From Griffiths, Faraday's law is given by: $$ \oint_C \mathbf{E}_{induced} \cdot d\mathbf{l} = - \iint_S \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{a} = - \frac{d \Phi}{dt} $$
On page 323, it states that if we are using this to compute the induced electric field, we are making a quasistatic assumption by assuming that the magnetic field $\mathbf{B}$ is "static enough" to use tools from magnetostatics. For example, an Amperian loop can be used above to compute the induced electric field.
I do not quite understand this statement, why are we making a quasistatic approximation when $\dfrac{\partial \mathbf{B}}{\partial t}$ clearly indicates that the magnetic field $\mathbf{B}$ is changing?
If we compare the defining equations of a pure Faraday field: $$ \nabla \cdot \mathbf{E}_{induced} = 0 \hspace{20mm} \nabla \times \mathbf{E}_{induced} = - \frac{\partial \mathbf{B}}{\partial t} $$
to a magnetostatic field: $$ \nabla \cdot \mathbf{B} = 0 \hspace{20mm} \nabla \times \mathbf{B} = - \mu_0 \mathbf{J} $$
Clearly to use the apparatus of magnetostatics, which demands that $\mathbf{J}$ be a constant vector (since magnetostatics applies only for steady currents), in our case of the induced electric field, we need to demand $\dfrac{\partial \mathbf{B}}{\partial t}$ to be a constant. This however does NOT imply that $\mathbf{B}$ is a constant (which is assumed in a quasistatic approximation).
What am I missing here?