tag:blogger.com,1999:blog-3153881528840143069.post2304247124277423908..comments2017-03-27T02:28:10.495+10:00Comments on Grant Trebbin: Off Axis Magnetic Field of a Circular Current LoopGrant Trebbinhttps://plus.google.com/104565984284177357788noreply@blogger.comBlogger31125tag:blogger.com,1999:blog-3153881528840143069.post-83902207676051233842016-11-14T23:21:08.052+10:002016-11-14T23:21:08.052+10:00I feel like this is a problem that is beyond solvi...I feel like this is a problem that is beyond solving with equations. You also have the problem of hysteresis in the material. This makes the problem somewhat nonlinear. The only thing I can think of is trying some of the software in this list https://en.wikipedia.org/wiki/List_of_finite_element_software_packages I like COMSOL, but don't have access to it anymore. Another resource I can recommend searching is Google Books. There are some good physics textbooks in there. Some are restricted. some aren't. Sorry I couldn't be more help.Grant Trebbinhttp://www.blogger.com/profile/13253301568478517009noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-33056708535015580222016-11-14T09:35:04.499+10:002016-11-14T09:35:04.499+10:00Yes this is the case, that's why I'm strug...Yes this is the case, that's why I'm struggling. Would you please recommend any reference? <br /><br />Thank you Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-33756412954783286462016-11-10T22:47:49.318+10:002016-11-10T22:47:49.318+10:00I'm glad the video helped you out. I'd li...I'm glad the video helped you out. I'd like to help but wouldn't know where to start. As soon as you start adding materials to the scenario it becomes complicated to the point that you really need to do numerical modelling.Grant Trebbinhttp://www.blogger.com/profile/13253301568478517009noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-75363440756763858062016-11-09T10:50:55.362+10:002016-11-09T10:50:55.362+10:00I mean of the axis :)I mean of the axis :)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-91633250793240116482016-11-09T10:38:39.754+10:002016-11-09T10:38:39.754+10:00you just made things really easy with your video, ...you just made things really easy with your video, Thank you.<br /><br />I'm really having a problem with calculating the field B for solenoid with core and I could not find any clear formula or code. Would you please help me with that? Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-36361368975162409112016-10-08T15:10:20.484+10:002016-10-08T15:10:20.484+10:00This comment has been removed by the author.D Monroehttp://www.blogger.com/profile/09526248982624312393noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-55872450410377744312016-09-07T19:58:11.746+10:002016-09-07T19:58:11.746+10:00Aah..I see now: one can simply rotate the coordina...Aah..I see now: one can simply rotate the coordinate system in order to generalise the solution. I get what you're saying about radial symmetry. Thanks.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-58118859245856522692016-09-07T17:54:39.535+10:002016-09-07T17:54:39.535+10:00Thanks for the compliment. I'm not sure about...Thanks for the compliment. I'm not sure about that, I may not have understood your question correctly (I'm slightly confused about why you mention the angle between vec(OP) and the x-axis) but I don't think so. Working things out in the x-z plane makes things simple to calculate and then the radial symmetry generalises it for all of space. Technically can calculate the answer for any location of P, but it's easier to do it when it's in the x-z plane. It's been a while since I've read this so I'm very rusty. Sorry I couldn't give you a definitive answer.Grant Trebbinhttp://www.blogger.com/profile/13253301568478517009noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-7942129542114288792016-09-07T03:38:42.308+10:002016-09-07T03:38:42.308+10:00Your video is absolutely fantastic, and is the ONL...Your video is absolutely fantastic, and is the ONLY source on the internet I have found that finds the magnetic field off-axis using the magnetic vector potential--and with such ease. One question though. You find the vector potential first for all points in the x-z plane, one can generalise this for points in all of space by substituting $\phi'$ with $\phi'-\phi_1$, where $\phi_1$ is the angle that $\vec{OP}$ makes with the x-axis, can't one?Unknownhttp://www.blogger.com/profile/12299195813070589503noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-25061027819638651042016-04-13T19:36:25.023+10:002016-04-13T19:36:25.023+10:00Well spotted. It should be k^2 or m. Everything s...Well spotted. It should be k^2 or m. Everything still works though. If I remember correctly the documentation for the software I was using got it wrong and I carried it through. It's a bit hard to change now, but I'll add a note to the post. Thanks for reading.Grant Trebbinhttp://www.blogger.com/profile/13253301568478517009noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-90372800078736926552016-04-13T19:12:46.277+10:002016-04-13T19:12:46.277+10:00Is your k the standard k^2 = 4*r*a/(z^2 + (a + r)^...Is your k the standard k^2 = 4*r*a/(z^2 + (a + r)^2) and if so shouldn't the formula use EllipticE(k^2) and EllipticK(k^2)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-15631773121583444592016-02-09T22:10:45.387+10:002016-02-09T22:10:45.387+10:00thanks manthanks manAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-68839415777842036662015-12-04T23:18:41.446+10:002015-12-04T23:18:41.446+10:00I've placed a worked solution for the in plane...I've placed a worked solution for the in plane magnetic field of a circular current loop here<br /><br />http://www.grant-trebbin.com/2015/12/in-plane-magnetic-field-of-current-loop.html<br /><br />It starts with the Biot Savart law and progresses from there.Grant Trebbinhttp://www.blogger.com/profile/13253301568478517009noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-20675764800172430932015-11-29T23:29:40.785+10:002015-11-29T23:29:40.785+10:00Now I see what you're getting at. I just trie...Now I see what you're getting at. I just tried to work it out and got lost. One thing that might help is that you only need to calculate the field along the x or y axis. Due to rotational symmetry this is valid for all points the same distance from the loop axis. This means you can set x or y in your equation to 0. From what I can tell, there will still be elliptic integrals in the solution.