Force acting on a particle between a tube and a wire (elektrodynamics)

In summary, the conversation discusses a problem with a counting tube for a particle accelerator and the calculation of the electric field and force acting on a particle in the tube. After some discussion and calculations, it is determined that the ratio of the electric field between the wire and tube is 400, with the wire having a stronger field due to its smaller radius. The conversation ends with a clarification about the radius of the tube being a typo and a solution to the problem.
  • #1
fara0815
45
0
Hello physics community!

After working on this problem for more than 4 hours and reading about the topic in different books, I decided to ask here for help since I do not seem to have a clue.

"A counting tube for a particle accelerator consists of a thin-walled metal tube and a wire that goes along the tube's center axis. The tube has a radius of 12 mm and the wire of 12 X 10^-6 m. Between the tube and the wire there is a current of 1000 volts.

a) Calculate the factor of which the force acting on a particle increases on its way from the tube's wall to the wire.
Answer: 400 times

I have problems to make a connection between the 1000 volts and the electric field with which I could calculate a force (according to Gauss).

What could I start with?

Any help will be appreciated!
 
Physics news on Phys.org
  • #2
What can you say bout the field between the wire and the pipe? How does the field vary with distance?
 
  • #3
mh, I guess that in the tube's center there is no field at all since all the field lines eliminate each other in the center. So only the wire's field is acting?
The wire can also being considert a point with the electric field of
[tex]E=\frac{q}{4\pi e} \frac{1}{r^2}[/tex]

Am I getting closer?
 
  • #4
I was thinking more that the field is uniform. How does the electric field vary with respect to distance in an uniform field?

Also, just a slight correct; I'm sure you meant a potential of 1000V;
Between the tube and the wire there is a current of 1000 volts
 
  • #5
I am really sorry but I am not making any progress. If the field is uniform, it does not change its intensity in respect to distance, right ?

What am I missing?
 
  • #6
By dividing the electrical potential by the radius, I get a electric field of 83333 V/m.
Is this part of the way I have to take ?
 
  • #7
I still cannot figure it out :(
 
  • #8
Nobody has an idea?
 
  • #9
I think I have got it!

Since the electric field is uniform, the equation to get the electric field is:
[tex]E= \frac{U}{r}[/tex] where r is the distance to the particle.
[tex]r_{tube}=0,012m[/tex] and [tex]r_{wire}=3 x 10^-5m[/tex]
[tex]E_{tube}=\frac{1000V}{0,012m}= 83333.3 Vm[/tex] and for the wire
[tex]E_{wire}=\frac{1000V}{3 x 10^-5m}= 33333333.3 Vm[/tex]
for the ratio you do:
[tex]\frac{E_{wire}}{E_{tube}}= 400.000[/tex]

I think that is it !
 
  • #10
fara0815 said:
I think I have got it!

Since the electric field is uniform, the equation to get the electric field is:
[tex]E= \frac{U}{r}[/tex] where r is the distance to the particle.
[tex]r_{tube}=0,012m[/tex] and [tex]r_{wire}=3 x 10^-5m[/tex]
[tex]E_{tube}=\frac{1000V}{0,012m}= 83333.3 Vm[/tex] and for the wire
[tex]E_{wire}=\frac{1000V}{3 x 10^-5m}= 33333333.3 Vm[/tex]
for the ratio you do:
[tex]\frac{E_{wire}}{E_{tube}}= 400.000[/tex]

I think that is it !
Looks good to me :smile: . Just one question however, in your initial problem you stated that the radius of the tube was 12 X 10^-6 m, was this a typo? (I was wondering why tmy numbers didn't make sense :confused: )
 
  • #11
Hootenanny said:
Looks good to me :smile: . Just one question however, in your initial problem you stated that the radius of the tube was 12 X 10^-6 m, was this a typo? (I was wondering why tmy numbers didn't make sense :confused: )

Oh, I am sorry. You are right, it is a typo. It is supposed to be 30 x 10^-6m :)
 

Related to Force acting on a particle between a tube and a wire (elektrodynamics)

1. What is the force acting on a particle between a tube and a wire?

The force acting on a particle between a tube and a wire is known as the Lorentz force, which is the combined effect of the electric and magnetic forces acting on the particle. This force is perpendicular to both the electric and magnetic fields and is given by the equation F = q(E + v x B), where q is the charge of the particle, E is the electric field, v is the velocity of the particle, and B is the magnetic field.

2. How does the distance between the tube and wire affect the force on the particle?

The distance between the tube and wire can affect the force on the particle in a few ways. First, if the distance is increased, the electric field between the two may decrease, resulting in a weaker electric force on the particle. Additionally, the magnetic field may also be affected by the distance, which could change the direction or strength of the force on the particle.

3. Can the force on the particle be controlled by changing the properties of the tube or wire?

Yes, the force on the particle can be controlled by changing the properties of the tube or wire. For example, the strength of the electric and magnetic fields can be altered by changing the charge or current in the wire, respectively. Additionally, the material of the tube or wire may also affect the force on the particle.

4. What is the role of the particle's charge in the force between the tube and wire?

The particle's charge plays a crucial role in the force between the tube and wire. The electric force in the Lorentz force equation is directly proportional to the charge of the particle, meaning that a higher charge will result in a stronger force on the particle. Additionally, the direction of the force is also dependent on the charge of the particle, with positive and negative charges experiencing forces in opposite directions.

5. How is the force on the particle affected by the velocity of the particle?

The velocity of the particle affects the force on the particle in two ways. First, the velocity is used in the cross product in the Lorentz force equation, meaning that a higher velocity will result in a stronger force. Additionally, the direction of the velocity also plays a role in determining the direction of the force, as seen in the v x B term in the equation. Therefore, changing the velocity of the particle can alter the magnitude and direction of the force acting on it.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
616
  • Introductory Physics Homework Help
2
Replies
43
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
1K
Replies
27
Views
2K
  • Introductory Physics Homework Help
Replies
28
Views
1K
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
6K
Back
Top