Rowan-Classes/5th-Semester-Fall-2023/EEMAGS/Notes/Chapter2.md

Chapter 2

Maxwell's Equations in Differential Form

\int \vec{E} \cdot d\vec{s} = \frac{Q_{enc}}{\varepsilon_0} \xleftrightarrow{\text{divergence}} \nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0} \int \vec{B} \cdot d\vec{s} = 0 \xleftrightarrow{} \nabla \cdot \vec{B} = 0 \int \vec{E} \cdot d\vec{l} = -\frac{d}{dt}\vec{B} \xleftrightarrow{\text{stokes}} \nabla \times \vec{E} = -\frac{d}{dt}\vec{B} \int \vec{B} \cdot d\vec{l} = \mu_0\left( \vec{J} + \varepsilon_0 \frac{d\vec{E}}{dt}\right) \xleftrightarrow{} \nabla \times \vec{B} = \mu_0 J + \mu_0 \varepsilon_0 \frac{\partial \vec{E}}{\partial t}

Before Maxwell (only true for magnetostatics):

\nabla \times \vec{B} = \mu_0\vec{J} \nabla \cdot (\nabla \times \vec{B}) = \mu_0(\nabla \cdot \vec{J}) \nabla \cdot \vec{J} = 0

For changing magnetic fields:

\nabla \cdot \vec{J} = -\frac{\partial}{\partial t} \rho_v \nabla \cdot \vec{J} + \frac{\partial}{\partial t} \rho_v = 0

Example 2.1

\vec{J} = e^{-x^2}\hat{x}

Find the time rate of change of charge densisty at x=1:

\nabla \cdot \vec{J} = -\frac{\partial}{\partial t} \rho_v \frac{\partial \rho}{\partial t} = -\nabla \cdot \vec{J} = -\frac{d}{dx}e^{-x^2} = 2xe^{-x^2} \therefore 2 e^{-1}

Example 2.2

For some spherical current density:

\vec{J} = \frac{J_0 e^{-t/\tau}}{\rho}\hat{\rho}

Find the total current that leaves the surface of radius, t = \tau:

I = \int \vec{J} \cdot d\vec{s} = 4\pi a^2 \left(\frac{J_0 e^{-t/\tau}}{a}\right) I = 4\pi a J_0 e^{-1}

Find \rho_v(\rho, t):

\nabla \cdot \vec{J} = -\frac{\partial \rho}{\partial t} = \frac{1}{\rho^2} \frac{\partial}{\partial \rho}(\rho^2 J)

Plug in for J:

\nabla \cdot \vec{J} = \frac{1}{\rho^2} \frac{\partial}{\partial \rho}\left(\rho^2 \frac{J_0 e^{-t/\tau}}{\rho}\right) = \frac{J_0}{\rho^2} e^{-t/\tau} = -\frac{\partial}{\partial t} \rho_v

Solve for \rho_v:

\rho_v = \int -\frac{J_0}{\rho^2}e^{-t/\tau} dt \rho_v = \frac{J_0}{\rho^2} e^{-t/\tau} \tau

Wave Propagation

\nabla \times \vec{E} = -\frac{\partial}{\partial t} \vec{B} \nabla \times \vec{B} = \mu_0 (\vec{J} + \varepsilon_0 \frac{\partial}{\partial t} \vec{E}) \nabla \cdot \vec{E} = \frac{\rho_v}{\varepsilon_0} \nabla \cdot \vec{B} = 0

Take the curl of the first equation:

\nabla \times (\nabla \times \vec{E}) = -\frac{\partial}{\partial t}(\nabla \times \vec{B}) \nabla (\nabla \cdot \vec{E}) - \nabla^2\vec{E} = -\frac{\partial}{\partial t}(\nabla \times \vec{B})

Substitute the second equation into the one directly above:

\nabla(\nabla \cdot \vec{E}) - \nabla^2\vec{E} = -\frac{\partial}{\partial t} \left( \mu_0 \vec{J} + \mu_0\varepsilon_0 \frac{\partial}{\partial t}\vec{E} \right)

Homogenious vector wave for $\vec{E}$-fields

Lets say we are considering wave propagation in a source free region (\rho_v =0).

