tag:blogger.com,1999:blog-3153881528840143069.post2304247124277423908..comments2019-04-22T14:54:31.296+10:00Comments on Grant Trebbin: Off Axis Magnetic Field of a Circular Current LoopUnknownnoreply@blogger.comBlogger36125tag:blogger.com,1999:blog-3153881528840143069.post-44876931686680042752017-08-31T19:25:37.409+10:002017-08-31T19:25:37.409+10:00I think I was a bit (actually, very) wrong in my f...I think I was a bit (actually, very) wrong in my first commment. I said: "the magnetic field created by the current along the loop at any point of the loop itself is not zero. In the case of a circular loop, as is our case, the magnetic field at a point of the loop itself is not zero regardless what the plots may indicate". Well, the plots, your plot 'In plane magnetic field of coil with radius of 1.55 m and current of 1 Amp' precisely says that the magnetic at an exact point of the loop itself (when r=1.55) is infinite. I did the calculations myself and the result I got was infinite, not zero like I said it was. I just didn't loot at the plots close enough!<br /><br />This should not come as a surprise, since filiforms circuits are idealizations, like point charges are (this might be discussed). Real circuits have some volume and in that case I think the magnetic field along the axis of the circuit would probably be zero or approximately zero.<br /><br />Lenz's law is usually stated like that, but as you suggest, that statement is a bit misleading, because the induced magnetic field never manages to keep the magnetic flux in the loop constant. The example I gave of a conducting bar on conducting rails is an example of that.<br />Alfonso Díez Calvohttps://www.blogger.com/profile/06027242234964207021noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-59759289959621896462017-08-25T15:10:02.635+10:002017-08-25T15:10:02.635+10:00Thanks for the compliment. It's an interesting...Thanks for the compliment. It's an interesting question. I have to admit that I haven't come across this problem before and am not sure of the answer. I'd always thought the magnetic field from the induced current was an indirect statement of Lenz's law. From what I remember Lenz's law is usually stated as "The induced magnetic field inside any loop of wire always acts to keep the magnetic flux in the loop constant". I'm not sure that's right though. I might need to think about this some more.Grant Trebbinhttps://www.blogger.com/profile/13253301568478517009noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-8967339529095303652017-08-23T23:30:23.884+10:002017-08-23T23:30:23.884+10:00Hello from Spain. First of all, thank you, Grant, ...Hello from Spain. First of all, thank you, Grant, for your video and work.<br /><br />I was revisiting the typical problem of a conducting bar on conducting rails forming a complete loop with resistance R, in the presence of an external field B, and I noticed that in most books authors neglect the magnetic field created by the current around the loop itself. After doing some computatios, one can see that this magnetic field is neglegible compared to that of the external field after. But what's really remarkable is the fact that the magnetic field created by the current along the loop at any point of the loop itself is not zero. In the case of a circular loop, as is our case, the magnetic field at a point of the loop itself is not zero regardless what the plots may indicate. Why do you think this fact is so often ignored? Excuse my English if there were any mistakes.Alfonso Díez Calvohttps://www.blogger.com/profile/06027242234964207021noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-60746243955684269642017-04-11T00:00:28.836+10:002017-04-11T00:00:28.836+10:00First of all the problem with python sounds like s...First of all the problem with python sounds like something to do with numerical stability. It sounds like your equations are right but there is something wrong with the implementation.<br /><br />That equation is strange. I've never seen it before but it pops up when you search for "short solenoid inductance" I'm finding references to something called Wheeler's formula from the early 1900's as well. I'll have another look tomorrow when I have some time.Grant Trebbinhttps://www.blogger.com/profile/13253301568478517009noreply@blogger.comtag:blogger.com,1999:blog-3153881528840143069.post-44013233034340891572017-04-09T09:10:12.203+10:002017-04-09T09:10:12.203+10:00Hi Grant, thank you for your great work!
I have a...Hi Grant, thank you for your great work! <br />I have a similar script in Python for calculating the inductance of large coil (not a long solenoid). Now the thing that bothers me is that the B-field becomes essentially huge, if not infinite when the current filament is approached. The magnetic flux I get varies greatly depending on the step size in the script, since the closer it gets the current, the greater the flux gets. Now this doesn't happen in real world, flux and inductance is finite for a current loop. Obviously copper wire (and more so, 500 of them) is not infinitely thin filament. <br />I have found couple of calculators online for inductance, and they are in good agreement with my script (when using reasonable step size). These calculators seem to be using something else than Biot-Savart's law. I'm sure it is derived from Maxwell's equations, but I just can't find the derivation anywhere. My question is, if you could perhaps point me to the right direction how I could solve the problem with infinitely large magnetic field near the wire (in a coil), and also if you know how this approximation for inductance is derived: L ~ N^2*R µ_0 (ln(8R/a)-2), where R is radius of coil, a is radius of wire, N is number of loops and ~ should be "is approximately" For example <a href="https://www.eeweb.com/toolbox/coil-inductance" rel="nofollow">here</a>. <br /><br />Only cases I can find are the calculations for magnetic forces in plasma streams in toroidal Tokamak reactors. Of course I can see a slight resemblance with my coil there ;). <br /><br />This form of equation pops up for example in: <a href="https://books.google.fi/books?id=ZEn5FC3LNqgC&pg=PA112&lpg=PA112&dq=magnetic+pressure+ln(8R/a)&source=bl&ots=DRxS7UISpg&sig=UetkorqhwFwbPSoL4MfpgzwL0Do&hl=fi&sa=X&ved=0ahUKEwiVxP6S25XTAhVEIJoKHR3lAeoQ6AEILDAC#v=onepage&q=magnetic%20pressure%20ln(8R%2Fa)&f=false" rel="nofollow">This book</a> and Wikipedia <a href="https://en.wikipedia.org/wiki/Magnetic_pressure" rel="nofollow"> Magnetic pressure</a>. I am still having hard time understanding where this all comes from.<br /><br /><br />PetePetrihttps://www.blogger.com/profile/01911327450384004828noreply@blogger.comtag: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 Trebbinhttps://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 Trebbinhttps://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 Monroehttps://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 Trebbinhttps://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?Unknownhttps://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 Trebbinhttps://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 Trebbinhttps://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 Trebbinhttps://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.georgehttps://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.georgehttps://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 Trebbinhttps://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?georgehttps://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.jpggeorgehttps://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 Trebbinhttps://www.blogger.com/profile/13253301568478517009noreply@blogger.com