Symmetry of line integral of $B$ over circle at infinity with finite number of wires cross plane perpendicularly I am studying problems which look like this:


Some number of wires perpendicular to plane, some line segment (here x-axis is taken), find integral along line segment (here x-axis)

In answer to the question specifying a subcase where I took the above picture from, Floris writes:

Take a straight line from - to + infinity, then a semicircle to get back. The integral of the semicircle is exactly half the integral if you went all the way around the circle.

I am not seeing how one could rigorously justify this. Could someone elaborate?
 A: I think the easiest approach to justify the calculations is to actually just directly evaluate the integral due to a single wire, then sum the results. If you have a single wire at $(0,y_0)$ with current $I$ flowing along the positive $z$-axis, then
\begin{align}
\int_{\Bbb{R}_x}\mathbf{B}\cdot d\mathbf{l}&=\frac{\mu_0I}{2}\cdot\text{sign}(y_0).
\end{align}
($\Bbb{R}_x$ denotes the $x$-axis) You can calculate this simply by writing out the definition of the line integral and plugging in the known $B$-field; the result is a pretty trivial integral. Since E&M is a linear theory, you can individually add up the separate $B$-field contributions if you have multiple wires; this gives the final answer of $\frac{\mu_0}{2}(2+1-4-5)=-3\mu_0$.

The other approach is to use a careful contour-integral argument by deforming the contour. For the sake of concreteness, assume $y_0>0$. For each $R>0$,

*

*let $\gamma_R$ be the curve from $[-R,R]$ on the $x$-axis.

*let $\sigma_{R}$ be the semicircular arc having center $(0,0)$, from $(R,0)$ to $(-R,0)$ (i.e the upper semicircle going counter clockwise)

*let $\delta_{R,+}$ be $(0,y_0)+\sigma_R$, i.e shift up the contour $\sigma_R$ so that it has center $(0,y_0)$.

*let $\delta_{R,-}$ be the 'lower half' of $\delta_{R,+}$, again oriented in the counter clockwise sense.

Draw these four contours! Then, the following equations hold (insert $\mathbf{B}\cdot d\mathbf{l}$ in each case)
\begin{align}
\int_{\gamma_R}+\int_{\sigma_R}=\mu_0I,\quad\text{and}\quad \int_{\delta_{R,+}}=\int_{\delta_{R,-}}=\frac{\mu_0I}{2}.
\end{align}
The first is obvious from Gauss' law, and the second is obvious from Gauss' law and cylindrical symmetry about the wire's position $(0,y_0)$. The non-trivial technical part is showing that
\begin{align}
\lim_{R\to \infty}\int_{\sigma_R}=\lim_{R\to \infty}\int_{\delta_{R,+}}.
\end{align}
This should intuitively make sense: both $\sigma_R$ and $\delta_{R,+}$ are "upper semi-circles" of radius $R$, the only difference is that their centers are slightly off-center and the $B$ field decays at infinity, so we should expect the difference between these contour integrals becomes negligible. Usually this step of deforming the contour is trivial because the fields decay very rapidly at infinity, but in this case, we only have a $\frac{1}{r}$ decay in the $B$-field, which is a little unnerving. I leave it to you to verify the details here.
So, putting together these facts, we get $\int_{\Bbb{R}_x}\mathbf{B}\cdot d\mathbf{l}=\mu_0I-\frac{\mu_0I}{2}=\frac{\mu_0I}{2}$, just as in the previous case (which was honestly easier for this particular example).
A: Part 1:
There are probably better ways of showing this, however we need to prove that at large distances, the field is cylindrically symettric
Let's start off by defining the simplest case, 2 current elements of equal magnitude either side of the origin, separated by a distance of 2
$\vec{B} =[\frac{I\mu_{0}}{2\pi\sqrt{x^2+(y-1)^2}} + \frac{I\mu_{0}}{2\pi\sqrt{x^2+(y+1)^2}}]\hat \phi  $
Convert to cylindrical coordinates by expanding and substituting the definition of y
=$[\frac{I\mu_{0}}{2\pi\sqrt{1+r^2-2r\sin(\phi})} + \frac{I\mu_{0}}{2\pi\sqrt{1+r^2+2r\sin(\phi})}]\hat \phi$
Let's look at the denominator
$1+r^2+2r\sin{\phi}$
Factor out $r^2$
$r^2[\frac{1}{r^2} + 1 + \frac{2\sin{\phi}}{r}]$
When $r$ is very large,  the left and right components dissappear, namely the part that is dependant on $\phi$
What is left, is that at large distances,this denominator term is now
$r^2$
Under the square root it becomes $r$
Doing the same for the other part of the field gives the same answer.
We have now proven that at large distances the field is cylindrically symettric  as it is only dependant on r and not $\phi$
Part 2:
The answer is that the $\vec{B}$ Field is cylindrically symettric.
Each $\vec{B} \cdot \vec{dl}$ element on a circular path is the same
therefore half of a circle is half of the line integral of the full circle, a 1/3 the line integral of a circle, is 1/3 the line integral of the full circle
This method works for all distributions like this for all magnitudes.
I think a better way to show this would.be to use a surface current of variable extension, and showing that aslong as the extension is finite, it will be cylindrically symettric (you may find some problems with my previous method, showing it doesnt hold for infinite number of currents)
