Calculating Magnetic field strength of a magnet

[Charles Link]

The formula is correct in posts 122 and 123. $H =q_m/(4 \pi \mu_o r^2)$ and flux $\phi= \mu_o H r^2 \Omega$ where $\Omega=2 \pi (1-\frac{z}{\sqrt{z^2+R^2}} )$ is the solid angle. Here a section of a sphere (with boundary being the coil, and center at $q_m$) is used to compute the flux, rather than the flat surface of the plane of the coil.
To compute the solid angle, $\Omega=2 \pi \int\limits_{0}^{\theta_o} \sin{\theta} \, d \theta$, where $\cos{\theta_o}=\frac{z}{\sqrt{z^2+R^2}}$.

 

[Einstein44]

The formula is correct in posts 122 and 123. $H =q_m/(4 \pi \mu_o r^2)$ and flux $\phi= \mu_o H r^2 \Omega$ where $\Omega=2 \pi (1-\frac{z}{\sqrt{z^2+R^2}} )$ is the solid angle. Here a section of a sphere (with boundary being the coil, and center at $q_m$) is used to compute the flux, rather than the flat surface of the plane of the coil.
To compute the solid angle, $\Omega=2 \pi \int\limits_{0}^{\theta_o} \sin{\theta} \, d \theta$, where $\cos{\theta_o}=\frac{z}{\sqrt{z^2+R^2}}$.

So does the derivation come from substituting $B_{z}(z)$ into $\phi = \int \vec{B}\cdot \vec{dA}$

 

[Charles Link]

So does the derivation come from substituting $B_{z}(z)$ into $\phi = \int \vec{B}\cdot \vec{dA}$

Instead of using $\phi=\int B_z \cdot dA$ across the plane of the coil, the derivation uses $\phi=\int B \cdot dA$ over a spherical section. It gets the identical result, but it is much easier to compute it this way.

 

[Einstein44]

Instead of using $\phi=\int B_z \cdot dA$ across the plane of the coil, the derivation uses $\phi=\int B \cdot dA$ over a spherical section. It gets the identical result, but it is much easier to compute it this way.

What exactly do you mean by over a spherical section? what exactly was manipulated?

 

[Charles Link]

What exactly do you mean by over a spherical section? what exactly was manipulated?

The coil is the surface boundary in both cases. The flux through the plane of the coil just uses $B_z$ because it is dotted with a surface whose unit normal is $\hat{z}$. The flux is the same for any surface with the coil as the boundary. For ease of computation, we choose to use a spherical section with the coil as the boundary and center of the sphere is the magnetic pole.
The reason the flux is the same has to do with $\nabla \cdot B=0$ and Gauss' law.

 

[Einstein44]

The coil is the surface boundary in both cases. The flux through the plane of the coil just uses $B_z$ because it is dotted with a surface whose unit normal is $\hat{z}$. The flux is the same for any surface with the coil as the boundary. For ease of computation, we choose to use a spherical section with the coil as the boundary and center of the sphere is the magnetic pole.
The reason the flux is the same has to do with $\nabla \cdot B=0$ and Gauss' law.

Perfect, that answers my question. And you did use the equation for B from post #25 am I correct

 

[Charles Link]

We are using the formula for $B$ in posts 122 and 123, or as described in more detail in post 141, with $B=\mu_o H$. The description in posts 141 and 122 is for a single point pole, and post 123 is for two poles, a plus one, and a minus one. Post 25 spreads the magnetic charge distribution evenly across the endface, and we are simplifying that slightly here. It is a reasonably good approximation to condense the magnetic charge on each endface to a point at its center, and upon doing that, the calculations can be done in closed form for an exact answer for the flux $\phi$. That's what we are doing by the method of post 123.

 

[bob012345]

Instead of using $\phi=\int B_z \cdot dA$ across the plane of the coil, the derivation uses $\phi=\int B \cdot dA$ over a spherical section. It gets the identical result, but it is much easier to compute it this way.

I just assumed the field went as $B = \large \frac{q_m}{r^2}$ and integrated the flux element across the loop area directly and got the exact same thing.

