Cap'n Refsmmat Posted April 8, 2010 Posted April 8, 2010 This is ridiculous. post #23: http://www.scienceforums.net/forum/showpost.php?p=556655&postcount=23 [math] \mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2} [/math] Cap'n Refsmmat: ...gives us the force on a piece of wire -- its units are Newtons, not Newtons per meter. * Ampere's force law IS in N/m. Fm =MEANS= F/1meter * That equation given by BIPM is in Newtons (N) and that is NOT Ampere's force law but Biot-Savart-law and Loerntz force as demonstrated above. Please learn to read. You were talking about this equation: [math] B = \frac{\mu_0 I}{2\pi r} [/math] and that is the equation my comment is relevant to. The above equation has units of Newtons per meter. (Well, the proper form of it does. The numerator should have the currents of both wires in it, not just one wire, but I believe it was set up for a situation where the wires have equal currents.) Incidentally, it is Ampere's force law. Take a look on Wikipedia or on this site. Incidentally, that link plugs the numbers in and solves for the force on one wire. You might be interested.
Mr Skeptic Posted April 8, 2010 Posted April 8, 2010 Can you print out the complete code for the complete equation? -- There is no 't', no time, as I tried to explain at the beginning. I'm the only one here who actually got any result that makes sense from that equation, and I did it in three different ways, here are two: And if you had read what I said, you'd know that t is not time -- it is a generic variable. Also, if you use l(t) = t (ie a straight line) then it turns out equal to a regular integral, and if you have your function as a conservative force then the path doesn't matter (both give the same answer). If you have a closed loop, you will easily notice how very different it is to do a path integral or not. TWO PARALLEL WIRES, ampere unit setup: I1=I2= 1A; r= 1m But what are C1 and C2? Are you using infinitely long straight wires, or 1 m straight wires, or a loop the shape of the milky way? You can't get an answer without defining your paths. As for your results, you must not be using infinitely long wires because if so you are getting the wrong results: http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law
ambros Posted April 8, 2010 Author Posted April 8, 2010 (edited) These are not paths. They are direction vectors. The paths are C1 and C2. Direction vectors defining direction of the current, hence defining the position of wires, which instead of being two are now one wire with twice the el. current. -- That explains it, you're simply calculating DOUBLE magnetic in one wire. That term next to cross product equals to one, do you not see it is the same above and below? You can not evaluate upper half of it without lower, the s1 and s2 can not be different above from below, so that term equals one (+/-). Edited April 8, 2010 by ambros
darkenlighten Posted April 8, 2010 Posted April 8, 2010 You are dreaming. Go ahead, show us that post where we can see how you put REAL VARIABLES of: I1= 1A, I2= 1A and r=1m, in THIS EQUATION: [math] \mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2} [/math] What post, what number? What say you? postcount 47 for the correct derivation of the B field (just look at the scanned text) postcount 59 Shows the straight forward method postcount 122 showing you how the equation from BIPM was derived.
ambros Posted April 8, 2010 Author Posted April 8, 2010 Please learn to read. You were talking about this equation: [math] B = \frac{\mu_0 I}{2\pi r} [/math] and that is the equation my comment is relevant to. The above equation has units of Newtons per meter. (Well, the proper form of it does. The numerator should have the currents of both wires in it, not just one wire, but I believe it was set up for a situation where the wires have equal currents.) Incidentally, it is Ampere's force law. Take a look on Wikipedia or on this site. Huh?? That is magnetic field, there is no force there. How could you be talking about N or N/m in relation to magnetic field equation? -- What is your point then, what are you saying, can you summarize what is it you believe is wrong and what is correct? It is 1Tesla = 1N/A*m.
