Calculating Magnetic field strength of a magnet

  • I
  • Thread starter Einstein44
  • Start date
  • #176
Charles Link
Homework Helper
Insights Author
Gold Member
5,224
2,508
I cannot disagree more vehemently as we have previously discussed. It is a confusing sideshow IMHO.
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.
Hopefully we can get back on track, but I think it was useful to address the matter in some detail. :)
 
Last edited:
  • #177
Charles Link
Homework Helper
Insights Author
Gold Member
5,224
2,508
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?
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.
 
  • #178
bob012345
Gold Member
1,057
345
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.
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## .
 
  • Like
Likes Einstein44 and Charles Link
  • #179
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
19,483
10,248
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)
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}.$$
It's at the heart of this problem in fact! There is no scalar potential of the electric field and no voltage but an EMF (which is a pretty bad old name, I like the German word "Ringspannung" much more, but I don't know whether there's a English translation for that).
 
  • Like
Likes Einstein44, hutchphd and Charles Link
  • #180
hutchphd
Science Advisor
Homework Helper
3,952
3,120
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
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
 
  • Like
Likes Einstein44, bob012345 and Charles Link
  • #181
Delta2
Homework Helper
Insights Author
Gold Member
4,590
1,874
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)
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}.$$
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.
 
  • #182
bob012345
Gold Member
1,057
345
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
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.
 
  • #183
bob012345
Gold Member
1,057
345
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.
How is this related to the ##emf## of magnets falling through loops?
 
  • #184
Delta2
Homework Helper
Insights Author
Gold Member
4,590
1,874
How is this related to the ##emf## of magnets falling through loops?
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.
 
  • #185
bob012345
Gold Member
1,057
345
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.
Probably the derivation for the Maxwell equations based on microscopic considerations of electrons and nuclei in Jackson section 6.7 Second edition.
 
  • #186
hutchphd
Science Advisor
Homework Helper
3,952
3,120
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.
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.....
 
  • Like
  • Informative
Likes Vanadium 50 and Einstein44
  • #187
bob012345
Gold Member
1,057
345
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.....
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.
 
  • Like
  • Informative
Likes Einstein44 and hutchphd
  • #188
Charles Link
Homework Helper
Insights Author
Gold Member
5,224
2,508
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.
 
  • #189
116
28
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.
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.
 
  • Like
Likes Charles Link
  • #190
Charles Link
Homework Helper
Insights Author
Gold Member
5,224
2,508
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. :)
 
  • Like
Likes Einstein44, bob012345 and hutchphd
  • #191
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
19,483
10,248
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.
Just for curiosity: Why should you get penalised for sharing your experimental results?
 
  • #192
116
28
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. :)
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.
 
  • Like
Likes vanhees71 and Charles Link
  • #193
116
28
Just for curiosity: Why should you get penalised for sharing your experimental results?
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.
 
  • Like
Likes vanhees71 and Charles Link
  • #194
Charles Link
Homework Helper
Insights Author
Gold Member
5,224
2,508
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.
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.
 
Last edited:
  • Like
Likes vanhees71 and Einstein44
  • #195
116
28
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.
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...
 
  • Like
Likes vanhees71 and Charles Link
  • #196
Charles Link
Homework Helper
Insights Author
Gold Member
5,224
2,508
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...
@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. :)

additional note: posts 138-139 are also of interest here. The ## M ## of ## B=\mu_o H+M ## makes things slightly more complex in computing the flux. From a theoretical standpoint, when the derivative is taken on @hutchphd 's formula (post 120) for the flux, it gives exactly what we are looking for, as discussed in post 139. This makes the solution of posts 120-123 all the better, and we really couldn't ask for more. If you were to compare your results to those that are predicted here, I think you would find that they match to a "T".
 
Last edited:
  • Like
Likes Einstein44, vanhees71, bob012345 and 1 other person
  • #198
rude man
Homework Helper
Insights Author
Gold Member
7,973
836
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?
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.

To find the equivalent amperian current you could experimentally determine the magnetic dipole moment ## \bf \mu ## of your magnet, probably best by determining the torque ## \bf \tau ## exerted on the magnet by an external known ## \bf B ## field. ## \bf \tau = \bf \mu \times \bf B ##. For the equivalent solenoid, ## \mu = i ~ \times ~area~ of~ the ~ solenoid~ \times number~ ot~ turns ##., i = solenoid current = equivalent amperian current.
 
  • #199
Charles Link
Homework Helper
Insights Author
Gold Member
5,224
2,508
@rude man This one we pretty much solved. We made it a joint effort. The solution that we found to work is found in posts 120-123. See also post 196 above.
 

Related Threads on Calculating Magnetic field strength of a magnet

Replies
6
Views
5K
  • Last Post
Replies
5
Views
7K
Replies
5
Views
2K
  • Last Post
Replies
2
Views
1K
Replies
1
Views
3K
Replies
1
Views
17K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
1
Views
1K
Top