How to Determine Proton Movement in Electric and Magnetic Fields?

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SUMMARY

This discussion focuses on determining the motion of a proton in the presence of electric and magnetic fields. The proton, with mass mp and charge e, is subjected to an electric field E and Earth's magnetic field B. To maintain a straight trajectory while moving upward, the electric field must be oriented in the positive x-direction to counteract the magnetic force, which acts in the negative x-direction due to the right-hand rule. The relevant equations include the Lorentz force equation F = q(v × B) and the relationship E = Fe/q.

PREREQUISITES
  • Understanding of the Lorentz force equation F = q(v × B)
  • Familiarity with the right-hand rule for determining force directions
  • Basic knowledge of electric fields and their relationship to force E = Fe/q
  • Concept of vector fields in physics
NEXT STEPS
  • Study the Lorentz force and its implications on charged particles in electric and magnetic fields
  • Learn about the right-hand rule and its application in determining force directions
  • Research vector fields and their properties in electromagnetism
  • Explore resources on hyperphysics for deeper insights into electric and magnetic forces
USEFUL FOR

Students studying electromagnetism, particularly those tackling problems involving charged particles in electric and magnetic fields, as well as educators seeking to clarify these concepts for their students.

  • #61
physicsbhelp said:
yes but the particle gets curved because it leaves that box thingy. and is moved by the force in a circular path
Correct. The path is circular, and it would travel in a circular path until it hit the side of the box.

The radius of the path is the cyclotron radius.


Since we established that the magnetic force is initially in the -x direction, the path must be circular in the counterclockwise direction.
 
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  • #62
exactly isn't that what i drew!?

and i have no clue what the cyclotron radius is haha

it is the first picture i drew, but more circular?
 
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  • #63
The proton moves in a circular path immediately upon leaving the box. If it had a small radius of curvature, then the proton would impact the top of the box. If it had a large rad. of curv. it would hit the side.

See hint I provided, and look at the figure.
 
  • #64
but can you draw a sample because i can't really understand.
 
  • #66
so i draw three lines? to the left
 
  • #67
physicsbhelp said:
so i draw three lines? to the left

The three lines could be three different speeds, mass or charges in the same magnetic field.

The radius is the ratio of the linear momentum (mv) and the product of q and B, i.e.

r = (mv)/(qB).

If v, q, and B are constant then r is simply a linear function or m and that is how one can separate particles of different mass (but with same v and q)
 
  • #68
h/o
let me draw another figure and post it so you can see if it is right
 
  • #69
physicsbhelp said:
h/o
let me draw another figure and post it so you can see if it is right

Just draw a circle, such that the circumference is tangent at the exit of the box and goes counterclockwise to the top or side of the box.
 
  • #70
physicsbhelp said:
so even theoughtthe problem says draw it after the proton leaves the box it can go three ways.
What do you mean three ways?

This is just a sketch!

If you drew it to scale, then you could use the knowledge of m, q, v, B to draw an exact path.

Your sketch is more or less correct. If it was a fast proton, or if B was very strong and the box relatively large, the path would be a semi-circle into the top of the box. If the proton was moving very slow or the B was very weak, then the proton would travel a path between a semi-circle and 3/4 of a circle into the side of the box, or perhaps more.
 
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  • #71
so should i draw the really slow one.
 
  • #72
You could draw one for low v (or high B) and one for high v (or low B).

Another way to look at this is r = (m/q)*(v/B). Note the (m/q) or more traditionally (q/m) is constant for very low (non-relativistic) speeds.
 
  • #73
Thank You So Much!
 

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