Motion of a charged particle in magnetic field

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SUMMARY

The discussion focuses on the motion of a charged particle in a magnetic field, specifically analyzing a particle with charge -q and mass m entering a magnetic field B at point A with speed v1 at angle alpha and exiting at point C with speed v2 at angle beta. Key conclusions include that alpha equals beta, v1 equals v2, and the time spent in the field is given by t=(2m(pi-alpha))/qB. The Lorentz force equation, F=qvB, is crucial for understanding the particle's motion, which follows a circular path due to the magnetic force acting as centripetal force.

PREREQUISITES
  • Understanding of Lorentz force and its vector form
  • Knowledge of circular motion principles in physics
  • Familiarity with the concept of centripetal force
  • Ability to resolve vector components in physics problems
NEXT STEPS
  • Study the derivation and applications of the Lorentz force equation
  • Learn about the motion of charged particles in uniform magnetic fields
  • Explore the relationship between velocity, angle, and magnetic force in particle motion
  • Investigate the effects of varying angles of entry on the trajectory of charged particles
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of charged particles in magnetic fields will benefit from this discussion.

Abhimessi10
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Homework Statement


A particle of charge -q and mass m enterd magnetic field B at A with speed v1 at an angle alpha and leaves the field at C with v2 at angle beta.Then,

a)alpha=beta
b)v1=v2
c)particle remains in the field for time t=(2m(pi-alpha))/qB
d)All of these

Please see image below

https://ibb.co/iLMrie

Abhimessi10fig.jpg

Homework Equations


F=qvB

The Attempt at a Solution


Really don't know how to do it.Tried resolving components.
 

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Abhimessi10 said:

Homework Statement


A particle of charge -q and mass m enterd magnetic field B at A with speed v1 at an angle alpha and leaves the field at C with v2 at angle beta.Then,

a)alpha=beta
b)v1=v2
c)particle remains in the field for time t=(2m(pi-alpha))/qB
d)All of these

Please see image below

https://ibb.co/iLMrie

Homework Equations


F=qvB

The Attempt at a Solution


Really don't know how to do it.Tried resolving components.[/B]
Resolving components will not get you very far. Other than F=qvB, what else do you know about the motion of particles in magnetic fields? Can you draw the path connecting the entry and exit points of the particle?
 
kuruman said:
Resolving components will not get you very far. Other than F=qvB, what else do you know about the motion of particles in magnetic fields? Can you draw the path connecting the entry and exit points of the particle?

I don't know to draw the path for particles entering at general angles.just the case when angle=90, 0, and in between.
 
Abhimessi10 said:
I don't know to draw the path for particles entering at general angles.just the case when angle=90, 0, and in between.
OK. What kind of path do you get in these cases? Why is the path like that and like something else?
 
kuruman said:
OK. What kind of path do you get in these cases? Why is the path like that and like something else?
Because of magnetic force acting as centripetal force?
 
Abhimessi10 said:
Because of magnetic force acting as centripetal force?
Yes., and if the magnetic force is centripetal, what kind of path does the particle describe for as long as the force remains centripetal?
 
kuruman said:
Yes., and if the magnetic force is centripetal, what kind of path does the particle describe for as long as the force remains centripetal?

Depends on the angle velocity makes with the magnetic field.On a general basis i can say circular-kinda motion.
 
  • #10
Abhimessi10 said:
Depends on the angle velocity makes with the magnetic field.On a general basis i can say circular-kinda motion.
Look at your drawing. The magnetic field is directed into the screen and the velocity vector is in the plane of the screen. What do you think the angle between the magnetic field and the velocity is?
 
  • #11
kuruman said:
Look at your drawing. The magnetic field is directed into the screen and the velocity vector is in the plane of the screen. What do you think the angle between the magnetic field and the velocity is?
The sin alpha component of velocity is only perpendicular
and cos alpha would remain same?
 
  • #12
Abhimessi10 said:
The sin alpha component of velocity is only perpendicular
Imagine placing your index finger perpendicular to the face of an analog clock and keeping it fixed. What is the angle between your finger and the seconds hand at any time t? The seconds hand is like the velocity vector that changes direction in the plane of the screen and your index finger is like the magnetic field the direction of which is fixed.
 

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