Cap'n Refsmmat Posted April 8, 2010 Posted April 8, 2010 You are not making sense that is magnetic field there is no force? How could you be talking about N or N/m in relation to magnetic field equation? -- What is your point then, can you summarize what is it you believe is wrong and what is correct? Whoops. Sorry, that's the equation for the magnetic field around an infinite wire. There's no length because it's for an infinitely long wire. If you want to calculate the magnetic field for a finite wire, use Biot-Savart.
ambros Posted April 8, 2010 Author Posted April 8, 2010 (edited) postcount 47 for the correct derivation of the B field (just look at the scanned text)postcount 59 Shows the straight forward method postcount 122 showing you how the equation from BIPM was derived. darkenlighten: - I've put in real variables/constants to show you how you get 2 x 10^-7 N per unit length. In first two links there is no even mention of that equation, but this: [math]\mathbf{B}_2 = \frac{\mu_0 I_2}{2\pi s} \hat{\phi} ; \mathbf{F}_1 = I_1 \int{d\mathbf{l} \times \mathbf{B}_2} \Rightarrow F_1 = \frac{\mu_0 I_1 I_2}{2\pi s} \mathbf{\hat{r}} [/math]per unit length ...third link has this equation at the very, not used to do any calculation or plug in any values in. [math] \mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2} [/math] Well done, Alice, a man with no shame. Edited April 8, 2010 by ambros
darkenlighten Posted April 8, 2010 Posted April 8, 2010 Oh my!!! Open your eyes man.... Read the last post again. The equation from the BIPM did not just mysteriously come into existence. I showed you how it came to be!!!! Using the Lorentz Force Law and Biot-Savart Law and combining them you get what you are so fond of. And I have solved that. Edit: Your original post was more funny
Cap'n Refsmmat Posted April 8, 2010 Posted April 8, 2010 Hmm. This thread would be a whole lot easier if we could establish one or two big details. So, ambros, I have a challenge for you, that I'd like to use to better understand your position. Perhaps I can figure out how you're doing this integration this way. So, here's the situation. I have two 1m lengths of wire. One of these is curled into a loop of 1m circumference. They are arranged something like this: Ignore the multiple loops -- I just wanted a diagram that gives a general idea. We have one wire going vertically and another looping around that wire. The loop is exactly in the middle of the vertical wire, 0.5m from the bottom and 0.5m from the top. There is one amp of current going from the top to the bottom of the center wire. There is one amp of current going counterclockwise (as viewed from above) in the loop. What is the net force on the loop due to the magnetic field? Use 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] Then I can understand exactly how you're doing this.
ambros Posted April 8, 2010 Author Posted April 8, 2010 Oh my!!! Open your eyes man.... Read the last post again. The equation from the BIPM did not just mysteriously come into existence. I showed you how it came to be!!!! Using the Lorentz Force Law and Biot-Savart Law and combining them you get what you are so fond of. And I have solved that. Edit: Your original post was more funny Hahaa. That's one of the first things I told you in that thread. AND, THIS IS HOW TO DO IT WITH REAL NUMBERS: TWO PARALLEL WIRES, ampere unit setup: I1=I2= 1A; r= 1m *DO NOT MODIFY EQUATIONS, USE THEM & CALCULATE: No.3 ================================================ [math]B= \frac{\mu_0*Q*v \times \hat{r}}{4\pi*|r|^2}[/math] [math]=> \frac{4\pi*10^-7_{N/A^-2}*1_C*1_{m/s} \times 1}{4\pi*1_{m^2}}[/math] [math]=> \frac{10^-7_{N/A^-2}*1_A*1_m }{1_{m^2}} \ =OR= \ \frac{10^-7_{N/A^-2}*1_{C/s}}{1_m}[/math] [math]=> \frac{10^-7_{N/A}}{1_m} \ = \ 10^-7 \ N/A*m[/math] [math]F = Q*v \times B[/math] [math]=> 1_C*1_{m/s} \times 10^-7_{N/A*m}[/math] [math]=>1_{C/s}*1_m \times 10^-7_{N/A*m} \ =OR= \ 1_A*1_m \times 10^-7_{N/A*m} \ = \ 10^-7 \ N[/math] Can you do that, can you point mistake? Oh my!!! Open your eyes man.... Read the last post again. The equation from the BIPM did not just mysteriously come into existence. I showed you how it came to be!!!! Using the Lorentz Force Law and Biot-Savart Law and combining them you get what you are so fond of. And I have solved that. Edit: Your original post was more funny darkenlighten: - I've put in real variables/constants to show you how you get 2 x 10^-7 N per unit length. Go away liar.
darkenlighten Posted April 8, 2010 Posted April 8, 2010 Big manz on the internetz. But really, you are not integrating right and have the wrong B field. That is your mistake. I did it correctly. There is not just 1 charge in a wire, you need integrate properly in order to solve for a wire. And I never argued that you couldn't use the Biot-Savart Law, but for an infinite wire, it is heck alot easier to solve using Ampere's Law.
