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 Author Topic: Cyclotron  (Read 298376 times) 0 Members and 2 Guests are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: January 29, 2004, 09:39:30 pm » posted from:,,Satellite Provider

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The magnetic force for charge particle (with charge q) moving with velocity
in a magnetic field is

The magnetic force is perpendicular to both the velocity and magnetic field ,

The power done by the magnetic force
So the magnetic force will not change the kinetic energy of the charge particle,

It only change the direction of the velocity for the charge particle.

If the velocity is in the same direction of the magnetic field,
then there is no magnetic force acting on the charge particle.

For charge particle moving in a uniform magnetic field ,
The velocity can be represented as ,
is a component of velocity
along B, which will not change.

The particle advances along while it moves in a circle in the plane formed by and.

The resulting trajectory forms a spiral with its axis along .
, where R is the radius.

So we have momentum
And the angular velocity of the circular motion
Which is independent of the velocity of the charge particle!( is the cyclotron frequency)

The constancy of the cyclotron frequency led to a device called cyclotron.
This java applet let you play with cyclotron.

This is a top view of the region of a cyclotron in which the particle circulate.

is a picture of a real cyclotron.

The two hollow D-shaped objects (open on their straight edges) are made of copper sheet.
These dees, as they are called, form part of an electrical oscillator, which establishes an alternating potential difference across the gap between them.

The dees are immersed in a magnetic field whose direction is into the plane of the screen,

Suppose that a proton, starts from the blue dot near the center of the cyclotron, initially moves toward a negatively dee.
It will accelerate toward this dee and will enter it.

Once inside, it is "screened" from electric fields by the copper walls of the dee.
The magnetic field is not screen by the (nonmagnetic) copper dee, so the proton moves in the circle path.

Assume that at the instant the proton emerges into the center gap (again) from first dee,
The accelerating potential difference has changed sign.
Thus the proton again faces a negatively charged dee and is again accelerated.

This process continues, the circulating proton always being in step with the oscillations of the dee potential, until the proton spiral out of the edge of the dee system.
The frequency of the electric oscillator must match the cyclotron frequency.

If the gap between the dees is very small , the two frequency is the same.
What if the gap is not small? You will need to adjust the frequency of the electric oscillator.

Enter the value of the frequency ratio (oscillator frequency/cyclotron frequency) into the textfield.
( Do not forget to hit RETURN button after you change the frequency ratio)

Click the red dot near the electric oscillator and drag it up/down to change the voltage of the oscillator.
Click Start to start the animation,
Click right mouse button to pause, click right mouse button again to resume.

The velocity of the charge particle is represented by yellow line.
The force acting on the particle is represented by red line.
Vy-Vx (velocity) of the charge particle is also shown to the right.
Click Clear to erase the trace of the trajectory.
Click Reset for default values.

For an electron : the charge , mass ,
If the energy of the electron is 1eV (accelerated under 1V biased voltage) and the magnetic field is 1T(Tesla),
The cyclotron frequency (It is so big!)

The velocity of the electron (it is moving so fast!)

If the velocity is perpendicular the magnetic field, the radius of the circular motion
( it is so small!)
If the velocity has a small component along the magnetic (say 1%),
The electron will move along the magnetic field line with velocity
Image how the electron moves under the above conditions!

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 « Embed this message Reply #1 on: January 30, 2004, 11:00:21 am » posted from:,,Satellite Provider

Subject: problems with cyclotron java applet
Date: Thu, 4 Jun 1998 19:37:38 +0200
From: "Pietro Diviacco" <diviacco@unige.it>
To: <hwang@phy03.phy.ntnu.edu.tw>
My name is Pietro Diviacco. I am an Italian surgeon and live and work in Genova, Italy.
I tried to play Your applet "Cyclotron" but there was some problems.
Some of the images (*.gif 7 and 14) were not available and the applet was not working.
In any case Your applets are very good and interesting (the working ones, of course).
Thank You in advance for fixing the problems.
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 « Embed this message Reply #2 on: January 30, 2004, 11:25:54 am » posted from:,,Satellite Provider

