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Moving electron in a uniform magnetic field

  1. Apr 10, 2008 #1
    Recently, in a test, I had a question involving a magnetic field and an electron. I've attached the diagram. The "X" denote the magnetic field, moving inwards.

    So, there's a uniform magnetic field, moving into the plane of the picture, and theres an electron passing over that plane with its motion parallel to the plane. It is shown by (1)

    The question is, what happens to the shape of the path of the electron when the speed of the electron is increased? Does its motion match (2), or (3)???

    At first, I did the question by intuition. I thought that since the speed has increased, it'll have less change in direction per unit time. So I chose (2).

    But later, the formula F = Bqv was pointed out to me, where F=Force, B=Magnetic Field strengthm, q=charge on the electron, v=velocity of the electron. According to this, (3) should be the correct path!

    So what is the correct new path and why? Because I still feel it is (2), even though the formula proves otherwise.
     

    Attached Files:

  2. jcsd
  3. Apr 10, 2008 #2

    tiny-tim

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    … angular velocity ω is constant …

    Hi zinc79! :smile:

    Any charged object moving perpendicular to a uniform magnetic field will move in a circle.

    Objects with the same mass and the same charge in the same field will move with the same angular velocity, ω.

    In other words, they will all complete a full circle in the same time, with the faster ones of course having to move in larger circles (to go a distance 2πr where r is larger, you need more speed!).

    The reason is that the centripetal force needed to keep to a circle of radius r, which you are probably familiar with in the form m.v^2/r or m.r.ω^2, can also be written m.ω.v … and since the force is proportional to q.v, that means that m.ω/q is constant (independent of speed)! :smile:
     
  4. Apr 10, 2008 #3
    I see, so although the force has increased, the radius still increases?

    What's with all this dependence and independence anyway? I never seem to make sense of it, or make it out on my own. If you hadn't told me that the angular velocity remains the same (which I didn't know), and if I was using m.v^2/r, how would I tell that the radius is dependent on the force, i.e. whether r would increase/decrease? (because v^2 is there to increase/decrease the centripetal force in correspondence with the force Bqv on the electron when its speed increases/decreases)

    And could you explain WHY the angular velocity is constant? How do we explain its independence when it doesn't cancel off in the force equations on the electron?
     
  5. Apr 10, 2008 #4

    tiny-tim

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    Hi zinc79! :smile:

    Because you would have m.v^2/r = qBv;

    divide both sides by v, and you get v/r = qB/m, which is a constant.

    In other words, v/r is constant, and so r is proportional to v, which is proportional to the force.

    (And also v/r = ω, the angular velocity, so ω is constant.)
    Sorry, I'm not following that. :confused:
     
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