Cap'n Refsmmat Posted April 4, 2010 Posted April 4, 2010 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? Nope. I'm saying that if you use the equation they give, you will get the result I showed. Ampere's law works the same way: [math]\int_C \vec{B} \cdot d\vec{l} = \mu_0 I[/math] The magnetic field around a wire is circular around it. We choose the line integral to go in one circle around the wire, at a radius R from the wire. [math]\int_C \vec{B} \, d\vec{l} = \int_C \vec{B} R \, d\theta = \int_0^{2\pi} \vec{B} R \, d\theta = 2\pi R B = \mu_0 I[/math] [math]B = \frac{\mu_0 I}{2 \pi R}[/math] Physics works!
swansont Posted April 4, 2010 Posted April 4, 2010 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? Not at all. What everyone here has said is that you aren't applying the Biot-Savart law correctly, which is why you are getting the wrong result. The Biot-Savart page doesn't show the equation for a long wire because it doesn't show the solution for a long wire. The Ampere's force law page does show the force of two long wires, and it's a 1/r relationship (which is what you get if you combine the magnetic field equation with the force equation, F=ILxB) 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? How can you find the field from A and the force it exerts at B, without angles? =======A=============|<---dl1--->|====================>> I1 | | |r | | =====================|<---dl2--->|========B===========>> I2
darkenlighten Posted April 4, 2010 Posted April 4, 2010 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. No its not clear to you, because this is incorrect. The length contributes and yes there is such a variable [math] \int d\mathbf{l} [/math]. You claim to understand vector calculus but you have been wrong about this the whole time. At point "P" away from the wire there is a magnitude of the magnetic field that is caused by each segment of wire, not just by the segment below it, otherwise it would make it act no different than one moving charge, which is obviously not true. For example: take multiple moving charges equally spaced apart traveling at the same speed (a steady current) and confine them in 1 dimension (a straight line). They all have their own magnetic field, but if you look at a point in space, the total magnetic field at that point will be due to the superposition of all of the moving charges magnetic fields. Since their magnetic fields depend on distance away, this correlates to the length of a wire. 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. Technically it is more than one field, due to the example above (about multiple moving charges). 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. That is not what it means when it says the forces between the wires. Is one force not left and the other right, hence one negative and one positive? If you sum them, you will get zero. 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. No you are misunderstanding what it means "the force between the wires". What it means is the force that one wire feels due to the external magnetic field of the other wire. Is there any question that I ignored or did not answer directly? It is you who still needs to provide some reference for those equations supposedly derived from Maxwell's, or you could simply say that you were mistaken. Once again I have already provided reference, but you refuse to accept it, why?
ambros Posted April 4, 2010 Author Posted April 4, 2010 [math]B = \frac{\mu_0 I}{2 \pi R}[/math] Physics works! That's mathematics. This is physics: -"The ampere is that constant current which' date=' if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to [b']2 * 10^–7 newton per metre of length[/b]." r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 --------------------------------------------- B1= ?; B2 = ?; F(total)= ? If that equation is PHYSICALLY correct you ought to be able to actually use it on this case scenario and predict the correct result of this experiment. Will you please demonstrate whether you can indeed obtain the correct result with your equation?
Cap'n Refsmmat Posted April 4, 2010 Posted April 4, 2010 Using the definition of force on a wire: For a straight wire of length L in a uniform magnetic field, the contributions simply add to give a force of: [math]F = ILB \, \sin \alpha[/math] So, let's apply what we've learned. I = 1 amp R = 1 meter [math]B = \frac{\mu_0 I}{2\pi R} = 2 \times 10^{-7}[/math] Now, if we convert the quoted equation to force per unit length: [math]\frac{F}{L} = IB \, \sin \alpha[/math] and consider that the magnetic field must be perpendicular to the wire, so [imath]\sin \alpha[/imath] is 1, it's very clear that [math]\frac{F}{L} = 2 \times 10^{-7} N/m[/math] Good enough for you?
