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Bug introduced in 7.0, persisting through 13.2.

So i am trying to plot the magnetic vector field surrounding a "flat conducting cable" (don't know the proper english expression).

This is the "code" (there are some constants in the analytic formulas, which I have excluded for the sake of clarity):

 VectorPlot[{Piecewise[{{ArcTan[y/(x + 3)] - (Pi - ArcTan[y/(x - 3)]), 

 Abs[x] < 3}},

 ArcTan[y/(x + 3)] - ArcTan[y/(x - 3)]],

 Log[Sqrt[(y^2 + (x + 3)^2)]/Sqrt[(y^2 + (x - 3)^2)]]},

 {x, -10, 10}, {y, -10, 10}]

The error message I get is the following:

Part::partw: Part 1 of {} does not exist. >>

I am not sure whether Piecewise can be used with VectorPlot at all since I haven't managed to find a syntactic, so that is probably where the problem lies. (I have managed to plot the separate functions, basically I am experiencing problems implementing Piecewise into the code.)

Thank you very much!

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A vector plot needs two components:

 VectorPlot[
   Piecewise[
  {
   {{ArcTan[y/(x + 3)] - (Pi - ArcTan[y/(x - 3)]), y}, 
    Abs[x] < 3},
   {{ArcTan[y/(x + 3)] - ArcTan[y/(x - 3)], y}, 
     Log[Sqrt[(y^2 + (x + 3)^2)]/Sqrt[(y^2 + (x - 3)^2)]]}
   }
            ],
   {x, -10, 10}, {y, -10, 10}
           ]

enter image description here

Of course, you will need to include the proper formula for the $x$ and the $y$ components of your vector in the ranges you state.

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  • $\begingroup$ Thank you for the answer, but it has left me even more confused as I don't see how it fits with the syntax for the Piecewise or VectorPlot functions at all ... First let me ask you why did you put the y at the end of the first two formulas and nothing at the end of the last one? (actually the first two were defining x in ranges bordered by 3 and -3 and the third one was for y, but that doesn't really matter) $\endgroup$
    – HudPesjan
    Commented Apr 30, 2015 at 13:00
  • $\begingroup$ @HudPesjan A VectorPlot is (of course) a plot of vectors, which here must have two components. Your original posting had just a scalar (one component). My answer included two components (placed within {}), and I arbitrarily introduced a second component whose value was $y$. As I stated, you must write both components of your vector in your function. The conditional statement, however, can be one-dimensional, as indeed all of the above are. $\endgroup$
    – David G. Stork
    Commented Apr 30, 2015 at 14:56
  • $\begingroup$ Your answer makes perfect sense, but I am not sure I understand what is going on in the code. My questions would be firstly: Have you included both, the x and y components inside the Piecewise function and wether it is neccesesary to do so? And secondly: If I have understood correctly, what you have written inside the Piecewise function is: { {component (for x?), y}, condition (for x?)}, { {component (for y?), y}, condition (for y?)}, It kinda doesn't look like the suggested Piecewise syntax, so I don't see what you really did there. $\endgroup$
    – HudPesjan
    Commented May 3, 2015 at 9:39
  • $\begingroup$ @HudPesjan It is indeed necessary to include a list of two elements as your function—the $x$ and the $y$ components of your vector field. I just put in a simple $y$... you must determine the formula for the $x$ component and $y$ component of your vector. The condition which determines which vector function to use must evaluate to True or False, and thus can be a function of one variable, two variables, or whatever. $\endgroup$
    – David G. Stork
    Commented May 3, 2015 at 15:01

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