According to Math. Res. Lett. 7 (2000), no. 1, 123–132 (https://arxiv.org/abs/math/9909052) , if $g\ge 2$ is an integer and $f(x)$ is a complex polynomial of degree $2g+1$ without multiple roots, has coefficients in a (sub)field $K$, is irreducible over $K$ and its Galois group over $K$ is the full symmetric group $\mathbf{S}_{2g+1}$ then the jacobian $J(C_f)$ of the genus $g$ hyperelliptic curve $$C_f: y^2=f(x)$$ has no nontrivial endomorphisms, i.e., its endomorphism ring $\mathrm{End}(J(C_{f}))$ coincides with the ring of integers $\mathbb{Z}$; in particular, its Picard number is $1$.

For example, let $t_1, \dots t_{2g+1}$ be complex numbers that are algebraically independent over the field $\mathbb{Q}$ of rational numbers. Let us consider the field $L=\mathbb{Q}(t_1, \dots, t_{2g+1})$ generated by all $t_i$'s and let $K$ be its subfield of symmetric functions in $t_1, \dots t_{2g+1}$. Then $$f_t(x):=\prod_{i=1}^{2g+1}(x-t_i) \in K[x]$$
and the Galois group of $f(x)$ over $K$ is $\mathbf{S}_{2g+1}$. This implies that $\mathrm{End}(J(C_{f_t}))=\mathbb{Z}$ and the Picard number of $J(C_{f_t})$ is $1$.