### Author Topic: Cyclotron  (Read 396266 times)

#### Zahraa

• Newbie
• Posts: 11
##### Re: Cyclotron
« Reply #15 on: October 21, 2010, 09:52:37 pm »
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 ...

#### Fu-Kwun Hwang

• Hero Member
• Posts: 3062
##### Re: Cyclotron
« Reply #16 on: October 21, 2010, 10:10:11 pm »
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{qvec{E}}{m}\$
And you can calculate how the charge will move (velocity,displacement)from acceleration.

#### Zahraa

• Newbie
• Posts: 11
##### Re: Cyclotron
« Reply #17 on: October 22, 2010, 02:13:18 am »
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{qvec{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 ??

#### Zahraa

• Newbie
• Posts: 11
##### Re: source code
« Reply #18 on: October 22, 2010, 02:28:24 am »

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!

#### Fu-Kwun Hwang

• Hero Member
• Posts: 3062
##### Re: Cyclotron
« Reply #19 on: October 22, 2010, 09:55:45 am »
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

#### Zahraa

• Newbie
• Posts: 11
##### Re: Cyclotron
« Reply #20 on: October 22, 2010, 12:57:17 pm »
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 ??

#### ahmedelshfie

• Hero Member
• Posts: 930
##### Re: Cyclotron
« Reply #21 on: October 22, 2010, 05:17:38 pm »
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.

#### Fu-Kwun Hwang

• Hero Member
• Posts: 3062
##### Re: Cyclotron
« Reply #22 on: October 22, 2010, 05:47:01 pm »
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?

#### Zahraa

• Newbie
• Posts: 11
##### Re: Cyclotron
« Reply #23 on: October 22, 2010, 06:31:18 pm »
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.

#### Fu-Kwun Hwang

• Hero Member
• Posts: 3062
##### Re: Cyclotron
« Reply #24 on: October 22, 2010, 10:30:39 pm »
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.

#### Zahraa

• Newbie
• Posts: 11
##### Re: Cyclotron
« Reply #25 on: November 18, 2010, 01:00:18 am »
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 ??

#### Fu-Kwun Hwang

• Hero Member
• Posts: 3062
##### Re: Cyclotron
« Reply #26 on: November 18, 2010, 01:03:15 pm »
Just apply the Lorentz's force \$vec{F}=qvec{v}   imesvec{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.

#### Zahraa

• Newbie
• Posts: 11
##### Re: Cyclotron
« Reply #27 on: November 18, 2010, 07:08:24 pm »
Just apply the Lorentz's force \$vec{F}=qvec{v}   imesvec{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

#### Fu-Kwun Hwang

• Hero Member
• Posts: 3062
##### Re: Cyclotron
« Reply #28 on: November 18, 2010, 09:45:55 pm »
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.

#### Zahraa

• Newbie
• Posts: 11
##### Re: Cyclotron
« Reply #29 on: November 19, 2010, 02:17:30 am »
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"