Distance of particle drop in magnetic field (probably geometry)

In summary: CLqQrBggASACIAMgBCAFIAYgBygBKAIoAygH/hgbzOISA7LFxF96Sk24ynqGuBtMLIhtL66D0AEny48I?size_mode=5In summary, the problem involves finding the distance that a particle will drop due to a magnetic field in a system of three plate capacitors. The particle is released from the first plate with a known charge and voltage, and when it reaches the second plate it is no longer accelerating. The
  • #1
pollytheparrot
11
0

Homework Statement


IZi8obz.png


Basically there are 3 plate capacitors, with a distance a between the first two and b between the last two. In between the first two plates is voltage V. A particle of charge q is released from the first plate, and when it leaves the second plate it is no longer accelerating. There is a magnetic field of B in the second part, and then a hole of radius r at the third plate, find the distance that the particle drops due to the magnetic field. I neglected to outline this specifically in the picture, but basically it is the distance of the bottom black line and could be called "d"

Homework Equations


U=K
F=qv x B
Sum F = ma
a=v^2/r
E=qV

The Attempt at a Solution


[/B]
First I had to find the velocity of the particle at the point of the first plate, and did this using conservation of energy.

U=K
qV=.5mv^2
v=sqrt(2qV/m)

Second, is finding the radius of the circle in the second part, using F=ma and angular acceleration

Then Sum F = ma, only force is magnetic
qv x B = ma

qv x B = m(v^2/R)

R=m*v^2 / q*v*B

R=m*(2qV/m) / q*B*sqrt(2qV/m)

R=2V/Bsqrt(2qV/m)

At this point I have the radius R of the circular path that the magnetic field induces, but I'm totally lost at this point. I'm assuming it's just geometry of some fashion. Really would appreciate help.
 
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  • #2
Because my picture was pretty potato quality, here is a new version. Also I found a simplification of R, where it is mv^2/qvB, leading to mv/qB, which is m*sqrt(2qV/m) / qB

https://photos-1.dropbox.com/t/2/AABdnf-4ewEkuSufX_wYb_F1zWxQY_4lbn-mX5OImOW3dg/12/51054650/jpeg/32x32/1/1436472000/0/2/2015-07-09%2013.36.16.jpg/CLqQrBggASACIAMgBCAFIAYgBygBKAIoAygH/LLxgaJo81ulvhjEAskZsAlZZK8XGg4aTgh9X-QeV0m4?size=1024x768&size_mode=2
 
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  • #3
Welcome to PF!

Yes, it's just geometry. Your red circular path looks a little sloppily drawn. Try to make it more accurate. In particular, make sure you have the center of the circular path placed properly and make sure that the circular path passes through the second plate such that the tangent line to the path is in the correct direction.

I presume that there is a little hole in the second plate for the charge to pass through and that the distance between the second and third plates is known.

Your expression for R looks good, but you can simplify it a bit.
 
  • #4
TSny said:
Welcome to PF!

Yes, it's just geometry. Your red circular path looks a little sloppily drawn. Try to make it more accurate. In particular, make sure you have the center of the circular path placed properly and make sure that the circular path passes through the second plate such that the tangent line to the path is in the correct direction.

I presume that there is a little hole in the second plate for the charge to pass through and that the distance between the second and third plates is known.

Your expression for R looks good, but you can simplify it a bit.

Perhaps my second post answered these questions? I'm assuming that the center of the circle is in line with the second (middle) plate
 
  • #5
OK, I did not see your second picture. I see that b represents the distance between the second and third plates and that you have the direction of the path at the second plate in the right direction (horizontal). I'm still not too sure where you are placing the center of the circular path.
 
  • #6
pollytheparrot said:
I'm assuming that the center of the circle is in line with the second (middle) plate

Yes.
 
  • #7
In your second picture, you have drawn a radius R. Try making that the hypotenuse of a right triangle.
 
  • #8
TSny said:
OK, I did not see your second picture. I see that b represents the distance between the second and third plates and that you have the direction of the path at the second plate in the right direction (horizontal). I'm still not too sure where you are placing the center of the circular path.

That's still an assumption on my part. Between the second and third plates, the only force that should be acting on the particle is the magnetic force, which is F=q*v x B, and that it should move in a circle, and reasoning (albeit crappy explanation it is) thought it should be where it is, is there a different spot it should be?
 
  • #9
pollytheparrot said:
That's still an assumption on my part. Between the second and third plates, the only force that should be acting on the particle is the magnetic force, which is F=q*v x B, and that it should move in a circle, and reasoning (albeit crappy explanation it is) thought it should be where it is, is there a different spot it should be?

You are correct that the center of the circle should be in line with the second plate. But neither of your pictures seems to show this. Looks to me that you have the center a little bit to the right of the second plate.
 
