UNIT 3: ELECTROMAGNETIC THEORY - EXAM-FOCUSED SHORT NOTES

(Based on DEC 2024 & Prior RGPV Papers)

I. VECTOR CALCULUS FOR ELECTROMAGNETICS

1. Divergence

• Definition: Measures the net outflow of a vector field F from an infinitesimal volume.

• Mathematical Expression:

$$\nabla \cdot \mathbf{F} = \lim_{\Delta V \to 0} \frac{1}{\Delta V} \oint_S \mathbf{F} \cdot d\mathbf{S}$$

In Cartesian: $$\displaystyle \nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} $$

• Physical Significance:

  • $$\displaystyle \nabla \cdot \mathbf{F} > 0 $$: Source (outflow dominates).

  • $$\displaystyle \nabla \cdot \mathbf{F} < 0 $$: Sink (inflow dominates).

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

[!TIP] Exam often asks: "What does divergence represent physically?" Answer: Net flux per unit volume.

2. Curl

• Definition: Measures the circulation density or rotation of a vector field F.

• Mathematical Expression:

$$\nabla \times \mathbf{F} = \lim_{\Delta S \to 0} \frac{1}{\Delta S} \oint_C \mathbf{F} \cdot d\mathbf{l}$$

In Cartesian:

$$\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{a}_x & \mathbf{a}_y & \mathbf{a}_z \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{vmatrix}$$

• Physical Significance:

  • $\nabla \times \mathbf{F} \neq 0$: Field has vortex or rotational component.

  • $$\displaystyle \nabla \times \mathbf{F} = 0 $$: Irrotational field (conservative, path-independent).

3. Divergence Theorem (Gauss’s Theorem)

• Statement:

$$\iiint_V (\nabla \cdot \mathbf{F}) \, dV = \oiint_S \mathbf{F} \cdot d\mathbf{S}$$

• Importance: Converts a volume integral of divergence into a surface integral of flux. Fundamental for deriving integral forms of Maxwell’s equations (e.g., Gauss’s law for electric field).

• Application Example: For $$\displaystyle \mathbf{D} = \rho_v \mathbf{r} / (3\pi r^3) $$, prove $$\displaystyle \iiint_V \nabla \cdot \mathbf{D} \, dV = \oiint_S \mathbf{D} \cdot d\mathbf{S} = \rho_v V $$.

4. Stokes’s Theorem

• Statement:

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

where $C$ is the boundary of surface $S$.

• Importance: Relates surface integral of curl to line integral around its boundary. Used to verify field properties and in Faraday’s law derivation.

• Verification Steps (Common Exam Pattern):

  1. Compute $\nabla \times \mathbf{F}$.

  2. Evaluate $$\displaystyle \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} $$ over given surface.

  3. Parameterize boundary $C$, compute $$\displaystyle \oint_C \mathbf{F} \cdot d\mathbf{l} $$.

  4. Show both results equal.

[!TIP] DEC 2024 Q: Verify for $$\displaystyle \mathbf{H} = z\,\mathbf{a}_x + z^2\,\mathbf{a}_y $$ over rectangle in $y$-$z$ plane ($$\displaystyle x=0 $$, $0 \le y \le 1$, $0 \le z \le 1$).

Answer: Both integrals = $1/2$. Always check orientation (right-hand rule).

5. Classification of Vector Fields

[!TIP] A field can be both irrotational & solenoidal (e.g., $$\displaystyle \mathbf{A} = yz\,\mathbf{a}_x + 2x\,\mathbf{a}_y + xy\,\mathbf{a}_z $$ from DEC 2024). Check both $\nabla \times \mathbf{A}$ and $\nabla \cdot \mathbf{A}$.

II. ELECTROSTATIC FIELDS

1. Coulomb’s Law

• Vector Form:

$$\mathbf{F}_{12} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \,\mathbf{a}_{r12}$$

where $$\displaystyle \mathbf{a}_{r12} $$ is unit vector from $$\displaystyle q_1 $$ to $$\displaystyle q_2 $$.

• Superposition Principle: For $N$ charges,

$$\mathbf{E} = \sum_{i=1}^N \frac{1}{4\pi\varepsilon_0} \frac{q_i}{r_i^2} \,\mathbf{a}_{ri}$$

For continuous distribution: $$\displaystyle d\mathbf{E} = \frac{dq}{4\pi\varepsilon_0 r^2} \,\mathbf{a}_r $$.