\nabla \cdot \vec{E} = \frac{\rho_v}{\varepsilon_0} =0 \vec{J} = -\frac{\partial}{\partial t} \rho_v = 0

Now we have:

-\nabla^2\vec{E} = -\mu_0\varepsilon_0 \frac{\partial}{\partial t} \vec{E} \therefore \nabla^2\vec{E}-\mu_0\varepsilon_0 \frac{\partial}{\partial t}\vec{E} = 0

Homogeneous vector wave for $\vec{B}$-fields

\nabla^2\vec{B} - \mu_0\varepsilon_0 \frac{\partial^2}{\partial t^2}\vec{B} = 0

Phasor representations

In general, all fields and their sources vary as a function of position and time. If the time variations are sinusoidal, with angular frequency, \omega, then each of their quantities can be represented by a time independent phasor.

\vec{E}(x,y,z,t) = \Re\{\vec{E}(x,y,z)e^{j \omega t}\} \vec{E}(r,t) = \Re\{\hat{E}(r) e^{j \omega t}\} \vec{B}(r,t) = \Re\{\hat{B}(r) e^{j \omega t}\}

\hat{E} and \hat{B} are the complex time variations in vector form. (\hat{E} = \hat{\vec{E}}) .

If \hat{E}(r) = E_0 e^{j \theta},

\vec{E}(r, t) = \Re\{E_0 e^{j \theta} e^{j \omega t}\} = E_0\cos(\omega t + \theta)

Consider a plane wave in the x-direction only (E_y = E_z = 0), and does ont vary the x or y direction (\frac{\partial}{\partial x}\hat{E} = \frac{\partial}{\partial y}\hat{E} = 0). Assume free space for propagation (\hat{J} = \hat{\rho}_v = 0).

\nabla \times \vec{E} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \hat{E}_x & \hat{E}_y & \hat{E}_z \end{vmatrix} = -j\omega (\hat{B}_x\hat{x} + \hat{B}_y \hat{y} + \hat{B}_z \hat{z} \therefore \begin{Bmatrix} -\frac{\partial}{\partial z}\hat{E} = -j\omega\hat{B}_x \\ \frac{\partial}{\partial z}\hat{E} = -j\omega\hat{B}_y \\ 0 = -j\omega\hat{B}_z \end{Bmatrix}

Similarly, Ampere's Law:

\begin{Bmatrix} -\frac{\partial \hat{B}_y}{\partial z} = j\omega\varepsilon_0\mu_0\hat{E}_x \\ \frac{\partial \hat{B}_x}{\partial z} = j\omega\varepsilon_0\mu_0\hat{E}_y \\ 0 = j\omega\varepsilon_0\mu_0\hat{E}_z\end{Bmatrix}

Therefore, B_x and E_y are related to each other. B_x acts as the source for generating E_y, and E_y acts as the source for generating B_x. Solving for \hat{E}_x and \hat{B}_y:

\frac{\partial^2 \hat{E}_x}{\partial z^2} = -j\omega\frac{\partial \hat{B}_y}{\partial z} = -\omega^2\mu_0\varepsilon_0\hat{E}_x \therefore \frac{\partial^2}{\partial z^2}\hat{E}_x + \mu_0\varepsilon_0\omega^2\hat{E}_x = 0

The general solution:

\hat{E}_x = \hat{C}_1 e^{-j \beta_0 z} + \hat{C}_2 e^{j \beta_0 z}

Where \hat{C}_1 and \hat{C}_2 are complex.

\therefore \hat{E}_x = \hat{E}_m^+ e^{-j \beta_0 z} + \hat{E}_m^- e^{j \beta_0 z}

Where E_m^+ and E_m^- are wave amplitudes (volts per meter).

The phase constant:

\beta_0 = \omega \sqrt{\mu_0 \varepsilon_0} = \frac{2\pi}{\lambda} = \frac{2\pi f}{C}

Real time form of the solution:

E_x(z,t) = \Re\{\hat{E}_x e^{j \omega t}\} = \Re\{E_m^+ e^{j(\omega t - \beta_0 z)} + E_m^- e^{j(\omega t + \beta_0 z)}\} \therefore E_x(z,t) = E_m^+ \cos(\omega t - \beta_0 z) + E_m^- \cos(\omega t + \beta_0 z)

If E_m^+ = E_m^-, the magnitude and direction is E_0.

\therefore E_x(z,t) = E_0\cos(\omega t \pm \beta_0 z)

Where:

\omega = 2\pi f

v_0 is the phase velocity

The phase velocity

v_0 = \frac{1}{\sqrt{\mu \varepsilon}}

In a vacuum:

v_0 = c

Also

\hat{B}_y = \frac{\hat{E}_x}{c}

Common \vec{E} and \vec{B} field ratio is:

\mu_0 \hat{H}_y = \frac{\hat{E}_x}{c} \therefore \frac{\hat{E}_x}{\hat{H}_y} = \mu_0 c = \frac{\mu_0}{\sqrt{\mu_0 \varepsilon_0}} = \sqrt{\frac{\mu_0}{\varepsilon_0}} = \eta_0 \approx 120\pi = 377\Omega

Where \eta_0 is the intrinsic wave impedance.

