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Magnetic Fields and electron beam

  1. Apr 13, 2004 #1
    I have a few problems that I'm not sure how to do (not sure what equations to use), can anyone help me?

    1) At a certain location, the horizontal component of the earth's magnetic field is 2.3 x 10^-5 T, due north. A proton moves eastward with just the right speed, so the magnetic force on it balances its weight. Find the speed of the proton.

    2) The electrons in the beam of a television tube have a kinetic energy of 3.23 x 10^-15 J. Initially, the electrons move horizontally from west to east. The vertical component of the earth's magnetic field points down, toward the surface of the earth, and has a magnitude of 1.85 x 10^-5 T. What is the acceleration of an electron?

    3) A magnetic field has a magnitude of 0.0011 T, and an electric field has a magnitude of 3.9 x 10^3 N/C. Both fields point in the same direction. A positive 2 micro-C charge moves at a speed of 3.7 x 10^6 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.

    4) A beam of protons moves in a circle of radius 0.36 m. The protons move perpendicular to a 0.44-T magnetic field. (a) What is the speed of each proton? (b) Determine the magnitude of the centripetal force that acts on each proton.
     
  2. jcsd
  3. Apr 13, 2004 #2

    turin

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    Open your book. Look at the formulas that use the magnetic field. If this option fails, then you need to know that Fmagnetic = q v x B, where that x is not a variable, but the cross-product operator. Then, draw a free-body diagram for the problem and put the forces on it.
     
  4. Apr 13, 2004 #3
    I figured out the last one but I'm still not sure of the other 3...
     
  5. Apr 13, 2004 #4

    turin

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    Draw your free body diagram. There will be two force arrows pointing in opposite directions. The condition that I have quoted tells you the relationship of the lengths of the arrows. One of the arrows depends on v. You need to decide how it depends on v, and then put in the v that satisfies the above condition.




    I don't get it. It seems like you also need to know the N-directed component of B.




    You will get 2 force vectors. Vector addition will give you the resultant.
     
  6. Apr 14, 2004 #5
    No, you don't, because the cross product of two parallel vectors is 0 so the horizontal component of the magnetic field will not affect the electrons, at least initally (and it is my understanding that the problem asks to find the inital acceleration of the electrons).
     
  7. Apr 14, 2004 #6
    For question 3 I was going to use the equation FE=Eq but I'm not sure how to get q. Do I multiply the 2 micro C with 1.6x10^-19?
     
  8. Apr 14, 2004 #7
    The charge q is 2μc, the field E is 3.9 x 10^3 N/C. Multiply them to find the electric force.
     
  9. Apr 14, 2004 #8
    Not sure I'm following. Do I multiply (3.9x10^3)x(2μc) which would equal 7800 N?
     
  10. Apr 14, 2004 #9
    Yes, but it's 7.8x10-3N...
     
  11. Apr 14, 2004 #10
    Ok, for question 3 here's the work I've done:

    FE=Eq ---> (3.9x10^3)(2μc) = 7.8x10^-3N
    Fm=qVxB ---> (7.8x10^-3N)(3.7x10^6)x(0.0011 T) = 31.746
    (7.8x10^-3N)^2 + (31.746)^2 = 1007.808577

    But the program is saying it's incorrect, where did I go wrong?
     
  12. Apr 14, 2004 #11
    For starters, I get Fm = 8.14x10-3N. Why are you using the electric force for q when calculating the magnetic force? Just use the charge, 2μc. Also, the answer is the root of (FE2 + Fm2).
     
  13. Apr 14, 2004 #12
    I ended up getting the correct answer of 0.11269 N but I still don't understand how you got 8.14x10-3. Not sure if I'm entering it correctly in my calculator but here's how I entered it:

    (7.8x10^-3)x(3.7x10^6)x(0.0011) = 31.746 on my calc
     
  14. Apr 14, 2004 #13
    Again, why are you using 7.8x10-3 for q? q is 2μc.
     
  15. Apr 14, 2004 #14
    I guess I completely overlooked that. I think I've spent too much time on this prob. Thanks for your help though! Also, quick question regarding the second problem. I know I should use the equation a=F/mp. How do I solve for F since it's in Joules?
     
  16. Apr 14, 2004 #15
    From the kinetic energy of the electrons you can find their velocity, then find the magnetic force using qvB, and finally divide that by their mass to get the acceleration.
     
  17. Apr 14, 2004 #16

    turin

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    Just because it is horizontal doesn't make it parallel to the velocity vector of the electrons. The horizontal plane is 2-D.

    - The electrons are moving in the E direction.

    - The N directed projection of the B-field cannot be determined from problem #2.

    - The N direction and the E direction are not parallel; they are perpendicular.

    Therefore the force cannot be determined. Please explain what I am doing wrong or what I have misinterpretted.
     
    Last edited: Apr 14, 2004
  18. Apr 14, 2004 #17

    Doc Al

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    ambiguity in problem #2

    I think you are exactly right, turin. It's a sloppily worded problem. I presume that whoever devised the question just messed up.
     
  19. Apr 15, 2004 #18

    turin

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    That doesn't surprise me, but then what is the problem really telling us? I guess the most natural assumption from the wording of the problem would be that the N directed field is zero? That just seems strange, since we all learn from kindergarten that the magnetic field has a N-S direction. Any ideas?
     
  20. Apr 15, 2004 #19
    You are turin, I did not think of that. Seems like the wording has been a bit sloppy, but it can probably be understood that they are asking for the vertical acceleration. Otherwise there'd be no way to solve it. :smile:
     
  21. Apr 15, 2004 #20

    turin

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    You meant "the N-S acceleration," of course.
     
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