## Quasistatic Assumption At relatively low frequencies where $\frac{\omega}{2\pi}\ll \frac{c}{2X}$ ($\omega =$ frequency, $X =$ size of the system, $c =$ speed of light) we can assume that the speed of light is infinite: $c = \frac{1}{\sqrt{ \mu_{0}\epsilon_{0} }}\approx \infty$ This means that either $\mu_{0}=0$ or $\epsilon_{0}=0$. This leads us to one of the following: #### Electroquasistatic (EQS) assumption - electric fields dominate (capacitors, electronics, etc) - $\mu_{0}=0$ - $\frac{ \partial \vec{B} }{ \partial t } \approx 0$ #### Magnetoquasistatic (MQS) assumption - magnetic fields dominate (motors, actuators, etc) - $\epsilon_{0}=0$ - $\frac{ \partial \vec{E} }{ \partial t } \approx 0$ | | EQS | MQS | | ------- | ------------------------------------------------------------------ | --------------------------------------------------------------------------- | | E-Gauss | $\vec{\nabla}\cdot \vec{E}=\frac{\rho}{\epsilon_{0}}$ | ${} \vec{\nabla}\cdot\vec{E}=\frac{\rho}{\epsilon_{0}}$ | | M-Gauss | ${} \vec{\nabla}\cdot \vec{B}=0 {}$ | ${} \vec{\nabla}\cdot \vec{B}=0 {}$ | | Faraday | $\vec{\nabla}\times \vec{E}=0$ | ${} \vec{\nabla}\times \vec{E}=-\frac{ \partial \vec{B} }{ \partial t } {}$ | | Ampere | ${} \nabla\cdot \vec{J}+\frac{ \partial \rho }{ \partial t }=0 {}$ | $\vec{\nabla} \times \vec{B} = \mu_{0} ~\vec{J}$ | ### Solving EQS Systems Let $E = -\nabla \Phi$ [^1] where $\Phi$ is the electric potential scalar field (only differences in $\Phi$ are measurable). We cannot define a unique electric potential in MQS since $\nabla \times E\neq 0$. Then the line integral $\int _{A}^{B} E\, dl=\underbrace{ \Phi_{A}-\Phi_{B} }_{ \text{Voltage} }$ is path-independent and thus $\vec{E}$ is a conservative field. Kirchoff's Voltage Law (KVL) follows from this. Then, if $\rho=0$, we have Laplace's Equation $\nabla^{2}\Phi=0$. Otherwise, we have Poisson's Equation $\nabla^{2}\Phi=-\frac{\rho}{\epsilon_{0}}$ If we know the charge distribution, we can calculate $\phi$ using the **superposition integral**: $ \Phi = \iiint_{V'} \frac{\rho}{4\pi \epsilon_{0}\lvert \vec{r}-\vec{r}' \rvert } ~dV' $ ### Solving MQS Systems We let $\vec{B} = \nabla \times \vec{A}$ [^2] where $\vec{A}$ is the magnetic vector potential (also immeasurable). This only partially defines $\vec{A}$, since a vector field is specified by both a divergence and a curl. We choose one of the following 'gauges': - Coulomb Gauge (useful for MQS): $0 = \nabla \cdot \vec{A}$ - Lorenz Gauge (useful for dynamics): $\nabla\cdot \vec{A}=-\epsilon_{0}\mu_{0} \frac{ \partial \Phi }{ \partial t }$ The Coulomb Gauge gives us the superposition integral for magnetic vector potential: $ \vec{A}=\iiint_{V'}\frac{\mu_{0}\vec{J}(\vec{r}')}{4\pi \lvert \vec{r}-\vec{r}' \rvert } dV' $ Note that each contribution to $\vec{A}$ is parallel to the $\vec{J}$ that sources it. In 2D ($\vec{A}=A_{z}\hat{z}$), lines of constant $A_{z}$ are the $\vec{B}$-field lines. From this, we derive the **Biot-Savart Law**: $ \vec{B} = \nabla \times \vec{A} = \iiint_{V'} \frac{\mu_{0}\vec{J}(\vec{r}')\times(\vec{r}-\vec{r}')}{4\pi \lvert \vec{r}-\vec{r}' \rvert^{3} } ~dV' $ [^1]: Since $\nabla \times \vec{E}=0$ and we have the identity $\nabla \times(\nabla f) \equiv 0$ for scalar fields $f$ [^2]: Since $\nabla\cdot \vec{B}=0$ and we have the identity ${} \nabla \cdot (\nabla \times \vec{A})\equiv0 {}$ for vector fields $\vec{A}$