darkenlighten Posted March 30, 2010 Posted March 30, 2010 No. I already stated that was an example for a changing B field. Stop being a fool and read back through. I am done.
Cap'n Refsmmat Posted March 30, 2010 Posted March 30, 2010 Guys, there's no need to be confrontational. Slow down, please.
darkenlighten Posted March 30, 2010 Posted March 30, 2010 I'm sorry for being rash. This is just kind of frustrating, so I will probably stop here.
ambros Posted March 30, 2010 Author Posted March 30, 2010 (edited) All you had to do is to solve the simplest example we started with. No wonder you got angry. Regarding your example, those equations are wrong and you did not provide any reference for them. Hate to break it to you but [math]\hat{r}[/math] is a polar coordinate.. http://en.wikipedia.org/wiki/Biot%E2%80%93Savart_law [math]\mathbf{B} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2}[/math] * [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 * dl is a vector' date=' whose magnitude is the length of the differential element of the wire, and whose direction is the direction of conventional current In our case dl is unit vector too, we are again not doing any integrals as we are again looking at the single point in a single instant in time, it is called "infinitesimal element of wire length". *** THE SIMPLEST EXAMPLE THERE IS: [img']http://dev.physicslab.org/img/2162109e-eb6e-4193-8e52-9d141f928e92.gif[/img] Q1: Magnitude of E filed at distance 'r' can change with varying current? Q2: Magnitude of E filed at distance 'r' can change with changing B field? Q3: Magnitude of E filed at distance 'r' can change with moving magnets? darkenlighten: - "for a changing current there will be an electric field around the wire" - "...E field exists due to a changing B Field" You could just as well said pigs have wings. Wires do not change their electrical neutrality around them due to any changes in current, motion of any magnets or due to change of any magnetic fields. Voltage, my young friend, you are not talking about any "time-varying E field" as there is no such thing, you are talking about "potential difference" and "electric current" - distribution and electron kinetics, these particular effects are not measured at any point AROUND the wire, and these you should not call "E field". [math] B®= \frac{\mu_0 I d\mathbf{l} \times \mathbf{r}}{4\pi r^2} = \frac{\mu_0 I sin(angle)}{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] You are wrong. So lets take a look at the derivation and as to why you are wrong: Ay, caramba! Concentrate on your part and provide solution via Maxwell's equations, if you can. In the very next message I put back that term in Biot-Savart law you were complaining about to avoid silly arguments and embarrass you further... P P' | | r= distance(AP) | | |angle= AW-AP = 90 ----------------A---------------------B-------wire W---> Of course if we place 'probe' at point P we are measuring the magnitude in relation to wire at point A. It is ridiculous that someone would try to measure field potential around the wire at some distance 'r' by placing the instrument at point P and thinking it will measure the magnitude in relation to point/segment B that is further away from the instrument, that's just utter nonsense - it's equivalent of trying to measure your own body temperature by placing the thermometer not up yours, but up someone else's... armpit. P |\ | \ r' | \ |r \ | \ ----------------A-----B-----------------wire W---> In my example I assume laws of logic still hold and people are sane. In my example when I say distance 'r', it, OF COURSE, means distance A-P, the SHORTEST DISTANCE. No sane person would ever measure the DISTANCE from a straight wire at any other angle but 90 degrees. Distance is the shortest path between two points, ok? Do you not see what is 'angle', what is 'dl' and what is 'r'? Steady 1 ampere, what is E® & B®? Edited March 30, 2010 by ambros
Klaynos Posted March 30, 2010 Posted March 30, 2010 ambros, I'd suggest you re read darkenlighten's posts, clearly and carefully.