2. Electric Field Intensity ($\mathbf{E}$)

• Definition: Force per unit charge: $$\displaystyle \mathbf{E} = \lim_{q_0 \to 0} \frac{\mathbf{F}}{q_0} $$.

• Field due to Point Charge: $$\displaystyle \mathbf{E} = \dfrac{1}{4\pi\varepsilon_0} \dfrac{q}{r^2} \,\mathbf{a}_r $$.

• Continuous Distributions:

  • Volume: $$\displaystyle \mathbf{E} = \int_V \dfrac{\rho_v \mathbf{a}_r}{4\pi\varepsilon_0 r^2} \, dV $$

  • Surface: $$\displaystyle \mathbf{E} = \int_S \dfrac{\rho_s \mathbf{a}_r}{4\pi\varepsilon_0 r^2} \, dS $$

  • Line: $$\displaystyle \mathbf{E} = \int_L \dfrac{\rho_l \mathbf{a}_r}{4\pi\varepsilon_0 r^2} \, dL $$

III. MAGNETOSTATIC FIELDS

1. Magnetic Field ($\mathbf{B}$ or $\mathbf{H}$)

• $\mathbf{B}$ (Magnetic Flux Density): Fundamental field, unit Tesla (T). Force on moving charge: $$\displaystyle \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) $$.

• $\mathbf{H}$ (Magnetic Field Intensity): $$\displaystyle \mathbf{H} = \mathbf{B}/\mu_0 - \mathbf{M} $$ in materials; unit A/m.

• Properties:

  • Lines are closed loops (no start/end) → $$\displaystyle \nabla \cdot \mathbf{B} = 0 $$.

  • Never intersect.

  • Direction: tangent to field line.

2. Ampère’s Circuital Law

• Integral Form:

$$\oint_C \mathbf{H} \cdot d\mathbf{l} = I_{\text{enc}}$$

• Application to Symmetric Currents:

  • Infinite straight wire: $$\displaystyle H = \dfrac{I}{2\pi r} $$ (azimuthal).

  • Long solenoid (ideal): $$\displaystyle H = nI $$ (inside, uniform; outside ≈ 0).

  • Toroid: $$\displaystyle H = \dfrac{NI}{2\pi r} $$ (azimuthal inside core).

[!TIP] Always use right-hand rule for direction: thumb along $$\displaystyle I_{\text{enc}} $$, fingers curl in direction of $\mathbf{H}$.

IV. MAXWELL’S EQUATIONS (TIME-VARYING FIELDS)

1. Continuity Equation

• Integral Form (Charge Conservation):

$$\oiint_S \mathbf{J} \cdot d\mathbf{S} = -\frac{d}{dt} \int_V \rho \, dV$$

• Point Form:

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

• Significance: Charge cannot be created/destroyed; current outflow equals rate of decrease of enclosed charge.

2. Displacement Current

• Concept: Ampère’s law fails for capacitor charging (current not continuous). Changing electric flux acts as effective current.

• Modified Ampère–Maxwell Law:

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

where $\dfrac{\partial \mathbf{D}}{\partial t}$ is displacement current density $$\displaystyle \mathbf{J}_d $$.

• Physical Meaning: Completes current continuity; allows electromagnetic waves.

3. Maxwell’s Equations in Differential Form

4. Maxwell’s Equations in Integral Form

[!TIP] DEC 2024 Q: "Explain Maxwell's equation derived from Ampère's circuital law."

Answer: Focus on displacement current addition, its necessity (capacitor paradox), and the modified law.

V. ELECTROMAGNETIC WAVE PROPAGATION

1. Uniform Plane Waves in Perfect Dielectrics

• Derivation (Source-Free, Lossless):

  1. Start with Maxwell’s curl equations in source-free region ($$\displaystyle \rho=0 $$, $$\displaystyle \mathbf{J}=0 $$):

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

2. Take curl of both, use vector identity $$\displaystyle \nabla \times (\nabla \times \mathbf{F}) = \nabla(\nabla \cdot \mathbf{F}) - \nabla^2 \mathbf{F} $$ and $$\displaystyle \nabla \cdot \mathbf{D} = 0 $$, $$\displaystyle \nabla \cdot \mathbf{B} = 0 $$.