ambros Posted April 4, 2010 Author Posted April 4, 2010 Once again I have already provided reference, but you refuse to accept it, why? No, I'm not talking about Biot-Savart law and imaginary scenarios, I'm talking about these four monsters that you supposedly DERIVED FROM MAXWELL'S EQUATIONS: Have you derived these from Maxwell's equations and will you provide some reference finally? Merged post follows: Consecutive posts mergedUsing the definition of force on a wire: So, let's apply what we've learned. I = 1 amp R = 1 meter [math]B = \frac{\mu_0 I}{2\pi R} = 2 \times 10^{-7}[/math] Now, if we convert the quoted equation to force per unit length: [math]\frac{F}{L} = IB \, \sin \alpha[/math] and consider that the magnetic field must be perpendicular to the wire, so [imath]\sin \alpha[/imath] is 1, it's very clear that [math]\frac{F}{L} = 2 \times 10^{-7} N/m[/math] Good enough for you? I like the "F/L" term, but unfortunately that's not complete, just like darkenlighten's solution. Why are you calculating only the force for one wire, do you think that is what is meant by "force between conductors"? -- I do not know what to say, where is that guy 'Bignose' - Maths Expert? You see, even if that was supposed to be the complete result for this scenario, that equation would still contradict the real one, so do you think this equation is wrong: http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law and consider that the magnetic field must be perpendicular to the wire, so [math]\sin \alpha[/math] is 1 That's right.
Cap'n Refsmmat Posted April 4, 2010 Posted April 4, 2010 I like the "F/L" term, but unfortunately that's not complete, just like darkenlighten's solution. Why are you calculating only the force for one wire, do you think that is what is meant by "force between conductors"? -- I do not know what to say, where is that guy 'Bignose' - Maths Expert? You see, even if that was supposed to be the complete result for this scenario, that equation would still contradict the real one, so do you think this equation is wrong: http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law It is the complete result for this scenario. Each wire feels that force -- the same force, because each wire carries the same current. So there is a force of that magnitude between the two conductors. Using the Wikipedia article you just linked, the version of Ampere's force law that applies for two infinitely long, straight parallel wires is: [math] F_m = 2 k_A \frac {I_1 I_2 } {r}[/math] [math]F_m = \frac{\mu_0}{2\pi} \frac{1 \cdot 1}{1} = 2 \times 10^{-7} N / m[/math] Same answer. You do realize that it's possible for two equations to give the same result, don't you? They just approach the problem from different sides. The fact is that if you apply the Biot-Savart law to an infinitely long wire, you'll get a simple formula, as I have posted before. Just do the integral. Why must you make this so complicated?
darkenlighten Posted April 4, 2010 Posted April 4, 2010 No, I'm not talking about Biot-Savart law and imaginary scenarios, I'm talking about these four monsters that you supposedly DERIVED FROM MAXWELL'S EQUATIONS: Have you derived these from Maxwell's equations and will you provide some reference finally? Okay, though I've repeated myself. The first 2 equations, the E field I have there can be ignored, since we have already corrected that saying that the E field is zero, E = 0. The B field there is correct with the reference of the scanned pages of Griffith's text showing the full derivation using the Biot-Savart Law. I used Maxwell's equations to derive everything there, I don't need a reference using Maxwell's because my math is not wrong, so the reference I have is to show I have the correct answer. As far as changing B field scenario, I do not have a reference since that was a scenario I made up, but since you do not believe the previous answer, you will not believe the last 2 I gave here. Which you apparently had a problem with a changing Magnetic field creating an Electric field...which is silly.
