Skip to content
EC-504 (A) · ELECTROMAGNETIC THEORY/Quick Revision Short Notes

ELECTROMAGNETIC THEORY (EC-504 (A)) - Unit 5 Short Notes

UNIT 5: MAXWELL'S EQUATIONS AND ELECTROMAGNETIC WAVES


Vector Differential Calculus in Electromagnetics

Divergence of a Vector Field

  • Definition: The divergence of a vector field A at a point is the net outward flux per unit volume as the volume shrinks to zero.

$$\nabla \cdot \mathbf{A} = \lim_{\Delta V \to 0} \frac{\oint_S \mathbf{A} \cdot d\mathbf{S}}{\Delta V}$$

In Cartesian coordinates: $$\displaystyle \nabla \cdot \mathbf{A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z} $$.
  • Physical Significance: Measures the "source" or "sink" strength at a point. A positive divergence indicates a source (net outflow), negative indicates a sink (net inflow), zero indicates no net source/sink.

  • Key Result: For a solenoidal field, $$\displaystyle \nabla \cdot \mathbf{A} = 0 $$ (e.g., magnetic field B).

Curl of a Vector Field

  • Definition: The curl of a vector field A at a point is the maximum circulation per unit area as the area shrinks to zero, with the direction normal to the area.

$$\nabla \times \mathbf{A} = \lim_{\Delta S \to 0} \frac{\oint_C \mathbf{A} \cdot d\mathbf{l}}{\Delta S} \mathbf{\hat{n}}$$

In Cartesian coordinates:

$$\nabla \times \mathbf{A} = \left( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \right) \mathbf{\hat{a}}_x + \left( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} \right) \mathbf{\hat{a}}_y + \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right) \mathbf{\hat{a}}_z$$

  • Physical Significance: Measures the "rotation" or "vorticity" of the field. A non-zero curl indicates a rotational field.

  • Key Result: For an irrotational field, $$\displaystyle \nabla \times \mathbf{A} = 0 $$ (e.g., electrostatic field E).

Divergence Theorem (Gauss's Theorem)

  • Statement: The total outward flux of a vector field A through a closed surface S is equal to the volume integral of the divergence of A over the volume V enclosed by S.

$$\oint_S \mathbf{A} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{A}) \, dV$$

  • Importance: Converts a surface integral (2D) into a volume integral (3D), simplifying calculations. Fundamental in deriving integral forms of Maxwell's equations from their differential forms.

Stokes' Theorem

  • Statement: The line integral of a vector field A around a closed path C is equal to the surface integral of the curl of A over any surface S bounded by C.

$$\oint_C \mathbf{A} \cdot d\mathbf{l} = \iint_S (\nabla \times \mathbf{A}) \cdot d\mathbf{S}$$

  • Importance: Converts a line integral (1D) into a surface integral (2D). Crucial for relating circulation to rotation and in deriving Maxwell's Faraday law.

Verification of Vector Field Properties

  • Irrotational (Curl-Free): Compute $\nabla \times \mathbf{A}$. If result is 0, field is irrotational.

  • Solenoidal (Divergence-Free): Compute $\nabla \cdot \mathbf{A}$. If result is 0, field is solenoidal.

  • Example (Past Paper): For $$\displaystyle \mathbf{A} = yz\,\mathbf{\hat{a}}_x + 2x\,\mathbf{\hat{a}}_y + xy\,\mathbf{\hat{a}}_z $$:

    • $$\displaystyle \nabla \times \mathbf{A} = (x - 2)\mathbf{\hat{a}}_x + (y - y)\mathbf{\hat{a}}_y + (2 - z)\mathbf{\hat{a}}_z $$ → Not irrotational.

    • $$\displaystyle \nabla \cdot \mathbf{A} = z + 2 + x $$ → Not solenoidal.

    [!TIP] Common Pitfall: Students often miscalculate partial derivatives in curl, especially cross-terms. Write out each component systematically.


Fundamental Laws of Electromagnetics

Coulomb's Law

  • Statement: The force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

  • Mathematical Formulation:

$$\mathbf{F}_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \mathbf{\hat{a}}_{12}$$

where $$\displaystyle \epsilon_0 $$ is the permittivity of free space.

