Chapter 1 — The Electromagnetic Model

Key Theory — Chapter 1

Condensed from Cheng, Field and Wave Electromagnetics, §1-1 through §1-3. Read this before attempting the problems below.


1-1   What is Electromagnetics?

Electromagnetics is the study of the effects of electric charges at rest and in motion. There are two kinds of charges — positive and negative — and both are sources of an electric field. Moving charges constitute a current, which gives rise to a magnetic field.

A field is a spatial distribution of a quantity, which may or may not be a function of time. A time-varying electric field is always accompanied by a magnetic field, and vice versa — the two are coupled, forming an electromagnetic field. Under certain conditions, time-dependent electromagnetic fields produce waves that radiate from the source.

Why fields, not just circuits? Circuit theory is a restricted special case of electromagnetic theory, valid only when the dimensions of a network are much smaller than the wavelength (quasi-static regime). Two situations where circuit theory fails:

  1. A monopole antenna on a walkie-talkie appears as an open circuit to circuit theory, yet it clearly radiates — only a non-uniform current flowing along the open-ended conductor (predicted by field theory) explains transmission.
  2. An electromagnetic wave incident on a conducting wall with a small aperture produces fields on the far side, even at points not directly behind the hole — a diffraction/shielding problem circuit theory cannot describe.

1-2.1   Deductive vs. Inductive Approach

Two approaches exist in developing a scientific subject:

Cheng uses the deductive approach because it is more elegant and develops the subject in an orderly way.

Three steps to build any theory on an idealized model

  1. Define the basic quantities germane to the subject.
  2. Specify the rules of operation (the mathematics) for those quantities.
  3. Postulate the fundamental relations (axioms/laws), based on experimental observation.

Familiar example — circuit theory: basic quantities are \(V, I, R, L, C\); rules of operation are algebra, ODEs, and Laplace transforms; fundamental postulates are Kirchhoff's voltage and current laws.

For electromagnetics: Chapter 1 covers step 1; Chapter 2 covers step 2 (vector algebra & calculus); step 3 is introduced in three substeps — Chapters 3 (electrostatics), 6 (magnetostatics), and 7 (time-varying fields/Maxwell's equations).


1-2.2   Source Quantities

Quantities in the electromagnetic model fall into two categories: source quantities (the charges and currents that produce fields) and field quantities (the fields themselves).

Electric charge and the charge of an electron

Electric charge is a fundamental property of matter and exists only in positive or negative integer multiples of the elementary charge \(e\):

\[ e = 1.60 \times 10^{-19} \ \text{C} \qquad \text{(1-1)} \]

Principle of conservation of charge — like conservation of momentum, this is a fundamental postulate of physics: charge can neither be created nor destroyed. The algebraic sum of positive and negative charges in an isolated system remains unchanged. This principle is expressed mathematically by the equation of continuity (§5-4), and Kirchhoff's current law is simply an assertion of it applied to a junction.

Volume, surface, and line charge densities

Although charge is discrete microscopically, electromagnetic effects of large aggregates are well described by smoothed-out density functions defined as point functions of space coordinates:

\[ \rho = \lim_{\Delta v \to 0} \frac{\Delta q}{\Delta v} \quad (\text{C/m}^3) \qquad \text{(1-2)} \]
\[ \rho_s = \lim_{\Delta s \to 0} \frac{\Delta q}{\Delta s} \quad (\text{C/m}^2) \qquad \text{(1-3)} \]
\[ \rho_\ell = \lim_{\Delta \ell \to 0} \frac{\Delta q}{\Delta \ell} \quad (\text{C/m}) \qquad \text{(1-4)} \]

"Small enough" to give accurate variation yet "large enough" to contain many discrete charges — e.g. a cube of side 1 μm has volume \(10^{-18}\ \text{m}^3\) and still contains about \(10^{11}\) atoms.

Current and current density

Current is the rate of change of charge with respect to time:

\[ I = \frac{dq}{dt} \quad (\text{C/s} = \text{A}) \qquad \text{(1-5)} \]

Current flows through a finite area, so it is not a point function. Two vector point functions are defined to describe current density:


1-2.3   The Four Fundamental Field Quantities

Electromagnetics has four fundamental vector field quantities, all point functions of space (and time, in the dynamic case):

Symbol Unit
Electric: Electric field intensity E V/m
Electric flux density (displacement) D C/m²
Magnetic: Magnetic flux density B T (= V·s/m²)
Magnetic field intensity H A/m
Table 1-1 — Fundamental Electromagnetic Field Quantities

Which field do you need, and when?

Static vs. dynamic coupling: under static, steady, or stationary conditions the two pairs \(\{\mathbf{E},\mathbf{D}\}\) and \(\{\mathbf{B},\mathbf{H}\}\) are independent. In time-dependent cases they are coupled — a time-varying \(\mathbf{E}\)/\(\mathbf{D}\) produces \(\mathbf{B}\)/\(\mathbf{H}\), and vice versa.