Subject: Re: Regarding your Cyclotron Applet.
Date: Wed, 6 Jan 1999 14:32:25 +0000 (WET)
From: Joao Manuel Henriques <jmh@camoes.rnl.ist.utl.pt>
To: Fu-Kwun Hwang <hwang@phy03.phy.ntnu.edu.tw>
Hi Again.
First let me apologise for not responding sooner, but I was on vacation.
Second, I would like to thank you for having sent the Source Code, it
really helped alot...
Now it is my turn to help you out (if U want to), I made another program
that simulated: ' Attwood's Machine', since you dont have it on the list I
could send you a copy of it ..(it is in C though...)
Let my know if you want it or not...BTW it also uses RK4, but it looks

Anyway thank you very much.

On Fri, 18 Dec 1998, Fu-Kwun Hwang wrote:

>
>
> Joao Manuel Henriques wrote:
>
> > Hi.
> > I am a student of the University of Lisbon and I was a assigned a task to
> > simulate the cyclotron.
> > It has been really hard to make something that actually resembled a
> > cyclotron, but i still cant make the electron spin according to the
> > frequency of the Voltage.
> >
> > Here is what I need to know:
> > -I calculated a formula that calculates Theta, but the center of the
> > circular movement is constant. In other words the Ray of the circunference
> > increases, but the center doesnt change like it should.
> > Can you tell me the formula to theta?
> >
>
> This is not the way I did it. There is a better way to do it:
> If you know the velocity V and force (F=qVXB) at time t,
> you will be able to calculate velocity at t+dt;
> If you know the position and velocity of a particle at time t,
> you will be able to calculate next position at time t+dt.
> Repeat the above loop, you will know the next position and velocity at next
> moment t+dt;
>
>
> >
> > There is something else I want to ask.
> > Is there a possibility of you sending me the source code?
> > I know this is an unusual request, but it would really make me understand
> > how the program really works. My source code is so complex that not even I
> > can understand it, and above all it doesnt simulate the cyclotron right.
> > I would be very gratefull if you could send it. I promise I wouldn't
> > change the code and say that the program is mine. All I need to know is
> > how you did it.
> >
> > Well, please e-mail me back with a response, preferably one with the
> > reply i want to get.
> >
> > Thank you for your attention.
> >
> >
> > Joao Henriques
> > http://camoes.rnl.ist.utl.pt/~jmh
>
> The source code is attached with this e-mail. But you will need to know about
> the
> Runge-Kutta method to understand the rk4 class.
>
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arika8178
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 « Embed this message Reply #3 on: January 28, 2005, 03:34:40 pm » posted from:Kajang,Negeri Sembilan,Malaysia

I'm so interested with your achievement in physics especially in cyclotron. I'm confuse with cyclotron mathematicals equation. How to used in Runga-Kutta Method? Anybody can help me?
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Fu-Kwun Hwang
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 « Embed this message Reply #4 on: February 12, 2005, 02:04:22 pm »

This is a two dimension simulation, so there are components for the force.
Runga-Kutta Method solve first order differential equation.
So force equation (second order ) need to be break into two first order equation.
Change $\frac{d^2 \vec{V}}{dt^2}=q \vec{V}\times\vec{B}$
into
$\frac{d\vec{x}}{dt}= V$ and $\frac{d\vec{V}}{dt}=q \vec{V}\times\vec{B}$
For the above two equations, each one have two components.
So you will have 4 equations for Runga-Kutta Method solver. :-)
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arika8178
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 « Embed this message Reply #5 on: May 13, 2005, 09:05:40 am » posted from:Kajang,Negeri Sembilan,Malaysia

Hi!
I had receive cyclotron java applet from your webside. I find all what i want except the source code. Actually, i would like to learn your complete cyclotron java applet programming. I really want to know how you derive all cyclotron variables and display the cyclotron simulation. Until now, i can't do it!

I really need your help. I would like to request your source code. I know it's ridiculous but I promise I could't change the code and say that the program is mine. All I need to know is how you do it.