The flux element was $B_z dA = B cos(\theta) 2\pi \rho d\rho$ where $cos(\theta) = \large \frac{z}{\sqrt{(z^2 + \rho^2)}}$.

 

[Charles Link]

I just assumed the field went as $B = \large \frac{q_m}{r^2}$ and integrated the flux element across the loop area directly and got the exact same thing.

The flux element was $B_z dA = B cos(\theta) 2\pi \rho d\rho$ where $cos(\theta) = \large \frac{z}{\sqrt{(z^2 + \rho^2)}}$.

For extra detail, the $B=\frac{q_m}{4 \pi r^2}$, but yes, the integral will give the exact same thing. :)

 

[Einstein44]

I just assumed the field went as $B = \large \frac{q_m}{r^2}$ and integrated the flux element across the loop area directly and got the exact same thing.

The flux element was $B_z dA = B cos(\theta) 2\pi \rho d\rho$ where $cos(\theta) = \large \frac{z}{\sqrt{(z^2 + \rho^2)}}$.

Why are you not using an integral?
And where does the $2\pi$ come from since the area of the circle is $\pi r^{2}$

 

[bob012345]

Why are you not using an integral?
And where does the $2\pi$ come from since the area of the circle is $\pi r^{2}$

I did use an integral but I was just showing the differential flux element before the integration occurs. The magnetic field is symmetric in cylindrical coordinates around the $z$ axis so I used a differential element in the plane of the loop which is $2\pi \rho$ in circumference and $d\rho$ thick because $B_z$ is constant over that differential element.

 

[Charles Link]

Flux $\phi(z)=\int\limits_{0}^{2 \pi} \int\limits_{0}^{R} (\frac{q_m}{4 \pi r^2}) \frac{z}{\sqrt{z^2+\rho^2}} \rho \, d \rho \, d \phi$, where $r^2=z^2+\rho^2$. The $2 \pi$ comes from integrating over $\phi$.

 

[Einstein44]

@bob012345 @Charles Link
I have a question about the experiment, when I am using bare copper, the coils are not allowed to touch am I correct? Because I have seen this being done on YouTube with the coils touching, but I am unsure if this is correct. If this is not allowed, would insulated wire be a good solution?

 

[bob012345]

@bob012345 @Charles Link
I have a question about the experiment, when I am using bare copper, the coils are not allowed to touch am I correct? Because I have seen this being done on YouTube with the coils touching, but I am unsure if this is correct. If this is not allowed, would insulated wire be a good solution?

The wire needs to be insulated or its one thick loop instead of many loops.

 

[cmb]

@bob012345 @Charles Link
I have a question about the experiment, when I am using bare copper, the coils are not allowed to touch am I correct? Because I have seen this being done on YouTube with the coils touching, but I am unsure if this is correct. If this is not allowed, would insulated wire be a good solution?

You might have been seeing enamelled wire, which is insulated but appears copper-coloured.

You will get bigger EMF for smaller clearances between magnet and wire, so it is worth reducing the clearances.

 

[bob012345]

If there were one solid loop with uninsulated wire then it seems to me one could at most pick up half the $emf$ if the contact points were exactly opposite on the loop. Less if closer according to the ratio of the short path to the total circumference.

 

[Einstein44]

If there were one solid loop with uninsulated wire then it seems to me one could at most pick up half the $emf$ if the contact points were exactly opposite on the loop. Less if closer according to the ratio of the short path to the total circumference.

I finished my experiment (using insulated wire), this time I had very little fluctuations so overall some good results. Although the best fit curve turned out to not be fully linear, but slightly curved. Ill have to have a look into that.

If anyone is interested in seeing these results, you can send me your email and I will share a google sheets with you (I cannot post this on here since I will be using this data, but since some of you were asking to see it I am more than happy to show it to you.)

 

[hutchphd]

If there were one solid loop with uninsulated wire then it seems to me one could at most pick up half the $emf$ if the contact points were exactly opposite on the loop. Less if closer according to the ratio of the short path to the total circumference.

