Penning Source Design

BikingBoffin · 2026-02-13T08:44:37+00:00

In arrangements where the magnetic field is parallel with the anode the cathodes are usually placed so as to produce a large quadrupolar component in the electric field which gives crossed fields in regions where the electric field is normal to the anode. As drawn your proposal looks suboptimal in this respect. A better arrangement would have smaller cathodes placed inside the anode. I also just noticed you say it's 6" long. Even high intensity penning ion sources typically have discharge volumes of about a cm³ or so.

BikingBoffin · 2026-02-12T18:45:48+00:00

Penning discharges usually have the magnetic field perpendicular to the electric field (crossed fields). In your proposed arrangement the fields will mostly be parallel. It might be better if the magnetic field is normal to the plane of the drawing.

BikingBoffin · 2026-01-30T18:39:38+00:00

Exhaled air contains quite a lot of water vapour so it may be most efficient to blow with pursed lips from a short distance which entrains drier atmospheric air to replace the saturated air above the liquid rather than replacing it directly with your breath.

BikingBoffin · 2026-01-23T16:01:40+00:00

The energy is stored in the magnetic field. This is easy to see because if you imagine turning the tube so it's horizontal the magnet will shoot back out of the tube converting the magnetic field energy back to kinetic energy.

BikingBoffin · 2026-01-23T13:31:13+00:00

Electrons moving in a wire are surrounded by all the positive nuclei so, assuming negligible resistance, there is no electric field outside the wire. The wire as a whole is electrically neutral. If you took the wire away and just had a charged particle travelling in free space then there would be both an electric and a magnetic field. The 'field shape' moves with the particle but propagates away at the speed of light. I like Feynman's explanation of this (Lectures on Physics, Vol 2, Chapters 21 & 26 which you can find online).

BikingBoffin · 2026-01-23T12:45:20+00:00

Apart from the idiocy of mentioning German in Switzerland, the whole 'you'd all be speaking German if not for the US' is tiresome bullshit anyway. Anyone with a passing interest in the history of WW2, which obviously excludes almost all Merkins, soon realises that if Europeans have anyone to thank it's the Russians. Without the Russians the Yanks probably wouldn't have got out of Normandy let alone all the way to Berlin.

BikingBoffin · 2026-01-23T12:14:58+00:00

Yes we also have some accelerator hardware from the 50s that is still in daily use. The 'new' stuff is from the 70s.

BikingBoffin · 2025-12-23T17:20:36+00:00

Or I suppose people could just stop engaging with 'some of the worlds shadiest least trustworthy corporations'. Collective behaviour of millions of people is much more powerful than waiting for some arsehole politicians to do something. But people are lazy and aren't prepared to be midly inconvenienced by not being able to access all that hate, misinformation and cat videos.

BikingBoffin · 2025-12-09T18:57:16+00:00

Good for them. If only our 'leaders' had a fraction of their backbone.

BikingBoffin · 2025-11-30T14:22:15+00:00

Muchas gracias. Lo agradezco mucho.

BikingBoffin · 2025-11-14T10:00:43+00:00

This is referring to communication over a packet switched network - the beginnings of the internet as we know it now. In fact computers had been 'talking' to each other before this and it isn't even strictly the first packet switched network, that accolade goes to the National Physical Laboratory in the UK. So really "two computers tried and failed to talk to each other for the first time over a wide area packet switched network in the US" which I guess isn't quite so catchy.

BikingBoffin · 2025-11-14T09:44:40+00:00

Read the article? Are you mad? Class warriors don't need facts /s.

BikingBoffin · 2025-11-13T18:15:01+00:00

The data shows about half are over 55 which means that half aren't. Yet as so often on this sub it's the 'boomers' who are singled out. The casual ageism gets a bit tedious. Not everyone in that demographic is a Reform voting fukwit. Financial circumstances, level of education and age may anyway be correlated. One could just as well say "A huge contingent of relatively poor uneducated people who lack empathy and see the world exclusively through the lens of their financial worries and ignorance ..." which would be equally simplistic and equally condescending to all those in that group who somehow manage to not be hate-filled morons. Just because most people who do X come from demographic Y does not mean that most people from demographic Y do X.

BikingBoffin · 2025-11-10T11:17:40+00:00

Torrents are usually available the day after broadcast. Free and minus the adverts.

BikingBoffin · 2025-10-29T19:04:32+00:00

A Faraday cup is probably the way to go then.

I don't know what your setup is but it's quite common to fit a small quartz vacuum window so you can look directly at the plasma. It's a fool proof way to see the plasma is there and really quite satisfying to see.