ambros Posted April 4, 2010 Author Posted April 4, 2010 It is the complete result for this scenario. Each wire feels that force -- the same force' date=' because each wire carries the same current. So there is a force of that magnitude between the two conductors. [/quote'] Your wording is very ambiguous and you are not using proper notation. Force is always calculated between TWO objects (fields), so there is always [math]F_{12}[/math] and [math]F_{21}[/math]. You also need to make clear all your other notation, like what do you mean by [math]F_m[/math], that usually represent the total magnitude of two forces, i.e. [math]F_{12} + F_{21}[/math]. Cap'n Refsmmat: - "Each wire feels that force" 1.) There is force [math]F_{12}[/math] and force [math]F_{21}[/math], yes? Cap'n Refsmmat: - "So there is a force of that magnitude between the two conductors." 2.) What magnitude? Just the force of one wire, say: [math]F_{12}[/math], gives the complete result of the force BETWEEN TWO conductors? -OR- Do we need the total magnitude, say: [math]F_m = F_{12} + F_{21}[/math], alternatively for this particular case: [math]2 * F_{12}[/math]? Using the Wikipedia article you just linked, the version of Ampere's force law that applies for two infinitely long, straight parallel wires is: [math] F_m = 2 k_A \frac {I_1 I_2 } {r}[/math] [math]F_m = \frac{\mu_0}{2\pi} \frac{1 \cdot 1}{1} = 2 \times 10^{-7} N / m[/math] Same answer. You do realize that it's possible for two equations to give the same result, don't you? They just approach the problem from different sides. That is not the equation I gave. Look again, and pay attention to "4Pi" and "r^2". The Ampere's force law has nothing to do with your solution using Lorentz force. The formula I gave is Lorentz force and Biot-Savart law combined: [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2} [/math] http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law *** TWO PARALLEL WIRES: r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 --------------------------------------------- F(between two wires) = ? [math]\mathbf{F}_{12} = \frac{\mu_0}{4\pi} \frac{1}{r^2} = 10^{-7} N/m[/math] [math]\mathbf{F}_m = \mathbf{F}_{12} + \mathbf{F}_{21} = 2* \mathbf{F}_{12} = 2 * 10^{-7} N/m[/math] That is how it is done. -- If you were to calculate the displacement between these two wires, would you calculate that with just one force or would you include the both forces to calculate the total displacement due to attraction? The fact is that if you apply the Biot-Savart law to an infinitely long wire, you'll get a simple formula, as I have posted before. Just do the integral. Why must you make this so complicated? The length of the wire has absolutely no impact, nor is considered in any of these equations. Length of the wire is not the same thing as segment "dl". -- In most cases these wires are supposed to be infinite anyway, but not because the length matters in any way to the magnitude of the B field in relation to distance, only to avoid dealing with the edges of the magnetic field "cylinder". [math] F_m = 2 k_A \frac {I_1 I_2 } {r}[/math] [math]F_m = \frac{\mu_0}{2\pi} \frac{1 \cdot 1}{1} = 2 \times 10^{-7} N / m[/math] Same answer. You do realize that it's possible for two equations to give the same result, don't you? They just approach the problem from different sides. I gave you completely different equation, plus we were talking about your solution where you calculate B field and force with the Lorentz force. But, in any case, that is very strange way to make that derivation. You are skewing the meanings and the value of the multiplier and constant. [math]F_m = 2* \frac{\mu_0}{4\pi} \frac{1}{1} = 2 *10^{-7}N/m[/math]
swansont Posted April 4, 2010 Posted April 4, 2010 Your wording is very ambiguous and you are not using proper notation. Force is always calculated between TWO objects (fields), so there is always [math]F_{12}[/math] and [math]F_{21}[/math]. You also need to make clear all your other notation, like what do you mean by [math]F_m[/math], that usually represent the total magnitude of two forces, i.e. [math]F_{12} + F_{21}[/math]. No that is not normally meaningful. [math]F_{12} = F_{21}[/math] by virtue of Newton's third law, and the force on an object is what determines its motion. Merged post follows: Consecutive posts merged The length of the wire has absolutely no impact, nor is considered in any of these equations. Length of the wire is not the same thing as segment "dl". -- In most cases these wires are supposed to be infinite anyway, but not because the length matters in any way to the magnitude of the B field in relation to distance, only to avoid dealing with the edges of the magnetic field "cylinder". No, this is not correct. The magnitude of the field at any point is the result of contributions from the length of the wire. A shorter wire will give you a weaker field. This is why the BIPM definition of current references infinite wires.