Constitutive relations

Material (medium) properties determine the relations between \(\mathbf{E}\) and \(\mathbf{D}\), and between \(\mathbf{B}\) and \(\mathbf{H}\). These relations are called the constitutive relations of the medium. In free space (see §1-3):

\[ \mathbf{D} = \epsilon_0 \mathbf{E} \qquad \text{(1-7)} \]
\[ \mathbf{H} = \frac{1}{\mu_0}\, \mathbf{B} \qquad \text{(1-8)} \]

1-3.1   The SI System

SI (Système International d'Unités) is a rationalized MKSA system — four base units, from which every other unit in electromagnetics is derived:

Quantity Unit Abbreviation
Length meter m
Mass kilogram kg
Time second s
Current ampere A
Table 1-2 — Fundamental SI Units

All units in Table 1-1 are derived:

"Rationalized" means the factor \(4\pi\) has been absorbed into the constitutive constants so that it does not appear in Maxwell's equations themselves (though it does appear in many derived relations such as Coulomb's law).


1-3.2   The Three Universal Constants

Free space (vacuum) is characterized by three universal constants, which are not independent of one another:

\[ c \cong 3 \times 10^{8} \ \text{m/s} \quad \text{(velocity of light in vacuum)} \qquad \text{(1-6)} \]
\[ \mu_0 = 4\pi \times 10^{-7} \ \text{H/m} \quad \text{(permeability of free space)} \qquad \text{(1-9)} \]
\[ \epsilon_0 = \frac{1}{c^2 \mu_0} \cong \frac{1}{36\pi}\times 10^{-9} \cong 8.854 \times 10^{-12} \ \text{F/m} \qquad \text{(1-11)} \]

In SI, \(\mu_0\) is defined exactly as \(4\pi\times 10^{-7}\) H/m (a choice of the unit system, not an experimentally measured quantity). The speed of light is fixed by the modern definition of the meter. The permittivity \(\epsilon_0\) is then derived from the master identity

\[ c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} \qquad \text{(1-10)} \]

which links electromagnetism to optics — the velocity of light is a consequence of the electric and magnetic properties of vacuum.

Constant Symbol Value Unit
Velocity of light (free space) c 3 × 10⁸ m/s
Permeability of free space μ₀ 4π × 10⁻⁷ H/m
Permittivity of free space ε₀ (1/36π) × 10⁻⁹ F/m
Table 1-3 — Universal Constants in SI Units

Chapter 1 at a Glance

Problems and Solutions


Review Questions (Tier 1)

R.1-1● EasyTier 1?
Review: What is electromagnetics? → Answer
R.1-1 Answer↑ Question

Study of the effects of electric charges at rest and in motion. Static charges produce \(\vec{E}\); moving charges (currents) produce \(\vec{B}\); time-varying fields couple as electromagnetic waves.

R.1-2● EasyTier 1?
Review: Describe two phenomena that cannot be adequately explained by circuit theory. → Answer
R.1-2 Answer↑ Question

(1) Open-circuit antenna that nevertheless radiates (e.g., monopole on a walkie-talkie). (2) Diffraction of an EM wave through a small aperture in a conducting wall.

R.1-3● EasyTier 1?
Review: What are the three essential steps in building an idealized model for the study of a scientific subject? → Answer
R.1-3 Answer↑ Question

(1) Define basic quantities. (2) Specify rules of operation (the mathematics). (3) Postulate fundamental relations (axioms) verified against experiment.

R.1-4● EasyTier 1?
Review: What are the four fundamental SI units in electromagnetics? → Answer
R.1-4 Answer↑ Question

Meter (m), kilogram (kg), second (s), ampere (A).

R.1-5● EasyTier 1?
Review: What are the four fundamental field quantities in the electromagnetic model? What are their units? → Answer
R.1-5 Answer↑ Question

\(\vec{E}\) — V/m; \(\vec{D}\) — C/m\(^2\); \(\vec{B}\) — T (V·s/m\(^2\)); \(\vec{H}\) — A/m.

R.1-6● EasyTier 1?
Review: What are the three universal constants in the electromagnetic model, and what are their relations? → Answer
R.1-6 Answer↑ Question

\(c=3{\times}10^8\,\text{m/s}\) (free-space light speed); \(\mu_0=4\pi{\times}10^{-7}\,\text{H/m}\); \(\epsilon_0=1/(36\pi){\times}10^{-9}\,\text{F/m}\). Relation: \(c=1/\sqrt{\mu_0\epsilon_0}\).

R.1-7● EasyTier 1?
Review: What are the source quantities in the electromagnetic model? → Answer
R.1-7 Answer↑ Question

Electric charge \(q\) (or charge density \(\rho_v\)) and current density \(\vec{J}\).


Ex 1.1●● MediumTier 1?
Convert Charge Units

Convert 1 coulomb to statcoulombs (ESU) and abcoulombs (EMU).

→ Solution
Ex 1.1 Solution↑ Problem
1 C = 3×10⁹ statcoulombs (ESU/Gaussian)
\[1 C = 0.1 abcoulombs (EMU)\]
Origin: In CGS-ESU, F = q₁q₂/r², whereas SI uses F = q₁q₂/(4πε₀r²).
The factor 3×10⁹ comes from c = 2.998×10¹⁰ cm/s ≈ 3×10¹⁰ cm/s.

Ex 1.2●● MediumTier 1?
Speed of Light in Different Unit Systems

Express c in m/s, cm/s, ft/ns, and km/s.

→ Solution
Ex 1.2 Solution↑ Problem
\[c = 2.998\times 10^{8} m/s \approx 3\times 10^{8} m/s\]
\[c = 2.998\times 10^{10} cm/s \approx 3\times 10^{10} cm/s\]
\[c = 2.998\times 10^{5} km/s \approx 3\times 10^{5} km/s\]
c = 0.9836 ft/ns ≈ 1 ft/ns (useful for digital circuit delays)
c = 186,282 miles/s ≈ 186,000 miles/s
Practical note: a signal travels ~1 foot per nanosecond in free space
(~6 inches/ns in PCB traces with εᵣ ≈ 4).

