What Forces Affect an Electron in a Particle Accelerator?

Click For Summary
The discussion focuses on the forces affecting an electron in a particle accelerator, specifically using Gauss' and Ampere's laws to determine electric (E) and magnetic (B) fields inside and outside an electron beam. The Lorentz force equation is applied to analyze the force experienced by an electron at the beam's edge, with the magnetic force directed towards the center of the beam and perpendicular to its velocity. There is confusion regarding the balance of forces, particularly whether the repulsive electric force counteracts the magnetic force to maintain a straight trajectory. Additionally, the Large Hadron Collider (LHC) is mentioned, with clarification that it primarily uses protons rather than electrons, and that electrons in such a context would travel close to the speed of light. The discussion emphasizes the need for a deeper understanding of the forces at play in particle accelerators.
tarkin
Messages
13
Reaction score
0

Homework Statement


[/B]
Consider an electron beam traveling with velocity v. The total current of the beam is I.The beam is of uniform charge density and has radius R.

(a) Find E and B at r<R and r>R using Gauss' and Ampere's laws. ( This part is fine.)

(b) Consider an electron in the beam at r=R. What is the magnitude and direction of the force it experiences?
What would the force on the electron be if it were traveling in the Large Hadron Collider?

Homework Equations



Lorentz force equation: F = q(E + v x B)

The Attempt at a Solution



From the right hand grasp rule, the electron beam should generate a magnetic field curling around. If we think of the beam as directed into the page, the B field will curl anticlockwise.

The motion of an electron in the beam should be perpendicular to the magnetic field,
so the equation just becomes F=qE +qvB, and then just sub in the answers from part a, is this correct?

What's confusing me is the direction the force will be in. Using the right hand rule, I'm finding that, for one electron, the magnetic force should be directed towards the centre of the beam, perpendicular to it's current velocity. Is this correct? does the repulsive electric force balance this so the beam still goes in a straight line?
If so, wouldn't the total force on the electron just be zero? I feel like I'm definitely missing something here...

And for the short Large Hadron Collider question, I'm unsure what to say exactly. I know it uses magnets to direct the beam in a circular path, what more should I say here?Thanks in advance for any help!
 
Physics news on Phys.org
tarkin said:
From the right hand grasp rule, the electron beam should generate a magnetic field curling around. If we think of the beam as directed into the page, the B field will curl anticlockwise.
Yes.

The motion of an electron in the beam should be perpendicular to the magnetic field,
so the equation just becomes F=qE +qvB, and then just sub in the answers from part a, is this correct?
Yes

What's confusing me is the direction the force will be in. Using the right hand rule, I'm finding that, for one electron, the magnetic force should be directed towards the centre of the beam, perpendicular to it's current velocity. Is this correct?
That's correct.
does the repulsive electric force balance this so the beam still goes in a straight line?
Use your results from part (a) to see if the forces balance.

And for the short Large Hadron Collider question, I'm unsure what to say exactly. I know it uses magnets to direct the beam in a circular path, what more should I say here?
I'm not sure about this part of the question. Seems like they should give you more information. I don't think the LHC uses electron beams. Maybe they just want you to assume that in the LHC the electrons would travel very close to the speed of light.
 
Last edited:
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

Replies
3
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
34
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K