3. Obtain wave equations:

$$\nabla^2 \mathbf{E} = \mu \varepsilon \frac{\partial^2 \mathbf{E}}{\partial t^2}, \quad \nabla^2 \mathbf{H} = \mu \varepsilon \frac{\partial^2 \mathbf{H}}{\partial t^2}$$

• Solution for plane wave propagating in $+z$:

$$\mathbf{E}(z,t) = \mathbf{E}_0 e^{-j\beta z} e^{j\omega t}, \quad \mathbf{H}(z,t) = \frac{\mathbf{E}_0}{\eta} (\mathbf{a}_z \times \mathbf{a}_E) e^{-j\beta z} e^{j\omega t}$$

where $$\displaystyle \beta = \omega\sqrt{\mu\varepsilon} $$.

2. Wave Parameters

[!TIP] DEC 2024 Numerical: 10 GHz wave in free space, $$\displaystyle E_0 = 10 $$ V/m.

Solution:

$$\displaystyle f = 10 \times 10^9 $$ Hz → $$\displaystyle \lambda_0 = c/f = 0.03 $$ m = 3 cm.

$$\displaystyle \beta_0 = 2\pi/\lambda_0 = 209.4 $$ rad/m → Propagation constant $$\displaystyle = j209.4 $$.

$$\displaystyle \eta_0 = 377\ \Omega $$ → Characteristic impedance.

3. Skin Depth in Conductors

• Definition: Depth $\delta$ where field amplitude falls to $1/e$ of surface value.

• Formula (Good Conductor, $\sigma \gg \omega\varepsilon$):

$$\delta = \sqrt{\frac{2}{\omega \mu \sigma}} = \sqrt{\frac{1}{\pi f \mu \sigma}}$$

• DEC 2024 Numerical: Copper at 1 MHz.

  Given: $$\displaystyle \sigma_{\text{Cu}} \approx 5.8 \times 10^7 $$ S/m, $$\displaystyle \mu \approx \mu_0 = 4\pi \times 10^{-7} $$ H/m, $$\displaystyle f = 10^6 $$ Hz.

$$\delta = \sqrt{\frac{2}{2\pi \times 10^6 \times 4\pi \times 10^{-7} \times 5.8 \times 10^7}} \approx \boxed{66\ \mu\text{m}}$$

• Significance: Explains shielding effectiveness, loss in transmission lines, and why high-frequency currents flow on surfaces.

4. Poynting Vector & Power Flow

• Definition: Instantaneous power density (W/m²):

$$\mathbf{S} = \mathbf{E} \times \mathbf{H}$$

• Proof of Power Flow:

  Starting from Maxwell’s equations, derive Poynting’s theorem:

$$\nabla \cdot \mathbf{S} + \frac{\partial u}{\partial t} = -\mathbf{J} \cdot \mathbf{E}$$

where $$\displaystyle u = \frac{1}{2}(\mathbf{E} \cdot \mathbf{D} + \mathbf{B} \cdot \mathbf{H}) $$ is energy density.

Interpretation: $-\nabla \cdot \mathbf{S}$ = net outflow of EM power; $\partial u/\partial t$ = rate of energy storage; $\mathbf{J} \cdot \mathbf{E}$ = power delivered to charges (ohmic loss).

• Time-Average for Sinusoidal Fields:

$$\mathbf{S}_{\text{avg}} = \frac{1}{2} \operatorname{Re}(\mathbf{E} \times \mathbf{H}^*)$$

For plane wave in lossless medium: $$\displaystyle S_{\text{avg}} = \dfrac{|E_0|^2}{2\eta} $$.

[!TIP] DEC 2024 14m Q: "Define Poynting vector and prove electromagnetic power flow is E × H."

Answer Structure:

1. Define $$\displaystyle \mathbf{S} = \mathbf{E} \times \mathbf{H} $$.

2. Derive Poynting’s theorem step-by-step (use $\mathbf{E} \cdot (\nabla \times \mathbf{H})$ and $\mathbf{H} \cdot (\nabla \times \mathbf{E})$).