ambros Posted March 30, 2010 Author Posted March 30, 2010 (edited) ambros, I'd suggest you re read darkenlighten's posts, clearly and carefully. What are you talking about, please explain yourself. --- EXAMPLE Q1 ------------------------------------------------------ Wire along x-axis has steady current of 1 ampere, solve for E® and B®. darkenlighten: Do you mean to say this is correct answer? Are these equations valid at all, is that correct derivation of Maxwell's equations? - Do you also mean to agree with him about Biot-Savart law and deny that actually this is the correct expression, where both the full and simplified version can be applied to this problem: Can you solve this example? Can you CLEARLY write down the two equations derived from Maxwell's that actually can solve E® and B® for this scenario, and can you please use CORRECT NOTATION or point some reference? --- CHANGING CURRENT --------------------------------------------------- darkenlighten: Do you mean to say these equations are correct? Do you mean to agree this is true: - "for a changing current there will be an electric field around the wire... E field exists due to a changing B Field"? Edited March 30, 2010 by ambros
darkenlighten Posted April 2, 2010 Posted April 2, 2010 --- EXAMPLE Q1 ------------------------------------------------------ Wire along x-axis has steady current of 1 ampere, solve for E® and B®. darkenlighten: [ATTACH]2452[/ATTACH] [ATTACH]2453[/ATTACH] Do you mean to say this is correct answer? Are these equations valid at all, is that correct derivation of Maxwell's equations? - Do you also mean to agree with him about Biot-Savart law and deny that actually this is the correct expression, where both the full and simplified version can be applied to this problem: So as stated before, we already determined that [math] \mathbf{E} = 0 [/math] and the B field I gave there is correct. This is not correct, reason being is that you are dealing with a wire and all of the pieces contribute towards the total magnetic field, therefore there has to be an integral to sum up all of the pieces, which also means that [math] d\mathbf{l} \times \mathbf{r} [/math] is not 1 but what is shown in my previous post with the reference from Griffith's text. So why do you refuse to believe a highly reputable source? Can you solve this example? Can you CLEARLY write down the two equations derived from Maxwell's that actually can solve E® and B® for this scenario, and can you please use CORRECT NOTATION or point some reference? --- CHANGING CURRENT --------------------------------------------------- darkenlighten: Do you mean to say these equations are correct? Do you mean to agree this is true: - "for a changing current there will be an electric field around the wire... E field exists due to a changing B Field"? I've already gave enough information for why these are correct. I am using correct notation, as cylindrical coordinates are best, if you are not familiar with them, read about them and it will make things easier. Also I think the underlying issue is that you do not believe that a changing B field creates an electric field??? Which I don't understand considering that's electric motors work. A changing magnetic field induces an emf (electromotive force) which is a vector potential (a voltage) that corresponds to an electric field.
ambros Posted April 2, 2010 Author Posted April 2, 2010 (edited) Le Bureau international des poids et mesures: -"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." [math]B®= \frac{\mu_0 I d\mathbf{l} \times \mathbf{r}}{4\pi r^2} = \frac{\mu_0 I}{4\pi r^2}[/math] [math]F= I\int d\boldsymbol{\ell}\times \mathbf{B} = I \mathbf{L} \times \mathbf{B}[/math] r= 1m; dl= 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 * dl x B2 = 1 * 1 x 10^-7 = 10^-7 F(2-1)= I2 * dl x B1 = 1 * 1 x 10^-7 = 10^-7 --------------------------------------------- F(net) = 2 * 10^-7 N/m :: Check and mate. Edited April 2, 2010 by ambros
darkenlighten Posted April 2, 2010 Posted April 2, 2010 On the contrary: You would still be wrong if that meant the net force, but the net force here is zero, and that is not what it means when saying the force between the wires. And you are off by a half on each and since you are working with a unit of 1m it being r^2 or r will not make a difference for this example. This is the correct answer: [math] \mathbf{B}_2 = \frac{\mu_0 I_2}{2\pi s} \hat{\phi} ;[/math] [math] \mathbf{F}_1 = I_1 \int{d\mathbf{l} \times \mathbf{B}_2} [/math] [math]\Rightarrow F_1 = \frac{\mu_0 I_1 I_2}{2\pi s} \mathbf{\hat{r}} [/math] per unit length [math] \mu_0 = 4\pi\times 10^{-7};[/math] [math] I_1 = I_2 = 1 A;[/math] [math] s = 1m [/math] [math]\Rightarrow F_1 = 2\times 10^{-7}N[/math] per unit length One force is negative one force is positive, look at your own diagram. So [math] F_{net} = F_1 + F_2 = 2\times 10^{-7} - 2\times 10^{-7} = 0[/math] Please answer me again when I ask: Why do you not trust my source? Also, where you got the picture from even states I am correct: http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law
