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Magnetic Field Problem! HELP NEEDED

  • #51
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now i have to : Determine the speed of the proton while in the box if it continues to move vertically upward. Express your answer in terms of the fields and the given quantities.

to find velocity do i use the formula Fm=qv*B
 
  • #52
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but how can i express myh anwer in terms of the fields when i am not given anything about charge.
do i somehow need to incorporate the proton.

or should i use a different formula.
 
  • #53
Doc Al
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You need to solve for the speed in terms of both fields. Set up an equation for the net force on the charge in terms of the fields. (What's the net force on the charge?)
 
  • #54
Astronuc
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Fm=Fe

qvB=qE??
Correct. Note that the charge cancels. Charge by the way is a scalar quantity.

Now solve for v.
 
  • #55
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thanks
 
  • #56
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this is for part c) On the figure above, sketch the path of the proton after it leaves the box.
 
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  • #57
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which one is it? i drew all three of them, i just don't know which way im supposed to draw it.
 
  • #59
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too complicated words? i am not sure what you mean, i looked at the hint.
wait why can't you tell me because i drew those three image.s
 
  • #60
Astronuc
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too complicated words? i am not sure what you mean, i looked at the hint.
wait why can't you tell me because i drew those three image.s
trajectory is the path. What path does a charged particle of constant speed take in a uniform magnetic field?
 
  • #61
Astronuc
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yes but the particle gets curved becuase 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.
 
  • #62
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exactly isnt 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
Astronuc
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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
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but can you draw a sample cuz i can't really understand.
 
  • #66
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so i draw three lines? to the left
 
  • #67
Astronuc
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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
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h/o
let me draw another figure and post it so you can see if it is right
 
  • #69
Astronuc
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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
Astronuc
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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
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so should i draw the really slow one.
 
  • #72
Astronuc
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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
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Thank You So Much!
 

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