That is not correct unless the wire is closed. The loop will instead simply consist of the half circle and the leads to the voltmeter. See Prof Walter Lewin's famous lecture on induced EMF

 

[bob012345]

That is not correct unless the wire is closed. The loop will instead simply consist of the half circle and the leads to the voltmeter. See Prof Walter Lewin's famous lecture on induced EMF

In that comment I assumed the wire is a closed loop. Not like this;

But like this only the wires are uninsulated;

 

[hutchphd]

The measurement will depend upon how you close the loop with your meter. Even for a closed loop (although it is a bit more complicated). Prof Lewin will convince you.

 

[bob012345]

The measurement will depend upon how you close the loop with your meter. Even for a closed loop (although it is a bit more complicated). Prof Lewin will convince you.

I see your point. So, if there is a solid closed loop, there is an $emf$ around the loop. Is it possible to tap only a portion of that induced voltage say if the leads went straight out on either side to infinity? Given that the current through the leads to the voltmeter is ≈ zero and there has to be a current in the closed loop, it seemed reasonable to me it could be tapped. I was thinking of it like a rheostat.

 

[hutchphd]

But think about traversing the loop with the two leads from the DVM. When the leads are close together do you read zero or the full loop EMF? How about when they are 180 deg apart ? Is it +EMF/2 ? Maybe -EMF/2?
Even with the leads to infinity you got to attach the DVM in a particular way eventually and the result will depend.
The answer I like best is that Kirchhoff simply doesn't work when the are time dependent B fields. There is no single valued voltage.
Prof Lewin says there were several EE faculty (at MIT) who thought this was a trick! Not easy (I never really understood the issue until Lewin lectures...although I knew there was something amiss in my understanding)

 

[bob012345]

But think about traversing the loop with the two leads from the DVM. When the leads are close together do you read zero or the full loop EMF? How about when they are 180 deg apart ? Is it +EMF/2 ? Maybe -EMF/2?
Even with the leads to infinity you got to attach the DVM in a particular way eventually and the result will depend.
The answer I like best is that Kirchhoff simply doesn't work when the are time dependent B fields. There is no single valued voltage.
Prof Lewin says there were several EE faculty (at MIT) who thought this was a trick! Not easy (I never really understood the issue until Lewin lectures...although I knew there was something amiss in my understanding)

Wow! I watched the video and I did find it fascinating. Thanks. So, non-conservative fields...

I was just thinking earlier today that in a single closed loop if you attached voltmeter leads at two random point on the loop how would it know what the voltage should read because it seemed that it would be path dependent. I resolved that by thinking it was like a rheostat whereby it would choose the shorter path. Now, it seems it is kind of like a rheostat but which path depends on which way the meter is. So doing two measurements would always yield the total $emf$ if you subtract them. Based on this it seems putting the leads 180° apart would give half the value of the $emf$ since $R_1$ is equal to $R_2$ but the sign would depend on the meter but as prof. Lewin says one just uses Lenz's law to figure the actual sign.

Prof. Lewin has a great lecture style.

 

[Charles Link]

Wow! I watched the video and I did find it fascinating. Thanks. So, non-conservative fields...

I was just thinking earlier today that in a single closed loop if you attached voltmeter leads at two random point on the loop how would it know what the voltage should read because it seemed that it would be path dependent. I resolved that by thinking it was like a rheostat whereby it would choose the shorter path. Now, it seems it is kind of like a rheostat but which path depends on which way the meter is. So doing two measurements would always yield the total $emf$ if you subtract them. Based on this it seems putting the leads 180° apart would give half the value of the $emf$ since $R_1$ is equal to $R_2$ but the sign would depend on the meter but as prof. Lewin says one just uses Lenz's law to figure the actual sign.

Prof. Lewin has a great lecture style.

We have a couple of recent threads on the Physics Forums that discuss Professor Lewin's paradox in detail.

See https://www.physicsforums.com/threads/how-to-recognize-split-electric-fields-comments.984872/
You might also find the "link" in post 2 of this thread of interest, where a couple of us worked a tricky homework problem involving a changing flux in the loop. We finally got an answer that we all agreed upon towards the very end of what was a rather lengthy series of posts.