Thanks!
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Fu-Kwun Hwang
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 « Embed this message Reply #6 on: May 13, 2005, 09:20:47 am »

However, it was written many years ago with JDK 1.0.2
If you want to program similar java applet. I would suggest you try to learn EJS (Easy Java Simulation) tool. It will generate the java aource code for you, but you need to provide the physics model.
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mma943
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 « Embed this message Reply #7 on: December 18, 2006, 10:53:52 pm »

Thank u for evrey things

-Can I knew a long of particl"s track  insid Dees?

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minhtue
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 « Embed this message Reply #8 on: October 15, 2007, 11:26:15 am » posted from:Singapore,,Singapore

Hi,
I'm Vo Thanh Minh Tue, Nus High School, Singapore. The Java Applet for the cyclotron is very interesting. I note that there are different meanings of the term "cyclotron frequency" in physics text. In some American textbooks (such as Serway's Physics for Scientist and Engineering and Hyper Physics Forum), cyclotron frequency refers to the angular frequency of the square wave of electric field applied between the dees. Hence, the angular frequency of the electric field matches the angular velocity of the particle: w=qB/m.
Thank you very much,
Vo Thanh Minh Tue.
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Fu-Kwun Hwang
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 « Embed this message Reply #9 on: October 15, 2007, 10:10:03 pm »

If the device was built to accelerate the electron in the cyclotron, the frequency has to be matched with the frequency for the electron to move around in the device. It is the frequency of the power supply should match the cyclotron frequency if the magnetic field is fixed (and normally this is the case). So I would define "cyclotron frequency" as qB/(2*pi*m) which is a property for the device, instead of the frequency for the electric field applied between the dees, which could be changed from the power generator.
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minhtue
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 « Embed this message Reply #10 on: October 16, 2007, 06:03:41 pm » posted from:Singapore,,Singapore

Thank you very much.
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crownabhisek
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 « Embed this message Reply #11 on: February 17, 2009, 02:15:26 am » posted from:Bhubaneswar,Orissa,India

Read the first few lines, it says, the magnetic field "charges" the direction. It should be "changes"
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lookang
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http://weelookang.blogspot.com

 « Embed this message Reply #12 on: February 17, 2009, 08:11:35 am » posted from:SINGAPORE,SINGAPORE,SINGAPORE

It will be easier for Fu-Kwun Hwang to see what you mean by using the "quote" and edit out the rest away

So the magnetic force will not change the kinetic energy of the charge particle,

It only charge the direction of the velocity for the charge particle.

If the velocity is in the same direction of the magnetic field,
then there is no magnetic force acting on the charge particle.

Yes, crownabhisek  is right, there appears to be a spelling error Thanks! crownabhisek
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Fu-Kwun Hwang
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 « Embed this message Reply #13 on: February 17, 2009, 03:09:25 pm »

Quote
the magnetic field "charges" the direction. It should be "changes"
It is fixed now. Thank you!
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dannydesiliva
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 « Embed this message Reply #14 on: September 22, 2009, 01:19:59 pm » posted from:Amritsar,Punjab,India

I'm trying to build a small cyclotron and am thus searching for a magnet. I've been looking for a neodythium disc magnet that I could have cut in half to act as the two 'D' magnets in the cyclotron. The size of the magnet needs be somewhere around 12" in diameter but I can't seem to find anyplace that sells them around this size. If anyone knows where I could find some magnets or is willing to sell some of their own that would be great.
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Zahraa
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 « Embed this message Reply #15 on: October 21, 2010, 08:52:37 pm » posted from:,,Syrian Arab Republic

hello...

I would like to thank you for this nice cyclotron

I am a Syrian student in Damascus university and I have to do simulation for the cyclotron
I am facing a problem with physical study , I don't know how to study the motion of the charged particle in the gap between the tow dee , I wonder how the motion will be in the gap circular? or right straight ? and why ?