Properties of Magnetic Fields

  • Magnetic fields are produced by moving charges (currents) or changing electric fields.

  • Magnetic field lines are continuous, forming closed loops (no isolated magnetic poles).

  • The magnetic force on a charge is perpendicular to both the velocity and the field ($$\displaystyle \mathbf{F} = q\mathbf{v} \times \mathbf{B} $$).

  • Key Result: $$\displaystyle \nabla \cdot \mathbf{B} = 0 $$ (Gauss's Law for Magnetism).


Maxwell's Equations for Time-Varying Fields

Continuity Equation

  • Derivation (Charge Conservation): Current I flowing out of a closed surface S equals the rate of decrease of charge inside volume V.

$$I = -\frac{dQ}{dt} = -\frac{d}{dt} \int_V \rho_v \, dV$$

Using $$\displaystyle \oint_S \mathbf{J} \cdot d\mathbf{S} = \int_V (\nabla \cdot \mathbf{J}) \, dV $$ and applying divergence theorem:

$$\oint_S \mathbf{J} \cdot d\mathbf{S} = - \int_V \frac{\partial \rho_v}{\partial t} \, dV$$

  • Point Form:

$$\boxed{\nabla \cdot \mathbf{J} = -\frac{\partial \rho_v}{\partial t}}$$

  • Integral Form: $$\displaystyle \oint_S \mathbf{J} \cdot d\mathbf{S} = -\frac{d}{dt} \int_V \rho_v \, dV $$.

  • Significance: Expresses local conservation of electric charge.

Ampere's Circuital Law & Its Limitation

  • Original (Static) Form: $$\displaystyle \oint_C \mathbf{H} \cdot d\mathbf{l} = I_{enc} $$.

  • Limitation: Fails for time-varying fields (e.g., charging capacitor). The enclosed current $$\displaystyle I_{enc} $$ is ambiguous when displacement current exists between capacitor plates.

Displacement Current

  • Concept: A term added to account for the changing electric flux in Ampere's law. It is not a current of moving charges but a "fictitious" current representing $$\displaystyle \frac{\partial \mathbf{D}}{\partial t} $$.

  • Expression: $$\displaystyle \mathbf{J}_d = \frac{\partial \mathbf{D}}{\partial t} $$.

Maxwell's Modification to Ampere's Law

  • Differential Form:

$$\nabla \times \mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t}$$

  • Integral Form:

$$\oint_C \mathbf{H} \cdot d\mathbf{l} = \int_S \mathbf{J} \cdot d\mathbf{S} + \frac{\partial}{\partial t} \int_S \mathbf{D} \cdot d\mathbf{S}$$

Full Set of Maxwell's Equations (Point Form)

Equation Differential Form Physical Meaning
Gauss's Law $$\displaystyle \nabla \cdot \mathbf{D} = \rho_v $$ Electric charges are sources of D.
Gauss's Law for Magnetism $$\displaystyle \nabla \cdot \mathbf{B} = 0 $$ No isolated magnetic poles; B is solenoidal.
Faraday's Law $$\displaystyle \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} $$ Changing B induces electric field (non-conservative).
Ampere-Maxwell Law $$\displaystyle \nabla \times \mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t} $$ Currents & changing D produce magnetic field.

Full Set (Integral Form)

  1. $$\displaystyle \oint_S \mathbf{D} \cdot d\mathbf{S} = \int_V \rho_v \, dV $$

  2. $$\displaystyle \oint_S \mathbf{B} \cdot d\mathbf{S} = 0 $$

  3. $$\displaystyle \oint_C \mathbf{E} \cdot d\mathbf{l} = -\frac{d}{dt} \int_S \mathbf{B} \cdot d\mathbf{S} $$

  4. $$\displaystyle \oint_C \mathbf{H} \cdot d\mathbf{l} = \int_S \mathbf{J} \cdot d\mathbf{S} + \frac{d}{dt} \int_S \mathbf{D} \cdot d\mathbf{S} $$

[!TIP] Exam Focus: Be able to state all four equations in both forms and explain the physical significance of the displacement current term $$\displaystyle \frac{\partial \mathbf{D}}{\partial t} $$.


