The complex number $z$ satisfies the relations $|z|\le 2$ and $\dfrac{\pi}{6}\le \arg z \le \dfrac{\pi}{2}$. Illustrate both of these relations on a single Argand diagram as the region $R$.

1. Determine the rightmost vertex of the region $R$ to establish its position relative to the origin of the locus $l$.

The region $R$ is bounded by $|z| \le 2$ and $\frac{\pi}{6} \le \arg z \le \frac{\pi}{2}$.
The vertex of $R$ with the maximum real part is at the intersection of the arc $|z| = 2$ and the ray $\arg z = \frac{\pi}{6}$:
$z = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = \sqrt{3} + i$
Let this point be $B(\sqrt{3}, 1)$.
Thus, for all points $z \in R$, $\text{Re}(z) \le \sqrt{3}$ and $\text{Im}(z) \ge 0$.
The locus $l$ is a half-line emanating from $A(\sqrt{3}, 0)$ with an angle $k$ to the positive real direction.
Notice that $A$ is vertically below $B$, meaning the vector $\overrightarrow{AB}$ goes straight up, corresponding to the angle $k = \frac{\pi}{2}$.

2. Analyze the sweep of the half-line $l$ to find the geometric range of $k$ resulting in no intersection.

For $k = \frac{\pi}{2}$, the half-line passes directly through $B(\sqrt{3}, 1)$, so it intersects $R$.
As $k$ increases such that $\frac{\pi}{2} < k \le \pi$, the half-line points upwards and leftwards. Since it starts from $x=\sqrt{3}$ below $R$, it must cross the boundary segment of $R$ connecting the origin to $B$, thus intersecting $R$.
For $0 \le k < \frac{\pi}{2}$, the half-line points to the right ($x > \sqrt{3}$). Since all points in $R$ have $x \le \sqrt{3}$, there is no intersection.
For $-\pi < k < 0$, the half-line points into the lower half-plane ($y < 0$). Since $R$ is strictly in the upper half-plane ($y \ge 0$), there is no intersection.
Therefore, $l$ and $R$ have no common point when $-\pi < k < \frac{\pi}{2}$.