Greetings,
In the hopes of reducing the amount of collisions that occur between my grid and the incoming deuterons, I have been searching fusor.net and other sites for potential manners to optimize grid transparency and deflect 40 keV deuterons.
After doing so, I came across a few sources, one of which included an old post by Jeroen Vriesman back in 2015 (see here: viewtopic.php?f=18&t=10265&p=68448&hilit=SmCo#p68448).
In the post linked above, the OP talks more about how he desires to use this for reducing the current pull of the fusor. However, it has occurred to me that the magnetic field produced by the magnets may be used to deflect the trajectories of deuterons in the slightest fashion to reduce collisions with the grid.
Thus, I propose that a typical grid configuration comprised of two loops made of nickel-coated samarium cobalt may be used to reduce collisions with the grid. The new grid would be made up entirely of this SmCo magnetic material (beyond the nickel layer), thus, the high voltage would be applied to it. The grid would look somewhat like the item below:
The apparatus that I am using has a -40 kV 400 W power source (the likes of which are still being modified). The magnetic SmCo grid will have an outer diameter of 1.25” and shall be placed in a stainless steel hemisphere of a 6” outer diameter. At the moment, I am not planning on using an ion gun, but its addition may prove tempting in the future.
If I succeed in getting this grid operational, I am planning to move onto more geometrically transparent designs (such as those modeled by the University of Wisconsin-Madison). Thus, I will be able to determine if the use of magnetic materials (SmCo magnets) are able to decrease the number of collisions with the inner grid wires.
Thus, I was curious if the individuals on this forum (who are far more experienced and knowledgeable on this topic than I) have any input on my design and whether or not the use of magnets would do any useful work to prevent deuterons from colliding with the grid.
If it happens to be that my proposed magnets are of no use for deuterons deflection, I was curious if they might be useful for some other purpose (i.e. decreasing current draw). Thank you.
Sincerely,
Joshua Guertler
SmCo Grid for Deuteron Deflection
-
- Posts: 198
- Joined: Fri Apr 21, 2017 10:59 am
- Real name: Joshua Guertler
- Rich Feldman
- Posts: 1471
- Joined: Mon Dec 21, 2009 6:59 pm
- Real name: Rich Feldman
- Location: Santa Clara County, CA, USA
Re: SmCo Grid for Deuteron Deflection
Interesting idea, Joshua. Reminds me of original polywell fusors (?) with ring-shaped electromagnet coils inside.
Issue 1. How do you propose to orient your permanent magnets? All north poles up? All north poles away from the sphere center? Each arc magnetized along its length, or transversely in the plane of the spherical surface?
How will the resulting static magnetic field be oriented in space outside the magnets? How is it supposed to deflect moving ions away from hitting the grid? Do you understand the basic f = q times (v cross B) formula for force on a charged particle in a magnetic field?
There's an inherent magnetic field from the HV current in any fusor, but its magnitude is negligible compared to, say, the geomagnetic field, or the weakest intentionally permanent magnets.
Issue 2. How do you propose to cool the grid, to keep high energy magnetic materials from overheating? SmCo is much more tolerant than NdFeB, but its useful temperature range is all lower than the useful temperature range of ordinary soldering irons.
Issue 1. How do you propose to orient your permanent magnets? All north poles up? All north poles away from the sphere center? Each arc magnetized along its length, or transversely in the plane of the spherical surface?
How will the resulting static magnetic field be oriented in space outside the magnets? How is it supposed to deflect moving ions away from hitting the grid? Do you understand the basic f = q times (v cross B) formula for force on a charged particle in a magnetic field?
There's an inherent magnetic field from the HV current in any fusor, but its magnitude is negligible compared to, say, the geomagnetic field, or the weakest intentionally permanent magnets.
Issue 2. How do you propose to cool the grid, to keep high energy magnetic materials from overheating? SmCo is much more tolerant than NdFeB, but its useful temperature range is all lower than the useful temperature range of ordinary soldering irons.
