7. In the Cartesian coordinate system $x O y$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $x^{2}-\frac{y^{2}}{b^{2}}=1(b>0)$, respectively. A tangent line is drawn from point $F_{1}$ to the circle $x^{2}+y^{2}=1$, intersecting the left and right branches of the hyperbola at points $A$ and $B$, respectively. If $F_{2} B=A B$, then the value of $b$ is $\qquad$
7. $1+\sqrt{3}$.

From the given and the definition of the hyperbola, we have
$$
\begin{array}{l}
A F_{1}=B F_{1}-A B=B F_{1}-B F_{2}=2, \\
A F_{2}=2+A F_{1}=4 .
\end{array}
$$

As shown in Figure 3, let $A B$ be tangent to the circle at point $T$.
Then, in the right triangle $\triangle O T F_{1}$,
$$
O T=1, O F_{1}=c, T F_{1}=b \text {. }
$$

Thus, $\cos \angle F_{2} F_{1} A=\frac{b}{c}$.
In $\triangle A F_{1} F_{2}$, by the cosine rule, we have
$$
\begin{array}{l}
\cos \angle F_{2} F_{1} A=\frac{F_{1} F_{2}^{2}+A F_{1}^{2}-A F_{2}^{2}}{2 F_{1} F_{2} \cdot A F_{1}} \\
=\frac{c^{2}-3}{2 c} .
\end{array}
$$

Thus, $c^{2}-3=2 b$.
Also, from $c^{2}=1+b^{2}$, we get
$$
b^{2}-2 b-2=0
$$
$\Rightarrow b=1 \pm \sqrt{3}$ (the negative value is discarded).