\therefore \hat{E}_x = \hat{H}_y \eta_0 \therefore \hat{H}_y = \frac{\hat{E}_x}{\eta_0}

Therefore, both \vec{H} and \vec{E} fields are in phase.

In general:

\hat{H} = \frac{\hat{k}}{\eta_0} \times \hat{E} \hat{E} = -\eta_0 \hat{k} \times \hat{H}

Example 2.3

\vec{E} = 50\cos(10^8t + \beta_0 x)\hat{y} \text{[v/m]} E_y = E_0\cos(\omega t + \beta x)

E_y propagates in the -\hat{x} direction.

\beta_0 = \omega \sqrt{\mu_0 \varepsilon_0} = \frac{\omega}{c} = \frac{10^8}{3 \times 10^8} = \frac{1}{3}

What is the time it takes to travel a distance of \frac{\lambda}{2}?

\omega = 2\pi f = \frac{2\pi}{T} T = \frac{2\pi}{\omega} \therefore t_{\lambda/2} = \frac{T}{2} = \frac{\pi}{\omega} = \frac{\pi}{10^8} \approx 31.42\text{[ns]}

The refractive index of a medium is given by the ratio of c and v_p:

c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} v_p = \frac{1}{\sqrt{\mu \varepsilon}} n = \frac{c}{v_p} = \sqrt{\frac{\mu \varepsilon}{\mu_0 \varepsilon_0}} = \sqrt{\mu_r \varepsilon_r}

For a non-magnetic material:

\mu_r \approx 1 \therefore n = \sqrt{\varepsilon_r}

Polarization

$$\begin{rcases} \nabla \cdot \vec{E} = {\rho \over \varepsilon_0} \ \nabla \cdot \vec{B} = 0 \end{rcases} \text{Gauss}$$

$$\begin{rcases} \nabla \times \vec{E} = -{\partial \over \partial t} \vec{B} \end{rcases} \text{Faraday}$$

$$\begin{rcases} \nabla \times B = \mu_0 \vec{J} + \mu_0 \varepsilon_0{\partial \over \partial t}\vec{E} \end{rcases} \text{Ampere}$$

Polarization of a uniform plane wave describes the shape and olcus of the tip of the $\vec{E}$-field vector in the plane orthoganal to the direction of propagation. There are two pairs of \vec{E} and \vec{B} fields (\hat{E}_x, \hat{H}_y) and (\hat{E}_y, \hat{H}_x). When both pairs are present, we can evaluate polarization of plane waves. The $\vec{E}$-field has components in the \hat{x} and \hat{y} directions and travels in \hat{z}.

\hat{E} = (\hat{E}_x \hat{x} + \hat{E}_y \hat{y})e^{-j \beta z} \hat{E}_x = \lvert \hat{E}_{x_0} \rvert e^{-j \beta a} \hat{E}_y = \lvert \hat{E}_{y_0} \rvert e^{-j \beta b}

Locus

The shape the tip of the $\vec{E}$-field vector traces out while in motion.

Phase

Typically defined relative to a reference point such as z=0 or t=0 or some combination.

Polarization Characteristics

Linear polarization

\hat{E}_x and \hat{E}_y have the same phase angle: a = b, so the x and y components of the $\vec{E}$-field will be in phase.

\hat{E} = (\lvert \hat{E}_x \rvert \hat{x} + \lvert \hat{E}_y \rvert \hat{y} ) e^{-j( \beta z - a)}

In real time:

\hat{E} = (\lvert \hat{E}_x \rvert \hat{x} + \lvert \hat{E}_y \rvert \hat{y} ) \cos(\omega t - \beta z + a)

As the wave continues to propagate in the \hat{z} direction, the $\vec{E}$-field vector maintains its direction with angle, \theta, with respect to the y-axis.

\tan(\theta) = {\lvert \hat{E}_x \rvert \over \vert \hat{E}_y \rvert}

When z=0, the \vec{E} field is given by:

\vec{E}(0, t) = (\lvert \hat{E}_x \rvert \hat{x} + \lvert \hat{E}_y \rvert \hat{y} ) \cos(\omega t + a) \vec{E}(0, 0) = (\lvert \hat{E}_x \rvert \hat{x} + \lvert \hat{E}_y \rvert \hat{y} ) \cos(a)

Elliptical Polarization

\hat{E}_x and \hat{E}_y have different phase angles. \vec{E} is no longer in one plate.