All models are wrong; some models are useful. -- George Box
- Trent Carter
- Posts: 23
- Joined: Tue Dec 20, 2016 3:38 pm
- Real name: Trent Carter
- Location: Melbourne, Florida
- Contact:
Re: SmCo Grid for Deuteron Deflection
The concept wont work, but you should still do it as you will learn a lot. Major scientific advancements were the end result of experimentation without merit. Heck Nikola Tesla did not even believe in electrons!
1. Your SmCo will get very hot very quickly and will reach the point where its dipoles can flip on their own, resulting in magnet-no-more
2. The SmCo will trap the plasma in pretty plasma rings. Take some pics, it will be cool, but wont last long. See (1)
3. The Deuterons wont deflect much while traveling quite a handsome chunk of the speed of light, accelerated by your 40keV. However your electrons at 1/4000 the mass of the p+n will deflect more. However electrostatic forces are already in your favor there.
4. If you switch to a 2D cathode then you could use an external helmholtz coil to reduce the electrons hitting the chamber by turning them in the B field, much like a magnetrons magnets. You could do some research there. But does not fit your theory.
5. Most commercial / research grade fusion devices uses magnets. Mostly electromagnets, even more often high temp superconductive ceramics at >> Teslas, but its a start.
Good luck
Trent
1. Your SmCo will get very hot very quickly and will reach the point where its dipoles can flip on their own, resulting in magnet-no-more
2. The SmCo will trap the plasma in pretty plasma rings. Take some pics, it will be cool, but wont last long. See (1)
3. The Deuterons wont deflect much while traveling quite a handsome chunk of the speed of light, accelerated by your 40keV. However your electrons at 1/4000 the mass of the p+n will deflect more. However electrostatic forces are already in your favor there.
4. If you switch to a 2D cathode then you could use an external helmholtz coil to reduce the electrons hitting the chamber by turning them in the B field, much like a magnetrons magnets. You could do some research there. But does not fit your theory.
5. Most commercial / research grade fusion devices uses magnets. Mostly electromagnets, even more often high temp superconductive ceramics at >> Teslas, but its a start.
Good luck
Trent
-
- Posts: 1850
- Joined: Wed Apr 21, 2004 10:29 pm
- Real name: John Futter
- Contact:
Re: SmCo Grid for Deuteron Deflection
Trent
Protons deuterons do not get anywhere relativistic with 40kV acceleration
from memory when i did work this out years ago somewhere around 1000km/sec
all of this due to the difference between electron mass and amu mass
Protons deuterons do not get anywhere relativistic with 40kV acceleration
from memory when i did work this out years ago somewhere around 1000km/sec
all of this due to the difference between electron mass and amu mass
- Trent Carter
- Posts: 23
- Joined: Tue Dec 20, 2016 3:38 pm
- Real name: Trent Carter
- Location: Melbourne, Florida
- Contact:
Re: SmCo Grid for Deuteron Deflection
Yes, 1% is not "near" C, however its way too fast for a magnet to have much affect. I had to make a quick XLS on this, as I recalled it being around 50% of C, but that was an electron at 40keV. I stand corrected.
Correction: I used the mass of a proton, obviously the Deuteron is just over 2x the mass of a proton. So more like 0.65% C. updated image below
Correction: I used the mass of a proton, obviously the Deuteron is just over 2x the mass of a proton. So more like 0.65% C. updated image below
Last edited by Trent Carter on Thu Sep 05, 2019 7:41 pm, edited 2 times in total.
- Rich Feldman
- Posts: 1471
- Joined: Mon Dec 21, 2009 6:59 pm
- Real name: Rich Feldman
- Location: Santa Clara County, CA, USA
Re: SmCo Grid for Deuteron Deflection
Let's do a SWAG without looking up any coefficients on the Internet.
Rest mass of electron is about 500 keV. Proton is about 1800 times more massive (900 MeV); deuteron is twice that.
Velocity of deuteron with kinetic energy of 40 keV:
m * v^2 / 2 = 40/1800000 m * c^2.
v^2/c^2 = 4/90000 = 1/22500
v/c = 1/150
v = 2000 km/s. Nice how the answer is a round number. 2 mm per nanosecond.
An electron at 40 keV, neglecting relativity, is 3600 times less massive, so 60 times faster.
v/c = 0.4.
v = 120,000 km/s. In a nanosecond it travels 120 mm, comparable to the scale of our fusors.