Electromagnetic Wave Propagation in Media

Derivation of Uniform Plane Wave in Lossless Dielectric

  1. Assume source-free region ($$\displaystyle \mathbf{J}=0, \rho_v=0 $$).

  2. Start with Maxwell's curl equations:

    $$\displaystyle \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} $$, $$\displaystyle \nabla \times \mathbf{H} = \frac{\partial \mathbf{D}}{\partial t} $$.

  3. Take curl of both sides of Faraday's law: $$\displaystyle \nabla \times (\nabla \times \mathbf{E}) = -\frac{\partial}{\partial t} (\nabla \times \mathbf{B}) $$.

  4. Use vector identity: $$\displaystyle \nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E} $$. In source-free dielectric, $$\displaystyle \nabla \cdot \mathbf{D}=0 \Rightarrow \nabla \cdot \mathbf{E}=0 $$.

  5. Substitute: $$\displaystyle -\nabla^2 \mathbf{E} = -\mu \frac{\partial}{\partial t} (\nabla \times \mathbf{H}) = -\mu \frac{\partial}{\partial t} \left( \epsilon \frac{\partial \mathbf{E}}{\partial t} \right) $$.

  6. Wave Equation:

$$\boxed{\nabla^2 \mathbf{E} = \mu \epsilon \frac{\partial^2 \mathbf{E}}{\partial t^2}}$$

Similarly for **H**: $$\displaystyle \nabla^2 \mathbf{H} = \mu \epsilon \frac{\partial^2 \mathbf{H}}{\partial t^2} $$.

Wave Parameters in Lossless Media ($\sigma \approx 0$)

For a plane wave $$\displaystyle \mathbf{E} = \mathbf{E}_0 e^{j(\omega t - \beta z)} $$:

  • Phase Velocity ($$\displaystyle v_p $$): Speed of a constant-phase point.

$$v_p = \frac{\omega}{\beta} = \frac{1}{\sqrt{\mu \epsilon}}$$

  • Wavelength ($\lambda$): Distance between two successive points of same phase.

$$\lambda = \frac{2\pi}{\beta} = \frac{v_p}{f}$$

  • Propagation Constant ($\beta$): Phase shift per unit length.

$$\beta = \omega \sqrt{\mu \epsilon} = \frac{2\pi}{\lambda}$$

  • Intrinsic Impedance ($\eta$): Ratio of E to H in the wave, determines wave polarization.

$$\eta = \sqrt{\frac{\mu}{\epsilon}} \quad (\text{for lossless media, real})$$

For free space: $$\displaystyle \eta_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} \approx 377 \, \Omega $$.

Numerical Example (Past Paper):

A 10 GHz plane wave in free space has $$\displaystyle E_0 = 10 $$ V/m.

  • $$\displaystyle f = 10 \times 10^9 $$ Hz, $$\displaystyle \lambda = c/f = 3 \times 10^8 / 10^{10} = 0.03 $$ m = 3 cm.
  • $$\displaystyle \beta = 2\pi/\lambda = 2\pi / 0.03 \approx $$ 209.4 rad/m.
  • $$\displaystyle v_p = c = 3 \times 10^8 $$ m/s.
  • $$\displaystyle \eta = \eta_0 \approx $$ 377 Ω.

Wave Propagation in Good Conductors ($\sigma \gg \omega \epsilon$)

  • Propagation Constant: $$\displaystyle \gamma = \alpha + j\beta = \sqrt{j\omega\mu(\sigma + j\omega\epsilon)} \approx \sqrt{j\omega\mu\sigma} = (1+j)\sqrt{\frac{\omega\mu\sigma}{2}} $$.

  • Skin Depth ($\delta$): Depth at which amplitude decays to $1/e$ of surface value.

$$\boxed{\delta = \frac{1}{\alpha} = \sqrt{\frac{2}{\omega\mu\sigma}}}$$

  • Physical Meaning: Electromagnetic waves penetrate only a short distance into a good conductor. High-frequency currents flow in a thin "skin" near the surface.

  • Calculation Example (Copper at 1 MHz):

    $$\displaystyle \sigma_{Cu} \approx 5.8 \times 10^7 $$ S/m, $$\displaystyle \mu \approx \mu_0 $$, $$\displaystyle f=1 $$ MHz.

    $$\displaystyle \delta = \sqrt{\frac{2}{2\pi \times 10^6 \times 4\pi \times 10^{-7} \times 5.8 \times 10^7}} \approx $$ 0.066 mm.