\hat{E} = \hat{x} \lvert \hat{E}_x \rvert e^{j (a - \beta z)} + \hat{y} \lvert \hat{E}_y \rvert e^{j(b - \beta z)}

Where:

E_x = \lvert \hat{E}_x \rvert \cos(\omega t - a + \beta z)

E_y = \lvert \hat{E}_y \rvert \cos(\omega t - b + \beta z)

If a=0 and b = {\pi \over 2}:

\vec{E}_x(z,t) = \lvert \hat{E}_x \rvert \cos(\omega t - a + \beta z) \vec{E}_y(z,t) = \lvert \hat{E}_y \rvert \cos(\omega t - b + \beta z)

Circular Polarization

\hat{E}_x and \hat{E}_y have the same magnitude with a phase angle difference of {\pi \over 2}.

Example

Determine the real-valued $\vec{E}$-field.

\hat{E}(z) = -3j\hat{x} e^{-j\beta z} \vec{E}(z, t) = -3j\hat{x} e^{-j \beta z} e^{j\omega t} 3 \hat{x} e^{-j \beta z} e^{j \omega t} e^{-\pi \over 2} 3 \cos\left(\omega t- \beta z - {\pi \over 2}\right)

Example

What are E_x and E_y

\hat{E}(z) = (3\hat{x} + 4\hat{y})e^{j\beta z} \vec{E}(z, t) = (3\hat{x} + 4\hat{y})e^{j\beta z}e^{j\omega t} E_x = 3\cos(\omega t + \beta z) E_y = 4\cos(\omega t + \beta z)

Example

\hat{E}(z) = (-4\hat{x} + 3\hat{y})e^{-j\beta z} \hat{E}(z, t) = (-4\hat{x} + 3\hat{y})e^{-j\beta z}e^{j\omega t} E_x = 4\cos(\omega t - \beta z + \pi) E_y = 3\cos(\omega t - \beta z)

Non-Sinusoidal Waves

Analytical solution of a 1-D traveling wave.

{\partial^2 \over \partial z^2}\vec{E} - {1\over c^2}{\partial^2 \over \partial t^2}\vec{E} = 0

D'Alemberts Solution

\vec{E}(z, t) = E(z - ct) + E'(z + ct)

Show that the function, F(z - ct) = F_0 e^{-(z+ct)^2} is a solution of the wave equation.

Let \gamma = z - ct, and {\partial \gamma \over \partial z} = 1:

F(z-ct) = F_0 e^{-\gamma^2} F'(z + ct) = 0 {\partial F \over \partial z} = {\partial F \over \partial \gamma} {\partial \gamma \over \partial z} = F_0 e^{-\gamma^2}(-2\gamma) {\partial^2 F \over \partial z^2} = {\partial \over \partial z}\left({\partial F \over \partial \gamma}{\partial \gamma \over \partial z}\right) G = \left({\partial F \over \partial \gamma}{\partial \gamma \over \partial z}\right)

Don't forget chain rule:

{\partial G \over \partial \gamma}{\partial \gamma \over \partial z} = {\partial \over \partial z}\left({\partial F \over \partial \gamma}{\partial \gamma \over \partial z}\right){\partial \gamma \over \partial z} {\partial \over \partial \gamma} - 2\gamma F_0 e^{-\gamma^2} = -2F_0 {\partial \over \partial \gamma} \gamma e^{-\gamma^2}

By product rule:

-2F_0 (\gamma (-2\gamma e^{-\gamma^2}) + e^{-\gamma^2}) F_0 (4\gamma^2 e^{-\gamma^2} - 2e^{-\gamma^2}) = F_0(-2 + 4\gamma^2)e^{-\gamma^2} ={\partial^2 \over \partial \gamma^2}F \left({\partial \gamma \over \partial z}\right)^2 \therefore {\partial^2 \over \partial t^2}F = {\partial^2 F \over \partial \gamma^2} \left({\partial \gamma \over \partial t}\right)^2 = F_0(-2 + 4\gamma^2)e^{-\gamma^2}(C^2)

This solution satisfies:

{\partial^2 \over \partial z^2}\vec{E} - {1\over c^2}{\partial^2 \over \partial t^2}\vec{E} = 0