ambros Posted April 8, 2010 Author Posted April 8, 2010 Hmm. This thread would be a whole lot easier if we could establish one or two big details. So, ambros, I have a challenge for you, that I'd like to use to better understand your position. Perhaps I can figure out how you're doing this integration this way. So, here's the situation. I have two 1m lengths of wire. One of these is curled into a loop of 1m circumference. They are arranged something like this: Ignore the multiple loops -- I just wanted a diagram that gives a general idea. We have one wire going vertically and another looping around that wire. The loop is exactly in the middle of the vertical wire, 0.5m from the bottom and 0.5m from the top. There is one amp of current going from the top to the bottom of the center wire. There is one amp of current going counterclockwise (as viewed from above) in the loop. What is the net force on the loop due to the magnetic field? Use 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] Then I can understand exactly how you're doing this. y | |____ x I like examples! I'm in bed already so I will not plug in any numbers now, but tomorrow this is what I will do: 1 vector in Y direction to represent straight wire, and 1 vector in x direction representing 'tangential' vector or "dl" segment of the loop. I would then ask you about the precision you require and according to that I would split that loop into number of segments. I could even make it be square or a triangle for a start. The more accuracy you want, the more segments I will make and smaller they will be. It is like drawing a "circle" with 3 lines, 4 lines, 5 lines.... 19, 21... 50 lines, and that's where it starts to look like a real circle. The size of "dl" is a matter of resolution desired and is used to approximate geometry. It's very similar how we actually draw circles or any other curved paths on computer screen, as there are no real curves, but all is made of little line segments "dl". The smaller they are, the better the smoothness, precision and accuracy.
Cap'n Refsmmat Posted April 8, 2010 Posted April 8, 2010 I'm not too worried about precision; it's the method that matters. Try just a square or something simple. And please, everyone else, let's wait to see what ambros does before jumping in with comments on how to do it.
darkenlighten Posted April 8, 2010 Posted April 8, 2010 I challenge you this ambros: Obtain the exact fields of both wires. Then obtain the force felt on each wire from the external magnetic fields. I believe this will be best to understand what is going on is to be exact, no approximations. I will do the same and provide reference and compare.
Cap'n Refsmmat Posted April 8, 2010 Posted April 8, 2010 I challenge you this ambros: Obtain the exact fields of both wires. Then obtain the force felt on each wire from the external magnetic fields. I believe this will be best to understand what is going on is to be exact, no approximations. I will do the same and provide reference and compare. No, don't worry about it, as I have it all calculated out. I'm just curious to see what ambros has. Let's keep this as simple as possible for the moment.
ambros Posted April 8, 2010 Author Posted April 8, 2010 (edited) But really, you are not integrating right and have the wrong... Stop repeating someone else's opinion. What do you imagine how words "is wrong" can have impact on me, and what it can possibly contribute to the discussion? So lame, pathetic... go away! Merged post follows: Consecutive posts mergedNo, don't worry about it, as I have it all calculated out. I'm just curious to see what ambros has. Let's keep this as simple as possible for the moment. Do you want to see if my result will be zero or see how I do integrals? Are you actually have experimental measurements or you're going to use Maxwell's equations, or something like that? You are not planing to use the same equation as me, do you? -- That's all fine, but it would be much better if you can find example with some experimental measurements. Edited April 8, 2010 by ambros Consecutive posts merged.
Cap'n Refsmmat Posted April 8, 2010 Posted April 8, 2010 Do you want to see if my result will be zero or see how I do integrals? Both. Are you actually have experimental measurements or you're going to use Maxwell's equations, or something like that? I've solved the problem using the same equation. I just want to see how you do it. Incidentally, I've changed your mind. Please give the exact value through the formula.
darkenlighten Posted April 8, 2010 Posted April 8, 2010 Stop repeating someone else's opinion. What do you imagine how words "is wrong" can have impact on me, and what it can possibly contribute to the discussion? So lame, pathetic... go away! Hmm I'm sure I was first to say you were wrong, but what do I know its not like I've spent the last 7 months at The Ohio State University taking an Electrodynamics course, nah no not me, I'm not the Engineering Physics student. Oh wait...
