Question

Help Please,
I would like to calculate the lengths of tubes in the old linac wideroe structure.

Answer 1

the voltage in the structure oscilates V=v0cos(wt)

Answer 2

This looks as though it's about synchronising the tube/electrode voltages with the electron trajectories to get the electrons to be at a required point at a required time. I seem to remember words like "bunching" and "catching" in connection with controlling streams/currents of electrons.
Linac IS LINear ACcelerator (imaginative lot these people) as I now see from the enclosed picture.

Answer 3

so, match the transit time of the electrons in the tubes with the frequency of the ac ? (sort of a wave made up of two bits - "time AC voltage", and space - length of tube ?)

Answer 4

the particles are accelerated in the gaps and because the gained speed the length of the tubes becames longer.. I think I am missing some equations if someone could reffer

Answer 5

this looks like it's "VELOCITY MODULATION" in a jazzed up Klystron microwave oscillator ?

Answer 6

it's just one of the first accelerators from 1928

Answer 7

As the electrons get faster (and without getting into spec relativity) they'll travel further in the time period of the AC ? So, you need progressively longer tubing to keep on "giving them a kick" ?

Answer 8

yes, they recieve the kick and thus travell longer each time. I am missing some equations as to connect the Voltage applied - the force - with the length traveled

Answer 9

I may be able to wade through a book or two here, but it'll take a bit of time (I'm actually equation/algebra weary at the moment !). So, it depends on how long you've got, who else joins in the "fun", and whether or not it can be pieced/banged together.
As for the force, is that related to the equation
F=Eq -> F=(V/d) x e ?

Answer 10

@radar ??

Answer 11

Sorry but I am no help here, while I have had experience with klystrons, magnetrons, and amplitrons, they all were utilizing beams consisting of electrons and never did I get involved with their actual structure. The amplitron was used as a "after-burner" being driven by a magnetron and was a cross field device that was fail-safe, in that if it failed, the maggie would still provide the radar power but greatly reduced power.

Will come back and see what else has been added.......interesting!

Answer 12

no help? :((
It shouldn't be very difficult- some equation as to match the energy gained from the voltage with the speed gained each time

Answer 13

**It shouldn't be very difficult**

agreed, provided you are willing to look into it for yourself

Answer 14

For LINAC, the regions within the electrodes are field free and ions drift at constant velocity within the electrons. Since an AC voltage is applied across each gap, the ions must spend half the period of the AC cycle within every electron, else the phase will synchronize and the ions are slowed down instead of being boosted.

Let the voltage be \(V=V_{0}\cos{\omega t}\). For an ion with initial speed \(v_{0}\), charge \(q\) and mass \(m\), after the \(n\mathrm{th}\) electrode it has the kinetic energy: \[\frac{1}{2}mv_{n}^{2}=\frac{1}{2}mv_{0}^{2}+nqV\] and thus the speed of the electron in the \(n\mathrm{th}\) electrode is\[v_{n}=\sqrt{v_{0}^{2}+\frac{2nqV}{m}}\] Each ion must spend half the period \(T=\frac{2\pi}{\omega}\) om the electrode, thus the electrode length \(L\) must be: \[L_{n}=\frac{T}{2}v_n=\frac{\pi}{\omega}\sqrt{v_{0}^{2}+\frac{2nqV}{m}}\] Thus for the first electrode, \(n=0\) and the length \(L_0=\frac{\pi}{\omega}v_{0}\)

Answer 15

great answer! If you could please explain how you recieved nqV the potential energy

Answer 16

qV is related to work done as in
E=F/q, F=Eq, E=V/d, F=V/d q, F.d = Vq. (final exp is work)
Don't know about the n term..

Answer 17

@IrishBoy123 Bravo

Answer 18

@amirw81 Note that I am using the non-relativistic expression for kinetic energy expression for simplicity. Note that @osprey is right: when the ion is between electrons, it experiences a potential difference of \(V\), so it gains the energy \(qV\).

When it leaves the first electrode and reaches the second electrode, it gains an additional \(qV\) of kinetic energy. When it leaves the second electrode and reaches the third electrode, it gains another \(qV\) of energy. The \(n\) term simply captures this increase in kinetic energy as it is boosted.

Answer 19

@Festinger In slang english it gets kicks up the backside on a velocity modulated timing perhaps ?

Answer 20

I don't know what the speeds of electrons are in a LINAC, but I usually think as a rule of thumb that below about half the speed of light is "low" speed, and thus non relativistic. Mind you, an improved rule of thumb on the relativity speed, and an idea of the actual speeds might be "interesting".
It might also be worth "thinking" about LINACs doing electrons as opposed to protons and what would happen if that were "thought" about. On the other hand we might all be asleep by now.

Answer 21

10x, didn't think about adding up the energy qV (using the n).
if it interests you we were given some data-E=100eV energy at start and to calculate for particles of e, P and gold. But helped me a lot so thanks again

Answer 22

100eV sounds pretty low. But the post refers to an old linac, I think. I'm thinking that the speeds are sub relativistic.
Bear in mind, possibly, that CERN is so big that it crosses geographical borders and goes round in a circle, as do the particles (protons) being accelerated. That is very prob relativistic, and the particles that it "generates" are relativistic too.

Answer 23

15MeV gives speed of 0.9995c from an MIT linac. That is relativistic.
Source - "Special Relativity" AP French, MIT