Seeking the minimum distance between the rings.
Posted: Fri Jun 22, 2012 8:15 pm
This seems to be a common problem. What is the ideal minimum distance between each of the rings? Bussard offers us some hints as to a solution. Below is a quote from his IAF paper. The gyroradius of the electron is what could be needed. I did that calculation here. I would be interested in trying to “balance” the central plane B field with the main axes field. To get the gyroradius, you start by assuming some distance between the rings. From this, the gyroradius of the electron can be found using the method outlined below. The goal would be to see if the initial assumption matched the calculated result.
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“…The spacing between coils should be such that the central plane B field is approximately the same as that of the B field on main face axes. Typically, this may be at minimum the order of a few (5-10) electron gyro radii at the inter-corner field strength, but not larger than this...”
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Find the electron gyroradius:
Imagine an electron. Now, put a magnetic field on it. The electron will not move in any direction. Now “kick” the electron. The electron experiences a Lorentz force. This causes the electron to move in a circle. The direction (counterclockwise or clockwise) will depend on the direction of the magnetic field. The radius of this motion is the electron gyroradius. You can calculate it if you know the strength of the magnetic field and if you know the velocity of the electron perpendicular to that field.
Find the magnetic field:
I can find the magnetic field strength. I solved it for the November 2005, WB-6 tests. I assumed that the minimum distance between the rings was 4.25”, center to center. This was an educated guess. I am going to apply this method and see what minimum distance is needed. If this guess disagrees with the calculated value, either the guess is wrong or the method is flawed. Bussard estimated that at the corners the field was 70 to 100 Gauss [IAF, Page 6]. I calculated 89. Here is how I found it. During those four tests Bussard stated: there was 4,000 amps of current running through each ring [IAF, Page 12]. The “corner” is a point equidistant and centrally located between the three rings, about 5.2” from each ring. Locally, I treated the rings like they were straight wires. Then I used the Biot-Savart law to find the magnetic field. I accounted for 3 rings. The answer was 89 Gauss. This seemed reasonable and nicely agreed with Bussards' estimate.
Find the electron velocity via Energy:
I can also calculate the velocity of the electron at the corner. It is about 5.6% the speed of light. But I cannot know what percentage of this is parallel or perpendicular to the magnetic field. Hence, I cannot outright find the gyroradius. This velocity can be solved by assuming the total energy of an electron is constant. From Bussards paper, he estimated the average electron had 2,500 eV of energy in the center of the device [IAF, Page 12]. The ions need ~10,000 eV to fuse, so this seems reasonable. Incidentally, Rider would argue here. Rider argues energy transfer happens so fast that everything would have to be either 2,500 eV or 10,000 eV. He argues you cannot have two temperatures. Someone should look into this.
Energy In The Center:
I will assume the dead center of the Polywell has nearly no potential energy. This is for the electrons only. The ions see a completely different energy map. This energy has two parts: contributions from the magnetic and electric fields. To have no electric potential means that the electron is surrounded by uniform electrostatic charge in all directions. Additionally, the key to the Polywell is a no magnetic fields in the center, so there is no magnetic potential, either. Hence, the electron has only kinetic energy. This means the electron is moving about 9.9% the speed of light. As the electron moves from the center to the corner, it goes up in potential energy. This is because it is moving into regions where there are denser magnetic fields and lower electric fields. To move, the electron must transfer its kinetic energy into potential energy. If we assume the total energy is constant, then if we know the potential energy at the corner, we will know the electron's kinetic energy. From this we can find the velocity of the electron at the corner.
Surprisingly, the magnetic potential energy at the corner, is tiny. The reason for this is the magnetic moment of an electron is so small. The magnetic moment I used 9.28E-26 Joules/Tesla. There are lots of magnetic moments in the world. There are several moments for the electrons; the value is different for each place an electron could be found and different ways the electron spins. It is not an easy value to solve/estimate. Nominally, the moment is some multiple of the Bohr magneton (9.87E-24 J/T). The magnetic potential energy is the cross product of the moment and the field. This is why the magnetic moment is important. If the moment is higher the magnetic potential energy is higher. But, even if the moment was 3 times the Bohr value and the moment lined up in the highest energy state – the magnetic potential energy would still be tiny. The the range of possible values runs from E-29 to E-25. This is so, so, much smaller than the electric potential - that I just ignored the magnetic component.
The electric potential is easy to find. Remember this is for the electrons, not the ions. Bussard stated that there was a 10 Kv drop in the middle of the machine. I applied Gausses law and found that there were 5.5E11 electrons in the center. I treat this like a point charge. The corner is 0.46 meters away. So the electric potential energy from a point charge at this distance is 1,691 eV. This number makes sense. The electric potential is like a valley between two mountains. At dead center, your in the valley. Move a small distance away and the electron is repulsed from the other electrons in the cloud. This is the mountain peak. The single charge is sitting next to a big pile of electrons, and is repulsed by the columbic force. It is in a dense electric field at that point. So the electron flies away from the center. when it reaches the corner it has 809 eV in kinetic energy. This means the electron is traveling about 5.6% the speed of light at the corner.
Using Some Assumptions:
The overall velocity can be estimated from the kinetic energy. However, it is unknown how much of that runs perpendicular to the B field. I thought about trying to use the Lorentz force. The force will tell you some of the behaviors of the electron at the corner. The problem is you need to know the velocity before you can apply that equation. Dead end. If you watch simulations by HappyJack and Indrek, you can see that the electron is mostly moving back and forth up the field lines. Hence you would expect the perpendicular velocity to be less than 50%. I picked, 50% 30%, 10% and 2% and here is the minimum distance.
