How Does Particle Displacement Affect Momentum in a Magnetic Field?

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SUMMARY

The discussion focuses on calculating the momentum of a charged particle in a uniform magnetic field, specifically addressing the relationship between particle displacement and momentum. The participant derived the momentum equation as p = Bqaz, where B is the magnetic field strength, q is the charge, a is the horizontal displacement, and z indicates the direction of momentum. Concerns were raised regarding the vertical displacement, d, and its influence on the momentum calculation, leading to the conclusion that while the magnetic field alters the direction of momentum, it does not affect its magnitude.

PREREQUISITES
  • Understanding of classical mechanics principles, particularly momentum and forces.
  • Familiarity with magnetic fields and the Lorentz force equation.
  • Knowledge of vector calculus, specifically cross products.
  • Basic understanding of kinematics, including displacement and velocity.
NEXT STEPS
  • Study the Lorentz force and its implications on charged particle motion in magnetic fields.
  • Learn about the derivation and application of the right-hand rule in vector cross products.
  • Explore the concept of circular motion of charged particles in magnetic fields.
  • Investigate the effects of varying magnetic field strengths on particle trajectories.
USEFUL FOR

Physics students, educators, and professionals in fields related to electromagnetism and particle physics will benefit from this discussion, particularly those focusing on the dynamics of charged particles in magnetic fields.

darkpsi
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Homework Statement


A particla of charge q enters a region of uniform magnetic field B (pointing into the page). The field deflects the particle a distance d above the original line of flight. In terms of a, d, B, and q, find the momentum of the particle.


Homework Equations


F=dp/dt
Fmag=(vXB)q
I put v=vy and B=-Bx


The Attempt at a Solution


I started by setting the two above equations equal to each other.
dp/dt=(vXB)q
I took the cross product and got
dp=Bvqzdt
and since v=ds/dt
dp=Bq(ds)z
and since the displacement was from when the particle entered the field to when it would be leaving, I said it went from 0 to a and
p=Bqaz

This is the answer I got but I was just concerned because it didn't include the value that the particle was vertically displaced by the magnetic force, d. Is this answer right, or does the vertical displacement appear somewhere that I've overlooked?
EDIT: I just remembered that the direction of the velocity changes because of the magnetic force. So instead of the displacement being just a wouldn't it now be √(a2+d2 ?
 
Last edited:
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It is not possible to figure out what is going on without a diagram defining quantities a and d with respect to the particle's trajectory. Whatever the case may be, the magnetic field changes the direction of the particle's momentum but not its magnitude, so if the question asks you to find the momentum. All you need to provide is the new direction and that you can probably figure out from the diagram.
 
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