ambros Posted April 4, 2010 Author Posted April 4, 2010 (edited) No that is not normally meaningful. [math]F_{12} = F_{21}[/math] by virtue of Newton's third law, and the force on an object is what determines its motion. Uh, ah, there is a whole another discussion there, about "equal" and "opposite" direction of those magnetic forces, but it is not necessary to go there in order to resolve these simple scenarios, fortunately, so I'll skip that. http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}[/math] So, you are saying this equation is wrong after all? EDIT: ...I just realized the internet is actually full of web-pages to support the crazy formula. That's simply insane because in the same time all the pages talking specifically about Biot-Savart law and current carrying wires will agree with me and make clear what is the actual equation for that particular case, such as all the Wikipedia articles on the subject. How can some crazy derivation make those two equal? One equation must be wrong, which one? Wiki: - "The Biot–Savart law is used to compute the magnetic field generated by a steady current, for example through a wire... The equation in SI units is: [math]\mathbf{B} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2}[/math] -not this-> [math]B = \frac{\mu_0 I}{2\pi R}[/math] The length of the wire has absolutely no impact, nor is considered in any of these equations. Length of the wire is not the same thing as segment "dl". No, this is not correct. The magnitude of the field at any point is the result of contributions from the length of the wire. A shorter wire will give you a weaker field. This is why the BIPM definition of current references infinite wires. Is the "wire length" to you the same thing as wire segment "dl"? If a shorter wire give weaker field, then infinite wire gives infinite field? Edited April 4, 2010 by ambros
swansont Posted April 4, 2010 Posted April 4, 2010 (edited) http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}[/math] So, you are saying this equation is wrong after all? No. I don't think anyone has claimed that this equation is wrong. EDIT: ...I just realized the internet is actually full of web-pages to support the crazy formula. That's simply insane because in the same time all the pages talking specifically about Biot-Savart law and current carrying wires will agree with me and make clear what is the actual equation for that particular case, such as all the Wikipedia articles on the subject. How can some crazy derivation make those two equal? One equation must be wrong, which one? They are both correct. Wiki: - "The Biot–Savart law is used to compute the magnetic field generated by a steady current, for example through a wire... The equation in SI units is: [math]\mathbf{B} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2}[/math] -not this-> [math]B = \frac{\mu_0 I}{2\pi R}[/math] When you solve the integral (properly) for the case of the infinite wire, these two equations are equivalent. Is the "wire length" to you the same thing as wire segment "dl"? If a shorter wire give weaker field, then infinite wire gives infinite field? No wire length is not the same as dl. [math]L = \int dl[/math] I'm sorry, I mis-stated what I meant to say. A smaller wire gives you a smaller force; I was referring to the BIPM statement, which talks about the force between wires. It's the results of the Biot-Savart law tells you that the field is weaker for a shorter length of wire. An infinite wire gives a finite field, [math]B = \frac{\mu_0 I}{2\pi r}[/math] Edited April 4, 2010 by swansont
ambros Posted April 4, 2010 Author Posted April 4, 2010 (edited) [math]\mathbf{B} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} =equals ??=> B = \frac{\mu_0 I}{2\pi R}[/math] They are both correct. When you solve the integral (properly) for the case of the infinite wire' date=' these two equations are equivalent. ...It's the results of the Biot-Savart law tells you that the field is weaker for a shorter length of wire. An infinite wire gives a finite field, [math']B = \frac{\mu_0 I}{2\pi r}[/math] I solved it couple of times by now, there is nothing to integrate as Cap'n Refsmmat demonstrated with his "F/L" term - both of these: "dl x r" are unit vectors, there are no any angles but 90 degree, and dB(r, i=const)=0, so it is linear function. -- It's as if that crazy equation on the right is product of some "proper integration", it is not. It is derivation according to some fictional scenario that has no bearing in reality: Our scenario looks like this: =====================|----dl1--->|====================>> i1 | | |r | | alpha= 90 =====================|----dl2--->|====================>> i2 ...hence, the correct "shortcut" is this: [math] \mathbf{B} = \frac{\mu_0 q \mathbf{v}}{4\pi} \times \frac{\mathbf{\hat r}}{r^2} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} =wires=> \frac{\mu_0 I}{4\pi r^2} [/math] ...why? We must always remember what Cap'n Refsmmat told us: - "magnetic field must be perpendicular to the wire, so sin(alpha) is 