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The book from which I have taken this picture claims that the force on the segment X in the first case is less than th force acting on it in the second case. I don't think this is true unless one carries out the full calculation. The wire segment on the left causes a magnetic field that changes with each single point along the wire X in each case. So, is there a way in which I can make that comparison without actually currying out the full calculation?

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The $\bf B$ field for a wire of length L in the x direction is, \begin{equation} {\bf B}(x,y)=\frac{{\bf I\times{\hat j}}}{cy}\left[\frac{(L/2-x)}{\sqrt{L/2-x)^2+y^2}}+\frac{(L/2+x)}{\sqrt{L/2+x)^2+y^2}}\right]. \end{equation} You can derive from this that the field diminishes as x extends beyond L/2. This means there will be a weaker field acting on the second wire when they are displaced. Integrating I$\bf dl\times B$ over the second wire also shows that the force is weakens with displacement.

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  • $\begingroup$ By $\vec{I}$ do you mean $\vec{I}=IL\hat{n}$ $\endgroup$
    – Jack
    Commented Feb 28 at 19:24
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    $\begingroup$ $\bf I$ is in the x direction, so ${\bf I=}I{\bf\hat i}$ $\endgroup$ Commented Feb 28 at 19:30
  • $\begingroup$ I have a question. The magnetic field you posted above is for a segment whose center is at the origin, or whose left end starts at the origin? $\endgroup$
    – Jack
    Commented Feb 29 at 5:01
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    $\begingroup$ The center is at the origin. $x$ is the distance from the origin. $\endgroup$ Commented Feb 29 at 15:28

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