If $\mathcal{F}_b$ is an *invertible* sheaf and $B=\textrm{Spec} \mathbb{C}[\epsilon]/(\epsilon^2)$ (first-order deformations) then the obstruction theory for deforming $\mathcal{F}_b$ described in Angelo's answer becomes very explicit. Indeed, there is the following result, whose proof can be found in Sernesi's book "Deformation of algebraic schemes", p. 147.

Set $\mathcal{L}:=\mathcal{F}_b$, $X:=X_b$.

**THEOREM** Given a first-order deformation $\xi$ of $X$, there is a first-order deformation of $\mathcal{L}$ along $\xi$ if and only if

$\kappa(\xi) \cdot c(\mathcal{L})=0$.

Here $\kappa$ is the Kodaira-spencer map, $c$ is the first Chern class and "$\cdot$" denotes the composition

$H^1(X, T_X) \times H^1(X, \Omega^1_X) \to H^2(X, T_X \otimes \Omega^1_X) \to H^2(X, \mathcal{O}_X)$,

where the first arrow is induced by the cup-product and the second one by the duality pairing $T_X \otimes \Omega^1_X \to \mathcal{O}_X$.

Using this, one can prove for instance that if $X$ is an Abelian variety of dimension $g$ and $\mathcal{L}$ is an ample line bundle, then $\mathcal{L}$ extends along a subspace of $H^1(X, T_X)$ of dimension $g(g+1)/2$. For $g \geq 2$, $\mathcal{L}$ does not extend to the whole of $H^1(X, T_X)$ (not even to the first-order!), since the general deformation of $X$ is not projective.