Ex 1.3●● MediumTier 1?
Coulomb Force Between Two Charges

Find the force between two point charges Q₁ = Q₂ = 1 μC separated by r = 1 m.

→ Solution
Ex 1.3 Solution↑ Problem
\[F = Q_{1}Q_{2}/(4\pi \varepsilon _{0}r^{2})\]
\[= (10^{-6})^{2}/(4\pi \times 8.85\times 10^{-12} \times 1^{2})\]
\[= 10^{-12}/(1.113\times 10^{-10})\]
\[= 8.99\times 10^{-3} N \approx 9 mN (repulsive)\]
Using k = 1/(4πε₀) = 8.99×10⁹ N·m²/C²:
\[F = k Q_{1}Q_{2}/r^{2} = 8.99\times 10^{9} \times 10^{-12} = 8.99\times 10^{-3} N \checkmark \]

Ex 1.4●● MediumTier 1?
Relationship Between ε₀ and μ₀

Derive ε₀μ₀ = 1/c² and verify numerically.

→ Solution
Ex 1.4 Solution↑ Problem
From Maxwell's equations, the wave equation gives:
\[c = 1/\sqrt{\varepsilon _{0}\mu _{0}} \to \varepsilon _{0}\mu _{0} = 1/c^{2}\]
Numerical verification:
\[\varepsilon _{0} = 8.854\times 10^{-12} F/m\]
\[\mu _{0} = 4\pi \times 10^{-7} = 1.2566\times 10^{-6} H/m\]
\[\varepsilon _{0}\mu _{0} = 8.854\times 10^{-12} \times 1.2566\times 10^{-6} = 1.1127\times 10^{-17} s^{2}/m^{2}\]
\[1/c^{2} = 1/(2.998\times 10^{8})^{2} = 1/8.988\times 10^{16} = 1.1127\times 10^{-17} s^{2}/m^{2} \checkmark \]

Ex 1.5●● MediumTier 1?
Convert Current Units

Convert 1 ampere to statamperes and abamperes.

→ Solution
Ex 1.5 Solution↑ Problem
\[1 A = 3\times 10^{9} statamperes (ESU)\]
\[1 A = 0.1 abamperes (EMU)\]
Note: 1 abampere = 10 A (so 1 A = 0.1 abampere)
These follow from Q = I×t and the charge conversions in Problem 1.

Ex 1.6●● MediumTier 1?
Energy Density in an Electromagnetic Wave

A plane wave in free space has E₀ = 100 V/m. Find the electric and magnetic energy densities.

→ Solution
Ex 1.6 Solution↑ Problem
Electric energy density:
\[u_e = \varepsilon _{0}E_{0}^{2}/2 = 8.85\times 10^{-12} \times 10^{4}/2 = 4.43\times 10^{-8} J/m^{3}\]
\[H_{0} = E_{0}/\eta _{0} = 100/377 = 0.265 A/m\]
Magnetic energy density:
\[u_m = \mu _{0}H_{0}^{2}/2 = 4\pi \times 10^{-7} \times 0.0702/2 = 4.43\times 10^{-8} J/m^{3}\]
u_e = u_m ✓ (equal in a plane wave)
Total: u = ε₀E₀² = 8.85×10⁻⁸ J/m³

Ex 1.7●● MediumTier 1?
Power Density

For the wave in Problem 6, find the Poynting vector magnitude.

→ Solution
Ex 1.7 Solution↑ Problem
S = E₀ × H₀ = E₀²/η₀ (time-average = peak/2 for sinusoidal)
\[Peak: S_peak = 100 \times 0.265 = 26.5 W/m^{2}\]
\[Time-average: \langle S\rangle = E_{0}^{2}/(2\eta _{0}) = 10000/(2\times 377) = 13.26 W/m^{2}\]
Also: ⟨S⟩ = u_total × c / 2 = 8.85×10⁻⁸ × 3×10⁸/2 = 13.3 W/m² ✓

Ex 1.8●● MediumTier 1?
Dimensional Consistency of Maxwell's Equations

Verify ∇ × E = −∂B/∂t and ∇ × H = J + ∂D/∂t are dimensionally consistent.

→ Solution
Ex 1.8 Solution↑ Problem
\[\nabla \times E = -\partial B/\partial t:\]
\[LHS: [\nabla \times E] = [E]/[m] = (V/m)/m = V/m^{2}\]
\[RHS: [\partial B/\partial t] = [T]/[s] = (V\cdot s/m^{2})/s = V/m^{2} \checkmark \]
\[\nabla \times H = J + \partial D/\partial t:\]
\[LHS: [\nabla \times H] = (A/m)/m = A/m^{2}\]
\[RHS (J): [J] = A/m^{2} \checkmark \]
\[RHS (\partial D/\partial t): [D]/[t] = (C/m^{2})/s = A/m^{2} \checkmark \]
\[\nabla \cdot D = \rho _v:\]
\[[\nabla \cdot D] = (C/m^{2})/m = C/m^{3} = [\rho _v] \checkmark \]

Ex 1.9●● MediumTier 1?
Electric Field Unit Conversions

Convert E = 1 kV/m to (a) N/C, (b) V/cm, (c) force on a 1 μC charge.

