Do magnetic lenses do work on particles flowing through them?

Click For Summary
SUMMARY

Magnetic lenses effectively manipulate charged particles flowing through them by utilizing the Lorentz force, defined by the equation F = q v × B. The discussion confirms that while the magnetic field generated by the coil exerts a force on the charged particles, the instantaneous work done is zero, as shown by the equation P = v · (q v × B) = 0. This indicates that magnetic lenses can focus charged particle beams without performing work on them.

PREREQUISITES
  • Understanding of Lorentz force and its application in electromagnetism
  • Familiarity with the principles of magnetic fields and electric currents
  • Knowledge of vector calculus, particularly cross products
  • Basic concepts of particle physics and beam dynamics
NEXT STEPS
  • Study the principles of electromagnetic lens design in particle accelerators
  • Learn about the applications of magnetic lenses in electron microscopy
  • Explore the mathematical derivation of the Lorentz force law
  • Investigate the role of magnetic fields in controlling particle trajectories
USEFUL FOR

Physicists, electrical engineers, and students in advanced electromagnetism or particle physics who are interested in the application of magnetic fields in particle manipulation and beam focusing techniques.

BrandonBerchtold
Messages
46
Reaction score
6
Do magnetic lenses do work on charged particles flowing through them?

Intuitively I would think yes because the magnetic field produced by the electric current in the coil is applying a force on the particles flowing through the lens, so therefore an electric current (the beam of particles to be focused) flowing through the lens should apply an equal and opposite force on the current in the coil, right?
 
Physics news on Phys.org
BrandonBerchtold said:
Do magnetic lenses do work on charged particles flowing through them?
We can write the instantaneous force from the magnetic lens as ##\vec F = q \vec v \times \vec B##. The instantaneous work is ##P=\vec v \cdot \vec F##. So ## P=\vec v \cdot ( q \vec v \times \vec B) = 0##
 
  • Like
Likes   Reactions: BrandonBerchtold

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
984
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 12 ·
Replies
12
Views
1K