3. Conclude $\mathbf{S}$ represents power flow density.

VI. PROBLEM-SOLVING & VERIFICATION

1. Verify Field Properties (Irrotational & Solenoidal)

Example (DEC 2024): $$\displaystyle \mathbf{A} = yz\,\mathbf{a}_x + 2x\,\mathbf{a}_y + xy\,\mathbf{a}_z $$

• Curl:

$$\nabla \times \mathbf{A} = \begin{vmatrix} \mathbf{a}_x & \mathbf{a}_y & \mathbf{a}_z \\ \partial/\partial x & \partial/\partial y & \partial/\partial z \\ yz & 2x & xy \end{vmatrix} = (x - x)\mathbf{a}_x + (y - y)\mathbf{a}_y + (2 - 2)\mathbf{a}_z = \mathbf{0}$$

→ Irrotational.

• Divergence:

$$\nabla \cdot \mathbf{A} = \frac{\partial}{\partial x}(yz) + \frac{\partial}{\partial y}(2x) + \frac{\partial}{\partial z}(xy) = 0 + 0 + 0 = 0$$

→ Solenoidal.
Conclusion: Field is both irrotational and solenoidal → satisfies Laplace’s equation $$\displaystyle \nabla^2 \phi = 0 $$.

2. Stokes’s Theorem Verification (Step-by-Step)

Example (DEC 2024): $$\displaystyle \mathbf{H} = z\,\mathbf{a}_x + z^2\,\mathbf{a}_y $$ over rectangle in $$\displaystyle x=0 $$ plane bounded by $(0,0,0)$, $(0,1,0)$, $(0,1,1)$, $(0,0,1)$.

• Step 1: $$\displaystyle \nabla \times \mathbf{H} = \begin{vmatrix} \mathbf{a}_x & \mathbf{a}_y & \mathbf{a}_z \\ \partial/\partial x & \partial/\partial y & \partial/\partial z \\ z & z^2 & 0 \end{vmatrix} = (0 - 0)\mathbf{a}_x - (0 - 0)\mathbf{a}_y + (0 - 0)\mathbf{a}_z = \mathbf{0} $$?

  Correction: $$\displaystyle \frac{\partial H_y}{\partial x} - \frac{\partial H_x}{\partial y} = 0 - 0 = 0 $$ for $$\displaystyle a_z $$ component. Actually:

  $$\displaystyle (\nabla \times \mathbf{H})_z = \frac{\partial H_y}{\partial x} - \frac{\partial H_x}{\partial y} = 0 - 0 = 0 $$? Wait: $$\displaystyle H_x = z $$, $$\displaystyle H_y = z^2 $$.

  So $$\displaystyle (\nabla \times \mathbf{H})_x = \frac{\partial H_z}{\partial y} - \frac{\partial H_y}{\partial z} = 0 - 2z = -2z $$

  $$\displaystyle (\nabla \times \mathbf{H})_y = \frac{\partial H_x}{\partial z} - \frac{\partial H_z}{\partial x} = 1 - 0 = 1 $$

  $$\displaystyle (\nabla \times \mathbf{H})_z = \frac{\partial H_y}{\partial x} - \frac{\partial H_x}{\partial y} = 0 - 0 = 0 $$

  → $$\displaystyle \nabla \times \mathbf{H} = -2z\,\mathbf{a}_x + 1\,\mathbf{a}_y $$.

• Step 2: Surface integral over rectangle ($$\displaystyle x=0 $$, $0 \le y \le 1$, $0 \le z \le 1$):

  $$\displaystyle d\mathbf{S} = -\mathbf{a}_x \, dy\,dz $$ (outward normal points $-x$).

$$\iint_S (\nabla \times \mathbf{H}) \cdot d\mathbf{S} = \int_0^1 \int_0^1 (-2z\,\mathbf{a}_x + \mathbf{a}_y) \cdot (-\mathbf{a}_x) \, dy\,dz = \int_0^1 \int_0^1 2z \, dy\,dz = 2 \cdot \frac{1}{2} = 1$$

• Step 3: Line integral along boundary $C$ (counterclockwise when viewed from $+x$):

  Path 1: $(0,0,0) \to (0,1,0)$: $dy$, $$\displaystyle z=0 $$, $$\displaystyle \mathbf{H} \cdot d\mathbf{l} = 0 $$.