The correction for relativity, in the electron case, is on the order 8%. (ratio of kinetic energy to rest mass)
Now who wants to figure the radius of gyration in a magnetic field?
A starting point could be the numbers for a cyclotron, for someone who remembers the values.
Size, field strength, and particle energy.
As I warned Joshua, I think the temperature and magnet strength problems are not as fundamental as the geometric issues, for what he wanted to do. Trent had some specific suggestions along those lines.
[edit] 40 keV deuteron has same curve radius as 80 keV proton.
10 times smaller than in a cyclotron delivering 8 MeV protons or 4 MeV deuterons.
I figure radius of about 2 inches at B = 1 tesla, which is a much stronger field than Joshua's magnetized grid can produce in a space of that size.
Trent, want to check my numbers? I was surprised at the roundness of the answer (when given in inches).
Rest mass of electron is about 500 keV. Proton is about 1800 times more massive (900 MeV); deuteron is twice that.
Velocity of deuteron with kinetic energy of 40 keV:
m * v^2 / 2 = 40/1800000 m * c^2.
v^2/c^2 = 4/90000 = 1/22500
v/c = 1/150
v = 2000 km/s. Nice how the answer is a round number. 2 mm per nanosecond.
An electron at 40 keV, neglecting relativity, is 3600 times less massive, so 60 times faster.
v/c = 0.4.
v = 120,000 km/s. In a nanosecond it travels 120 mm, comparable to the scale of our fusors.
The correction for relativity, in the electron case, is on the order 8%. (ratio of kinetic energy to rest mass)
Now who wants to figure the radius of gyration in a magnetic field?
A starting point could be the numbers for a cyclotron, for someone who remembers the values.
Size, field strength, and particle energy.
As I warned Joshua, I think the temperature and magnet strength problems are not as fundamental as the geometric issues, for what he wanted to do. Trent had some specific suggestions along those lines.
[edit] 40 keV deuteron has same curve radius as 80 keV proton.
10 times smaller than in a cyclotron delivering 8 MeV protons or 4 MeV deuterons.
I figure radius of about 2 inches at B = 1 tesla, which is a much stronger field than Joshua's magnetized grid can produce in a space of that size.
Trent, want to check my numbers? I was surprised at the roundness of the answer (when given in inches).
All models are wrong; some models are useful. -- George Box
-
- Posts: 1850
- Joined: Wed Apr 21, 2004 10:29 pm
- Real name: John Futter
- Contact:
Re: SmCo Grid for Deuteron Deflection
thanks Rich
i'm out by a factor of two
not bad for an old git who did a few calcs 25 + years ago while designing implanters with mass selection magnets
i'm out by a factor of two
not bad for an old git who did a few calcs 25 + years ago while designing implanters with mass selection magnets
- Trent Carter
- Posts: 23
- Joined: Tue Dec 20, 2016 3:38 pm
- Real name: Trent Carter
- Location: Melbourne, Florida
- Contact:
Re: SmCo Grid for Deuteron Deflection
Mr Feldman; sure I will check your numbers...
I get 1.61 Inches. ...but pretty close to 2.
Magnetic forces apply a centripetal force F=mv^2/r. Then F=qvB
qvB=(mv^2)/r
r=mv/qB
v=Speed m=Mass q=Charge B=Magnetic field strength
Static:
Deuterium Mass: 3.35E-27
Deuterium Charge: 1.60E-19
Voltage: 40kV
B Field: 1 Tesla
Calculated:
Velocity: 1.96E+06
Radius: 1.61 In (40.912mm)
I get 1.61 Inches. ...but pretty close to 2.
Magnetic forces apply a centripetal force F=mv^2/r. Then F=qvB
qvB=(mv^2)/r
r=mv/qB
v=Speed m=Mass q=Charge B=Magnetic field strength
Static:
Deuterium Mass: 3.35E-27
Deuterium Charge: 1.60E-19
Voltage: 40kV
B Field: 1 Tesla
Calculated:
Velocity: 1.96E+06
Radius: 1.61 In (40.912mm)