# 物理代写|电磁学代写electromagnetism代考|PHYC20014

## 物理代写|电磁学代写electromagnetism代考|Quasi-Static Models

More general approximate models can be obtained by discriminating the time variations, respectively, of the electric field and the magnetic induction. Hence, after the scaling step in Maxwell’s equations in vacuum, that is, in Eqs. (1.107-1.110), if we suppose that
$$\bar{v} \overline{\bar{F}} \ll 1 \quad \text { and } \quad \frac{\bar{v}}{c} \frac{\bar{E}}{c \bar{B}} \approx 1,$$

we easily obtain that we may neglect the time derivative $\partial_t \boldsymbol{B}$ in Faraday’s law, whereas the coefficient of the time derivative $\partial_t \boldsymbol{E}$ in Ampère’s law is comparable to one. We then obtain the electric quasi-static model, which can be written in the physical variables $\boldsymbol{E}, \boldsymbol{B}$ as
\begin{aligned} &\operatorname{curl} \boldsymbol{E}=0, \ &\operatorname{div} \boldsymbol{E}=\frac{1}{\varepsilon_0} \varrho, \ &\operatorname{curl} \boldsymbol{B}=\mu_0 \boldsymbol{J}+\frac{1}{c^2} \frac{\partial \boldsymbol{E}}{\partial t}, \ &\operatorname{div} \boldsymbol{B}=0 . \end{aligned}

## 物理代写|电磁学代写electromagnetism代考|Darwin Model

Let us introduce another approximate model, also known as the Darwin model [90]. It consists in introducing a Helmholtz decomposition of the electric field as
$$\boldsymbol{E}=\boldsymbol{E}^L+\boldsymbol{E}^T$$
where $\boldsymbol{E}^L$, called the longitudinal part, is characterized by curl $\boldsymbol{E}^L=0$, and $\boldsymbol{E}^T$, the transverse part, is characterized by div $\boldsymbol{E}^T=0$. Starting from Maxwell’s equations in vacuum, one then assumes that $\varepsilon_0 \partial_t \boldsymbol{E}^T$ can be neglected in Ampère’s law: one neglects only the transverse part of the displacement current, whereas, in the quasi-static model, the total displacement current $\varepsilon_0 \partial_t \boldsymbol{E}$ is neglected. In this sense, it is a more sophisticated model than the quasi-static one. Moreover, it can be proven (see Sect. 6.4), by using the low frequency approximation (1.111) and the resulting dimensionless form of Maxwell’s equations, that this model yields a second-order approximation of the electric field and a first-order approximation of the magnetic induction.
The Darwin model in vacuum is written in the physical variables $\boldsymbol{E}, \boldsymbol{B}$ as
\begin{aligned} &\operatorname{curl} \boldsymbol{E}=-\frac{\partial \boldsymbol{B}}{\partial t}, \quad \operatorname{div} \boldsymbol{E}=\frac{\varrho}{\varepsilon_0}, \ &\operatorname{curl} \operatorname{curl} \boldsymbol{B}=\mu_0 \operatorname{curl} \boldsymbol{J}, \quad \operatorname{div} \boldsymbol{B}=0 . \end{aligned}
Then, if one uses the Helmholtz decomposition (1.120) with $\operatorname{div} \boldsymbol{E}^T=0$ and $\boldsymbol{E}^L=-\operatorname{grad} \phi$, we see that the three fields $\boldsymbol{B}, \boldsymbol{E}^T$ and $\phi$ solve three elliptic PDEs, namely (1.121) and
\begin{aligned} &-\Delta \phi=\frac{\varrho}{\varepsilon_0}, \ &\operatorname{curl} \boldsymbol{E}^T=-\frac{\partial \boldsymbol{B}}{\partial t}, \quad \operatorname{div} \boldsymbol{E}^T=0 . \end{aligned}

## 物理代写|电磁学代写electromagnetism代考|Quasi-Static Models

$$\bar{v} \overline{\bar{F}} \ll 1 \quad \text { and } \quad \frac{\bar{v}}{c} \frac{\bar{E}}{c \bar{B}} \approx 1,$$

$$\operatorname{curl} \boldsymbol{E}=0, \quad \operatorname{div} \boldsymbol{E}=\frac{1}{\varepsilon_0} \varrho, \operatorname{curl} \boldsymbol{B}=\mu_0 \boldsymbol{J}+\frac{1}{c^2} \frac{\partial \boldsymbol{E}}{\partial t}, \quad \operatorname{div} \boldsymbol{B}=0$$

## 物理代写|电磁学代写electromagnetism代考|Darwin Model

$$\boldsymbol{E}=\boldsymbol{E}^L+\boldsymbol{E}^T$$

$$\operatorname{curl} \boldsymbol{E}=-\frac{\partial \boldsymbol{B}}{\partial t}, \quad \operatorname{div} \boldsymbol{E}=\frac{\varrho}{\varepsilon_0}, \quad \operatorname{curl} \operatorname{curl} \boldsymbol{B}=\mu_0 \operatorname{curl} \boldsymbol{J}, \quad \operatorname{div} \boldsymbol{B}=0 .$$

$$-\Delta \phi=\frac{\varrho}{\varepsilon_0}, \quad \operatorname{curl} \boldsymbol{E}^T=-\frac{\partial \boldsymbol{B}}{\partial t}, \quad \operatorname{div} \boldsymbol{E}^T=0$$

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