  • #10
TSny said:
You are correct that the center of the circle should be in line with the second plate. But neither of your pictures seems to show this. Looks to me that you have the center a little bit to the right of the second plate.

Sorry, that is uncertainty and sloppiness in action there.

Is this what you meant by making a triangle? If you can find the hypotenuse created by the red line, using some kind of circle magic, you have the side b, and the last side should be d, correct? If this is the right course of action then what circle magic is done?

https://photos-1.dropbox.com/t/2/AACYhnNE5kOBe9cXiXds1vuFNx8SE31Le9ybSvxPspgdoA/12/51054650/jpeg/32x32/1/1436472000/0/2/2015-07-09%2013.51.02.jpg/CLqQrBggASACIAMgBCAFIAYgBygBKAIoAygH/hgbzOISA7LFxF96Sk24ynqGuBtMLIhtL66D0AEny48I?size_mode=5
 
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  • #11
Draw a right triangle with the hypotenuse as the solid line that you drew and labeled R. Forget the red line that you drew.
 
  • #12
TSny said:
Draw a right triangle with the hypotenuse as the solid line that you drew and labeled R. Forget the red line that you drew.

Surely it can't be that simple? d=√R^2-b^2 ?
 
  • #13
It looks kind of legit if you make that line and flip it around and upside down
 
  • #14
pollytheparrot said:
Surely it can't be that simple? d=√R^2-b^2 ?
Not quite that simple.
 
  • #15
pollytheparrot said:
It looks kind of legit if you make that line and flip it around and upside down

?
 
  • #16
Here's the right triangle I was suggesting.
 

Attachments

  • Circ deflec.png
    Circ deflec.png
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  • #17
TSny said:
Here's the right triangle I was suggesting.

So given your triangle, how would I use that to find d?

I was just assuming something like this. Don't know how picture got flipped.

https://photos-5.dropbox.com/t/2/AADxYk-aonmJAYOZcE2sYaAXx2xtx0obeTzHz0YEMyG5RA/12/51054650/jpeg/32x32/1/1436475600/0/2/2015-07-09%2014.07.38.jpg/CLqQrBggASACIAMgBCAFIAYgBygBKAIoAygH/DqJgHWIuRBypec8tKOV3Tk3svKkHv0AeV478BINibRc?size=1024x768&size_mode=2
 
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  • #18
https://photos-3.dropbox.com/t/2/AADWd9kWZiwMFH1Svwi-bbLvApEmtOjBYuDxYjJxCWD5JQ/12/2668640/jpeg/32x32/1/1436475600/0/2/2015-07-09%2021.04.34.jpg/CODwogEgASACIAMgBCAFIAYgBygBKAIoBw/mATI66H3HO_k1EIE84VJevokK97mR2xUrjAyfX-h5Yk?size_mode=5

Someone else worked it out like this, is this allowed? (Ignore the smaller red circle), where his d is my b, and his a is my d, and the triangle created by the circle's radius and "b" is radius - d, and then work from there?
 
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  • #19
Where R =d+x and x = Sqrt(R^2 - b^2), so d = R - Sqrt(R^2 - b^2) ?
 
  • #20
pollytheparrot said:
Where R =d+x and x = Sqrt(R^2 - b^2), so d = R - Sqrt(R^2 - b^2) ?
Looks good to me.
 
  • #21
TSny said:
Looks good to me.
Thanks for the help man! Really means a lot!

You keep doing you!
 

Related to Distance of particle drop in magnetic field (probably geometry)

1. What is the relationship between the distance of particle drop and the strength of the magnetic field?

The distance of particle drop is directly proportional to the strength of the magnetic field. This means that as the magnetic field becomes stronger, the distance the particle drops will also increase.

2. How does the geometry of the magnetic field affect the distance of particle drop?

The geometry of the magnetic field plays a significant role in determining the distance of particle drop. The shape and orientation of the magnetic field can either attract or repel the particle, resulting in a longer or shorter distance of drop.

3. What is the impact of the particle's mass on its distance of drop in a magnetic field?

The mass of the particle does not have a direct impact on its distance of drop in a magnetic field. However, a heavier particle may experience a stronger force from the magnetic field, resulting in a longer distance of drop.

4. Can the distance of particle drop in a magnetic field be altered by changing the velocity of the particle?

Yes, the velocity of the particle can affect the distance of drop in a magnetic field. A faster-moving particle will experience a greater force from the magnetic field, resulting in a longer distance of drop.

5. How can the distance of particle drop in a magnetic field be calculated?

The distance of particle drop in a magnetic field can be calculated using the equation d = v^2/2g, where d is the distance of drop, v is the velocity of the particle, and g is the acceleration due to gravity. This equation assumes that the particle is moving in a straight line and the magnetic field is uniform.

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