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Field region beams penetrate

  1. Dec 21, 2012 #1
    1. The problem statement, all variables and given/known data

    fieldregion_zps342a1109.jpg

    2. Relevant equations

    magnetic force = charge * velocity * magnetic field


    they ask for distance but this was the only relevant equation for this problem to corresponding section

    3. The attempt at a solution

    veloctity = 8.4* 10 ^ 6 m/s
    magnetic field = 190 Gauss
    perpendicular = sin (90) = 1

    magnetic force = (1.6 * 10 ^ -19 C) (8.4 * 10 ^ 6 m/s )(190 G)
     
  2. jcsd
  3. Dec 21, 2012 #2

    mfb

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    Which path do the electrons describe in the field?
    This will help to solve the problem.
     
  4. Dec 21, 2012 #3

    rude man

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    I would start by writing F= ma equations for all three axes. (Might two suffice?).
    Good problem.
     
  5. Dec 21, 2012 #4
    all 3 axes, even if z axis is not mentioned. we have to resolve components using a force diagram?

    okay here goes:
    F_x = m * a_x
    F_y = m * a_y
    F_z = m * a_z
     
  6. Dec 21, 2012 #5

    rude man

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    What I alluded to is that perhaps one of the axes never sees any motion. Since mag force is always perpendicular to the B field, which axis do you suppose that might be?

    PS certainly the z axis is in on this.
     
  7. Dec 21, 2012 #6
    magnetic force perpendicular to the magnetic field. the y -axis is perpendicular!
     
  8. Dec 21, 2012 #7

    mfb

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    Z-axis is the most important one...
    And please, do not use cartesian coordinates to solve equations of motions. This is just overkill for this problem.
     
  9. Dec 21, 2012 #8

    rude man

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    OK, now what are F_x, F_y and F_z in terms of q, m, velocities, etc? Also, write x_double-dot in lieu of a_x, etc.
     
  10. Dec 21, 2012 #9

    rude man

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    Request not granted :grumpy:. And the answer is in x, not z. :smile:
     
  11. Dec 21, 2012 #10

    rude man

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    Right. So there are only two equations, in x and in z.
     
  12. Dec 21, 2012 #11
    F_x is force in the x or force sub x(probably should have done that)

    you mean like this:

    F_subx = m* a..
    F_subz = m* a..
     
  13. Dec 21, 2012 #12

    rude man

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    No. Look at what I wrote.
    You need to introduce q, B, and velocity components in lieu of forces F_x etc.

    For a_x you should write d2x/dt2 etc.
     
  14. Dec 21, 2012 #13


    oh I see.

    charge* velocity* magnetic field = mass * (d^2 * x)/(d*t^2)

    may you explain how you got acceleration to be d^2 * x)/(d*t^2). I went through the chapter called amperes law where this question was associated with and don't see the equation like that of (d^2 * x)/(d*t^2)

    d as in distance, t as in time, and what is x and we need to find time?
     
    Last edited: Dec 21, 2012
  15. Dec 21, 2012 #14

    rude man

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    Question: have you had calculus incl. elementary differential equations? I never know what level of student I am dealing with.

    This is a pretty sophisticated problem, involving coupled motions. I.e. the initial x velocity generates a z velocity, and the z velocity in turn generates an x velocity. I wouldn't try to solve this in a non-calculus manner. Maybe mfb knows how ...
     
  16. Dec 21, 2012 #15
    i had two semester of calculus: differential and integral. taking 3rd and last semester of calculus in the spring
     
  17. Dec 21, 2012 #16

    rude man

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    OK. Then I can say that I meant d2x/dt2 to be "the second derivative of x with respect to time". Aka "acceleration in the x direction". Etc.

    My way you wind up with two second-order linear differential equations in x and z. Solve for x and there's your answer. The equation in x is the same you get from a spring-mass oscillator.

    (You're not asked for the z solution which is similar but not exactly the same as a spring-mass oscillator).
     
  18. Dec 21, 2012 #17
    oh i see. interesting.

    btw how come magnetic field is given in Gauss, the previous questions that involved with magnetic field were given in Tesla? Wouldn't Tesla have to be used?
     
    Last edited: Dec 21, 2012
  19. Dec 21, 2012 #18

    rude man

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    1 Tesla = 10,000 Gauss. Your teach is trying to make life even more miserable than it has to be ... tell him you won't do any more problems in any other than SI units... :smile:

    Actually, pure physicists tend to use the cgs system which includes the Gauss. Applied physicists and engineers gravitate towards the SI system, aka 'rationalized mks" system ("meter, kilogram, second").
    .
     
  20. Dec 21, 2012 #19
    he did teach us how to convert units super fast in the first physics lecture. he really want us to convert units in some assigned problems. but it's for practice which I totally needed but am getting better at. for this one it would be :

    90 Gauss * (1 Tesla/10,000 Gauss) = .009 Tesla.

    okay now to plug these values in:

    charge * velocity * magnetic field = (mass of an electron) (2nd derivative of position respect to time)

    (1.6 * 10 ^ -9 C) * (8.4 * 10 ^ 6 m/s) (.009 T) = (9.1 * 10 ^ -31 kg) (2nd derivative of position respect to time.

    well im left with acceleration given as a constant value of 1.3 * 10 ^ 26 m/s^2 . I could integrate this twice if this was a function so I can get distance but its just a constant value
     
    Last edited: Dec 21, 2012
  21. Dec 21, 2012 #20

    rude man

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    Right!
    Don't mess with numbers now. You need to get your equations first.

    So why not start with the z-axis. What does F = ma look like for it? Keep in mind that the velocity is in x and the B field is in y direction, so use the appropriate velocity component ...
     
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