→ Solution
Ex 1.9 Solution↑ Problem
(a) 1 V/m = 1 N/C (by definition, since V = J/C = N·m/C)
\[\to 1 kV/m = 1000 N/C\]
\[(b) 1 kV/m = 1000 V/m = 10 V/cm\]
\[(c) Force: F = qE = 10^{-6} \times 1000 = 10^{-3} N = 1 mN\]

Ex 1.10●● MediumTier 1?
Magnetic Field Unit Conversions

Convert B = 1 T to Gauss and find H in free space.

→ Solution
Ex 1.10 Solution↑ Problem
\[1 T = 10^{4} Gauss = 1 Wb/m^{2}\]
\[H in free space (\mu _{r} = 1):\]
\[H = B/\mu _{0} = 1/(4\pi \times 10^{-7}) = 7.958\times 10^{5} A/m\]
For B = 0.5 T (typical MRI magnet):
\[H = 3.98\times 10^{5} A/m\]
H in Oersteds: 1 A/m = 4π×10⁻³ Oe → H = 4997 Oe ≈ 5000 Oe

Ex 1.11●● MediumTier 1?
Impedance of Free Space

Derive η₀ = √(μ₀/ε₀) and verify it equals 120π Ω.

→ Solution
Ex 1.11 Solution↑ Problem
\[\eta _{0} = \sqrt{\mu _{0}/\varepsilon _{0}} = \sqrt{4\pi \times 10^{-7} / 8.85\times 10^{-12}}\]
\[= \sqrt{1.420\times 10^{5}}\]
\[= 376.7 \Omega \]
Alternative derivation:
\[\eta _{0} = \mu _{0}c = 4\pi \times 10^{-7} \times 3\times 10^{8} = 4\pi \times 30 = 120\pi \approx 376.99 \Omega \]
\[120\pi = 376.99 \Omega \checkmark \]
Useful: η₀ ≈ 377 Ω for quick calculations.

Ex 1.12●● MediumTier 1?
Verify c = 1/√(ε₀μ₀)

Calculate 1/√(ε₀μ₀) from the SI values of ε₀ and μ₀.

→ Solution
Ex 1.12 Solution↑ Problem
\[\varepsilon _{0} = 8.854\times 10^{-12} F/m\]
\[\mu _{0} = 4\pi \times 10^{-7} = 1.2566\times 10^{-6} H/m\]
\[\varepsilon _{0} \times \mu _{0} = 8.854\times 10^{-12} \times 1.2566\times 10^{-6} = 11.13\times 10^{-18} s^{2}/m^{2}\]
\[1/\sqrt{\varepsilon _{0}\mu _{0}} = 1/\sqrt{11.13\times 10^{-18}} = 1/(3.337\times 10^{-9}) = 2.998\times 10^{8} m/s = c \checkmark \]
This confirms: the values of ε₀ and μ₀ are not independent —
they are linked through the measured speed of light.

Ex 1.13●● MediumTier 1?
Energy Density Units

For E = 1 MV/m in free space, compute u_e in J/m³, erg/cm³, and eV/m³.

→ Solution
Ex 1.13 Solution↑ Problem
\[u_e = \varepsilon _{0}E^{2}/2 = 8.85\times 10^{-12} \times (10^{6})^{2}/2 = 8.85\times 10^{-12} \times 10^{12}/2 = 4.425 J/m^{3}\]
Convert to CGS: 1 J/m³ = 10 erg/cm³
\[u_e = 4.425 \times 10 = 44.25 erg/cm^{3}\]
Convert to eV/m³: 1 J = 1/(1.602×10⁻¹⁹) eV = 6.242×10¹⁸ eV
\[u_e = 4.425 \times 6.242\times 10^{18} = 2.76\times 10^{19} eV/m^{3}\]

Ex 1.14●● MediumTier 1?
Coulomb's Constant

Express k = 1/(4πε₀) in SI units and verify its numerical value.

→ Solution
Ex 1.14 Solution↑ Problem
\[k = 1/(4\pi \varepsilon _{0})\]
\[= 1/(4\pi \times 8.854\times 10^{-12})\]
\[= 1/(1.1127\times 10^{-10})\]
\[= 8.988\times 10^{9} N\cdot m^{2}/C^{2}\]
Also: k = c² × 10⁻⁷ = (3×10⁸)² × 10⁻⁷ = 9×10¹⁶ × 10⁻⁷ = 9×10⁹ N·m²/C²
Units: [k] = N·m²/C² = V·m/C = kg·m³/(A²·s⁴)

Ex 1.15●● MediumTier 1?
E/H Ratio in a Plane Wave

A plane wave in free space has H = 1 A/m. Find E, power density, and compare E/H to η₀.

→ Solution
Ex 1.15 Solution↑ Problem
\[E = \eta _{0} \times H = 377 \times 1 = 377 V/m\]
E/H = η₀ = 377 Ω (intrinsic impedance of free space)
Time-average Poynting vector:
\[\langle S\rangle = (1/2) E_{0} H_{0} = (1/2) \times 377 \times 1 = 188.5 W/m^{2}\]
\[Or: \langle S\rangle = \eta _{0}H_{0}^{2}/2 = 377/2 = 188.5 W/m^{2}\]
Also: ⟨S⟩ = E₀²/(2η₀) = 377²/(2×377) = 377/2 = 188.5 W/m² ✓

Chapter 2 — Vector Analysis

Key Theory — Chapter 2

Condensed from Cheng, Field and Wave Electromagnetics, §2-1 through §2-12. Read this before attempting the problems below.


