations give the same result for any length, and how about conspicuous absence of LENGTH variable in Ampere's law? Am I dreaming right now, or was this all a part of _your hallucination? Distances change, angles change and fields superimpose over length, right? Incidentally, the equation I gave will not give the correct results unless at least one wire is infinitely long. The L term is the length of only one wire, since the other wire is assumed to be infinitely long in the first integration. (Hence the choice of [imath]\theta_1[/imath] and [imath]\theta_2[/imath].)
ambros Posted April 6, 2010 Author Posted April 6, 2010 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]
Cap'n Refsmmat Posted April 6, 2010 Posted April 6, 2010 ambros, please point out the specific flaw in my equations (preferably pointing to one specific step), and of course the specific flaw in the four sources that I have cited that directly agree with me.
swansont Posted April 6, 2010 Posted April 6, 2010 The BIPM result has units of N/m. Dimensional analysis is the easiest check one can do to see if their solution is wrong. If the units are wrong, the solution is wrong.
Cap'n Refsmmat Posted April 6, 2010 Posted April 6, 2010 [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] So... if I have to integrate: [math]\int_0^2 x^2 \, dx[/math] ...your method would be to do this? [math]\int_0^2 x^2 \, dx = x^2 x[/math] Curious. Note that the integrals in the expressions you gave are [imath]\oint_{C_2}[/imath] -- note the C2. That means they're along paths, so you do the definite integral along the path. The paths in this case are the wires. You are not doing any definite integration. That is wrong.
ambros Posted April 7, 2010 Author Posted April 7, 2010 (edited) [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. Edited April 7, 2010 by ambros
darkenlighten Posted April 7, 2010 Posted April 7, 2010 So you do not understand what [math] q\mathbf{v} = I \int d\mathbf{l} [/math] means I'll go over it: First of all it's not going to be just q of one charge, but the charge density. So for a wire it is [math] \lambda [/math] therefore [math] \lambda\mathbf{v} = \lambda \frac{dx}{dt}[/math], now when you take it further you get [math] \frac{d\lambda}{dt} x [/math], this means the change in charge over time multiplied by the distance; hence [math] I \int d\mathbf{l} [/math] is the change in charge with respect to time multiplied by the length of the path. And you are not getting the right answer. It should be 2 x 10^-7 N/m on ONE wire.
ambros Posted April 7, 2010 Author Posted April 7, 2010 (edited) I did. BIPM wants force per unit length' date=' so I gave that. Look at the equations. I never divided by infinity, for one thing. [/quote'] No, BIPM *DEFINITION OF UNIT AMPERE* wants force per unit length, the equation is not given in relation to that experiment, but it just happens it is what defines it - indirectly - just like they gave Coulomb's force equation right above. That is the MOST GENERAL equation for MAGNETIC FORCE, and force is always in Newtons, just like the equation for ELECTRIC FORCE is always in Newtons too. In F(m), 'm' refers to "per meter". The correct way to denoted it would be: F/1m = 2Ka I^2/r My point is that your "dl" in the equation is a pointless fudge introduced to make the equation seem to work, not a real physical quantity. '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? Incidentally, the equation I gave will not give the correct results unless at least one wire is infinitely long. The L term is the length of only one wire, since the other wire is assumed to be infinitely long in the first integration. (Hence the choice of [imath]\theta_1[/imath] and [imath]\theta_2[/imath].) 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 mergedI count seven steps to get to that from the original equation' date=' not one. [/quote'] That makes it worse, why did not he show all the steps? That whole mess there is some attempt at derivation in relation to some specific imaginary scenario - just plug in the numbers and calculate, that's what equations are for. Sure it is. http://en.wikipedia.org/wiki/Ampere%27s_force_law For two thin, straight, infinitely long, stationary, parallel wires, the force per unit length one wire exerts upon the other in the vacuum of free space (emphasis added) The equation is only valid for infinite wires. 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 mergedambros, please point out the specific flaw in my equations (preferably pointing to one specific step), 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. and of course the specific flaw in the four sources that I have cited that directly agree with me. 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 mergedThe BIPM result has units of N/m. Dimensional analysis is the easiest check one can do to see if their solution is wrong. If the units are wrong, the solution is wrong. 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] Edited April 7, 2010 by ambros Consecutive posts merged.
swansont Posted April 7, 2010 Posted April 7, 2010 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] That's not "my" equation. Where did it come from? The equation that everyone who solves the equation properly gets is [math]F_m = \frac{\mu_0 I_1 I_2 }{2\pi R}[/math] which can only be used for infinite wires. You can't just multiply both sides by L and apply it to a finite-length system. You should not be getting 2 pi x 10^-7 N if you put a finite length in the equation. There' a reason the BIPM definition says the wires are infinite. If you use infinite wires, you have to express it as force/length, otherwise it diverges. We already established that your magnetic field solution has the wrong units. Why are you still insisting that you're doing this correctly?
ambros Posted April 7, 2010 Author Posted April 7, 2010 What would be the actual result if wires were indeed 1 meter long, less or more? We already established that your magnetic field solution has the wrong units. Why are you still insisting that you're doing this correctly? 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 mergedSo you do no... 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?
Cap'n Refsmmat Posted April 7, 2010 Posted April 7, 2010 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. Do you know what "radians" are? Merged post follows: Consecutive posts merged ---------------------------------------------------------------------------- \ theta L | \ | r \ | R \ | \ | --------------------------------------------------------A------------------ That's the triangle. You can see that the trig identities in post 72 work, assuming you know what radians are. (L is s in the textbook's example in 72.) As L goes to infinity in either direction, theta goes to either zero or pi.
darkenlighten Posted April 7, 2010 Posted April 7, 2010 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? This is just silly. You do realize that one force is negative and the other is positive, if you add them together, you get zero??? This is not what they mean by the force between the wires, they say that only because both wires are involved in making a force on ONE wire.
Cap'n Refsmmat Posted April 7, 2010 Posted April 7, 2010 This is just silly. You do realize that one force is negative and the other is positive, if you add them together, you get zero??? Or, if we're using vector forces, if you add opposing vectors with equal magnitude, you get the zero vector.
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