I hope I get help from you  because i don't have so much time for this study ...
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Fu-Kwun Hwang
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 « Embed this message Reply #16 on: October 21, 2010, 09:10:11 pm » posted from:,,Taiwan

There is an potential different between the gap which will form electric field $\vec{E}$
You can calculate acceleration $\vec{a}=\frac{\vec{F}}{m}=\frac{q\vec{E}}{m}$
And you can calculate how the charge will move (velocity,displacement)from acceleration.
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Zahraa
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 « Embed this message Reply #17 on: October 22, 2010, 01:13:18 am » posted from:,,Syrian Arab Republic

There is an potential different between the gap which will form electric field $\vec{E}$
You can calculate acceleration $\vec{a}=\frac{\vec{F}}{m}=\frac{q\vec{E}}{m}$
And you can calculate how the charge will move (velocity,displacement)from acceleration.

thx , but how can I study the effect of magnetic  field in the  gap ??
and how the charge particle  will move in the gap ?? circular motion or straight one ?
does it effect with the magnetic field ??

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Zahraa
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 « Embed this message Reply #18 on: October 22, 2010, 01:28:24 am » posted from:,,Syrian Arab Republic

I really need your help. I would like to request your source code. I know it's ridiculous but I promise I could't change the code and say that the program is mine. All I need to know is how you do it.

Thanks!
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Fu-Kwun Hwang
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 « Embed this message Reply #19 on: October 22, 2010, 08:55:45 am » posted from:,,Taiwan

The above simulation was created under the following assumption:

Only electric field exists between gap, there is no magnetic field between the gap.
And there is only uniform magnetic field in the two semicircular region.
The effect of the electric field between gap is to increase the energy of the charged particle (without changing direction).
And the effect of the magnetic field is to change direction of charged particle (without changing magnitude).

If you want to simulate magnetic field between the gap, you have to decide what kind of magnetic field in there.

You need to define your model before you can create an simulation.
The source code is available as attached file under the first message: cyclotron.java

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Zahraa
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 « Embed this message Reply #20 on: October 22, 2010, 11:57:17 am » posted from:,,Syrian Arab Republic

thank you again ...

but all the articles I have read about the cyclotron say : that the charged particles is effected with tow filed in the gap
magnetic and electric one ..

how ever, is there a cyclotron in the world with the  assumption  you said above  ??
if there is , where can I get pictures for it ??

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ahmedelshfie
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 « Embed this message Reply #21 on: October 22, 2010, 04:17:38 pm » posted from:SAO PAULO,SAO PAULO,BRAZIL

Hi Zahraa i want say that prof Hwang make all source code for all applets that he design free to all users,
By the way is see you from Syria im from Egypt.
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Fu-Kwun Hwang
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 « Embed this message Reply #22 on: October 22, 2010, 04:47:01 pm » posted from:,,Taiwan

thank you again ...

but all the articles I have read about the cyclotron say : that the charged particles is effected with tow filed in the gap
magnetic and electric one ..

Can you provide reference to the articles you read? Or copy the full text in detail?
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Zahraa
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 « Embed this message Reply #23 on: October 22, 2010, 05:31:18 pm » posted from:,,Syrian Arab Republic

thank you again ...

but all the articles I have read about the cyclotron say : that the charged particles is effected with tow filed in the gap
magnetic and electric one ..

Can you provide reference to the articles you read? Or copy the full text in detail?

see these pictures and notice that  the magnetic filed cross the gap.
 9.jpg (4.18 KB, 125x75 - viewed 22133 times.)  2.jpg (4.68 KB, 133x141 - viewed 22188 times.)  12.jpg (7.21 KB, 120x117 - viewed 22453 times.) Logged
Fu-Kwun Hwang
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 « Embed this message Reply #24 on: October 22, 2010, 09:30:39 pm » posted from:,,Taiwan

Normally the gap is very small so that the effect due to the magnetic field betwen the gap can be ignored.

For a cyclotron device, we usually only interested in the final beam came out at particular radius (with particular energy/momentum).
$p=mv=qBr$
The device was developed to accelerate charge particle (due to electric field between the gap) to high energy.
The detail of the trajectory usually is not so important.

