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Going from cylindrical to cartesian coordinates

  1. Oct 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi

    The expression for the magnetic field from an infinite wire is
    [tex]
    \boldsymbol B(r) = \frac{\mu_0I}{2\pi}\frac{1}{r} \hat\phi
    [/tex]
    which points along [itex]\phi[/itex]. I am trying to convert this into cartesian coordinates, and what I get is
    [tex]
    \boldsymbol B(x, y) = \frac{\mu_0I}{2\pi}\frac{1}{\sqrt{x^2+y^2}} \hat\phi
    [/tex]
    where
    [tex]
    \hat\phi = -\sin\phi \hat x + \cos\phi \hat y
    [/tex]
    I am trying to make a phase plot of this expression, so what I have done is to say that [itex]\phi = \arctan(y/x)[/itex], so [itex]\hat\phi = -\sin(\arctan(y/x))\hat x + \cos(\arctan(y/x))\hat y[/itex]. However I don't get the desired result. Have I missed something in my approach?
     
  2. jcsd
  3. Oct 19, 2012 #2
    sin (arctan a) and cos (arctan a) can be simplified. Let z = arctan a, that means tan z = a, and tan z = sin z/cos z = a, so you can express sin z and cos z in terms of a.
     
  4. Oct 19, 2012 #3
    Thanks, so I know that
    [tex]
    \frac{y}{x} = \frac{\sin(\phi)}{\cos(\phi)}
    [/tex]
    I can't see how this enables me to rewrite e.g. [itex]\sin(\arctan(y/x))[/itex].
     
  5. Oct 19, 2012 #4
    ## \sin (\arctan a) = \sin z ##. Since ##\tan z = \sin z/\cos z = a##, ##\sin^2 z = a^2 \cos^2 z = a^2(1 - \sin^2 z)##. So you can find ##\sin z## as a function of ##a##; ditto for ##\cos z##. Then substitute ## a = y/x ##.
     
  6. Oct 19, 2012 #5
    Ah, I see. So I get
    [tex]
    \sin z = \frac{a}{\sqrt{1+a^2}} \\
    \cos z = \frac{1}{\sqrt{1+a^2}}
    [/tex]
    But I still have my original problem: That when I plot B using these for negative x, then I don't see the correct magnetic field. I thought that I was perhaps missing a term [itex]\pi/2[/itex], but that didn't solve it either.
     
  7. Oct 19, 2012 #6
    What do you get and what is your expectation?
     
  8. Oct 19, 2012 #7
    I have attached a plot of what I see (the axes are (x, y), the current 1A and the units on the axis in meters), it is called "negative_x". If I only plot for positive x-values I get "positive_x", and there I see what I expect (as shown here, on the top: http://www.netdenizen.com/emagnet/solenoids/frommaxwellonly.htm).

    My code in Mathematica for plotting is:

    VectorPlot[(mu0/2 pi)*
    current*(1/(x^2 + y^2)^(1/2))*{-(y/x)/(1 + y^2/x^2)^(1/2),
    1/(1 + y^2/x^2)^(1/2)}, {x, -0.01, 0.01}, {y, -0.01, 0.01}]
     

    Attached Files:

  9. Oct 19, 2012 #8
    You should be able to simply the formula very significantly. Note that ## \frac 1 {\sqrt {1 + y^2/x^2}} = \frac x {\sqrt {x^2 + y^2}} ##, and the radical in the denominator nicely couples with that in the common factor. But even then your formula is not wrong, I am not sure why Mathematica does not plot it correctly.
     
  10. Oct 19, 2012 #9
    I don't know either. Strange, but nice to know that I have the correct exprssion. Thanks!
     
  11. Oct 19, 2012 #10
    OK, I just plotted it in MatLAB, and it *isn't* correct. For x<0 the y-coordinates all have to change sign. So the expression is not correct.

    EDIT: I have attached the plot.
     

    Attached Files:

    Last edited: Oct 19, 2012
  12. Oct 19, 2012 #11
    Have you tried the simplified formula as I suggested?
     
  13. Oct 19, 2012 #12
    Yes, it didn't change anything. It shouldn't either, since it is just a different way of expressing it.
     
  14. Oct 20, 2012 #13
    Your formula is ## (\mu_0/2 \pi)
    I \frac 1 {(x^2 + y^2)^{1/2}} \left(\frac {-y/x } {(1 + y^2/x^2)^{1/2}},
    \frac 1 {(1 + y^2/x^2)^{1/2}} \right)##

    Observe that the y-component is always positive, which is incorrect. If you transform it the way I suggested, you will get ## (\mu_0/2 \pi) I \frac 1 {x^2 + y^2} \left(-y, x\right)##, which restores the correct sign.
     
  15. Oct 21, 2012 #14
    thanks! I must have made an error somewhere then when I tried
     
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