% of velocity perpendicular: Minimum distance:
====================================
50% 0.00136”
30% 0.00082”
10% 0.00027”
2% 0.00005”
This distance varies, but it is well below the 4.25” assumed at the start of this calculation.
====================
“…The spacing between coils should be such that the central plane B field is approximately the same as that of the B field on main face axes. Typically, this may be at minimum the order of a few (5-10) electron gyro radii at the inter-corner field strength, but not larger than this...”
====================
Find the electron gyroradius:
Imagine an electron. Now, put a magnetic field on it. The electron will not move in any direction. Now “kick” the electron. The electron experiences a Lorentz force. This causes the electron to move in a circle. The direction (counterclockwise or clockwise) will depend on the direction of the magnetic field. The radius of this motion is the electron gyroradius. You can calculate it if you know the strength of the magnetic field and if you know the velocity of the electron perpendicular to that field.
Find the magnetic field:
I can find the magnetic field strength. I solved it for the November 2005, WB-6 tests. I assumed that the minimum distance between the rings was 4.25”, center to center. This was an educated guess. I am going to apply this method and see what minimum distance is needed. If this guess disagrees with the calculated value, either the guess is wrong or the method is flawed. Bussard estimated that at the corners the field was 70 to 100 Gauss [IAF, Page 6]. I calculated 89. Here is how I found it. During those four tests Bussard stated: there was 4,000 amps of current running through each ring [IAF, Page 12]. The “corner” is a point equidistant and centrally located between the three rings, about 5.2” from each ring. Locally, I treated the rings like they were straight wires. Then I used the Biot-Savart law to find the magnetic field. I accounted for 3 rings. The answer was 89 Gauss. This seemed reasonable and nicely agreed with Bussards' estimate.
Find the electron velocity via Energy:
I can also calculate the velocity of the electron at the corner. It is about 5.6% the speed of light. But I cannot know what percentage of this is parallel or perpendicular to the magnetic field. Hence, I cannot outright find the gyroradius. This velocity can be solved by assuming the total energy of an electron is constant. From Bussards paper, he estimated the average electron had 2,500 eV of energy in the center of the device [IAF, Page 12]. The ions need ~10,000 eV to fuse, so this seems reasonable. Incidentally, Rider would argue here. Rider argues energy transfer happens so fast that everything would have to be either 2,500 eV or 10,000 eV. He argues you cannot have two temperatures. Someone should look into this.
Energy In The Center:
I will assume the dead center of the Polywell has nearly no potential energy. This is for the electrons only. The ions see a completely different energy map. This energy has two parts: contributions from the magnetic and electric fields. To have no electric potential means that the electron is surrounded by uniform electrostatic charge in all directions. Additionally, the key to the Polywell is a no magnetic fields in the center, so there is no magnetic potential, either. Hence, the electron has only kinetic energy. This means the electron is moving about 9.9% the speed of light. As the electron moves from the center to the corner, it goes up in potential energy. This is because it is moving into regions where there are denser magnetic fields and lower electric fields. To move, the electron must transfer its kinetic energy into potential energy. If we assume the total energy is constant, then if we know the potential energy at the corner, we will know the electron's kinetic energy. From this we can find the velocity of the electron at the corner.
Surprisingly, the magnetic potential energy at the corner, is tiny. The reason for this is the magnetic moment of an electron is so small. The magnetic moment I used 9.28E-26 Joules/Tesla. There are lots of magnetic moments in the world. There are several moments for the electrons; the value is different for each place an electron could be found and different ways the electron spins. It is not an easy value to solve/estimate. Nominally, the moment is some multiple of the Bohr magneton (9.87E-24 J/T). The magnetic potential energy is the cross product of the moment and the field. This is why the magnetic moment is important. If the moment is higher the magnetic potential energy is higher. But, even if the moment was 3 times the Bohr value and the moment lined up in the highest energy state – the magnetic potential energy would still be tiny. The the range of possible values runs from E-29 to E-25. This is so, so, much smaller than the electric potential - that I just ignored the magnetic component.
The electric potential is easy to find. Remember this is for the electrons, not the ions. Bussard stated that there was a 10 Kv drop in the middle of the machine. I applied Gausses law and found that there were 5.5E11 electrons in the center. I treat this like a point charge. The corner is 0.46 meters away. So the electric potential energy from a point charge at this distance is 1,691 eV. This number makes sense. The electric potential is like a valley between two mountains. At dead center, your in the valley. Move a small distance away and the electron is repulsed from the other electrons in the cloud. This is the mountain peak. The single charge is sitting next to a big pile of electrons, and is repulsed by the columbic force. It is in a dense electric field at that point. So the electron flies away from the center. when it reaches the corner it has 809 eV in kinetic energy. This means the electron is traveling about 5.6% the speed of light at the corner.
Using Some Assumptions:
The overall velocity can be estimated from the kinetic energy. However, it is unknown how much of that runs perpendicular to the B field. I thought about trying to use the Lorentz force. The force will tell you some of the behaviors of the electron at the corner. The problem is you need to know the velocity before you can apply that equation. Dead end. If you watch simulations by HappyJack and Indrek, you can see that the electron is mostly moving back and forth up the field lines. Hence you would expect the perpendicular velocity to be less than 50%. I picked, 50% 30%, 10% and 2% and here is the minimum distance.
% of velocity perpendicular: Minimum distance:
====================================
50% 0.00136”
30% 0.00082”
10% 0.00027”
2% 0.00005”
This distance varies, but it is well below the 4.25” assumed at the start of this calculation.