swansont Posted April 9, 2010 Posted April 9, 2010 I laugh at your hands waving through the air. -- Your equation is wrong because it does not have and reference to LENGTH which you insist is so important. -- If it is "mathematical disaster" than you should be able to demonstrate it by using MATHEMATICS, stop waving hands. Can you show you work? [math]B = \frac{\mu_0*I}{2\pi r} [/math] Where is the length, and where did 2 Pies go? Where are the freaking variable angles in your equation??? The work has already been shown. The wire is infinite in length. Angles were evaluated with the limits of the integral. This has been explained a number of times. http://www.scienceforums.net/forum/showpost.php?p=554213&postcount=47 Acting like you haven't seen the solution strains credulity. Merged post follows: Consecutive posts merged 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] But isn't the right answer 2 x 10^-7 ?
Bignose Posted April 9, 2010 Posted April 9, 2010 How about you first learn this: unit for time = "s" - second unit for length = "m" - meter unit for mass = "kg" - kilogram It doesn't matter what specific units are used, metric, English, etc. That's I used the generic t for time unit, l for length, and m for mass. Heck, look at the chart about SI units: http://en.wikipedia.org/wiki/International_System_of_Units, the "symbol" category of the chart under the heading Units. The answer should be the same no matter what units you use, too. F=ma doesn't care if the force is in the Newtons, kilograms-force, or pounds-force; doesn't care if the mass is kilograms, grams, slugs, or pounds-mass; doesn't care if a is in meters per time squared, nautical miles per year squared, or light years per picosecond squared. In the end, the quantity on the right hand side will always be a mass times a length over a time squared. The bigger point is that you can come up with an infinite number of combinations of dimensional quantities to result in a mass times a length over a time squared. [math]F_{wrong1} = mvd [/math] where v is a velocity and d is a displacement will also have the same units, but this equation will not give the right answer except by coincidence. [math]F_{wrong2} = mjt [/math] where j is the jerk (derivative of acceleration with respect to time) and t is the elapsed time is also wrong, but has the right units. [math]F_{wrong3} = \frac{\rho d^4 j}{v} [/math] where [math]\rho[/math] is the density. Again, correct units in the end, but way wrong equation. The moral of this story is that dimensional analysis is a great 1st step, but it is no where near the end. It weeds out all equations that have the wrong dimensions, but it doesn't tell you anything about how correct any of the many different equations with the right dimensions are.
ambros Posted April 9, 2010 Author Posted April 9, 2010 (edited) It doesn't matter what specific units are used' date=' metric, English, etc. [/quote'] Gibberish, that's how it is called if worded in English. In mathematics and physics it is called "ERROR - DOES NOT COMPUTE". It took me 10 minutes to figure out what in the world were you talking about and then I double checked to make sure I do not misinterpret you. After 15 minutes I realized you were wasting everyone's time by saying the simplest thing in the most incomprehensible way, and even managed to make a wrong conclusion. In mathematics and physics, correctness is paramount. Merged post follows: Consecutive posts mergedBut isn't the right answer 2 x 10^-7 ? How would you know? ---TWO PARALLEL WIRES: i1=i2= 1A' date=' 1m distance --- integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} cross {1,0,0} cross {0,-1,0} [math]\int_0^1( \int_0^1((1*10^-7 N) {1,0,0} \times ({1,0,0} \times {0,-1,0}))/1^2 dx) dx = {0, \ 1*10^-7 N, \ 0}[/math] ---TWO PERPENDICULAR WIRES (loop around wire): i1=i2= 1A, 1m radius --- integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} cross {0,1,0} cross {0,0,-1} [math]\int_0^1( \int_0^1((1*10^-7 N) {1,0,0} \times {0,1,0} \times {0,0,-1}))/1^2 dx) dx = {0, \ 0N, \ 0}[/math] ...ENTER THE CODE IN MATHEMATICA, HERE: http://www.wolframalpha.com/input/ Edited April 9, 2010 by ambros Consecutive posts merged.
