The thing we want to do here is make sure that when we measure the voltage from the coil, that we indeed get a good measurement. We got off on a tangent, but the proposed connected ring would not work for this measurement, and the reason, as you mentioned in post 158, is basically because of Professor Lewin's concepts.hutchphd said:I cannot disagree more vehemently as we have previously discussed. It is a confusing sideshow IMHO.
We certainly should try to avoid getting sidetracked any further.bob012345 said:But if we exactly canceled the ##E_m## field there would be no current. I'd like to see what Kirk McDonald says on the matter. Also, should we start a new thread?
So far we have been focusing on the ##emf## produced in the coil with the goal of measuring it. We have ignored the reaction force opposing the magnets fall. I suspect it is small enough to ignore but could be easily calculated since it is just the current in the loops and the field at the loops. In this case the radial field ##B_{\rho}## instead of ##B_z## .Charles Link said:We certainly should try to avoid getting sidetracked any further.
Just a couple comments on what the OP was asking back around post 153 on whether it is necessary to use insulated wire: The answer is yes. If the coil connects into a closed conductive ring at any point, there will be currents that circulate in the closed loop from the Faraday EMF from the magnet. These currents will generate an opposing magnetic field, and the EMF will be greatly reduced. The goal is to measure the EMF from the changing flux from just the magnet. For that we need to have insulated wire in the coil.
If you have an open loop (closed only by the high impedance voltmeter) the current produced by the EMF will be very small and hence the reaction force negligiblebob012345 said:So far we have been focusing on the emf produced in the coil with the goal of measuring it. We have ignored the reaction force opposing the magnets fall. I suspect it is small enough to ignore but could be easily calculated since it is just the current in the loops and the field at the loops. In this ca
No I don't think this is very accurate. The most accurate thing to say is that in the general case, the electric field has a scalar potential ##V## as well as a vector potential ##\vec{A}## that is it is $$\vec{E}=-\nabla V-\frac{\partial \vec{A}}{\partial t}$$vanhees71 said:The important point is that for time-dependent fields the electric field has no scalar potential, because it's curl doesn't vanish due to (SI units)
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}.$$
That makes sense and if one has a closed loop of many turns then one should expect a respectable reaction force like when a magnet falls through a conductive pipe.hutchphd said:If you have an open loop (closed only by the high impedance voltmeter) the current produced by the EMF will be very small and hence the reaction force negligible
How is this related to the ##emf## of magnets falling through loops?Delta2 said:No I don't think this is very accurate. The most accurate thing to say is that in the general case, the electric field has a scalar potential ##V## as well as a vector potential ##\vec{A}## that is it is $$\vec{E}=-\nabla V-\frac{\partial \vec{A}}{\partial t}$$
In the lorenz gauge, both V and A obey the non homogeneous wave equations. The source of V waves are the time varying charge densities, while the source of A waves are the time varying current densities.
It is related in a way that cannot be expressed by words...bob012345 said:How is this related to the ##emf## of magnets falling through loops?
Probably the derivation for the Maxwell equations based on microscopic considerations of electrons and nuclei in Jackson section 6.7 Second edition.Delta2 said:It is related in a way that cannot be expressed by words...
Seriously speaking I don't know how the wave equations can be expressed when we have a magnet that is producing a magnetic field, the V and A formulation is when magnetic field is produced by current density.
I think the number of turns doesn't matter (only the amount of copper). Suppose I have two turns of longer thinner wire (with half the cross-section) closed once. Then the resistance will quadruple (it is thinner and longer) so the current will be half. But there are now two loops so there is no net change in the "back" field. You think this through until we are sure. This is why a pipe works fine also I reckon...bob012345 said:That makes sense and if one has a closed loop of many turns then one should expect a respectable reaction force like when a magnet falls through a conductive pipe.
What you say is true, a coil of one turn of cross sectional area ##A## gives the same reaction force as two turns of area ##A/2##. I only meant that an ##N## turn loop is better than a one turn loop of the same wire if you wanted a measurable effect since reaction force goes as ##B N I## and ##I## is the same for one loop or many since the increase in resistance due to length is countered by the increase ##emf## due to the number of turns.hutchphd said:I think the number of turns doesn't matter (only the amount of copper). Suppose I have two turns of longer thinner wire (with half the cross-section) closed once. Then the resistance will quadruple (it is thinner and longer) so the current will be half. But there are now two loops so there is no net change in the "back" field. You think this through until we are sure. This is why a pipe works fine also I reckon...