2-1   Introduction — Why Vector Analysis?

Electromagnetic quantities are either scalars (charge, energy, potential) or vectors (\(\mathbf{E}\), \(\mathbf{B}\), current density). In three-dimensional space a vector relation is three scalar relations, so an efficient vector calculus is essential.

Three pillars of this chapter:

  1. Vector algebra — addition, dot and cross products.
  2. Orthogonal coordinate systems — Cartesian, cylindrical, spherical.
  3. Vector calculus — gradient, divergence, curl, and the integral theorems that tie them together.

The physical laws themselves are invariant under choice of coordinates; coordinates are chosen only to match the geometry of a given problem.


2-2   Vector Addition and Subtraction

A vector has magnitude and direction; a unit vector points without magnitude:

\[ \mathbf{A} = \mathbf{a}_A A, \qquad \mathbf{a}_A = \frac{\mathbf{A}}{|\mathbf{A}|} \qquad \text{(2-1,3)} \]

Addition is commutative and associative; subtraction is \(\mathbf{A}-\mathbf{B} = \mathbf{A}+(-\mathbf{B})\).

\[ \mathbf{A}+\mathbf{B} = \mathbf{B}+\mathbf{A} \qquad \text{(2-4)} \]
\[ \mathbf{A}+(\mathbf{B}+\mathbf{C}) = (\mathbf{A}+\mathbf{B})+\mathbf{C} \qquad \text{(2-5)} \]

2-3   Products of Vectors

Dot (scalar) product — yields a scalar:

\[ \mathbf{A}\cdot\mathbf{B} \triangleq AB\cos\theta_{AB} \qquad \text{(2-9)} \]
\[ \mathbf{A}\cdot\mathbf{A} = A^{2} \qquad \text{(2-10)} \]

Zero iff vectors are orthogonal; equals the projection of one onto the other times the other's magnitude. Commutative and distributive; not associative (triple dot is meaningless).

Cross (vector) product — yields a vector perpendicular to the plane of the operands, magnitude equals the area of their parallelogram:

\[ \mathbf{A}\times\mathbf{B} \triangleq \mathbf{a}_n\,|AB\sin\theta_{AB}| \qquad \text{(2-14)} \]
\[ \mathbf{B}\times\mathbf{A} = -\mathbf{A}\times\mathbf{B} \qquad \text{(2-15)} \]

Direction \(\mathbf{a}_n\) by right-hand rule. Cross product is anti-commutative, distributive, and not associative.

Scalar triple product — cyclic permutation leaves it unchanged (geometrically, volume of the parallelepiped):

\[ \mathbf{A}\cdot(\mathbf{B}\times\mathbf{C}) = \mathbf{B}\cdot(\mathbf{C}\times\mathbf{A}) = \mathbf{C}\cdot(\mathbf{A}\times\mathbf{B}) \qquad \text{(2-18)} \]

Vector triple product — the BAC-CAB rule:

\[ \mathbf{A}\times(\mathbf{B}\times\mathbf{C}) = \mathbf{B}(\mathbf{A}\cdot\mathbf{C}) - \mathbf{C}(\mathbf{A}\cdot\mathbf{B}) \qquad \text{(2-20)} \]

Parentheses matter — \((\mathbf{A}\times\mathbf{B})\times\mathbf{C}\) is a different vector. Division by a vector is undefined.


2-4   Orthogonal Coordinate Systems

A right-handed orthogonal system has base vectors \(\mathbf{a}_{u_1},\mathbf{a}_{u_2},\mathbf{a}_{u_3}\) satisfying:

\[ \mathbf{a}_{u_1}\times\mathbf{a}_{u_2} = \mathbf{a}_{u_3},\quad \mathbf{a}_{u_2}\times\mathbf{a}_{u_3} = \mathbf{a}_{u_1},\quad \mathbf{a}_{u_3}\times\mathbf{a}_{u_1} = \mathbf{a}_{u_2} \qquad \text{(2-21)} \]
\[ \mathbf{a}_{u_i}\cdot\mathbf{a}_{u_j} = \delta_{ij} \qquad \text{(2-22,23)} \]

Coordinates \(u_i\) may not themselves be lengths (angles, for instance). The metric coefficient \(h_i\) converts a coordinate change to a length:

\[ d\ell_i = h_i\,du_i \qquad \text{(2-29)} \]
\[ d\mathbf{\ell} = \mathbf{a}_{u_1}(h_1\,du_1) + \mathbf{a}_{u_2}(h_2\,du_2) + \mathbf{a}_{u_3}(h_3\,du_3) \qquad \text{(2-31)} \]
\[ dv = h_1 h_2 h_3\,du_1\,du_2\,du_3 \qquad \text{(2-33)} \]
Table 2-1 — Three Basic Orthogonal Coordinate Systems
Cartesian Cylindrical Spherical
(x, y, z) (r, φ, z) (R, θ, φ)
Base vectors aₓ, a_y, a_z a_r, a_φ, a_z a_R, a_θ, a_φ
Metric h₁, h₂, h₃ 1, 1, 1 1, r, 1 1, R, R sin θ
Differential volume dx dy dz r dr dφ dz R² sin θ dR dθ dφ