Mr Skeptic Posted April 9, 2010 Posted April 9, 2010 In mathematics and physics, correctness is paramount. Which is why when doing dimensional analysis I have avoided using named units and instead used generic units. Dimensional analysis gives you energy, force, mass, not Joules, ergs, Newtons, pounds, kilograms, stones, or any specific unit. The number in dimensional analysis is also irrelevant, which is why I never specified it when doing a dimensional analysis. Merged post follows: Consecutive posts mergedAs for your computation, it seems, as I expected, that you are using 1 m wires instead of infinitely long wires. Yes, having a different length of wire means you get a different result.
ambros Posted April 9, 2010 Author Posted April 9, 2010 (edited) Which is why when doing dimensional analysis I have avoided using named units and instead used generic units. Dimensional analysis gives you energy, force, mass, not Joules, ergs, Newtons, pounds, kilograms, stones, or any specific unit. The number in dimensional analysis is also irrelevant, which is why I never specified it when doing a dimensional analysis. Merged post follows: Consecutive posts mergedAs for your computation, it seems, as I expected, that you are using 1 m wires instead of infinitely long wires. Yes, having a different length of wire means you get a different result. You can not really comment on my result if you can not show how do you obtain your result. So what is your result, can you show your work here with equations and mathematics, or simply give a code for Mathematica? Let me see your solution for parallel and perpendicular wire, eh? I wonder if you get zero force with perpendicular wires or perpendicular loop around straight wire... actually, with infinity, you can hardly get anything else but infinity, which means - ERROR, DOES NOT COMPUTE. As for your computation, it seems, as I expected, that you are using 1 m That's good, because I have been repeatedly saying it... though I can do that in many different ways, as I also demonstrated previously. -- So, you are complaining that I use "1m"? But, how else do you expect me to tell you the force PER METER, if I do not integrate OVER ONE METER? The integral itself is what give us this another dimension and with it - a new UNIT. It is absolutely necessary to ingrate over 1m, because that is how ampere unit is defined - force BETWEEN TWO wires of 2*1-^-7 Newtons per meter of length. I solved the problem in five different ways now, you have the math of it all right in front of you. I do not want to argue as I think this can not be said and demonstrated any more clear than this, but if not, then tell me - what do I need to do, or show you, or demonstrate to convince you? Merged post follows: Consecutive posts merged Yes, having a different length of wire means you get a different result. Having "different mathematics" can make 2+7 = 34. Why do you think you can comment on any of it if you can not obtain any result from that equation? What is the point of guessing? Either equation is wrong or I made a mistake, which it is? CASE A: CASE B: CASE C: + - + - | I2= 1A | | I2= 1A | I2= 1A |===================| |==========| + ==================infinite >> - | | | |1m |1m |1m | | | |===================| |==========| + ==================infinite >> - | I1= 1A | | I1= 1A | I1= 1A |<------ 35m ------>| |<-- 1m -->| + - + - CASE A: [math]F_m = 2 k_A \frac {I_1 I_2 }{r} = 2* \frac{4\pi*10^-7_{N/A^2} *1_{A} *1_{A}}{4\pi*1_m} = 2*10^-7 \ N/m[/math] CASE B: [math]F_m = 2 k_A \frac {I_1 I_2 }{r} = 2* \frac{4\pi*10^-7_{N/A^2} *1_{A} *1_{A}}{4\pi*1_m} = 2*10^-7 \ N/m[/math] CASE C: [math]F_m = 2 k_A \frac {I_1 I_2 }{r} = 2* \frac{4\pi*10^-7_{N/A^2} *1_{A} *1_{A}}{4\pi*1_m} = 2*10^-7 \ N/m[/math] Different results like that above? Ampere's force law (above) is not the same as that other equation for F(12) given by the BIPM, though they both can produce these same results, depending on how one integrates. Edited April 9, 2010 by ambros Consecutive posts merged.
swansont Posted April 9, 2010 Posted April 9, 2010 Either equation is wrong or I made a mistake, which it is? You have made numerous mistakes. They have been pointed out to you, over and over again, but you have ignored the corrections.
ambros Posted April 9, 2010 Author Posted April 9, 2010 You have made numerous mistakes. They have been pointed out to you, over and over again, but you have ignored the corrections. You 're hallucinating my young friend. What you said is a lie, amusing and interesting lie, disgusting too. -- Can you say what step exactly you imagined might be incorrect in my calculation? Why can you not print out that solution in Mathematica code so everyone can see it online? Merged post follows: Consecutive posts merged No' date=' don't worry about it, as I have it all calculated out. I've solved the problem using the same equation. I just want to see how you do it. [/quote'] Ok, let's see that now, your turn.
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