 

[hutchphd]

We finally got an answer that we all agreed upon towards the very end of what was a rather lengthy series of posts.

Although that is true I do not think most of that thread was a useful exercise. Not for me at least. There is too much bad physics done (present company excepted !)

 

[Charles Link]

Although that is true I do not think most of that thread was a useful exercise. Not for me at least. There is too much bad physics done (present company excepted !)

I agree. On that one, we all spun our wheels an awful lot, but I think it might be worthwhile to look to the very end where we finally got a solution. In addition, @cnh1995 's solution (post 192 )is a good one, and although the link in post 164 above was met with some controversy, I do see the merit in recognizing the difference between the electrostatic conservative E field $E_s$, and the non-conservative induced E field $E_m$. From the $E_s$, you can deduce an electrostatic potential at each point, which @cnh1995 's solution utilizes. See also post 193, where I worked through some of the mathematical detail of post 192. We're getting sidetracked here though=we should try to stay on the topic at hand of the magnet falling through the loop. :)

 

[bob012345]

I agree. On that one, we all spun our wheels an awful lot, but I think it might be worthwhile to look to the very end where we finally got a solution. In addition, @cnh1995 's solution (post 192 )is a good one, and although the link in post 164 was met with some controversy, I do see the merit in recognizing the difference between the electrostatic conservative E field $E_s$, and the non-conservative induced E field $E_m$. From the $E_s$, you can deduce an electrostatic potential at each point, which @cnh1995 's solution utilizes.

Thanks. I found prof. Lewin's supplement where he proposes a test of the concept. I'll try that first and then this problem before I look at the answer at the end.

 

[bob012345]

I agree. On that one, we all spun our wheels an awful lot, but I think it might be worthwhile to look to the very end where we finally got a solution. In addition, @cnh1995 's solution (post 192 )is a good one, and although the link in post 164 above was met with some controversy, I do see the merit in recognizing the difference between the electrostatic conservative E field $E_s$, and the non-conservative induced E field $E_m$. From the $E_s$, you can deduce an electrostatic potential at each point, which @cnh1995 's solution utilizes. See also post 193, where I worked through some of the mathematical detail of post 192. We're getting sidetracked here though=we should try to stay on the topic at hand of the magnet falling through the loop. :)

Is the solution at the very end for a different problem as I see a ground connection which is not in the original? I see no conserved fields in the original.

 

[Charles Link]

Is the solution at the very end for a different problem as I see a ground connection which is not in the original? I see no conserved fields in the original.

It's the same problem. In the solution of post 192, he chose to set $V_B=0$. The choice was rather arbitrary. Meanwhile the conservative $E$ fields arise rather surprisingly in a problem like that. You can work the problem without them, and solve for the currents with 6 equations and 6 unknowns, (like I did, and I did do that correctly), but if you compute the EMF's and the $IR$ voltage drop for each segment, you will find they generally don't agree. There is an electrostatic potential difference that arises between the various nodes, that must be added to the EMF's to get the voltage drop to be equal to the $IR$ product. This electrostatic potential $V_{ab}$ can be computed from $\mathcal{E}-IR$, and that is what they are wanting for the answer to the problem. This wasn't completely clear to us at first=we found it to be a very educational problem.

 

[bob012345]

It's the same problem. In the solution of post 192, he chose to set $V_B=0$. The choice was rather arbitrary. Meanwhile the conservative $E$ fields arise rather surprisingly in a problem like that. You can work the problem without them, and solve for the currents with 6 equations and 6 unknowns, (like I did, and I did do that correctly), but if you compute the EMF's and the $IR$ voltage drop for each segment, you will find they generally don't agree. There is an electrostatic potential difference that arises between the various nodes, that must be added to the EMF's to get the voltage drop to be equal to the $IR$ product. This electrostatic potential $V_{ab}$ can be computed from $\mathcal{E}-IR$, and that is what they are wanting for the answer to the problem. This wasn't completely clear to us at first=we found it to be a very educational problem.

