Problem 163

Since $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\dfrac {x^{2}}{a^{2}}+ \dfrac {y^{2}}{b^{2}}=1(a > b > 0)$,
it follows that the eccentricity $0 < e < 1$, $F_{1}(-c,0)$, $F_{2}(c,0)$, $c^{2}=a^{2}-b^{2}$.
Let point $P(x,y)$, from $PF_{1} \perp PF_{2}$, we get $(x-c,y) \cdot (x+c,y)=0$, which simplifies to $x^{2}+y^{2}=c^{2}$.
By solving the system of equations $\begin{cases} x^{2}+y^{2}=c^{2} \\\\ \dfrac {x^{2}}{a^{2}}+ \dfrac {y^{2}}{b^{2}}=1\end{cases}$, and rearranging, we get $x^{2}=(2c^{2}-a^{2})\cdot \dfrac {a^{2}}{c^{2}}\geqslant 0$.
Solving for $e$, we find $e\geqslant \dfrac { \sqrt {2}}{2}$, and since $0 < e < 1$,
$\therefore \dfrac { \sqrt {2}}{2}\leqslant e < 1$.
Therefore, the correct choice is: $\boxed{B}$.
By setting point $P(x,y)$ and using $PF_{1} \perp PF_{2}$ to get $x^{2}+y^{2}=c^{2}$, and combining it with the equation of the ellipse, we can find the range of the eccentricity of the ellipse.
This question tests the method of finding the eccentricity of an ellipse, which is a medium-level problem. When solving, it is important to carefully read the problem and flexibly apply knowledge of ellipse properties and perpendicular lines.