Recent content by WhoWeAre

I was wondering if anyone could tell me about fiber lasers. I don't mean those used as transmitters on optical fiber systems; I mean lasers that contain optical fibers as components. I know that it can be done, but I don't know what the structure is like. What part does the fiber play? Is...

Hi Everyone, I'm working with Laplace Transforms and I don't know how to deal with the Laplace Transform of [(Sin(t))^2]. If anyone could help me that would be great. Thanks

Ya, that's pretty easy integral. But I guess you don't understand my question. Oh well, thanks for try'in guy. Later

Right. I'm saying that it seems to me that the area of the loop (b*a) should be in the final formula, but it is not. Only the width of the loop (b) is. Why?

Oh, I see, I think. Are you saying that dA is equal to b*dx, where dx is the height of the loop?

I don't see how to do that. It doesn't seem like I need to do that because I am integrating over a. That should take care of B in terms of d and a. But what happens to dA in Ampere's Law. Does it become just A because it is constant or does something else happen to it?

I should have asked: Why is only the width of the loop included in the final formula. Shouldn't there be a (width*height) in the final formula?

I'm confused. Here is the problem I'm working on. Imagine a long, straight wire crossing the screen or page horizontally. It is carrying a current i. Now imagine a rectangular loop underneath the wire a distance of d away. The loop has a height of a and a width of b. I am supposed to find...

I'm working with the equation B=((mu sub naught)*i)/(2*pi*R). I know that R is changing, but I don't know what dB would look like. I thought that R in the original equation would be dR in the new one, but it doesn't make sense to me that dR would be to the negative one power. Do you have any...