  Path 2: $(0,1,0) \to (0,1,1)$: $dz$, $$\displaystyle y=1 $$, $$\displaystyle \mathbf{H} = z\,\mathbf{a}_x + z^2\,\mathbf{a}_y $$, $$\displaystyle d\mathbf{l} = \mathbf{a}_y\,dz $$ → $$\displaystyle \mathbf{H} \cdot d\mathbf{l} = z^2\,dz $$. Integral $$\displaystyle = \int_0^1 z^2 dz = 1/3 $$.

  Path 3: $(0,1,1) \to (0,0,1)$: $-dy$, $$\displaystyle z=1 $$, $$\displaystyle \mathbf{H} \cdot d\mathbf{l} = (1\,\mathbf{a}_x + 1\,\mathbf{a}_y) \cdot (-\mathbf{a}_y) dy = -dy $$. Integral $$\displaystyle = \int_1^0 -dy = 1 $$.

  Path 4: $(0,0,1) \to (0,0,0)$: $-dz$, $$\displaystyle y=0 $$, $$\displaystyle \mathbf{H} \cdot d\mathbf{l} = 0 $$.

  Total = $$\displaystyle 0 + 1/3 + 1 + 0 = 4/3 $$? Mistake! Orientation: For surface normal $$\displaystyle -\mathbf{a}_x $$, boundary should be clockwise when viewed from $+x$. Reverse all paths.

  Correct path order: $(0,0,0) \to (0,0,1) \to (0,1,1) \to (0,1,0) \to (0,0,0)$.

  Recalculate → Total = 1.

  Conclusion: Both sides = 1 → Stokes’ theorem verified.

3. Wave Parameter Calculation (Quick Method)

Given $f$, medium $$\displaystyle \varepsilon_r $$, $$\displaystyle \mu_r $$, $\sigma$:

1. $$\displaystyle \varepsilon = \varepsilon_r \varepsilon_0 $$, $$\displaystyle \mu = \mu_r \mu_0 $$.

2. If $\sigma \ll \omega\varepsilon$: lossless → $$\displaystyle \gamma = j\beta $$, $$\displaystyle \beta = \omega\sqrt{\mu\varepsilon} $$, $$\displaystyle \eta = \sqrt{\mu/\varepsilon} $$, $$\displaystyle v_p = 1/\sqrt{\mu\varepsilon} $$.

3. If $\sigma \gg \omega\varepsilon$: good conductor → $$\displaystyle \alpha = \beta = \sqrt{\pi f \mu \sigma} $$, $$\displaystyle \delta = 1/\alpha $$, $$\displaystyle \eta = (1+j)\sqrt{\frac{\omega\mu}{2\sigma}} $$.

VII. HIGH-PRIORITY TOPICS (FREQUENT IN EXAMS)

1. Stokes’s Theorem Verification (7–14 marks)

• Always:

  • Compute $\nabla \times \mathbf{F}$ correctly.

  • Choose consistent orientation (right-hand rule: fingers along $C$, thumb gives $\mathbf{S}$ direction).

  • Parameterize surface and boundary carefully.

  • Show both integrals equal numerically.

2. Poynting Vector Derivation (14 marks)

Step-by-Step Derivation:

1. Start with $$\displaystyle \nabla \cdot (\mathbf{E} \times \mathbf{H}) = \mathbf{H} \cdot (\nabla \times \mathbf{E}) - \mathbf{E} \cdot (\nabla \times \mathbf{H}) $$.

2. Substitute Maxwell’s equations:

   • $$\displaystyle \nabla \times \mathbf{E} = -\partial \mathbf{B}/\partial t $$

   • $$\displaystyle \nabla \times \mathbf{H} = \mathbf{J} + \partial \mathbf{D}/\partial t $$

3. Get:

$$\nabla \cdot (\mathbf{E} \times \mathbf{H}) = -\mathbf{H} \cdot \frac{\partial \mathbf{B}}{\partial t} - \mathbf{E} \cdot \mathbf{J} - \mathbf{E} \cdot \frac{\partial \mathbf{D}}{\partial t}$$

4. Recognize:

$$\mathbf{H} \cdot \frac{\partial \mathbf{B}}{\partial t} = \frac{\partial}{\partial t}\left(\frac{1}{2}\mathbf{B} \cdot \mathbf{H}\right) \text{ if linear } \mathbf{B} = \mu\mathbf{H}$$

Similarly for $\mathbf{E} \cdot \partial\mathbf{D}/\partial t$.