1". [math] B®= \frac{\mu_0 I d\mathbf{l} \times \mathbf{r}}{4\pi r^2} = \frac{\mu_0 I sin(alpha)}{4\pi r^2} = \frac{\mu_0 I sin(90)}{4\pi r^2} = \frac{\mu_0 I * 1}{4\pi r^2} = \frac{\mu_0 I}{4\pi r^2} [/math] No. I don't think anyone has claimed that this equation is wrong. Well, it's a little bit wrong, so I will use THE EQUATION: - "the official documentation regarding definition of the ampere BIPM SI Units brochure, 8th Edition, PAGE 105": http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf STEP 1: small modifications... [math]F_{12} = \frac {\mu_0} {4 \pi} \oint_{C_1} \oint_{C_2} \frac {I_1*d \mathbf{l_1}\ \mathbf{ \times} \ (I_2*d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] STEP 2: pif, paf, puf... No wire length is not the same as dl. [math]L = \int dl[/math] [math]F_{12} = \frac {\mu_0} {4 \pi} \frac {I_1*L_1 \mathbf{ \times} \ (I_2*L_2 \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] -OR- [math]\frac {F_{12}}{L} = \frac {\mu_0} {4 \pi} \frac {I_1 \mathbf{ \times} \ (I_2 \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] STEP 3: solve for r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 [math]F_{12} = \frac {\mu_0} {4 \pi} = 10^{-7} N[/math] -OR- [math]\frac {F_{12}}{L} = \frac {\mu_0} {4 \pi} = 10^{-7} N/m[/math] Conclusion: - to get the correct result for the combined force between the two wires according to the definition of unit ampere, we need to use the multiplier '2', i.e. we need to include this force too: [math]F_{21}[/math] Edited April 4, 2010 by ambros
Cap'n Refsmmat Posted April 4, 2010 Posted April 4, 2010 Well, it's a little bit wrong, so I will use THE EQUATION: - "the official documentation regarding definition of the ampere BIPM SI Units brochure, 8th Edition, PAGE 105": http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf STEP 1: small modifications... [math]F_{12} = \frac {\mu_0} {4 \pi} \oint_{C_1} \oint_{C_2} \frac {I_1*d \mathbf{l_1}\ \mathbf{ \times} \ (I_2*d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] STEP 2: pif, paf, puf... [math]F_{12} = \frac {\mu_0} {4 \pi} \frac {I_1*L_1 \mathbf{ \times} \ (I_2*L_2 \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] -OR- [math]\frac {F_{12}}{L} = \frac {\mu_0} {4 \pi} \frac {I_1 \mathbf{ \times} \ (I_2 \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] STEP 3: solve for r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 [math]F_{12} = \frac {\mu_0} {4 \pi} = 10^{-7} N[/math] -OR- [math]\frac {F_{12}}{L} = \frac {\mu_0} {4 \pi} = 10^{-7} N / m[/math] Conclusion: - to get the correct result for the combined force between the two wires according to the definition of unit ampere, we need to use the multiplier '2', i.e. we need to include this force too: [math]F_{21}[/math] This is completely wrong. [math]\oint_{C_1}[/math] denotes an integral along the line, and you did not integrate along the line. You didn't even integrate, really. You said you know vector calculus. Do the line integrals! Merged post follows: Consecutive posts mergedI solved it couple of times by now, there is nothing to integrate as Cap'n Refsmmat demonstrated with his "F/L" term - both of these: "dl x r" are unit vectors, there are no any angles but 90 degree, and dB(r, i=const)=0, so it is linear function. -- It's as if that crazy equation on the right is product of some "proper integration", it is not. It is derivation according to some fictional scenario that has no bearing in reality: You'll notice that I did, in fact, compute the integral, before then using the result of the integral to get the correct answer.
ambros Posted April 4, 2010 Author Posted April 4, 2010 This is completely wrong. [math]\oint_{C_1}[/math] denotes an integral along the line, and you did not integrate along the line. You didn't even integrate, really. You said you know vector calculus. Do the line integrals! Yes I did, did you? Do you not see the L is "LINE INTEGRAL", 1m in length? I did the same thing as you and exactly what Swansont wrote: L = integral(dl) 1.) can you please say what exactly is wrong? 2.) please show how do you integrate this equation:
Cap'n Refsmmat Posted April 4, 2010 Posted April 4, 2010 Yes I did, did you? 1.) can you say what exactly is wrong? Yes. r is not a constant, but you treated it as one. 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. They're not all at the same distance away from A.) 2.) why do I keep getting the correct result? Because you fudged it and said "oh, I have to multiply by two!" 3.) please show how do you integrate this equation: I already showed you how I would solve this problem. Can you show me the source of this equation so I know what each variable is supposed to represent?