<br /><br />The problem is also discussed here.<br />https://www.physicsforums.com/threads/magnetic-field-in-plane-of-loop.232182/<br /><br />I might have another crack at it tomorrow.Grant Trebbinhttp://www.blogger.com/profile/13253301568478517009noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-12585900376268285622015-11-29T03:38:28.329+10:002015-11-29T03:38:28.329+10:00Thanks for the help Grant, I appreciate your answe...Thanks for the help Grant, I appreciate your answer, I should have clarified that the formula I gave http://i.imgur.com/tqLvtQI.jpg takes into account the speed of the electrons rather than the current of the loop with k=(μ0*q*v)/(8*Pi^2). I'm interested in a qualitative answer rather than a numerical approach , I'm trying to compare the force field F=Brωq that I got when using uniform rotating B fields in spherical coordinates to the answer from the biot-savart law but with the biot savart law I can't calculate the integrals properly to compare the answers, they should either cancel out or leave a term that I want to find out about.georgehttp://www.blogger.com/profile/17413285842878331991noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-63057994544493809502015-11-29T03:10:08.923+10:002015-11-29T03:10:08.923+10:00This comment has been removed by the author.georgehttp://www.blogger.com/profile/17413285842878331991noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-73589406498640149462015-11-28T20:58:03.801+10:002015-11-28T20:58:03.801+10:00I've had a go at answering your question in an...I've had a go at answering your question in an update to the main article. I've also added a small python script to show how I'd calculate the field in the plane of the current loop. I hope this can be helpful to you.Grant Trebbinhttp://www.blogger.com/profile/13253301568478517009noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-82163645155131372662015-11-28T19:47:54.760+10:002015-11-28T19:47:54.760+10:00By the way does the program you're using gives...By the way does the program you're using gives you the elliptic integrals or you have to calculate them yourself?georgehttp://www.blogger.com/profile/17413285842878331991noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-57118945842374629542015-11-28T19:44:28.642+10:002015-11-28T19:44:28.642+10:00I've put a similar integral into the program m...I've put a similar integral into the program mathematica and it just hangs. I want to calculate the B field at a point anywhere inside the plane of the current loop, the formula without using the magnetic potential I think is : http://i.imgur.com/tqLvtQI.jpggeorgehttp://www.blogger.com/profile/17413285842878331991noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-58270306162757327082015-11-26T23:33:46.685+10:002015-11-26T23:33:46.685+10:00Thanks for that link. It's a really good resou...Thanks for that link. It's a really good resource. E1(k) and E2(k) are elliptic integrals of the first and second kind.<br />It's not the standard notation of E(k) and K(k) but it all works out.Grant Trebbinhttp://www.blogger.com/profile/13253301568478517009noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-39143626525661977972015-11-26T22:04:45.241+10:002015-11-26T22:04:45.241+10:00I've found the formulas for every case : http:...I've found the formulas for every case : http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20010038494.pdfgeorgehttp://www.blogger.com/profile/17413285842878331991noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-60293038249316112552015-11-26T22:02:44.640+10:002015-11-26T22:02:44.640+10:00What is E1(k) and E2(k)?What is E1(k) and E2(k)?georgehttp://www.blogger.com/profile/17413285842878331991noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-60119620118580966722015-11-26T21:59:48.641+10:002015-11-26T21:59:48.641+10:00This comment has been removed by the author.georgehttp://www.blogger.com/profile/17413285842878331991noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-79984414196886866812015-11-26T21:35:18.926+10:002015-11-26T21:35:18.926+10:00This comment has been removed by the author.georgehttp://www.blogger.com/profile/17413285842878331991noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-12869013136977395962015-11-21T13:27:26.837+10:002015-11-21T13:27:26.837+10:00Thanks for the feedback. I have a few thoughts bu...Thanks for the feedback. I have a few thoughts but I'm not entirely clear about your homework problem. Your use of the word dipole makes me think that you might be talking about a loop antenna? Usually this is for calculating far field radiation patterns, but the answers don't work nearby because of approximations used.<br /><br />I suppose it's a similar process though, the integration I used treats the loop as a series of infinitesimally small current carrying wires. <br /> <br />I may be misunderstanding the problem you're tying to solve. If you could send a copy of the question I might be able to give a better answer. I'm curious if there's another way to solve it.Grant Trebbinhttp://www.blogger.com/profile/13253301568478517009noreply@blogger.com