# 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?

• in quasistatic approximation $|\partial \mathbf {D}/\partial t| << |\mathbf {J}|$, hence $curl \mathbf {B} \approx \mu_0 \mathbf {J}$ and ignore the problem that $div \mathbf {J}$ is not necessarily zero everywhere (e.g., in a capacitor). In quasistatics, Faraday's law of induction is valid. Commented Aug 4, 2020 at 8:59
• @hyportnex Yes i understand that the quasistatic approximation in Maxwell's equation is: $$\nabla \times \mathbf{B} = \mu_0 \left(\mathbf{J} + \frac{\partial \mathbf{E}}{\partial t} \right) \approx \mu_0 \mathbf{J}$$ In other words, it is to prevent couping between the $\mathbf{E}$ and $\mathbf{B}$. However in the case of $\mathbf{E}_{induced}$, I do not see how this approximation applies here. For example, if we consider a straight wire carrying a time varying current $I(t)$, then the "quasistatic approximation" of the magnetic field at a distance $a$ from the wire would be Commented Aug 4, 2020 at 9:10
• $\mathbf{B} = \dfrac{\mu_0 I}{2 \pi a}$ (Computed using an Amperian Loop). But where in this scenario did we make the approximation $\dfrac{\partial \mathbf{E}}{\partial t} \approx 0$? Commented Aug 4, 2020 at 9:10
• as i said there is no approximation on Faraday's law but we do assume that $\epsilon \partial |E|/\partial t << |J|$ where $|J| \approx I/A$ and A is then a typical (characteristic) "area" and $\epsilon$ is a characteristic permittivity of the system where quasistatics is assumed, say $A$ is a capacitor plate. Commented Aug 4, 2020 at 11:40

Considering the full Maxwell equations, without the magnetotatic approximation, we end up in two coupled differential equations for your problem $$\nabla \times \mathbf{E}_\mathrm{indued} = -\frac{\partial\mathbf{B}}{\partial dt}$$ $$\nabla \times \mathbf{H} = \mathbf{J} + \epsilon_0 \frac{\partial \mathbf{E}_\mathrm{indued}}{\partial t}\,.$$