ambros Posted April 2, 2010 Author Posted April 2, 2010 On the contrary: You would still be wrong if that meant the net force, but the net force here is zero, and that is not what it means when saying the force between the wires. And you are off by a half on each and since you are working with a unit of 1m it being r^2 or r will not make a difference for this example. This is the correct answer: [math] \mathbf{B}_2 = \frac{\mu_0 I_2}{2\pi s} \hat{\phi} ;[/math] [math] \mathbf{F}_1 = I_1 \int{d\mathbf{l} \times \mathbf{B}_2} [/math] [math]\Rightarrow F_1 = \frac{\mu_0 I_1 I_2}{2\pi s} \mathbf{\hat{r}} [/math] per unit length [math] \mu_0 = 4\pi\times 10^{-7};[/math] [math] I_1 = I_2 = 1 A;[/math] [math] s = 1m [/math] [math]\Rightarrow F_1 = 2\times 10^{-7}N[/math] per unit length One force is negative one force is positive, look at your own diagram. So [math] F_{net} = F_1 + F_2 = 2\times 10^{-7} - 2\times 10^{-7} = 0[/math] Ay, caramba! There are two forces there and they act on TWO DIFFERENT objects. The "net force", of course, is the one that will correctly calculate the relative displacement, and both of these two forces play part there, as illustrated, of course. Please answer me again when I ask: Why do you not trust my source? Do you not understand "distance ®" is the SHORTEST PATH? Do you not see what is 'angle', what is 'dl' and what is 'r'? Now, compare that with your source - it's not the same, and if it was then "your" result would be wrong. Your source does not talk about any case scenario I have given here, what it talks about is simply insane. Why don't you trust my source? http://en.wikipedia.org/wiki/Biot%E2%80%93Savart_law http://en.wikipedia.org/wiki/Magnetic_field http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law Also, where you got the picture from even states I am correct: http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law What part do you think states you are correct? Maybe this: - "The physical origin of this force is that each wire generates a magnetic field (according to the Biot-Savart law), and the other wire experiences a Lorentz force as a consequence."? Ha-ha. ...and your equations, that's just hideous. How did that happen? Because you did not PROVIDE ANY REFERENCE, and I'm talking about these four monsters that you supposedly DERIVED FROM MAXWELL'S EQUATIONS:
darkenlighten Posted April 2, 2010 Posted April 2, 2010 Ay, caramba! There are two forces there and they act on TWO DIFFERENT objects. The "net force", of course, is the one that will correctly calculate the relative displacement, and both of these two forces play part there, as illustrated, of course. That doesn't mean I can't consider this whole system, but that is hardly the point. There is only 1 Force acting on each object and that force is 2 x 10^-7 N/m derived as I have shown. Do you not understand "distance ®" is the SHORTEST PATH? Do you not see what is 'angle', what is 'dl' and what is 'r'? Now, compare that with your source - it's not the same, and if it was then "your" result would be wrong. Your source does not talk about any case scenario I have given here, what it talks about is simply insane. Um yea it does. It solved the infinite wire with steady current I using the Biot-Savart Law...And once again do you understand why you look at dl and r? Think of it this way, say you start with a piece of wire that is 1L long (that somehow has a steady current), it will have a certain magnetic field. Then look at another piece of wire that is 2L, it will have another magnetic field different than that of the wire of length 1L. If you agree with that, then you should realize that dl x r is not 1 and should refer to the correct derivation 14 posts before this. Why don't you trust my source? http://en.wikipedia.org/wiki/Biot%E2%80%93Savart_law http://en.wikipedia.org/wiki/Magnetic_field http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law What part do you think states you are correct? Maybe this: - "The physical origin of this force is that each wire generates a magnetic field (according to the Biot-Savart law), and the other wire experiences a Lorentz force as a consequence."? Ha-ha. While these sources are correct, you do not solve them correctly, which is the half the point here. And from the source http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law it shows what I am saying read it: [math]F_m = 2 k_A \frac {I_1 I_2 } {r} [/math] ...and your equations, that's just hideous. How did that happen? Because you did not PROVIDE ANY REFERENCE, and I'm talking about these four monsters that you supposedly DERIVED FROM MAXWELL'S EQUATIONS: Not quite.
