Recent content by Vapor88

Wait, I'm still not seeing this. Wouldn't that leave -(m \cdot \widehat{r}) \widehat{r} = -m_{r} \widehat{r} and -(m \cdot \widehat{\theta}) \widehat{\theta} = -m_{\theta} \widehat{\theta}

Okay, so if I set it up like this, everything is the same except the coordinate system. So I get -(m \bullet \widehat{r}) \widehat{r} = -m cos ( \theta ) \widehat{r} -(m \bullet \widehat{\theta}) \widehat{\theta} = m sin ( \theta ) \widehat{\theta} ...I don't know why the \bullet or hats...

Homework Statement Example: Problem 5.33 Show that the magnetic field of a dipole can be written in the following coordinate free form: Homework Equations http://teacher.pas.rochester.edu/PHY217/LectureNotes/Chapter5/LectureNotesChapter5138.jpg...

Homework Statement Find the magnetic flux of the rectangular loop. Show that \int B x dl is independent of path by calculating \oint B x dl Homework Equations The Attempt at a Solution First step was to solve for B from the wire, I did that using Biot-Savart Law for a...

I used the binomial expansion, not the Taylor expansion... They are different, right?

What am I planning on integrating? lol... I'm not really sure, I'm very, very shaky on these legendre polynomials, they're completely new to me, but everyone else in my class is familiar with them. Anyways, I see from there, when plugged into the potential equation, I get an r^-3, which is...

P_{0} = 1 P_{1} = cos ( \theta) P_{2} = \frac{3cos^2 ( \theta) - 1}{2} I'm fairly confident that this is correct. Are these my P_{l} values? And would a^2/r^2 be Pm? Does this mean that I'll have three integrals, all evaluated from -1 to 1, such that \int P_{l}...

Ok, here's what I know of Legendre plynml's from my notes They are integrals such that \int P_{l} (x) P_{m} (x)dx = 0 When l =/= m When l = m, that integral is equal to \frac{2}{2l+1} For this problem, I want to take x to be cos(theta)? Does that turn the first two terms...

Alrighty... After banging my head on the desk, I've got the following. V = \frac{q}{4 \pi r \epsilon_{o}} [- \frac{a^2}{r^2} + \frac{3a^4}{4r^4} + \frac{3a^2}{2r^2} cos^2( \theta)] Now... to implement those Legendre polynomials.

*facepalm* thanks!

It's not, but I keep reworking it and get that. here is the work for how I got there [1 - \frac{1}{2} ( \frac{a^2}{r^2} - \frac{a}{r} cos( \theta)) + \frac{3}{8} ( \frac{a^4}{r^4} - \frac{2a^3}{r^3} cos( \theta) + \frac{a^2}{r^2} cos^2( \theta))] Then you add that to...

Ok, so I believe this is correct, however, I'm still stumped as far as legendre polynomials go. I have no idea how to implement them. V = \frac{q}{4 \pi \epsilon_{o} r} * [ - \frac{a^2}{r^2} + \frac{3a^4}{4r^4} + \frac{3a^2}{4r^2} cos^2( \theta)] Given this, I can...

For [1 + \frac{a^2}{r^2}- \frac{a}{r} cos( \theta)]^{-1/2} I expand to get [1 - \frac{1}{2} ( \frac{a^2}{r^2} - \frac{a}{r} cos( \theta)) + \frac{- \frac{1}{2} (- \frac{1}{2} - 1)}{2!} ( \frac{a^4}{r^4} - \frac{2a^3}{r^3} cos( \theta) + \frac{a^2}{r^2} cos^2( \theta))] Which gets...

Ok, so I got a bit further in the problem, he wants us to use the first three terms of binomial expansion, I only did the first two to get my results above, but they did not need Legendre polynomials. Though I do not know why my answer is incorrect. If I use the first three terms of the...

I just worked further into the problem, tried using that expansion the same way, don't know if I can really do that though. But I took the a^2/r^2 + a/r cos (theta) = x In the end, I did get something over r^3, which makes sense because it's a quadrupole, but I missed the point of the...