Why is electromagnetic induction a quasistatic approximation? 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?
 A: In magnetostatics, the calculation of the magnetic field does not depend on the time-varying electric field (as you already mention). According to Faradays law, the time varying magnetic induction creates a time-varying electric field. Here, it is assumed that this induced time-varying electric field is too slow to contribute to the magnetic induction.
Considering the full Maxwell equations, without the magnetotatic approximation, we end up in two coupled differential equations for your problem
$$\nabla \times \mathbf{E}_\mathrm{indued} = -\frac{\partial\mathbf{B}}{\partial dt}$$
$$\nabla \times \mathbf{H} = \mathbf{J} + \epsilon_0 \frac{\partial \mathbf{E}_\mathrm{indued}}{\partial t}\,.$$
Hence, the induced electric field alters the magnetic field, which again alters the induced electric field, and so on. These are two coupled differential equations where the electric and magnetic fields depend on each other. Using Faradays law, the magnetostatic approximation is used which decouples the fields and simplifies the calculation.