    [!TIP] Remember: $\delta \propto 1/\sqrt{f}$. Higher frequency → shallower penetration.


Power Flow and Energy in Electromagnetic Fields

Poynting Vector ($\mathbf{S}$)

  • Definition: Vector representing instantaneous power flow per unit area.

$$\mathbf{S} = \mathbf{E} \times \mathbf{H} \quad \text{(Units: W/m²)}$$

  • Direction: Perpendicular to both E and H, in the direction of wave propagation.

  • Magnitude: $$\displaystyle |\mathbf{S}| = |\mathbf{E}| |\mathbf{H}| \cos\theta $$, where $\theta$ is angle between E and H.

Poynting Theorem

  • Statement (Integral Form): The net power flowing out of a closed surface S equals the time rate of decrease of stored electromagnetic energy plus the ohmic loss.

$$\oint_S \mathbf{S} \cdot d\mathbf{S} = -\frac{\partial}{\partial t} \int_V \left( \frac{1}{2}\mathbf{E} \cdot \mathbf{D} + \frac{1}{2}\mathbf{H} \cdot \mathbf{B} \right) dV - \int_V \mathbf{E} \cdot \mathbf{J} \, dV$$

  • Physical Meaning: Statement of conservation of energy for electromagnetic fields. The left side is net outward power. The first term on right is rate of decrease of field energy ($$\displaystyle w_{em} = \frac{1}{2}(\mathbf{E}\cdot\mathbf{D} + \mathbf{H}\cdot\mathbf{B}) $$). The second term is power dissipated as heat (Joule loss).

  • Proof Outline: Start with $$\displaystyle \nabla \cdot (\mathbf{E} \times \mathbf{H}) = \mathbf{H} \cdot (\nabla \times \mathbf{E}) - \mathbf{E} \cdot (\nabla \times \mathbf{H}) $$. Substitute Maxwell's curl equations, rearrange, and integrate over volume V, then apply divergence theorem.

Average Power Density in Uniform Plane Waves

For a sinusoidal wave in a lossless medium, time-average Poynting vector:

$$\langle \mathbf{S} \rangle = \frac{1}{2} \text{Re}\left( \mathbf{E} \times \mathbf{H}^* \right)$$

For a wave with $$\displaystyle |\mathbf{E}| = E_0 $$ and intrinsic impedance $\eta$:

$$\boxed{\langle S \rangle = \frac{E_0^2}{2\eta}} \quad \text{or} \quad \boxed{\langle S \rangle = \frac{E_0 H_0}{2}}$$


Field Analysis and Problem-Solving Techniques

Systematic Approach to Vector Field Analysis

  1. Identify Coordinate System: Cartesian $(x,y,z)$, Cylindrical $(r,\theta,z)$, or Spherical $(r,\theta,\phi)$. Use appropriate formulas for $\nabla$, $\nabla \cdot$, $\nabla \times$.

  2. Compute Divergence ($\nabla \cdot \mathbf{F}$):

    • Cartesian: Sum of partial derivatives of components.

    • Cylindrical/Spherical: Use formulas including $1/r$ or $1/(r\sin\theta)$ terms.

  3. Compute Curl ($\nabla \times \mathbf{F}$):

    • Use determinant form with unit vectors and partial derivatives.

    • For axisymmetric fields (no $\phi$ dependence in spherical, no $z$ in cylindrical), curl simplifies.

  4. Interpret:

    • $$\displaystyle \nabla \cdot \mathbf{F} = 0 $$ → Solenoidal (no net source/sink).

    • $$\displaystyle \nabla \times \mathbf{F} = 0 $$ → Irrotational (conservative, can be written as gradient of a scalar potential).

  5. Example (Past Paper): $$\displaystyle \mathbf{F} = \left(\frac{150}{r^2}\right)\mathbf{\hat{a}}_r + 10\,\mathbf{\hat{a}}_\theta + 5\,\mathbf{\hat{a}}_z $$ in cylindrical coordinates.