Cartesian (§2-4.1):

\[ \mathbf{A}\cdot\mathbf{B} = A_x B_x + A_y B_y + A_z B_z \qquad \text{(2-42)} \]
\[ \mathbf{A}\times\mathbf{B} = \begin{vmatrix} \mathbf{a}_x & \mathbf{a}_y & \mathbf{a}_z \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} \qquad \text{(2-43)} \]
\[ d\mathbf{\ell} = \mathbf{a}_x\,dx + \mathbf{a}_y\,dy + \mathbf{a}_z\,dz,\quad dv = dx\,dy\,dz \qquad \text{(2-44,46)} \]

Cylindrical (§2-4.2): \((r,\phi,z)\). Useful for line charges, long wires, coax. \(\mathbf{a}_r\times\mathbf{a}_\phi=\mathbf{a}_z\).

\[ d\mathbf{\ell} = \mathbf{a}_r\,dr + \mathbf{a}_\phi\,r\,d\phi + \mathbf{a}_z\,dz \qquad \text{(2-52)} \]
\[ dv = r\,dr\,d\phi\,dz \qquad \text{(2-54)} \]
\[ x = r\cos\phi,\quad y = r\sin\phi,\quad z = z \qquad \text{(2-62)} \]

Spherical (§2-4.3): \((R,\theta,\phi)\). \(R\) is distance from origin, \(\theta\) is polar angle from \(+z\), \(\phi\) is azimuth. Ideal for point sources and far-field antenna problems. \(\mathbf{a}_R\times\mathbf{a}_\theta=\mathbf{a}_\phi\).

\[ d\mathbf{\ell} = \mathbf{a}_R\,dR + \mathbf{a}_\theta\,R\,d\theta + \mathbf{a}_\phi\,R\sin\theta\,d\phi \qquad \text{(2-66)} \]
\[ dv = R^{2}\sin\theta\,dR\,d\theta\,d\phi \qquad \text{(2-68)} \]
\[ x=R\sin\theta\cos\phi,\ y=R\sin\theta\sin\phi,\ z=R\cos\theta \qquad \text{(2-69)} \]

Caution: in curvilinear systems the base vectors depend on position (\(\partial\mathbf{a}_r/\partial\phi = \mathbf{a}_\phi\), \(\partial\mathbf{a}_\phi/\partial\phi = -\mathbf{a}_r\)). Differentiation of a vector in such systems must account for this.


2-5   Line, Surface, Volume Integrals

The differential area is treated as a vector \(d\mathbf{s} = \mathbf{a}_n\,ds\) (outward normal for a closed surface, right-hand-rule normal for an open surface bounded by a contour).

• Line integral along C: ∫_C A · dℓ (work, circulation, voltage)
• Surface integral over S: ∫_S A · ds (flux of A through S)
• Closed forms: ∮_C, ∮_S (closed loop / closed surface)
• Volume integral: ∫_V f dv (total charge, total energy, …)

2-6   Gradient of a Scalar Field

The gradient is the vector pointing in the direction of the maximum spatial rate of increase of a scalar \(V\), with magnitude equal to that rate:

\[ \nabla V \triangleq \mathbf{a}_n \frac{dV}{dn} \qquad \text{(2-86)} \]
\[ dV = (\nabla V)\cdot d\mathbf{\ell} \qquad \text{(2-88)} \]

General orthogonal curvilinear form:

\[ \nabla V = \mathbf{a}_{u_1}\frac{1}{h_1}\frac{\partial V}{\partial u_1} + \mathbf{a}_{u_2}\frac{1}{h_2}\frac{\partial V}{\partial u_2} + \mathbf{a}_{u_3}\frac{1}{h_3}\frac{\partial V}{\partial u_3} \qquad \text{(2-93)} \]

Cartesian form — also defines the del operator:

\[ \nabla V = \mathbf{a}_x\frac{\partial V}{\partial x} + \mathbf{a}_y\frac{\partial V}{\partial y} + \mathbf{a}_z\frac{\partial V}{\partial z} \qquad \text{(2-94)} \]
\[ \nabla \equiv \mathbf{a}_x\frac{\partial}{\partial x} + \mathbf{a}_y\frac{\partial}{\partial y} + \mathbf{a}_z\frac{\partial}{\partial z} \qquad \text{(2-96)} \]

\(\nabla V\) is always perpendicular to surfaces of constant \(V\). In electrostatics we use \(\mathbf{E}=-\nabla V\) — the minus sign makes \(V\) increase opposite to \(\mathbf{E}\).


2-7   Divergence of a Vector Field

Divergence measures the net outward flux of \(\mathbf{A}\) per unit volume — the flow source density:

\[ \nabla\cdot\mathbf{A} \triangleq \lim_{\Delta v\to 0}\frac{\oint_S \mathbf{A}\cdot d\mathbf{s}}{\Delta v} \qquad \text{(2-98)} \]

Coordinate forms:

\[ \text{Cartesian:}\quad \nabla\cdot\mathbf{A} = \frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z} \qquad \text{(2-108)} \]
\[ \text{Cylindrical:}\quad \nabla\cdot\mathbf{A} = \frac{1}{r}\frac{\partial}{\partial r}(rA_r)+\frac{1}{r}\frac{\partial A_\phi}{\partial\phi}+\frac{\partial A_z}{\partial z} \qquad \text{(2-114)} \]
\[ \text{Spherical:}\quad \nabla\cdot\mathbf{A} = \frac{1}{R^{2}}\frac{\partial}{\partial R}(R^{2}A_R)+\frac{1}{R\sin\theta}\frac{\partial}{\partial\theta}(A_\theta\sin\theta)+\frac{1}{R\sin\theta}\frac{\partial A_\phi}{\partial\phi} \qquad \text{(2-113)} \]

A field with \(\nabla\cdot\mathbf{A}=0\) everywhere is solenoidal — flux lines close on themselves, no sources or sinks (e.g., magnetic flux density \(\mathbf{B}\)).