ambros Posted April 4, 2010 Author Posted April 4, 2010 Yes. r is not a constant' date=' but you treated it as one. [/quote'] Two parallel wires, 'r' IS constant by definition. "MY EQUATION": [math]\mathbf{B} = \frac{\mu_0 I}{4\pi r^2}[/math] "YOUR EQUATION": [math]\mathbf{B} = \frac{\mu_0 I}{2\pi R}[/math] Everything is constant here and the only difference in these equations is "4Pi" and "r^2", but since r is 1m in this example, that means the only difference is in 2Pi = 6.283. -- "My" equation treats everything just the same as "yours", except that "mine" is more accurate, especially with distances different than 1m. 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. They're not all at the same distance away from A.) Superposition again? These equations do not do that - this is not numerical integration or time integral, these are 'line integrals'. "MY EQUATION": [math]\mathbf{B} = \frac{\mu_0 I}{4\pi r^2}[/math] "YOUR EQUATION": [math]\mathbf{B} = \frac{\mu_0 I}{2\pi R}[/math] Where do you see anything like what you describe? Did you not say magnetic angle in wire must always be perpendicular, so how can distance change if wires are supposed to be parallel and magnetic fields perpendicular, what is there to change, and why would it change in "your" but not in "my" equation? Because you fudged it and said "oh, I have to multiply by two!" Please, I showed all the steps and we are talking about mathematics here, so if something does not compute then it does not compute, and you should be able to point exactly what and where. I already showed you how I would solve this problem. Can you show me the source of this equation so I know what each variable is supposed to represent? All the variables are visible from the diagram and given by the definition of ampere unit, alternatively explained in that Wikipedia article. -- You have not solved this one yet. We are now talking about the very specific equation, which is THE DEFINITION, this is it, to be or not to be: - "the official documentation regarding definition of the ampere BIPM SI Units brochure, 8th Edition, PAGE 105": http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf ...alternatively: If you mean to claim my integration is wrong, then again, you need to first solve it yourself so we all can see what is the difference, what is wrong and what is right. This is the moment of truth, you really can not get but one result if you do it correctly here, and if you do, then the result will be the same as mine. =====================|----dl1--->|====================>> I1 | | |r | | =====================|----dl2--->|====================>> I2 AMPERE UNIT: r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 ------------------------------------------------------------- [math]F_{12} = \frac {\mu_0} {4 \pi} \oint_{C_1} \oint_{C_2} \frac {I_1*d \mathbf{l_1}\ \mathbf{ \times} \ (I_2*d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] Pick your favorite form of this equation and solve it please.
Cap'n Refsmmat Posted April 4, 2010 Posted April 4, 2010 (edited) Two parallel wires, 'r' IS constant by definition. "MY EQUATION": [math]\mathbf{B} = \frac{\mu_0 I}{4\pi r^2}[/math] "YOUR EQUATION": [math]\mathbf{B} = \frac{\mu_0 I}{2\pi R}[/math] Everything is constant here and the only difference in these equations is "4Pi" and "r^2", but since r is 1m in this example, that means the only difference is in 2Pi = 6.283. -- "My" equation treats everything just the same as "yours", except that "mine" is more accurate, especially with distances different than 1m. No, your equation was a line integral, and r is not constant in it. R (the distance between the wires) is constant by definition. Superposition again? These equations do not do that - this is not numerical integration or time integral, these are 'line integrals'. "MY EQUATION": [math]\mathbf{B} = \frac{\mu_0 I}{4\pi r^2}[/math] "YOUR EQUATION": [math]\mathbf{B} = \frac{\mu_0 I}{2\pi R}[/math] Where do you see anything like what you describe? Did you not say magnetic angle in wire must always be perpendicular, so how can distance change if wires are supposed to be parallel and magnetic fields perpendicular, what is there to change, and why would it change in "your" but not in "my" equation? Ooh, "line integrals." Scary. Of course, I took vector calculus just last semester, and passed with an A... So, here's the differing distances: -------------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. Please, I showed all the steps and we are talking about mathematics here, so if something does not compute then it does not compute, and you should be able to point exactly what and where. Multiplying by two does not compute. You do not understand what the force "between" two wires even means - there is no multiplication by two required. For example, we can say that the force "between" two massive objects is given by Newton's law of gravitation, but that doesn't mean we multiply the result of the formula by 2. It means that each object experiences that force. All the variables are visible from the diagram and given by the definition of ampere unit, alternatively explained in that Wikipedia article. -- You have not solved this one yet. We are now talking about the very specific equation, which is THE DEFINITION, this is it, to be or not to be: - "the official documentation regarding definition of the ampere BIPM SI Units brochure, 8th Edition, PAGE 105": http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf Which Wikipedia article? Let me know where this double integral formula comes from. (edit: never mind, I found it. I'll see what I can do.) Merged post follows: Consecutive posts merged[math]F_{12} = \frac {\mu_0} {4 \pi} \oint_{C_1} \oint_{C_2} \frac {I_1*d \mathbf{l_1}\ \mathbf{ \times} \ (I_2*d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] Pick your favorite form of this equation and solve it please. Okay, I'll pick my favorite form from the Wikipedia article: 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 is [math]F_m = 2 k_a \frac{I_1 I_2}{r}[/math] [math]k_a = \frac{\mu_0}{4\pi}[/math] So... [math]F_m = \frac{2 \mu_0 I_1 I_2}{4 \pi r}[/math] [math]F_m = \frac{\mu_0 I_1 I_2}{2 \pi r}[/math] For [imath]I_1 = I_2 = 1[/imath] and [imath]r=1[/imath], [math]F_m = \frac{\mu_0}{2\pi} = 2 \times 10^{-7} \mbox{ N/m} [/math] Notice that this is the force exerted on one wire by the other, not the sum of the forces between both. Merged post follows: Consecutive posts merged "MY EQUATION": [math]\mathbf{B} = \frac{\mu_0 I}{4\pi r^2}[/math] Derive this from scratch (from Biot-Savart or Ampere) for me please. Edited April 4, 2010 by Cap'n Refsmmat Consecutive posts merged.
