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Particle in electromagnetic field

  • Thread starter mystmyst
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  • #1
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[PLAIN]http://img23.imageshack.us/img23/421/electromagneticquestion.png [Broken]


A positively charged particle, which charge [tex]q[/tex] and mass [tex]m[/tex], reached an area with an electric field and a magnetic field. Electric field E is at y>0 and its direction is -y. Magnetic field B is at y<0 and its direction is at -z (into the paper). The particle enters at y=h at intial speed [tex]v_0[/tex]

Question: Determine the time it takes the particle to reach y=0.

Attempt: I'm not sure what equations are relevant in this question. I know there's

[tex]E = k \frac{q}{r^2}[/tex] and

[tex]F = qE[/tex]

but what formula has time in it?

(this portion of the question only deals with the electric field. The other parts involve the magnetic field. I'm sure once I understand this section, I'll understand the rest of the parts)

Thanks!
 
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Answers and Replies

  • #2
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F = ma
a = f(t)
 
  • #3
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thanks

this is what I got but i'm still stuck:

[tex] r(t) = \frac{qE}{2m}t^2 + v_0 t + h = 0 [/tex]

how do I solve for t? Am I supposed to use the quadratic equation?

Did I use the correct equation? Also can I use E in general, or do I need to split it between Ex, Ey, and Ez?

Thanks!
 
  • #4
gneill
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thanks

this is what I got but i'm still stuck:

[tex] r(t) = \frac{qE}{2m}t^2 + v_0 t + h = 0 [/tex]

how do I solve for t? Am I supposed to use the quadratic equation?

Did I use the correct equation? Also can I use E in general, or do I need to split it between Ex, Ey, and Ez?

Thanks!
What can you say about the initial velocity v0?
 
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  • #5
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What can you say about the initial velocity v0?
I don't think I can say anything about the initial velocity v0.

the formula I used is

[tex] r(t) = \frac{qE}{2M}t^2 + v_0 t + r_0 [/tex] where [tex]v_0[/tex] is initial speed at t=0, and [tex]r_0[/tex] is initial placement at t=0. It is given that [tex]r_0 = h[/tex] and we know that r(t) = 0 at y=0... That's how I got

[tex]
r(t) = \frac{qE}{2m}t^2 + v_0 t + h = 0
[/tex]

So I have no idea what I can say about the initial speed. Can you please give me some more direction?

Thanks!

***btw, I'm probably gonna be checking this post every 5 minutes to see if someone responded. So if you respond chances are I'm going to respond back within 10 minutes of your response. So if you can just stay with me, I'd love you forever!
 
  • #6
gneill
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The diagram that you provided would seem to indicate that the initial speed was horizontally directed. If that is the case, then the y-component is zero (and the x-component is v0). What do you know about computing the time for an object to fall from a height h while accelerating at rate a?
 
  • #7
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Why would the y-component be zero? The particle was moving all along, it didn't start from rest.

and i totally forgot. What formula do I use to compute the time.
there's

F = -mg
x = x_0 + (v_0 cos-alpha)t and
y = y_0 + (v_0 sin-alpha)t - 1/2gt^2

(i forgot all my high school math...)
 
  • #8
gneill
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Why would the y-component be zero? The particle was moving all along, it didn't start from rest.

and i totally forgot. What formula do I use to compute the time.
there's

F = -mg
x = x_0 + (v_0 cos-alpha)t and
y = y_0 + (v_0 sin-alpha)t - 1/2gt^2

(i forgot all my high school math...)
What is the vertical component of the velocity of an object that is moving horizontally?

Anyways, if the diagram is not "faithful", and the particle enters with some unknown angle, then you''' have to treat v0 like a vector and break it into components.

vx = v0*cos([tex]\theta[/tex])
vy = v0*sin([tex]\theta[/tex])

Of course, this will make the subsequent mathematical expressions a bit messier. Are you certain that the particle is not known to be travelling strictly horizontally when it enters?

You have the basic equation of motion to work from (as you previously deduced):

y = y0 + vy*t +(1/2)*a*t2

Solve for t when y = 0. As you can see, it would be simpler if vy happened to be zero.
 
  • #9
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Thanks. I think you're right. Vy =0

but I don't get one last thing:

I now have the formula: y = 1/2at^2.
what do I do with the other formula I had:
[tex]

r(t) = \frac{qE}{2m}t^2 + v_0t+ h = 0

[/tex]
 
  • #10
gneill
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20,795
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Thanks. I think you're right. Vy =0

but I don't get one last thing:

I now have the formula: y = 1/2at^2.
what do I do with the other formula I had:
[tex]

r(t) = \frac{qE}{2m}t^2 + v_0t+ h = 0

[/tex]
If you think about it, they're the same formula where y --> h, v0 --> 0. Let me amend it slightly:

[tex]y = y_0 + \frac{1}{2} a t^2[/tex]
[tex]y - y_0 = \frac{1}{2} a t^2[/tex]
[tex]\Delta y = \frac{1}{2} a t^2[/tex]

Your desired [tex]\Delta[/tex]y is -h. You've already found a (be sure to assign appropriate sign), so you should be able to determine t.

Once you've got t, you can find the velocity at which the particle is going to enter the magnetic field region...
 
  • #11
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thanks, you helped me so much!
 

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