If you are really interested in it, you can assume there are the same uniform magnetic field between the gap,
and find out is it going to produce similar result.
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Zahraa
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 « Embed this message Reply #25 on: November 18, 2010, 12:00:18 am » posted from:,,Syrian Arab Republic

Hello again

I have started my simulation  of the cyclotron in Java  but Am facing a problem:
we know that the magnetics filed doesn't affect the magnitude of the velocity vector in the dees , but how can I simulate that in my  program if the accelerator vector != zero vector , I mean that how can I applied this role mathematically to prove that when we move from point to another one in the dee that the  magnitude of velocity vector is still constant, but the direction and the axis  of it is changed ??
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Fu-Kwun Hwang
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 : 1 users think this message is good27 Re: Cyclotron « Embed this message Reply #26 on: November 18, 2010, 12:03:15 pm » posted from:Taipei,T'ai-pei,Taiwan

Just apply the Lorentz's force $\vec{F}=q\vec{v}\times\vec{B}$ to your simulation.
For B field in z direction, velocity in x-y plane.
$F_x=q*v_y*B_z$
$F_y=-q*v_x*B_z$
or the differential equations are
$\frac{d^2v_x}{dt}=\frac{q}{m}v_y*Bz$
$\frac{d^2v_y}{dt}=-\frac{q}{m}v_x*Bz$

And you can check out whether $v=\sqrt{v_x*v_x+v_y*v_y}$ is a constant or not.
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Zahraa
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 « Embed this message Reply #27 on: November 18, 2010, 06:08:24 pm » posted from:,,Syrian Arab Republic

Just apply the Lorentz's force $\vec{F}=q\vec{v}\times\vec{B}$ to your simulation.
For B field in z direction, velocity in x-y plane.
$F_x=q*v_y*B_z$
$F_y=-q*v_x*B_z$
or the differential equations are
$\frac{d^2v_x}{dt}=\frac{q}{m}v_y*Bz$
$\frac{d^2v_y}{dt}=-\frac{q}{m}v_x*Bz$

And you can check out whether $v=\sqrt{v_x*v_x+v_y*v_y}$ is a constant or not.

thx a lot..
but  actually I  don't understand so much , you write (Fx with Vy) and (Fy with Vx) I don't know why??

I see that you have studied the physics by dropping the vectors in the axises but I studied it by vectors
i.e :
in the gap vectors will be like this in first picture , then I conclude the next velocity vector V2 ( I mean velocity vector in the next moment , I have divided the time into amounts and study the motion of the charged particle in every moment ) by adding the tow V1 and acc ,according to the equation of the direct motion ,since if we study the motion in a very little  moment the motion will be straight , then the new velocity vector will increase and this is really because we know that velocity is increase in the gap ...

but in the dee velocity must be constant as we know , but please look at the second picture ,where I conclude V2 by adding  V1 and acc vectors
it will not be constant never but we know that in the dee velocity vector is constant  ..this is my problem

I hope you understand me

 vectors_gap.jpg (7.71 KB, 332x306 - viewed 853 times.)  vectors_dee.jpg (13.14 KB, 370x310 - viewed 796 times.) Logged
Fu-Kwun Hwang
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 « Embed this message Reply #28 on: November 18, 2010, 08:45:55 pm » posted from:,,Taiwan

You are thinking about a constant acceleration acc acting for a finite time.
However, the acceleration is always perpendicular to the velocity vector.

The magnitude of the velocity will change if the acceleration is in the direction of the velocity.
Because the acceleration is always pendicular to the velocity so it only change the direction of the velocity.
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Zahraa
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 « Embed this message Reply #29 on: November 19, 2010, 01:17:30 am » posted from:,,Syrian Arab Republic

You are thinking about a constant acceleration acc acting for a finite time.
However, the acceleration is always perpendicular to the velocity vector.

The magnitude of the velocity will change if the acceleration is in the direction of the velocity.
Because the acceleration is always pendicular to the velocity so it only change the direction of the velocity.

thank you again ..
but is this a general physics rule ??
that
"the acceleration is always pendicular to the velocity so it only change the direction of the velocity"
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"Nothing in life is to be feared, it is only to be understood." ..."Marie Curie 1867-1934, Polish born French Physicist, Twice Nobel Prize Winner- Physics and Chemistry)"