swansont Posted April 4, 2010 Posted April 4, 2010 I did the same thing as you and exactly what Swansont wrote: L = integral(dl) You asked if L and dl were the same. But the Biot-Savart law equation contains dl X r, which is a cross product. When you integrate along the length, r varies and the angle varies, as a function of position along L. You are choosing only the element for which the angle is 90 degrees, which means you are ignoring most of the wire. It's flat-out wrong. Check your units. Are they the units for B?
darkenlighten Posted April 4, 2010 Posted April 4, 2010 I'm going to state this again...( I said this in a previous post) Hopefully this will understand why you cannot look at just 1 element of dl, but the sum of all dl's (hence an integral) At point "P" away from the wire there is a magnitude of the magnetic field that is caused by each segment of wire, not just by the segment below it, otherwise it would make it act no different than one moving charge, which is obviously not true. For example: take multiple moving charges equally spaced apart traveling at the same speed (a steady current) and confine them in 1 dimension (a straight line). They all have their own magnetic field, but if you look at a point in space, the total magnetic field at that point will be due to the superposition of all of the moving charges magnetic fields. Since their magnetic fields depend on distance away, this correlates to the length of a wire.
Cap'n Refsmmat Posted April 4, 2010 Posted April 4, 2010 AMPERE UNIT: r= 1m; dl1=dl2= 1m; I1=I2= 1A; µ0= 4π*10^-7 Also, dl is an infinitesimal value (hence the "d"). It cannot be 1m.
darkenlighten Posted April 4, 2010 Posted April 4, 2010 Also to make it clear: [math] q \mathbf{v} = q \frac{d\mathbf{x}}{dt} = \frac{dq}{dt}\mathbf{x} = I \int{d\mathbf{l}} [/math]
ambros Posted April 5, 2010 Author Posted April 5, 2010 (edited) Also' date=' dl is an infinitesimal value (hence the "d"). It cannot be 1m. [/quote'] It makes no difference, dl can indeed be 1m. Call it L or "wire length" if you like, then: L= 1m. Okay' date=' I'll pick my favorite form from the Wikipedia article: [math'] F_m = 2 k_a \frac{I_1 I_2}{r} [/math] Wrong, you picked up completely different equation. Amere's force law does not define the unit of ampere, but LORENTZ FORCE and BIOT-SAVART LAW, as is clear from the Wikipedia article and from THE DEFINITION given by BIPM SI Units brochure, 8th Edition, PAGE 105": http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf [math]F_{12} = \frac {\mu_0} {4 \pi} \oint_{C_1} \oint_{C_2} \frac {I_1*d \mathbf{l_1}\ \mathbf{ \times} \ (I_2*d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] You have to use the correct definition, try again. -- All the other comments are completely unnecessary, and though amusing I've responded several times to each of those, the truth will be clear to anyone who actually starts using these equations on practical case scenarios, based on physics and real world, not hallucinations and false mathematics. Derive this from scratch (from Biot-Savart or Ampere) for me please. B(r) | | |r | | angle=90 ======================|---dl1--->|====================>> I1 | | |<---L1--->| ...hence' date=' the correct "shortcut" is this: [math']\mathbf{B} = \frac{\mu_0 q \mathbf{v}}{4\pi} \times \frac{\mathbf{\hat r}}{r^2} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} =wires=> \frac{\mu_0 I}{4\pi r^2}[/math] ...why? We must always remember what Cap'n Refsmmat told us: - "magnetic field must be perpendicular to the wire, so sin(alpha) is 1". [math]B®= \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} = \frac{\mu_0 I sin(alpha)}{4\pi r^2} = \frac{\mu_0 I sin(90)}{4\pi r^2} = \frac{\mu_0 I * 1}{4\pi r^2} = \frac{\mu_0 I}{4\pi r^2}[/math] Do not agree - fine, just solve the unit of ampere scenario from above and you should realize what equation is more correct, just please stop making any more assertions before you actually solve those two parallel wires as that is the only thing that is 100% unambiguous here, unfortunately. Merged post follows: Consecutive posts mergedI'm going to state this again... What angle? Cap'n Refsmmat: - "magnetic field must be perpendicular to the wire, so sin(alpha) is 1". You are being ignored because your arrogance wasted my time. Do you see how is the force defined by BIPM SI Units brochure, 8th Edition, PAGE 105": http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf ?? Stop talking and start learning. Do you think you are able to use THE definition and solve for two parallel wires and unit of ampere? I'm only interested in physics, not crazy theories, silly insults and hand-waving. Edited April 5, 2010 by ambros Consecutive posts merged.
