An infinitely long wire lying along z-axis carries a current I, flowing towards positive z-direction. There is no other current. Consider a circle in x-y plane with centre at (2 meter, 0, 0) and radius 1 meter. Divide the circle into small segments and let $d\overrightarrow{\mathcal{l}}$  denote the length of a small segment in anticlockwise direction, as shown.

 

The maximum value of path integral  $\int \overrightarrow{B}\cdot d\overrightarrow{\mathcal{l}}$ of the total magnetic field $\overrightarrow{B}$ along the
perimeter of the given circle between any two points on the circle is $\frac{{\mu }_0\mathrm{{\rm I}}}{n}$ . Find $n$
 

Consider the figure shown. 

$C$ is the centre of the circle of diameter $d$ in the $x-y$ plane. $C$ is located at a distance $2d$ from the origin $O$ .

An infinite wire (lying along $z$axis) and carrying current along $z$ axis is shown as a dot at the origin $O$.

With centre at the origin $O$, draw a circle whose arc $LVM$ lies inside the circle with centre at  $C$. The arc $LVM$ subtends an angle $\theta$ at the origin $O$. There is symmetry in figure about $x$ axis.

Extend the lines $OL$ and $OM$ to intersect the circle with centre at $O$ at points $P$ and $N$ respectively. Arc $PUN$belongs to circle with centre at the origin $O$ and radius equal to $OP=ON$.

Consider the arc $MVL$. All points of it are at same distance from the wire. Magnitude of magnetic field at any point of it is same and it is tangential to the arc.

By Ampere’s circuital law, for the complete circle passing through $M,V,L$

$\oint \overrightarrow{B}·\overrightarrow{d}l=+{\mu }_{0}I$     ($\overrightarrow{B}$is parallel to $\overrightarrow{dl}$ at each point of the circle. Hence positive sign)

Since magnitude of magnetic field is radially symmetric and direction of the magnetic field is tangential to the circumference of circle passing through $M,V,L$, hence,

$\underset{arcMVL}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}\propto \left(arclengthMVL\right)\propto 2\theta$

Hence, $\underset{arcMVL}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}=\frac{{\mu }_{0}I}{2\pi }\times 2\theta =\frac{{\mu }_{0}I\theta }{\pi }$                      …(1)

Notice that $\left(\underset{arcMVL}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}\propto \theta \right)$. It means the greater the angle $\theta$, greater is the line integral.

Also notice, $\underset{arcMVL}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}=\underset{arcNUP}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}=\frac{{\mu }_{0}I\theta }{\pi }$    (Since, both arcs subtend same angle at point $o$).

Further, now consider closed loop $MVLTM$. Since there is no current passing through it,

$\underset{arcMVLTM}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}=0$

$\Rightarrow \underset{arcMVL}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}+\underset{arcLTM}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}=0$

$\Rightarrow \frac{{\mu }_{0}I\theta }{\pi }+\underset{arcLTM}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}=0$                   (using (1))

$\Rightarrow \underset{arcLTM}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}=-\frac{{\mu }_{0}I\theta }{\pi }$           …(2)

In a similar way,

$\underset{arcNUPSN}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}=\underset{arcNUP}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}+\underset{arcPSN}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}=0$

$\Rightarrow \frac{{\mu }_{0}I\theta }{\pi }+\underset{arcPSN}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}=0$

  $\Rightarrow \underset{arcPSN}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}=-\frac{{\mu }_{0}I\theta }{\pi }$         …(3)

Even though length of arc $PSN$ is greater than that of arc $LTM$, both give same value of line integral of magnetic field.

Now let us increase $\theta$. Then points $L,P$ approach each other on the circumference. Likewise points $M,N$ approach each other on the circumference.

Since $\underset{arcLTM}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}\propto \theta$, at maximum value of $\theta$, maximum line integral is obtained through each of the arcs $MVL,LTM,NUV,PSN$(all of which will be equal in magnitude).

What is the maximum value of $\theta$? At maximum value of $\theta$, $L,P$overlap and $M,N$ overlap. Lines $OL$ and $OM$ becomes tangents to the circle with centre at $C$. Triangle $CLO$ is a right-angled triangle with $\angle CLO=90°$

$\sin \theta =\frac{CL}{OC}=\frac{d}{2d}=\frac{1}{2}$

$\Rightarrow \sin \theta =\frac{\pi }{6}$

So, ${\left[\underset{arcLTM}{\overset{ }{\int }}\overrightarrow{B}·\overrightarrow{dl}\right]}_{\max }=-\frac{{\mu }_{0}I}{\pi }{\left(\theta \right)}_{\max }=-\frac{{\mu }_{0}I}{\pi }\frac{\pi }{6}=-\frac{{\mu }_{0}I}{6}$

Further, in the above analysis, as $\theta$ varies from $0 \mathrm{to} \frac{\mathrm{\pi }}{2}$, we cover all possible lengths of arcs of the circle with centre at $C$. There is no length of arc (between $0$ to $2\pi r$) that is left uncovered. Here is how!

We can see, as the value of $\theta$ grows from $0°$ to $90°$, the arc length $PSN$ grows at an average rate of five times more than that of arc $LTM$.