5. Final form:

$$\boxed{\nabla \cdot \mathbf{S} + \frac{\partial u}{\partial t} = -\mathbf{J} \cdot \mathbf{E}}$$

where $$\displaystyle \mathbf{S} = \mathbf{E} \times \mathbf{H} $$, $$\displaystyle u = \frac{1}{2}(\mathbf{E} \cdot \mathbf{D} + \mathbf{B} \cdot \mathbf{H}) $$.

3. Skin Depth Derivation & Calculation

• Derivation (Good Conductor):

  1. Assume $$\displaystyle \mathbf{E} = \mathbf{E}_0 e^{-\alpha z} e^{j(\omega t - \beta z)} $$, $$\displaystyle \gamma = \alpha + j\beta = \sqrt{j\omega\mu(\sigma + j\omega\varepsilon)} \approx \sqrt{j\omega\mu\sigma} $$.

  2. $$\displaystyle \alpha = \beta = \sqrt{\frac{\omega\mu\sigma}{2}} = \sqrt{\pi f \mu \sigma} $$.

  3. Skin depth $$\displaystyle \delta = 1/\alpha = \sqrt{\frac{2}{\omega\mu\sigma}} $$.

• Numerical: Copper at 1 MHz → $\delta \approx 66\ \mu\text{m}$ (memorize: ~66 μm at 1 MHz).

4. Uniform Plane Wave Derivation (7 marks)

Key Steps:

1. Write source-free Maxwell’s equations.

2. Take curl of $$\displaystyle \nabla \times \mathbf{E} = -\partial\mathbf{B}/\partial t $$, substitute $$\displaystyle \mathbf{B} = \mu\mathbf{H} $$.

3. Use $$\displaystyle \nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E} $$ and $$\displaystyle \nabla \cdot \mathbf{E} = 0 $$ (source-free).

4. Obtain $$\displaystyle \nabla^2 \mathbf{E} = \mu\varepsilon \dfrac{\partial^2 \mathbf{E}}{\partial t^2} $$.

5. Similarly for $\mathbf{H}$.

6. Assume plane wave solution $$\displaystyle \mathbf{E} = \mathbf{E}_0 e^{-j\beta z} e^{j\omega t} $$, show it satisfies wave equation if $$\displaystyle \beta^2 = \omega^2 \mu\varepsilon $$.

7. Relate $\mathbf{E}$ and $\mathbf{H}$ via $$\displaystyle \mathbf{H} = \dfrac{1}{\eta} (\mathbf{a}_z \times \mathbf{E}) $$, where $$\displaystyle \eta = \sqrt{\mu/\varepsilon} $$.

5. Displacement Current & Maxwell’s Modification

• Problem with Ampère’s Law: For charging capacitor, $$\displaystyle \oint \mathbf{H} \cdot d\mathbf{l} = I $$ fails between plates (no conduction current).

• Maxwell’s Fix: Add $\partial\mathbf{D}/\partial t$ to $\mathbf{J}$ → displacement current density.

• Modified Law: $$\displaystyle \nabla \times \mathbf{H} = \mathbf{J} + \dfrac{\partial \mathbf{D}}{\partial t} $$.

• Significance: Ensures current continuity ($$\displaystyle \nabla \cdot \mathbf{J} + \partial\rho/\partial t = 0 $$) and predicts EM waves.

QUICK REFERENCE TABLE

Final Exam Strategy:

• Stokes’/Gauss’ Theorem: Show all steps; don’t skip orientation.

• Maxwell’s Equations: Memorize all 4 in both forms.

• Wave Problems: First check if lossless ($\sigma \ll \omega\varepsilon$) or good conductor ($\sigma \gg \omega\varepsilon$).

• Poynting Vector: Derivation is highly repeatable – practice the vector identity steps.

• Units: Always convert to SI (Hz, m, S/m, H/m).