2
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ – hyportnex Aug 4 '20 at 8:59
  • $\begingroup$ @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 $\endgroup$ – D. Soul Aug 4 '20 at 9:10
  • $\begingroup$ $\mathbf{B} = \dfrac{\mu_0 I}{2 \pi a}$ (Computed using an Amperian Loop). But where in this scenario did we make the approximation $\dfrac{\partial \mathbf{E}}{\partial t} \approx 0$? $\endgroup$ – D. Soul Aug 4 '20 at 9:10
  • $\begingroup$ 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. $\endgroup$ – hyportnex Aug 4 '20 at 11:40
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.