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Circular magnetic feilds

  1. Aug 2, 2009 #1
    can circular magnetic fields accelerate charged particles linearly?
    is their any method to make a powerful circular magnetic field , other than using a wire with large diameter.
     
  2. jcsd
  3. Aug 2, 2009 #2

    ZapperZ

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    Forget about "circular magnetic field". Can you show any example where ANY geometry of magnetic field can accelerate charged particles?

    Zz.
     
  4. Aug 2, 2009 #3
    it can only change the direction
    what if it was time varying?
     
    Last edited: Aug 2, 2009
  5. Aug 2, 2009 #4

    jasonRF

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    Bingo!
     
  6. Aug 2, 2009 #5

    ZapperZ

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    Time varying implies the existence of E-field, so you could simply have circumvented all of this by simply saying that there is an E-field, couldn't you? Yet, you didn't. So is this really part of your original question?

    Zz.
     
  7. Aug 2, 2009 #6
    i am sorry if i am wrong , i am just doing my 10+2 . i got this doubt when i saw the field of a wire which is circular , if a current flowing in it can generate a circular magnetic field.Then why cant a circular magnetic field accelerate a charge linearly.
    PS:i don't know vector calculus
     
    Last edited: Aug 2, 2009
  8. Aug 2, 2009 #7

    ZapperZ

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    If this is a HW/Coursework question, it must be done in the HW/Coursework forum and you have to show what you have attempted.

    Zz.
     
  9. Aug 2, 2009 #8
    This is not a HW/Coursework question.Can u please tell me why is this not possible
     
  10. Aug 2, 2009 #9

    ZapperZ

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    And I'll come back to my original question to you in which you should forget about your circular field, and show me ANY instance where ANY static magnetic field can actually produce a charged particle acceleration. Even when you don't know vector calculus, you must know the Lorentz force law, don't you? If you don't, now you have something to google on.

    Zz.
     
  11. Aug 2, 2009 #10
    yes i know Lorentz force law but how do i represent a circular magnetic field in that equation.
    But the equation shows that force is always perpendicular for magnetic fields having only one direction.
     
  12. Aug 2, 2009 #11

    Cyosis

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    As ZapperZ suggests, if we can prove that no magnetic field can accelerate a particle linearly then surely there does not exist a circular magnetic field that can.

    Try to calculate the work that the Lorentz force does on a particle.
     
  13. Aug 2, 2009 #12
    work done is zero because force is perpendicular to the displacement , that is when the field is only in a single direction .
    i just want to know the path. . . how will charge move,if the their is a circular field like that of a wire,and charge was initially moving through the center of the axis of the field.
     
    Last edited: Aug 2, 2009
  14. Aug 2, 2009 #13

    Cyosis

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    Your question keeps changing and by now has nothing to do with the original question you asked. How the charged particle moves, amongst other things, depends on its approach. Does the particle travel parallel to the wire or perpendicular to the wire. Just draw the situation and use the right hand rule.
     
  15. Aug 2, 2009 #14

    jtbell

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    You apply the Lorentz force law at a particular instant of time, using the magnetic field at the point where the particle is located at that instant, and the velocity of the particle at that instant.

    As time passes, the particle moves from one point to another. Now you use the magnetic field at the new point, as well as the particle's new velocity. The resulting force will generally have a different magnitude and direction than before.

    In general, finding the path of the particle means solving a set of coupled differential equations, which is not trivial!
     
  16. Aug 2, 2009 #15
  17. Aug 2, 2009 #16

    ZapperZ

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    You still don't get it. If the magnetic field is always perpendicular to v and B, as stated in the Lorentz force equation, then no work can be done by the magnetic field onto the charge. This means that the charge gains no energy from the magnetic field and so, it gains no linear acceleration.

    The shape of the static magnetic field is irrelevant!

    So which part of this did you not understand?

    Zz.
     
  18. Aug 2, 2009 #17
    Hulkhogan-
    You have learned that because of the Lorentz force, the force F is orthogonal to both the magnetic field B and the particle velocity v. Now visualize this. 1) A large vertical magnetic field (like most cyclotrons). 2) An azimuthal velocity of the particle (around the circumference of the magnetic field). 3) A radial Lorentz force making the particle go around a circle in the magnetic field.

    But the Lorentz force cannot accelerate the particle. What can produce an azimuthal electric field in the vacuum chamber where the particle is traveling? Hint: Think about how transformers work.
    Bob S
    [Added text] When you have completed the above, now think of the following design, originally proposed by Mark Oliphant. Consider two solid copper conductors each 10 cm diameter and 30 cm apart. If each were carrying 1 million amps, uniformly distributed in the copper, in opposite directions, what happens to the magnetic field between them? Suppose now visualize the two conductors were moved so that they were now 10 cm apart (center to center), so that the copper overlapped. What is the current in the overlapping region? Suppose all the copper were removed in the overlapping region, how would the magnetic field change? What about the uniformity of the magnetic field? Now visualize removing all the copper in the overlapping region, and adding a vacuum chamber.
     
    Last edited: Aug 2, 2009
  19. Aug 2, 2009 #18

    jasonRF

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    Perhaps this is taking the thread too off-point, but your statement isn't quite right. Of course a magnetic field does no work. But, if the strength of the field is variable in the direction of the field, then perpendicular energy (from cyclotron motion) will be converted to parallel (linear!) energy. By "perpendicular" I mean perpendicular to B, and "parallel" is parallel to B. This is why charged particles trapped in the magnetosphere "bounce" back and forth between north and south poles of the Earth's magnetic field. Likewise, one mechanism by which ionospheric ions are ejected out into the magnetosphere is as follows: wave processes accelerate the ions perpendicular to the Earths magnetic field; the fact that the field has a gradient in the direction of the field changes the perpendicular energy into parallel energy; the ions get enough parallel energy to escape the gravitational field of the Earth and they end up in the magnetosphere. This is why we sometimes measure significant quantities of oxygen in the magnetosphere - not much oxygen is in the solar wind!

    If you are interested in this stuff, almost any plasma physics text will discuss this.

    I'm sure the original poster wasn't thinking about this, which is why I didn't bring it up earlier.
     
    Last edited: Aug 2, 2009
  20. Aug 2, 2009 #19

    jtbell

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    Does the field also exist at other radii from the axis of the circles you've drawn, so those circles are just a sample. That is, would a more complete diagram also include other circles concentric with the ones you drew?

    Is the particle going down the middle (the axis of the circles), or is it off-center but still parallel to the axis?
     
    Last edited: Aug 3, 2009
  21. Aug 2, 2009 #20
    I suspected this is where you were comming from. This is a picture of a steady, magnetic field surrounding a steady, non-accelerating current. The electric charges are not speeding up, but continuing at a constant velocity.
     
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