Cap'n Refsmmat

No, your equation was a line integral, and r is not constant in it. R (the distance between the wires) is constant by definition. 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. 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. 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 Okay, I'll pick my favorite form from the Wikipedia article: http://en.wikipedia.org/wiki/Ampere%27s_force_law [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 Derive this from scratch (from Biot-Savart or Ampere) for me please.

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.) Because you fudged it and said "oh, I have to multiply by two!" 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?

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 merged You'll notice that I did, in fact, compute the integral, before then using the result of the integral to get the correct answer.

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?

The forums are now open to all those with more than 50 posts, rather than 100 as before.

Using 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'm not just referring to SFN here. I mean a community website in general, one that's not just "friends" from real life like Facebook or Twitter. So, what would your ideal Internet community have? Would members be anonymous, or would they use real names, like on The WELL, a 25-year-old discussion forum? How would moderation work? Would bad members be banned immediately or be given a chance? Just throw some ideas out. I'm curious to see what people think.

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!

A great column from the Times: http://www.timesonline.co.uk/tol/comment/columnists/melanie_reid/article7083395.ece

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.

I'd presume it's because the force required to keep the particle going in a circle goes up with [imath]v^2[/imath]. Instead of increasing the force, using stronger magnetic fields, and losing more energy to synchrotron radiation, it's easier to increase the radius. Particles in the LHC do run multiple loops before collision. But there's a limit to what you can achieve by increasing the fields, due to synchrotron radiation. There's probably other factors I don't know about as well.

Yes, and there's an integral sign, so you sum over the entire line. That's what integrals do.

I just got out of the lab after five hours, so I can attest to this. Agh.

You only say that because you know he'd win.

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.

Brilliant idea! Now done. Enjoy the ponies!

Ah. Looks like WordPress doesn't obey your timezone settings, so it's using GMT or something. Hence posts from the future.

What do you mean? Timestamps are off?

Inside the templates, <if conditional="$post[username]==$bbuserinfo[username]">omgponies<else />$post[username]</if> Something like that.

It's a bit tricky when you can't see that everyone else has the same thing, so next time I'll just switch Pangloss and bascule or something.

In case the joke is lost here, everyone shoes up as omgponies. The ones with an avatar get an added bonus.

It's an honor bequeathed on only our best members.

It could be worse. I could've used teletubbies.

Nice avatar you have there.