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Convert vector-field from cylindrical to cartesian

  1. Jun 20, 2013 #1
    1. The problem statement, all variables and given/known data
    I have a vector field (which happens to be a magnetic field)
    H = -[itex]\frac{I }{2 \pi r}[/itex]u[itex]\varphi[/itex]
    u[itex]\varphi[/itex] is the unit vector

    which is in the cylindrical coordinate system with only the [itex]\varphi[/itex] component nonzero so it circles around the z-axis. r is the radius of the circle.
    the question is: write H in the cartesian coordinate system.


    2. Relevant equations
    i already know the answer through common sense but I'm not able to derive it myself:
    H = [itex]\frac{-I}{2 \pi \sqrt{x^2+y^2}}[/itex][[itex]\frac{y}{\sqrt{x^2+y^2}}[/itex] - [itex]\frac{x}{\sqrt{x^2+y^2}}[/itex]]


    3. The attempt at a solution
    now I know these relations:
    x = r Cos[[itex]\varphi[/itex]]
    y = r Sin[[itex]\varphi[/itex]]
    r = [itex]\sqrt{x^2+y^2}[/itex]
    but i can't produce the answer with these
     
    Last edited: Jun 20, 2013
  2. jcsd
  3. Jun 20, 2013 #2

    LCKurtz

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    If ##\theta## is the usual polar coordinate angle, then the unit vector in the ##r## direction is ##\langle \cos\theta, \sin\theta\rangle##. You want to rotate this vector ##90^\circ## to get a unit vector in the ##\theta## direction. So put ##\varphi = \frac \pi 2 +\theta## to get$$
    \hat u_\varphi = \langle \cos(\frac \pi 2 +\theta),\sin(\frac \pi 2 +\theta)\rangle =
    \langle \sin\theta,-\cos\theta\rangle = \langle \frac y r,-\frac x r\rangle =
    \langle \frac y {\sqrt{x^2+y^2}},-\frac x {\sqrt{x^2+y^2}}\rangle$$
     
  4. Jun 23, 2013 #3
    Thank you very much, this solves my question
     
  5. Jun 23, 2013 #4

    vela

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    The righthand side, as you've written it, isn't a vector. Typo? It also looks like you made a sign error. Note that ##\hat{u}_\varphi = (-\sin \phi, \cos \phi,0)##. It points in the direction that ##\vec{r}## moves if you increase ##\varphi## by a small amount.
     
  6. Jun 23, 2013 #5
    Yes indeed a typo :redface: forgot the unit vectors
    and H points in the minus [itex]\varphi[/itex] direction so i think it's alright
     
  7. Jun 23, 2013 #6

    vela

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    If it points in the ##-\hat{u}_\varphi## direction, you have a sign error.
     
  8. Jun 23, 2013 #7
    I don't see this error and my professor didn't see it as an error either when i first handed it in so could you please elaborate a bit more?
     
  9. Jun 23, 2013 #8

    vela

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    I'm not sure what's there to elaborate on. Your expression is equal to ##\frac{I}{2\pi r}\hat{u}_\varphi##, not ##-\frac{I}{2\pi r}\hat{u}_\varphi##.
     
  10. Jun 23, 2013 #9
    Well I plotted it in Mathematica and I rotates clockwise around the z-axis as expected so I still don't see the error
     
  11. Jun 23, 2013 #10

    vela

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    According to your expression and assuming I>0, when x=y=1, for example, you have H pointing in the -x and +y direction. That's not clockwise.
     
  12. Jun 23, 2013 #11
    Then it will point in the +x and -y direction because there is a minus sign in front of the entire vector
    H = [itex]\frac{-I}{2[itex]\pi[/itex]\sqrt{x^2+y^2}}[/itex][[itex]\frac{y}{\sqrt{x^2+y^2}}[/itex]ux + [itex]\frac{-x}{\sqrt{x^2+y^2}}[/itex]uy]
     
  13. Jun 23, 2013 #12

    vela

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    Did you actually plug x=y=1 into that expression? Because of the minus sign out front, H points in the -x and +y directions.
     
  14. Jun 23, 2013 #13
    I see you are right and my professor didn't even notice :biggrin:
     
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