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Explains how devices work, and how magnetic waves propagate. Maxwell’s equation tell you about divergence and curl of these fields.

This is not detailed enough to be really understandable, but hopeful makes you a bit curious about physics 8.02.

## Electric / Magnetic fields

The electric field \(\vec E\) is a vector field that tells you wat force will be exerted on a charged particle. It is responsible for the flow of electrons when you have a voltage difference, because it is a gradient of a potential (=electric voltage). The force $$ \vec F = q\,\vec E \nonumber $$

A magnetic field is a vector field that deflects the trajectory of a moving charged particle and make it rotate. The force $$ \vec F = q\,\vec v \times \vec B \nonumber $$

### Electric field

#### Gauss-Coulomb law

\(\rho\) here is electric charge density (=charge per unit volume), and \(\epsilon_0\) is a physical constant $$ \nabla\cdot\vec E = \frac{\rho}{\epsilon_0} \label{eq:max1a} $$

This says: the divergence of \(\vec E\) is caused by the electric charge.

This is a partial differential equation satisfied by the electric charge. Not very intuitive. What is more intuitive is what we get when we apply the divergence theorem

To find the flux of the electric field out of the closed surface \(S\), using equation \(\eqref{eq:max1a}\) $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \oiint_S \vec E\cdot d\vec S &= \iiint_D \rm{div}(\vec E)\,dV \\ &= \frac{1}{\epsilon_0} \iiint_D \rho\,dV \end{align*} \nonumber $$

If we now integrate the charge density of the entire regions, I get the total electric charge \(Q\) in the region \(D\) $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \oiint_S \vec E\cdot d\vec S &= \frac{Q}{\epsilon_0} \end{align*} \nonumber $$

It tells us how electric charges influence the electric field around them.

One application is capacitors that store energy using a voltage between two plates. The voltage is obtained by integrating the electric field from one plate to the other plate. The charges in the plates are what causes the electric field between the plates. That gives you a relationship between voltage and charge in capacitor. (refer to Physics 802.2 Electricity and Magnetism)

#### Faraday’s law

The curl of the electric field (a gradient field) is \(0\), but you can create voltage using magnetic fields. You have a failure of conservativity of the electric force if you have a magnetic field. $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \nabla\times\vec E = -\pdv{\vec B}{t} \label{eq:max2a} $$

Again, it is a strange partial differential equation relating these vector fields. To make sense of it, apply Stokes theorem to compute the work done y the electric field around a closed curve.

That means if we have a wire in there, and you want to find the voltage along the wire generated by the varying magnetic field.

An application is a transformer. It passes an input voltage through a loop of wire. Another loop of wire is intertwined. The magnetic field \(\vec C\) generated by alternating current, varies over time. That causes curl of the electric field \(nabla\times\vec F\), what will generate voltage over the intertwined loop.

The intertwined loop is a closed curve. The voltage generated in this circuit $$ \oint_C \vec E\cdot\,d\vec r = \iint_S (\nabla\times\vec E)\cdot d\vec S \nonumber $$

Combined with equation \(\eqref{eq:max2a}\) $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \oint_C \vec E\cdot\,d\vec r = \iint_S \left(-\pdv{\vec B}{t}\right) \cdot d\vec S \nonumber $$

If you don’t move the loop, and only the magnetic field changes then you can take the \(\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \pdv{\vec B}{t}\) out of the integral.

It says, if the magnetic field changes over time, out of nowhere, it creates an electric field.

### Magnetic field

#### div

The divergence of the magnetic field is \(0\) $$ \nabla\cdot\vec B = 0 $$

That is fortunate, otherwise you would run into trouble trying to understand surface independence when you apply Stokes’ theorem in here.

The curl of the magnetic field is caused by motion of charged particles $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \nabla\times\vec B = \mu_0\,\vec J + \epsilon_0\,\mu_0\pdv{\vec E}{t} $$

Here

- \(J\) is the vector current density. It measures the flow of electrically charged particles.
- \(\mu_0\) is a physics constant

This is how the transformer generates it magnetic field.