2-8   Divergence Theorem

Also known as Gauss's theorem:

\[ \int_V \nabla\cdot\mathbf{A}\,dv = \oint_S \mathbf{A}\cdot d\mathbf{s} \qquad \text{(2-115)} \]

Converts volume integral of a divergence into a closed surface integral of the vector — the workhorse behind integral forms of Gauss's law and charge conservation.


2-9   Curl of a Vector Field

Circulation is the line integral of \(\mathbf{A}\) around a closed contour. Curl is the vortex-source density:

\[ \text{Circulation} = \oint_C \mathbf{A}\cdot d\mathbf{\ell} \qquad \text{(2-124)} \]
\[ \nabla\times\mathbf{A} \triangleq \lim_{\Delta s\to 0}\frac{1}{\Delta s}\!\left[\mathbf{a}_n\oint_C \mathbf{A}\cdot d\mathbf{\ell}\right]_{\max} \qquad \text{(2-125)} \]

The direction of \(\mathbf{a}_n\) is that which maximizes the circulation per unit area (right-hand rule relative to \(d\mathbf{\ell}\)).

Determinant form in Cartesian coordinates:

\[ \nabla\times\mathbf{A} = \begin{vmatrix} \mathbf{a}_x & \mathbf{a}_y & \mathbf{a}_z \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix} \qquad \text{(2-136)} \]

General orthogonal curvilinear form:

\[ \nabla\times\mathbf{A} = \frac{1}{h_1 h_2 h_3}\begin{vmatrix} h_1\mathbf{a}_{u_1} & h_2\mathbf{a}_{u_2} & h_3\mathbf{a}_{u_3} \\ \dfrac{\partial}{\partial u_1} & \dfrac{\partial}{\partial u_2} & \dfrac{\partial}{\partial u_3} \\ h_1 A_{u_1} & h_2 A_{u_2} & h_3 A_{u_3} \end{vmatrix} \qquad \text{(2-137)} \]

A field with \(\nabla\times\mathbf{A}=0\) everywhere is irrotational (or conservative). A static electrostatic field is always irrotational.


2-10   Stokes's Theorem

\[ \int_S (\nabla\times\mathbf{A})\cdot d\mathbf{s} = \oint_C \mathbf{A}\cdot d\mathbf{\ell} \qquad \text{(2-143)} \]

Converts surface integral of a curl into a closed line integral around the bounding contour. The directions of \(d\mathbf{\ell}\) and \(d\mathbf{s}=\mathbf{a}_n\,ds\) obey the right-hand rule.

Dictionary. The two integral theorems convert dimensionality:

Divergence theorem: ∫_V (∇·A) dv = ∮_S A · ds (volume ↔ closed surface)
Stokes's theorem: ∫_S (∇×A)·ds = ∮_C A · dℓ (open surface ↔ its bounding loop)

2-11   Two Null Identities

Repeated del operations give two identities of deep importance for potential theory:

\[ \nabla\times(\nabla V) \equiv 0 \quad \text{(Identity I)} \qquad \text{(2-145)} \]
\[ \nabla\cdot(\nabla\times\mathbf{A}) \equiv 0 \quad \text{(Identity II)} \qquad \text{(2-149)} \]

Converse statements (crucial for EM):


2-12   Helmholtz's Theorem

A vector field is determined — to within an additive constant — if both its divergence and its curl are specified everywhere, and the field vanishes at infinity.

Classification of vector fields:

Example
1. ∇·F = 0 and ∇×F = 0 : static E in a charge-free region
2. ∇·F = 0 and ∇×F ≠ 0 : steady B in a current-carrying conductor
3. ∇×F = 0 and ∇·F ≠ 0 : static E in a charged region
4. ∇·F ≠ 0 and ∇×F ≠ 0 : E in a charged medium with time-varying B

Any such field decomposes into an irrotational part and a solenoidal part:

\[ \mathbf{F} = \mathbf{F}_i + \mathbf{F}_s = -\nabla V + \nabla\times\mathbf{A} \qquad \text{(2-160)} \]

This is the axiomatic foundation for all of Cheng's subsequent chapters — specifying both div and curl of \(\mathbf{E}\) (Chapter 3) or of \(\mathbf{B}\) (Chapter 6) fixes the entire field. That is why Cheng postulates div and curl rather than starting from Coulomb's law.


Chapter 2 at a Glance

Problems and Solutions


Review Questions (Tier 1)

R.2-1● EasyTier 1?
Review: Three vectors \(\mathbf{A}\), \(\mathbf{B}\), \(\mathbf{C}\) form the sides of a triangle (head-to-tail). What are \(\mathbf{A}+\mathbf{B}+\mathbf{C}\) and \(\mathbf{A}+\mathbf{B}-\mathbf{C}\)? → Answer
R.2-1 Answer↑ Question

Head-to-tail closure: \(\vec{A}+\vec{B}+\vec{C}=0\). Therefore \(\vec{A}+\vec{B}-\vec{C}=-2\vec{C}\).

R.2-2● EasyTier 1?
Review: Under what conditions is \(\mathbf{A}\cdot\mathbf{B}\) negative? → Answer
R.2-2 Answer↑ Question

When the angle between \(\vec{A}\) and \(\vec{B}\) exceeds \(90^{\circ}\) (\(\cos\theta<0\)).