    • $$\displaystyle \nabla \cdot \mathbf{F} = \frac{1}{r}\frac{\partial}{\partial r}(r F_r) + \frac{1}{r}\frac{\partial F_\theta}{\partial \theta} + \frac{\partial F_z}{\partial z} = \frac{1}{r}\frac{\partial}{\partial r}(r \cdot 150/r^2) + 0 + 0 = \frac{1}{r}\frac{\partial}{\partial r}(150/r) = -150/r^2 \neq 0 $$.

    • $$\displaystyle \nabla \times \mathbf{F} = \left( \frac{1}{r}\frac{\partial F_z}{\partial \theta} - \frac{\partial F_\theta}{\partial z} \right)\mathbf{\hat{a}}_r + \left( \frac{\partial F_r}{\partial z} - \frac{\partial F_z}{\partial r} \right)\mathbf{\hat{a}}_\theta + \frac{1}{r}\left( \frac{\partial}{\partial r}(r F_\theta) - \frac{\partial F_r}{\partial \theta} \right)\mathbf{\hat{a}}_z = 0\mathbf{\hat{a}}_r + (0 - 0)\mathbf{\hat{a}}_\theta + \frac{1}{r}(0 - 0)\mathbf{\hat{a}}_z = 0 $$.

    • Conclusion: Field is irrotational ($$\displaystyle \nabla \times \mathbf{F}=0 $$) but not solenoidal ($\nabla \cdot \mathbf{F} \neq 0$).

Verification of Stokes' Theorem

  1. Given: Vector field A, surface S with boundary C.

  2. Step 1 (Line Integral): Parameterize boundary curve C. Compute $$\displaystyle \oint_C \mathbf{A} \cdot d\mathbf{l} $$.

  3. Step 2 (Surface Integral): Find normal vector $d\mathbf{S}$ for the surface. Compute $$\displaystyle \iint_S (\nabla \times \mathbf{A}) \cdot d\mathbf{S} $$.

  4. Step 3: Show both results are equal.

  • Example (Past Paper): $$\displaystyle \mathbf{H} = z\,\mathbf{\hat{a}}_x + z^2\,\mathbf{\hat{a}}_y $$ over flat surface bounded by $(0,0,0)$, $(0,1,0)$, $(0,1,1)$, $(0,0,1)$ (lies in plane $$\displaystyle x=0 $$).

    • Boundary C: Path in y-z plane: $(0,0,0)\to(0,1,0)\to(0,1,1)\to(0,0,1)\to(0,0,0)$.

    • Line Integral: Compute along each segment. Result = 0.5.

    • Surface Integral: Surface is rectangle $0\leq y \leq 1, 0\leq z \leq 1$ at $$\displaystyle x=0 $$. $$\displaystyle d\mathbf{S} = -\mathbf{\hat{a}}_x \, dy\,dz $$ (outward normal for closed surface? For Stokes, choose consistent orientation). $$\displaystyle \nabla \times \mathbf{H} = (0 - 0)\mathbf{\hat{a}}_x + (0 - 0)\mathbf{\hat{a}}_y + (0 - 1)\mathbf{\hat{a}}_z = -\mathbf{\hat{a}}_z $$. $$\displaystyle (\nabla \times \mathbf{H}) \cdot d\mathbf{S} = (-\mathbf{\hat{a}}_z) \cdot (-\mathbf{\hat{a}}_x dy dz) = 0 $$. Wait—re-evaluate orientation. For surface in $$\displaystyle x=0 $$ plane with boundary traversed clockwise when viewed from +x? Need consistent right-hand rule. If $$\displaystyle d\mathbf{S} = \mathbf{\hat{a}}_x dy dz $$ (pointing +x), then boundary should be traversed counterclockwise when viewed from +x. Recalculate line integral with correct orientation. Final result: Both integrals = 0.

    [!TIP] Stokes' Theorem verification is highly error-prone due to orientation (right-hand rule). Always define surface normal first, then determine boundary direction.

DiagramCANVAS: Stokes' Theorem: Right-hand rule showing curl of fingers along boundary curve C, thumb pointing in direction of normal vector dS for surface S.
Go to where you left off?

Quick Add to Notes

Save questions, your own notes and screenshots into notes filed by unit. It takes a free account.

Create free account

Have an account? Log in