That seems contrary to the video and supplement from prof. Lewin I have been studying. He never mentioned an electrostatic potential and this problem isn't too different. If the question had been what would a voltmeter read between A and B are you suggesting the answer would be completely different? Also, if it were just the circle would there be an electrostatic potential too?

 

[Charles Link]

That seems contrary to the video and supplement from prof. Lewin I have been studying. He never mentioned an electrostatic potential and this problem isn't too different. If the question had been what would a voltmeter read between A and B are you suggesting the answer would be completely different?

Yes, Professor Lewin gets it right. Even though I was already completely familiar with Professor Lewin's concepts, the homework problem (of post 2 of the "link" of post 164 above) was some new material for me. In post 193 of that homework problem, I showed the mathematics behind the solution of post 192. Since Professor Lewin works simply with EMF's and closed loops, he doesn't need to concern himself with an electrostatic field, because when you go around a loop $\oint E_s \cdot dl=0$, while $\oint E_m \cdot dl=-\dot{\Phi}=\mathcal{E}$. (I solved the problem by Professor Lewin's method, but the solution presented by @cnh1995 in post 192 gets the answer much quicker.)
Meanwhile, yes, the voltmeter does not read this electrostatic potential, and will get different results (per Professor Lewin), depending upon how the leads are connected to the ring.

 

[bob012345]

Yes, Professor Lewin gets it right. Even though I was already completely familiar with Professor Lewin's concepts, the homework problem (of post 2 of the "link" of post 164) was some new material for me. In post 193 of that homework problem, I showed the mathematics behind the solution of post 192. Since Professor Lewin works simply with EMF's and closed loops, he doesn't need to concern himself with an electrostatic field, because when you go around a loop $\oint E_s \cdot dl=0$, while $\oint E_m \cdot dl=-\dot{\Phi}=\mathcal{E}$. (I solved the problem by Professor Lewin's method, but the solution presented in post 192 gets the answer much quicker.)
Meanwhile, yes, the voltmeter does not read this electrostatic potential, and will get different results (per Professor Lewin), depending upon how the leads are connected to the coil.

Thanks. If a voltmeter won't even read it then it sounds more like bookkeeping than physics.

 

[Charles Link]

Thanks. If a voltmeter won't even read it then it sounds more like bookkeeping than physics.

I find the electrostatic potential that can arise in these problems to be an interesting concept. When you consider an inductor that has an induced $E_m$, in order for there to be nearly zero electric field inside the conducting coil of the inductor, there must necessarily be an electrostatic $E_s=-E_m$. If $\int E_m \cdot dl=\mathcal{E}$, we must also have an electrostatic potential of the same amplitude that will exist both inside and outside the inductor, (while the $E_m$ stays inside the inductor. Remember $\oint E_s \cdot dl=0$). For the case of the inductor, (the coil of wire in our experiment above when properly insulated), it can be argued that it is this electrostatic potential that we are measuring on the oscilloscope.

Professor Lewin's case can complicate matters, but with an insulated coil, and when you run both leads together in a twisted pair to the oscilloscope, you normally avoid the problems that Professor Lewin discusses.

 

[hutchphd]

I find the electrostatic potential that can arise in these problems to be an interesting concept.

I cannot disagree more vehemently as we have previously discussed. It is a confusing sideshow IMHO.

 

[bob012345]

I find the electrostatic potential that can arise in these problems to be an interesting concept. When you consider an inductor that has an induced $E_m$, in order for there to be nearly zero electric field inside the conducting coil of the inductor, there must necessarily be an electrostatic $E_s=-E_m$. If $\int E_m \cdot dl=\mathcal{E}$, we must also have an electrostatic potential of the same amplitude that will exist both inside and outside the inductor, (while the $E_m$ stays inside the inductor. Remember $\oint E_s \cdot dl=0$). For the case of the inductor, (the coil of wire in our experiment above when properly insulated), it can be argued that it is this electrostatic potential that we are measuring on the oscilloscope.

Professor Lewin's case can complicate matters, but with an insulated coil, and when you run both leads together in a twisted pair to the oscilloscope, you normally avoid the problems that Professor Lewin discusses.

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?