CEE 355 Applied Electromagnetics Study Notes

Maxwell's Equations

Point Form and Integral Form

Maxwell's equations represent the fundamental laws relating electric and magnetic fields. The two main formats are the point form and integral form. Each equation captures physical phenomena in electromagnetism:

  1. Displacement Field Equation:
    ∇⋅D=ρ\nabla \cdot D = \rho
    This equation describes how the electricity is distributed over an area, where:

    • $D$ is the electric displacement field.

    • $
      ho$ is the volume charge density.

  2. Ampère's Law with Maxwell's Addition:
    ∇×H=J+∂D∂t\nabla \times H = J + \frac{\partial D}{\partial t}
    Here, this law connects the curl of the magnetic field with electric current density $J$ and changing electric field, represented by displacement current.

    • $H$ is the magnetic field intensity.

  3. Faraday’s Law of Induction:
    ∇×E=−∂B∂t\nabla \times E = -\frac{\partial B}{\partial t}
    This law states that changes in the magnetic field will induce an electric field.

    • $B$ is the magnetic flux density.

  4. Gauss's Law for Electricity:
    ∇⋅E=ρϵ0\nabla \cdot E = \frac{\rho}{\epsilon_0}
    This describes how electric charges produce electric fields.

  5. Gauss's Law for Magnetism:
    ∇⋅B=0\nabla \cdot B = 0
    This states that there are no magnetic monopoles; magnetic field lines are closed loops.

Integral Form

The integral form of Maxwell’s equations is used for calculations leading to field interactions over a bounded region.

  1. Displacement Field:
    ∫<em>SD⋅dS=∫</em>Vρ dV\int<em>S D \cdot dS = \int</em>V \rho \, dV

    • Where $S$ is the surface over which the relation is evaluated.

  2. Ampère's Law:
    ∫<em>CH⋅dl=I</em>enclosed\int<em>C H \cdot dl = I</em>{enclosed}

    • Where $C$ is the closed path of integration.

  3. Faraday's Law:
    ∫<em>CE⋅dl=−ddt∫</em>SB⋅dS\int<em>C E \cdot dl = -\frac{d}{dt}\int</em>S B \cdot dS

  4. Gauss's Law for Electricity:
    ∫<em>SE⋅dS=1ϵ</em>0∫Vρ dV\int<em>S E \cdot dS = \frac{1}{\epsilon</em>0} \int_V \rho \, dV

  5. Gauss's Law for Magnetism:
    ∫SB⋅dS=0\int_S B \cdot dS = 0

Electric Current and Electric Current Density

Definition and Representation

  • Current is defined as the flow of free charge, typically electrons in a conductive material such as metals where outer electrons are free to move under an electric field influence.

  • Current density, denoted as $J$, represents the distribution of current over a surface and is measured in units of A/m².

  • Mathematically defined for a surface $s$ as:
    I=∫SJ⋅dSI = \int_S J \cdot dS
    This equation implies that only the component of $J$ that is perpendicular to the surface contributes to the net current.

Relation to Electric Field

In conductive materials:
J=σEJ = \sigma E
Where:

  • $\sigma$ is the conductivity of the material (in S/m); it indicates how well the material conducts electricity.

  • $E$ is the electric field (in V/m).

Properties:

  • For many conducting materials, $J$ and $E$ are in the same direction, indicating that the materials are isotropic.

  • Conductivity is often independent of the electric field; such materials are linear.

Kirchhoff's Current Law (KCL)

  • States that the total current entering a junction must equal the total current leaving a junction, reflecting conservation of charge within a closed surface $s$. This law is typically valid for steady-state (non-time-varying) currents.

Examples and Applications

Example 1: Determining Volume Charge Density

Given electric flux density $D = 8xy \hat{a} + 4z \hat{b}$, determine:

  • Volume charge density ($\rho$):
    Use Gauss's law:
    ρ=∇⋅D=∂D<em>x∂x+∂D</em>y∂y+∂Dz∂z\rho = \nabla \cdot D = \frac{\partial D<em>x}{\partial x} + \frac{\partial D</em>y}{\partial y} + \frac{\partial D_z}{\partial z}

  • Total charge contained in the volume defined by:
    0 < x < 1, 0 < y < 1, 0 < z < 1
    Integrate the charge density over the specified volume.

Example 2: Resistance Calculation

  • Example: To determine the resistance of a 1-km length of #20 gauge copper wire:

    • Given conductivity $\sigma = 5.8 \times 10^7 \text{ S/m}$ and the area of the wire can be calculated using its gauge.

    • Apply Ohm's law:
      R=LσAR = \frac{L}{\sigma A}
      Where:

    • $R$ = resistance,

    • $L$ = length of wire,

    • $A$ = cross-sectional area.

Example 3: Capacitor with Lossy Dielectric

  • For a parallel-plate capacitor with area $A = 100 cm^2$, separated by distance $d = 5 mm$ and filled with a dielectric of conductivity $\sigma = 10 \text{ S/m}$, calculate the resistance:
    R=dσAR = \frac{d}{\sigma A}

Example 4: Coaxial Cable Resistance

  • The per-unit-length resistance of a coaxial cable with inner radius $a$ and outer radius $b$ filled with dielectric of conductivity $\sigma$ is described by:
    Rper−unit=12πσln⁡(ba)R_{per-unit} = \frac{1}{2 \pi \sigma} \ln \left( \frac{b}{a} \right)
    Evaluated for typical dimensions: $a = 16$ mils, $b = 58$ mils.

Example 5: Magnetic Field of a Wire

  • Using Ampère's Law, if a wire carries a current of 10,000 Amperes, to find the magnetic field at a distance of 100m:
    B=μ<em>0I2πrB = \frac{\mu<em>0 I}{2 \pi r} For $\mu0 = 4 \pi \times 10^{-7} \text{ T m/A}$.

Example 6: Magnetic Flux Density Calculation

  • For a 5-A current flowing in a wire along the z-axis, calculate at point $(2m, 0, 0)$ using:
    B=μ0I2πrB = \frac{\mu_0 I}{2 \pi r}

Summary of Key Formulas

  1. Ohm's Law:
    J=σEJ = \sigma E

  2. Resistance:
    R=LσAR = \frac{L}{\sigma A}

  3. Magnetic Field From a Wire:
    B=μ0I2πrB = \frac{\mu_0 I}{2 \pi r}

  4. Gauss's Law for Electricity:
    ∫<em>SE⋅dS=1ϵ</em>0∫Vρ dV\int<em>S E \cdot dS = \frac{1}{\epsilon</em>0} \int_V \rho \, dV

  5. Ampère's Law:
    ∫<em>CH⋅dl=I</em>enclosed\int<em>C H \cdot dl = I</em>{enclosed}

These notes encompass the basics of applied electromagnetics covering Maxwell's equations, electrics currents, and examples relevant to core concepts.