ambros Posted April 2, 2010 Author Posted April 2, 2010 (edited) Not quite. It is not only ridiculous but I believe it is even against forum rules to keep making wrong statement and false assertion while blatantly and repeatedly refusing to provide any REFERENCE. Have you derived these from Maxwell's equations and will you provide some reference finally? Merged post follows: Consecutive posts merged While these sources are correct' date=' you do not solve them correctly, which is the half the point here. [/quote'] FOR STRAIGHT WIRE WITH STEADY CURRENT: YOU SAY THIS IS CORRECT: I SAY THIS IS CORRECT: There is nothing to solve - just COMPARE. If those sources are correct, as you admitted, then that automatically means I'm correct and you are wrong. You're mumbling. PICK ONE! http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law it shows what I am saying read it: FOR STRAIGHT WIRE(s) WITH STEADY CURRENT: YOU SAY THIS IS CORRECT: I SAY THIS IS CORRECT: - "The physical origin of this force is that each wire generates a magnetic field (according to the Biot-Savart law), and the other wire experiences a Lorentz force as a consequence." Ampere's force law is unsuitable approximation, Ampere's circuital law is unsuitable approximation too, and that's why the first paragraph in that Wikipedia article is clear that this interaction is actually DEFINED by BIOT-SAVART LAW and LORENTZ FORCE, and not with any of Ampere's and Maxwell's equations. Why? Because they are *different*, and it should be obvious by now what equations are more *accurate*, PICK ONE! -- I will not discuss basic vector calculus with you, check some real world measurements. Edited April 2, 2010 by ambros Consecutive posts merged.
Cap'n Refsmmat Posted April 2, 2010 Posted April 2, 2010 FOR STRAIGHT WIRE WITH STEADY CURRENT: YOU SAY THIS IS CORRECT: I SAY THIS IS CORRECT: Are you blind? Drunk? There is nothing to solve - just COMPARE. If those sources are correct, as you admitted, then that automatically means I'm correct and you are wrong. You're mumbling. PICK ONE! According to Physics for Engineers and Scientists, 3rd ed., page 932, the magnitude of the magnetic field around a wire is [imath]B = \frac{\mu_0 I}{2\pi r}[/imath]. ambros' expression looks correct. Now, if you integrate your expression for an infinitely long wire, you'll probably receive the exact same result. Try it. There's no need to be so confrontational about it.
ambros Posted April 2, 2010 Author Posted April 2, 2010 (edited) According to Physics for Engineers and Scientists, 3rd ed., page 932, the magnitude of the magnetic field around a wire is [imath]B = \frac{\mu_0 I}{2\pi r}[/imath]. ambros' expression looks correct. Now, if you integrate your expression for an infinitely long wire, you'll probably receive the exact same result. Try it. There's no need to be so confrontational about it. Ok, I moderated myself and deleted provocations. There is nothing to integrate really, this is the formula, that's it. There is this one and there is the other one you wrote down - but one of them is wrong or simply does not apply to these same situations. I've given many different Wikipedia articles that all confirm what equation is correct and more appropriate. Does no one have some instruments handy to measure this and resolve the problem? Alternatively, I'll find some experimental data in some paper or somewhere. http://dev.physicslab.org/Document.aspx?doctype=3&filename=Magnetism_BiotSavartLaw.xml - "The Biot-Savart Law is much, much, much more accurate than Ampere's Law (as its applications involve fewer assumptions)." -- We should really talk how that equation can come from Maxwell's equations, I did not think I would debating Biot-Savart law as it is well documented, and I am even very well aware how that alternative equation came to be: This is wrong, I do not know what is this, it' crazy. DISTANCE to that segment dl is not the same R that is in the equation, but the real distance is 'r' in that scenario. They are thinking to be measuring "dl" at distance R, but they are actually measuring it at distance r. What will their instrument measure? It would measure the CLOSEST segment to that 'point'. In order to measure segment dl at distance R, then they surely need to move the instrument right next to dl, so the distance R points to the middle of the segment dl, and so that R is indeed the shortest path and real "distance" from that point to that segment. Edited April 2, 2010 by ambros
swansont Posted April 2, 2010 Posted April 2, 2010 There is nothing to integrate really, this is the formula, that's it. There is this one and there is the other one you wrote down - but one of them is wrong or simply does not apply to these same situations. I've given many different Wikipedia articles that all confirm what equation is correct and more appropriate. Does no one have some instruments handy to measure this and resolve the problem? Alternatively, I'll find some experimental data in some paper or somewhere. You do need to integrate — that's why there is an integral sign in the formula. All of the line elements contribute to the total field, so you have to count the contribution from each. This is wrong, I do not know what is this, it' crazy. DISTANCE to that segment dl is not the same R that is in the equation, but the real distance is 'r' in that scenario. They are thinking to be measuring "dl" at distance R, but they are actually measuring it at distance r. What will their instrument measure? It would measure the CLOSEST segment to that 'point'. In order to measure segment dl at distance R, then they surely need to move the instrument right next to dl, so the distance R points to the middle of the segment dl, and so that R is indeed the shortest path and real "distance" from that point to that segment. No, the result is correct. You need to account for the contribution at every point; they all add up. A short wire and a long wire carrying the same current will give you a different field.
