2. Given $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $P$ is a point on the ellipse such that $\angle F_{1} P F_{2}=90^{\circ}$. If the area of $\triangle P F_{1} F_{2}$ is 2, then the value of $b$ is $(\quad)$.
(A) 1
(B) $\sqrt{2}$
(C) $\sqrt{3}$
(D) 2
2. B.

Let $\left|P F_{1}\right|=m,\left|P F_{2}\right|=n$. Then
$$
\begin{array}{l}
m+n=2 a, \\
m^{2}+n^{2}=4 c^{2}=4\left(a^{2}-b^{2}\right), \\
\frac{1}{2} m n=2 .
\end{array}
$$

From the above three equations, we get $b^{2}=2$, i.e., $b=\sqrt{2}$.