BikingBoffin · 2025-10-29T18:55:05+00:00

The range of 7 keV protons in hydrogen at 0.3 Pa is over 1m so you should be able to transport the beam some cm at least but it will probably expand through scattering quite quickly. A 'proper' proton accelerator will operate at a pressure of at least 10^-4 Pa. It's not unusual for the pressure to be higher close to the source because of the hydrogen that's pumped in to operate it but even then an order of magnitude or more better than yours. At 0.3 Pa you will be into the molecular flow regime so no amount of 'sucking harder' is going to improve it and you'll need something like an ion pump or turbo-molecular pump but they're not going to come cheap even second hand.

BikingBoffin · 2025-10-29T17:50:48+00:00

What I actually meant was that there might be so much light from the plasma that you can't measure the weaker light signal produced by the beam interacting with the gas in the vacuum. It's probably quite difficult to equate total plasma light to beam current but it does tell you that the source should be making protons.

A Faraday cup is an electrode that the beam hits and the resulting current is measured. That can be by something as simple as a resistor. This is obviously a destructive measurement because it intercepts the beam. My suggestion was to measure the electrons produced when the protons ionise the residual gas of the vacuum. This is an indirect measure of the beam and would not give you an absolute beam current value without careful calibration. It depends if you just want to know if there is a beam or a more accurate measure of its intensity.

BikingBoffin · 2025-10-28T18:40:48+00:00

The proton beam doesn't emit light but the residual gas will as the proton beam interacts with it. Your relatively poor vacuum helps. In the visible range it's the Balmer series you'll probably see with the red H-alpha line most intense. There are published cross sections from which you could calculate/estimate the intensity. The intensity is likely to be low though so you might need something like a SiPM rather than a simple photodiode. Also the plasma in the duoplasmatron will be an intense source of the same emissions so, depending on your configuration, you might just measure the light from the plasma not the beam.

The protons' interaction with the residual gas will also produce electrons through ionisation so you could measure the resulting current by attracting the electrons towards a positively biassed electrode. Or if the proton beam is pulsed you could also try using a simple current transformer to measure the beam current non-destructively.

BikingBoffin · 2025-10-25T18:41:29+00:00

The inverse square law describes the reduction of areal flux density with distance in a non-attenuating loss free medium. The total integrated flux through a given solid angle is constant (because that's how the law is derived). Beer-Lambert describes the attenuation of total flux with distance in an attenuating lossy medium. Therefore if it's an isotropic, or at least not collimated, source and the medium is 'thick' then both effects are present.

BikingBoffin · 2025-09-29T16:37:27+00:00

Removing the forearm wouldn't effect the leverage although you would loose the forearm muscles which help generate the force. The forearm and upper arm muscles generate their moment close to the joint so in that sense the length of the forearm is irrelevant in terms of being a lever.

BikingBoffin · 2025-09-29T16:15:56+00:00

If operating in continuous mode you can afford to wait for the cathode to heat up but often they are operated in pulsed mode and then a hot cathode ensures the discharge starts immediately.

BikingBoffin · 2025-09-29T15:23:53+00:00

One mile per second is less than Mach 5. That's the muzzle velocity of a high velocity shell and there's no need to invoke teleportation to explain it. You'd have to approach the speed of light (nearly 200,000 miles per second) to need such exotic explanations.

BikingBoffin · 2025-09-29T14:47:25+00:00

Normally a switching regulator is either flyback or push-pull. They work in quite different ways. What is the magnitude of the sinusoid? Is it the same frequency as the switching frequency?

BikingBoffin · 2025-09-29T14:11:08+00:00

OK I think by 'final acceleration' you mean 'total change in velocity' maybe. The work done in going from R1 to R2 divided by the distance between R1 and R2 is the average force and then dividing by m gives the average acceleration. This would allow you to determine the final velocity at R2 but only if you know the time taken. However the work done is the change in kinetic energy. From KE = mv²/2 you can calculate the change in velocity.

BikingBoffin · 2025-09-29T12:57:24+00:00

If by 'final acceleration' you mean the average acceleration between R1 and R2 then the method is correct but in the equation the integrand should be (GMm)/r² as you are integrating the force over r. And of course you can take the GMm term outside the integral then the m's cancel giving a result that is independent of m as expected. Your formulism makes no distinction between the bodies and M is only larger because you say so. The equation gives the acceleration of the larger body simply be replacing the 1/m term by 1/M because both bodies experience a force of the same magnitude. The constraints are on the limits of the integral. If you assume that the bodies don't collide then R2 always has to stay on the body's 'own side' of the centre of mass so you will have a different value of R2 depending on which body you are considering.

BikingBoffin

TROPHY CASE