ambros Posted April 3, 2010 Author Posted April 3, 2010 (edited) You do need to integrate — that's why there is an integral sign in the formula. =====================|<---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? All of the line elements contribute to the total field' date=' so you have to count the contribution from each. [/quote'] http://en.wikipedia.org/wiki/Biot%E2%80%93Savart_law http://en.wikipedia.org/wiki/Magnetic_field http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law Maybe when you work with loops and their center, but otherwise no, not "all", but the contribution from only one very specific line element, given by the vector "dl". -- There is only one correct formula of Biot-Savart law in relation to distance, and that relation is inverse SQUARED and "dl x r", though regardless of that there is again only one set of equations that will give correct answer for the definition of the ampere unit, and it's those ones with "4Pi", not "2pi". Do you see your equation to represent Biot-Savart law anywhere in these articles about Biot-Savart law? And that's still Biot-Savart law in either case, the real question is what is the solution given by the Maxwell's equations? No, the result is correct. You need to account for the contribution at every point; they all add up. A short wire and a long wire carrying the same current will give you a different field. 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? Edited April 3, 2010 by ambros
Cap'n Refsmmat Posted April 3, 2010 Posted April 3, 2010 Maybe when you work with loops and their center, but otherwise no, not "all", but the contribution from only one very specific line element, given by the vector "dl". Yes, and there's an integral sign, so you sum over the entire line. That's what integrals do.
ambros Posted April 3, 2010 Author Posted April 3, 2010 (edited) Yes, and there's an integral sign, so you sum over the entire line. That's what integrals do. 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. Edited April 3, 2010 by ambros
swansont Posted April 3, 2010 Posted April 3, 2010 (edited) 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? No, it's not ok, it's wrong. If you want the contribution from dl it will be [math]\mathbf{dB} = \frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2}[/math] But all of the line elements can make a contribution, because B follows superposition, so you have to integrate over all of them In the long wire case, dl x r will give you a term proportional to r, which is why the final answer depends on 1/r and not 1/r^2 http://electron9.phys.utk.edu/phys136d/modules/m7/Ampere.htm Since steady currents always flow in closed loops, we need to integrate dB over the entire circuit to evaluate the net field B at point P. (This is a vector integral. The contributions dB from different sections add vectorially.) All sections of the loop contribute to B. But because of the inverse-square dependence on distance, the sections closest to P make the largest contributions. 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". In the other equation the integral has already been evaluated. Edited April 3, 2010 by swansont Added link and quote
ambros Posted April 3, 2010 Author Posted April 3, 2010 (edited) No' date=' it's not ok, it's wrong... But all of the line elements can make a contribution, because B follows superposition, so you have to integrate over all of them. In the long wire case, dl x r will give you a term proportional to r, which is why the final answer depends on 1/r and not 1/r^2 [/quote'] The proof is in the pudding. Le Bureau international des poids et mesures: -"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." r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 --------------------------------------------- B1= ? B2= ? F(total)= ? Merged post follows: Consecutive posts mergedIf you want the contribution from dl it will be [math]\mathbf{dB} = \frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2}[/math] 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 But all of the line elements can make a contribution, because B follows superposition, so you have to integrate over all of them 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. In the long wire case, dl x r will give you a term proportional to r, which is why the final answer depends on 1/r and not 1/r^2 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. http://electron9.phys.utk.edu/phys136d/modules/m7/Ampere.htm Since steady currents always flow in closed loops, we need to integrate dB over the entire circuit to evaluate the net field B at point P. (This is a vector integral. The contributions dB from different sections add vectorially.) All sections of the loop contribute to B. But because of the inverse-square dependence on distance, the sections closest to P make the largest contributions. 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)= ? In the other equation the integral has already been evaluated. 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. Edited April 3, 2010 by ambros Consecutive posts merged.
