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Finding F

  1. Apr 29, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img245.imageshack.us/img245/7535/38187200.th.jpg [Broken]

    I need to find F from the picture above

    2. Relevant equations



    3. The attempt at a solution

    My answer so far is:

    [tex] \frac{-y}{(x^2+y^2)} \vec{i} + \frac{x}{(x^2+y^2)} \vec{j} [/tex]

    is this correct?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 30, 2009 #2

    benorin

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    Homework Helper

    No. The magnitude is at a distance r from the z-axis should be 4r. recall that [tex]r=\sqrt{x^2+y^2}[/tex]. Note that your "answer so far" has the properties depicted except that it has magnitude 1/r at a distance r from the z-axis.
     
  4. Apr 30, 2009 #3
    so it should then be:

    [tex] \frac{-y}{4(x^2+y^2)} \vec{i} + \frac{x}{4(x^2+y^2)} \vec{j} [/tex]
     
  5. Apr 30, 2009 #4

    benorin

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    Does [tex]\sqrt{\left(\frac{-y}{4(x^2+y^2)}\right)^2 + \left(\frac{x}{4(x^2+y^2)}\right)^2}=4\sqrt{x^2+y^2}[/tex]?

    No. Look, [tex]-y\vec{i}+x\vec{j}[/tex] is a vector field whose vectors point in all the right directions, make them the proper length.
     
  6. Apr 30, 2009 #5
    I have no clue.... I've tried [tex] -4y\vec{i} + 4x\vec{j} [/tex] and it seems to be wrong as well
     
  7. Apr 30, 2009 #6

    HallsofIvy

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    As benorin said, your first attempt had magnitude 1/r and it should be 4r. For your second attempt you just divided by 4. That makes the magnitude 1/4r, not 4r. You need to multiply your first answer by 4r2= 4(x2+ y2).
     
    Last edited by a moderator: May 4, 2017
  8. Apr 30, 2009 #7
    so it's just simply:

    [tex]-4y\vec{i} + 4x\vec{j}[/tex]
     
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