I didn't use an oscilloscope, because that was too complicated to use, but instead I went for a voltage probe, which displays the voltage on the computer like an oscilloscope would. The formula you have derived were helpful for understanding, however I wanted to do this project on my own, instead of relying on other people's work, so I made some simplifying assumptions and derived another formula which shares some similarities with the one you derived.Charles Link said:a follow-up on this one: In post 157, the OP @Einstein44 says he finished the experiment. Could you @Einstein44 furnish a little more detail: Did you use an oscilloscope?, and did you get a curve similar to @bob012345 's of post 123? Hopefully you found the formulas derived by @hutchphd in posts 120 and 122 of much use. Could enjoy getting a little more feedback on your experimental results.
Just for curiosity: Why should you get penalised for sharing your experimental results?Einstein44 said:I didn't use an oscilloscope, because that was too complicated to use, but instead I went for a voltage probe, which displays the voltage on the computer like an oscilloscope would. The formula you have derived were helpful for understanding, however I wanted to do this project on my own, instead of relying on other people's work, so I made some simplifying assumptions and derived another formula which shares some similarities with the one you derived.
I am not sure if this is a good idea if I talk in more detail about the experimental results, as I am scared that I am going to get penalised for doing so, since I will be using those. If you want I can send you a private message and explain this to you.
Yes, thank you :) The results were good and pretty accurate I believe, so that turned out well. The voltage probe did the job, it was possible to set it to record data at higher speeds, so this was no problem. Thanks to everyone who contributed as well. I have learned a lot, which is always what I am aiming for.Charles Link said:For the voltage probe=perhaps an "a to d" converter, (analog to digital), if it has a fast enough speed, it would be just as good as an oscilloscope. It is ok if your project is being graded or judged. It is always fun to see some experimental results, but I am confident that if you did the experiment well, that you did get results similar to what we forecast in the posts around 123. Glad we were able to be of assistance. You do have a very interesting experiment. :)
The problem is this is a graded project and I am not sure if I would be allowed to share this with other people who could help with that. Also, I think that if I post my results on here, it would appear as plagiarism if an originality check is made (even though they are mine), so I just want to avoid it altogether just to be sure I am not making a mistake here.vanhees71 said:Just for curiosity: Why should you get penalised for sharing your experimental results?
From what I could tell, the available on-line literature on this topic was somewhat limited. You found a good "link" with post 25, but even that was somewhat difficult to work with. I think @hutchphd 's computation in post 120 is first-rate, and it could be part of standard textbook type literature on the topic. I believe you also posted a "link" to someone that had done a variation of this experiment, (Edit: I'm referring to post 110, and now I see @bob012345 posted the "link"), but the calculations in the "link" are not terribly advanced, and @hutchphd 's calculation actually fills in this hole that previously existed in the on-line literature. I think it should be ok to seek higher expertise when the topic is not treated in enough detail in the write-ups that you do find on-line.Einstein44 said:The problem is this is a graded project and I am not sure if I would be allowed to share this with other people who could help with that. Also, I think that if I post my results on here, it would appear as plagiarism if an originality check is made (even though they are mine), so I just want to avoid it altogether just to be sure I am not making a mistake here.
Edit: which is why I decided to do my own derivation... even though the one suggested on here was really good, I can't really use it since it was posted here, although it is still good to know other ways of solving these problems.
Yes, I think the same thing. Although I don't believe they would accept me taking the exact model he has done and use that, but maybe getting a little help perhaps is ok...Charles Link said:From what I could tell, the available on-line literature on this topic was somewhat limited. You found a good "link" with post 25, but even that was somewhat difficult to work with. I think @hutchphd 's computation in post 120 is first-rate, and it could be part of standard textbook type literature on the topic. I believe you also posted a "link" to someone that had done a variation of this experiment, (Edit: I'm referring to post 110, and now I see @bob012345 posted it), but his calculations were not terribly advanced, and @hutchphd 's calculation actually fills in this hole that previously existed in the on-line literature. I think it should be ok to seek higher expertise when the topic is not treated in enough detail in the write-ups that you do find on-line.
@hutchphd 's calculations with the poles along with computing the flux from the solid angle are really simple, but also really the ideal solution for this problem. I understand your reluctance to use it, but in the future, if anyone else comes up with the same experiment, it will be posts 120-123 of this thread that I will refer them to. In that sense, you also made a contribution to this by asking the right questions. :)Einstein44 said:Yes, I think the same thing. Although I don't believe they would accept me taking the exact model he has done and use that, but maybe getting a little help perhaps is ok...
Since the B field around your magnet is caused by amperian currents circulating on the sides of the magnet, if the magnet is a right circular cylindrical you may assume that the B field resembles that of a solenoid of same physical shape.Einstein44 said:I have been trying to calculate the magnetic flux thought a single loop of wire occurring from a magnet (meaning it has a nonuniform field), so I have the following equation:
Φ=∮BdAcosθ
Now my problem is that I do not know how to calculate the magnetic field strength (B)of that magnet (which has a cylindrical shape) in order to find out its flux. I have not seen a clear answer to this on the internet, although I think that is is possible to use the Biot Savart equation in some form to do this?