Cap'n Refsmmat Posted April 5, 2010 Posted April 5, 2010 Wrong, you picked up completely different equation. Amere's force law does not define the unit of ampere, but LORENTZ FORCE and BIOT-SAVART LAW, as is clear from the Wikipedia article and from THE DEFINITION given by BIPM SI Units brochure, 8th Edition, PAGE 105": http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf [math]F_{12} = \frac {\mu_0} {4 \pi} \oint_{C_1} \oint_{C_2} \frac {I_1*d \mathbf{l_1}\ \mathbf{ \times} \ (I_2*d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] http://en.wikipedia.org/wiki/Ampere%27s_force_law The best-known and simplest example of Ampère's force law, which underlies the definition of the ampere, the SI unit of current, is as follows: 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 is [math]F_m = 2k_A \frac{I_1 I_2}{r}[/math] It's still Ampere's force law. [math]B®= \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} = \frac{\mu_0 I sin(alpha)}{4\pi r^2} = \frac{\mu_0 I sin(90)}{4\pi r^2} = \frac{\mu_0 I * 1}{4\pi r^2} = \frac{\mu_0 I}{4\pi r^2}[/math] Do not agree - fine, just solve the unit of ampere scenario from above and you should realize what equation is more correct, just please stop making any more assertions before you actually solve those two parallel wires as that is the only thing that is 100% unambiguous here, unfortunately. I have shown you earlier how I derived "my" formula. Can you find me sources that agree with you explicitly, and state the magnetic field created by a wire as you do? I can provide many that agree that you're wrong. http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html http://www.pa.msu.edu/courses/1997spring/PHY232/lectures/ampereslaw/wire.html http://webphysics.davidson.edu/physlet_resources/bu_semester2/c14_long.html And my textbook, as cited earlier. Any you have that agree with you? Merged post follows: Consecutive posts mergedStop talking and start learning. Do you think you are able to use THE definition and solve for two parallel wires and unit of ampere? I'm only interested in physics, not crazy theories, silly insults and hand-waving. You took the words right out of my mouth.
ambros Posted April 5, 2010 Author Posted April 5, 2010 (edited) You asked if L and dl were the same. But the Biot-Savart law equation contains dl X r, which is a cross product. When you integrate along the length, r varies and the angle varies, as a function of position along L. You are choosing only the element for which the angle is 90 degrees, which means you are ignoring most of the wire. It's flat-out wrong. Check your units. Are they the units for B? I checked everything. Now, you have to check it yourself. -- There is no need to be wondering about units, both equations have the same units but there is only one valid relation to correctly describe the magnitude of magnetic field B, of straight current carrying wire, in relation to distance: [math]B® = \frac{\mu_0 I}{2\pi r}[/math] -OR- [math]B® = \frac{\mu_0 I}{4\pi r^2}[/math] ...and to find out which one is correct you have to use THE DEFINITION given by the -BIPM SI Units brochure, 8th Edition, PAGE 105- solve it for the given values that define the unit of ampere, and see for yourself which one can produce that exact result. Merged post follows: Consecutive posts merged http://en.wikipedia.org/wiki/Ampere%27s_force_law It's still Ampere's force law. I have shown you earlier how I derived "my" formula. Can you find me sources that agree with you explicitly' date=' and state the magnetic field created by a wire as you do? And my textbook, as cited earlier. Any you have that agree with you? [/quote'] No, those equations are different and you will only know that if you actually use it. I repeat, "your" formula is not the same as THE DEFINITION GIVEN by the BIPM SI Units brochure, 8th Edition, PAGE 105. You also said my integration of it was wrong, so you need to show what you believe is the "correct" integration by applying it to the two parallel wires and unit of ampere setup, and see if it will indeed give the same result as "your" formula. -- Prove your assertion, argument your opinion, demonstrate your theory and show your work. It's a few lines of math, can you do that please? Edited April 5, 2010 by ambros Consecutive posts merged.
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