swansont Posted April 3, 2010 Posted April 3, 2010 The proof is in the pudding. Le Bureau international des poids et mesures: -"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." I'm not sure what you hope to demonstrate with this. The number itself does not prove or disprove any of the equations, because 1m and (1m)^2 will have the same numerical value. I think this was already pointed out. 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. No, this is not correct. You find the contribution from a differential element, and you integrate to find the total value. This is basic calculus; an integral is a summation with infinitesimal contributions. You do not sum over all terms to find the contribution of one. 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. You have some serious misconceptions about this. 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. Nope. darkenlighten has already posted a scan from a (very good) textbook showing how the Biot-Savart law gives the field of a long wire, and sets up the integral. I'm not talking about any loops, but STRAIGHT WIRES. And I'm talking about the concept, which doesn't really care if the wire is straight. The differential element is an infinitesimal, so the concept of straight vs curved is immaterial for the contribution; the shape will tell you how to do the integral. 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. You didn't get a correct result. You've merely been insisting that you did. People with some physics knowledge have been pointing out that you are not correct.
Cap'n Refsmmat Posted April 3, 2010 Posted April 3, 2010 I can quote from my physics textbook (R is the distance to the closest point on the wire): A second configuration calculable with the Biot-Savart Law is the case of a finite length of straight wire, as shown in Fig. 29.39. We need to sum the contributions of the form (29.24) ([imath]dB = \frac{\mu_0 I}{4\pi} \frac{ds \, \sin \theta}{r^2}[/imath]); now, however, the quantities r, s, and [imath]\theta[/imath] all vary along the wire. If we want to sum the contributions, we must put them in terms of a single variable. It so happens that using the angle [imath]\theta[/imath] is easiest. From Fig. 29.39, we can relate [math]\tan \theta = R/(-s)[/math] and we can take the derivative of [imath]s = -R/\tan \theta[/imath] with respect to [imath]\theta[/imath] to obtain [math]ds = \frac{R}{\sin^2 \theta}d\theta[/math] Also from Fig 29.39, [imath]r = R/\sin \theta[/imath]. Substituting these relations into (29.24), we can write [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] Using [imath]\int \sin \theta \, d\theta = -\cos \theta[/imath], we obtain [math]B = \frac{\mu_0 I}{4\pi R} (\cos \theta_1 - \cos \theta_2)[/math] This equation can be used repeatedly to sum contributions from various straight segments of a conductor. Note also that in the case of an infinite wire, we have [imath]\theta_1 = 0[/imath] and [imath]\theta_2 = \pi[/imath], so [imath]\cos \theta_1 - \cos \theta_2 = 1 - (-1) = 2[/imath], and the result reduces to the field of an infinite wire, [imath]B = \mu_0 I / 2\pi R[/imath], as it must. Note the bolded segment. If you disagree, take it up with Dr. Markert, the author of the textbook. I work in his lab, actually.
ambros Posted April 3, 2010 Author Posted April 3, 2010 I'm not sure what you hope to demonstrate with this. The number itself does not prove or disprove any of the equations, because 1m and (1m)^2 will have the same numerical value. I think this was already pointed out. *** 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? You have some serious misconceptions about this. You didn't get a correct result. You've merely been insisting that you did. People with some physics knowledge have been pointing out that you are not correct. 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.
darkenlighten Posted April 3, 2010 Posted April 3, 2010 Two Questions ambros: 1) Do you think that a wire of length 1m with steady constant current I will have a different magnetic field than a wire of length 2m? 2) When you are solving for, what you call, the total magnetic force, why are you adding up the individual forces from 2 different objects and why is one of them not negative? Answering these straight forward will put us back on the correct path. If you choose to ignore them and not answer directly, this discussion is going to continue to be circular.
ambros Posted April 4, 2010 Author Posted April 4, 2010 (edited) I can quote from my physics textbook (R is the distance to the closest point on the wire): Note the bolded segment. If you disagree, take it up with Dr. Markert, the author of the textbook. I work in his lab, actually. 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 mergedTwo Questions ambros: 1) Do you think that a wire of length 1m with steady constant current I will have a different magnetic field than a wire of length 2m? 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. 2) When you are solving for, what you call, the total magnetic force, why are you adding up the individual forces from 2 different objects and why is one of them not negative? 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. If you choose to ignore them and not answer directly, this discussion is going to continue to be circular. 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. Edited April 4, 2010 by ambros Consecutive posts merged.
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