Consider a circular current-carrying loop of radius $R$ in the $x-y$ plane with its centre at the origin. Consider the line integral $\Im(L) = \left| \int_{-L}^{L} \vec{B} \cdot d\vec{l} \right|$ taken along the $z$-axis.
$(a)$ Show that $\Im(L)$ monotonically increases with $L$.
$(b)$ Use an appropriate Amperian loop to show that $\Im(\infty) = \mu_0 I$,where $I$ is the current in the wire.
$(c)$ Verify this result directly.
$(d)$ Suppose we replace the circular coil with a square coil of side $R$ carrying the same current $I$. What can you say about $\Im(L)$ and $\Im(\infty)$?

(N/A) The magnetic field $\vec{B}$ on the $z$-axis is directed along the $z$-axis. Thus,$\vec{B} \cdot d\vec{l} = B_z dz$. Since $B_z$ is always positive (or negative depending on current direction) along the $z$-axis,the integral $\int_{-L}^{L} B_z dz$ represents the area under the $B_z$ vs $z$ curve. As $L$ increases,the integral accumulates more area,so $\Im(L)$ monotonically increases.
$(b)$ Consider an Amperian loop consisting of the segment on the $z$-axis from $-L$ to $L$ and a large semi-circular arc of radius $r \to \infty$ in the $x-y$ plane. By Ampere's Law,$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$. As $r \to \infty$,the contribution from the arc vanishes because $B \propto 1/r^3$. Thus,$\int_{-L}^{L} B_z dz + 0 = \mu_0 I$,so $\Im(\infty) = \mu_0 I$.
$(c)$ The magnetic field on the axis of a circular loop is $B_z = \frac{\mu_0 I R^2}{2(z^2 + R^2)^{3/2}}$. Integrating from $-\infty$ to $\infty$: $\int_{-\infty}^{\infty} \frac{\mu_0 I R^2}{2(z^2 + R^2)^{3/2}} dz$. Let $z = R \tan \theta$,then $dz = R \sec^2 \theta d\theta$. The integral becomes $\frac{\mu_0 I}{2} \int_{-\pi/2}^{\pi/2} \cos \theta d\theta = \frac{\mu_0 I}{2} [\sin \theta]_{-\pi/2}^{\pi/2} = \mu_0 I$.
$(d)$ For a square coil,the symmetry is different,but the total flux linked or the line integral along the axis still follows the same topological constraints of Ampere's law. Thus,$\Im(\infty)$ remains $\mu_0 I$ due to the same enclosed current,while $\Im(L)$ will still monotonically increase.