tag:blogger.com,1999:blog-3153881528840143069.post2609662165629297438..comments2023-08-02T20:54:52.429+10:00Comments on Grant Trebbin: In Plane Magnetic Field of a Current LoopGrant Trebbinhttp://www.blogger.com/profile/13253301568478517009noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-3153881528840143069.post-33705075791582611082016-05-03T17:29:07.132+10:002016-05-03T17:29:07.132+10:00Thanks for the kind words. Maxima is a bit hard i...Thanks for the kind words. Maxima is a bit hard initially, but after you understand its quirks it's pretty easy. If you have any more insights into this problem, I'd be glad to link to any blog or website you may have.Grant Trebbinhttps://www.blogger.com/profile/13253301568478517009noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-65678481562618214262016-04-28T03:35:20.419+10:002016-04-28T03:35:20.419+10:00This is brilliant, mate! Like you, I was unable t...This is brilliant, mate! Like you, I was unable to find this problem worked out anywhere on the internet. I set up the integrals for the in-plane version of this problem last year as an advanced problem for one of my better undergrad students in E&M. It is great to see you thinking like a physicist: checking the simple cases at the origin and checking that your more general solution in all of space reduces to this one in the plane of the loop. From the axial symmetry, we know that the only thing B_z can depend upon in the plane is r, and your solution agrees. It was also nice to see that you found the same solution from the pre-internet era. I haven't learned Maxima yet, but I plan to start by making a dual plot of E_2(k)/(a-r) and E_1(k)/(a+r) to demonstrate that inside the loop (r 0 and outside (R>a) B_z < 0, as the Right-hand Rule demands this circulation sense around the current, and show their expected crossover at r = a. Again, fine work!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-40568698292375880182015-12-08T09:50:21.572+10:002015-12-08T09:50:21.572+10:00Ok I understand, I appreciate your help.Ok I understand, I appreciate your help.georgehttps://www.blogger.com/profile/17413285842878331991noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-28242866314354984382015-12-07T22:10:27.235+10:002015-12-07T22:10:27.235+10:00Unfortunately there's no easy way to do this. ...Unfortunately there's no easy way to do this. On the Wikipedia page for elliptic integrals you can see that they can be expressed as an infinite summation, however this isn't helpful. Before computers they were just looked up in a table similar to functions like sine and logarithms.<br /><br />Most maths software will have functions to calculate them. Octave, Matlab, Maxima, Python, all have functions to do this.Grant Trebbinhttps://www.blogger.com/profile/13253301568478517009noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-71727923979496371552015-12-07T21:11:20.249+10:002015-12-07T21:11:20.249+10:00Good job Grant and thanks. By the way, how would y...Good job Grant and thanks. By the way, how would you proceed with calculating the elliptic integrals from now on?georgehttps://www.blogger.com/profile/17413285842878331991noreply@blogger.com