R.2-3● EasyTier 1?
Review: Write \(\mathbf{A}\cdot\mathbf{B}\) and \(\mathbf{A}\times\mathbf{B}\) when (a) \(\mathbf{A}\parallel\mathbf{B}\); (b) \(\mathbf{A}\perp\mathbf{B}\). → Answer
R.2-3 Answer↑ Question

(a) Parallel: \(\vec{A}\cdot\vec{B}=AB\), \(\vec{A}\times\vec{B}=0\). (b) Perpendicular: \(\vec{A}\cdot\vec{B}=0\), \(|\vec{A}\times\vec{B}|=AB\).

R.2-4● EasyTier 1?
Review: Which of the following make sense? \((\mathbf{A}\cdot\mathbf{B})\mathbf{C}\), \(\mathbf{A}(\mathbf{B}\cdot\mathbf{C})\), \(\mathbf{A}\times\mathbf{B}\times\mathbf{C}\), \(\mathbf{A}/\mathbf{B}\), \(\mathbf{A}/\mathbf{a}_A\), \((\mathbf{A}\times\mathbf{B})\cdot\mathbf{C}\). → Answer
R.2-4 Answer↑ Question

Valid: \((\vec{A}\cdot\vec{B})\vec{C}\), \(\vec{A}(\vec{B}\cdot\vec{C})\), \((\vec{A}\times\vec{B})\cdot\vec{C}\). Invalid: \(\vec{A}\times\vec{B}\times\vec{C}\) (no associativity), \(\vec{A}/\vec{B}\), \(\vec{A}/\vec{a}_A\) (no vector division).

R.2-5● EasyTier 1?
Review: Does \((\mathbf{A}\cdot\mathbf{B})\mathbf{C} = \mathbf{A}(\mathbf{B}\cdot\mathbf{C})\)? → Answer
R.2-5 Answer↑ Question

No. The first lies along \(\vec{C}\), the second along \(\vec{A}\); they are equal only in special cases.

R.2-6● EasyTier 1?
Review: Does \(\mathbf{A}\cdot\mathbf{B}=\mathbf{A}\cdot\mathbf{C}\) imply \(\mathbf{B}=\mathbf{C}\)? → Answer
R.2-6 Answer↑ Question

No. Only the components of \(\vec{B}\) and \(\vec{C}\) along \(\vec{A}\) must agree; perpendicular components are unconstrained.

R.2-7● EasyTier 1?
Review: Does \(\mathbf{A}\times\mathbf{B}=\mathbf{A}\times\mathbf{C}\) imply \(\mathbf{B}=\mathbf{C}\)? → Answer
R.2-7 Answer↑ Question

No. Only components of \(\vec{B}\) and \(\vec{C}\) perpendicular to \(\vec{A}\) must agree; parallel components are free.

R.2-8● EasyTier 1?
Review: Given \(\mathbf{A}\) and \(\mathbf{B}\), find (a) the component of \(\mathbf{A}\) in the direction of \(\mathbf{B}\); (b) the component of \(\mathbf{B}\) in the direction of \(\mathbf{A}\). → Answer
R.2-8 Answer↑ Question

(a) \((\vec{A}\cdot\vec{B})/|\vec{B}|=A\cos\theta\). (b) \((\vec{A}\cdot\vec{B})/|\vec{A}|=B\cos\theta\).

R.2-9● EasyTier 1?
Review: What makes a coordinate system (a) orthogonal, (b) curvilinear, (c) right-handed? → Answer
R.2-9 Answer↑ Question

(a) Orthogonal: basis vectors mutually perpendicular at every point. (b) Curvilinear: coordinate surfaces are curved (e.g., spheres, cylinders). (c) Right-handed: \(\hat{u}_1\times\hat{u}_2=\hat{u}_3\) cyclically.

R.2-10● EasyTier 1?
Review: Given \(\mathbf{F}\) in orthogonal curvilinear coordinates, explain how to determine \(|\mathbf{F}|\) and \(\mathbf{a}_F\). → Answer
R.2-10 Answer↑ Question

\(|\vec{F}|=\sqrt{F_1^2+F_2^2+F_3^2}\); unit vector \(\hat{a}_F=\vec{F}/|\vec{F}|\).

R.2-11● EasyTier 1?
Review: What are metric coefficients? → Answer
R.2-11 Answer↑ Question

Scale factors \(h_i\) such that \(d\ell_i=h_i\,du_i\); they convert coordinate differentials to physical lengths. Examples: cylindrical \((1,\rho,1)\); spherical \((1,r,r\sin\theta)\).

R.2-12● EasyTier 1?
Review: Given \(P_1(1,2,3)\) and \(P_2(-1,0,2)\), write \(\overrightarrow{P_1P_2}\) and \(\overrightarrow{P_2P_1}\) in Cartesian coordinates. → Answer
R.2-12 Answer↑ Question

\(\overrightarrow{P_1P_2}=-2\hat{x}-2\hat{y}-\hat{z}\); \(\overrightarrow{P_2P_1}=2\hat{x}+2\hat{y}+\hat{z}\).

R.2-13● EasyTier 1?
Review: Write \(\mathbf{A}\cdot\mathbf{B}\) and \(\mathbf{A}\times\mathbf{B}\) in Cartesian coordinates. → Answer
R.2-13 Answer↑ Question

\(\vec{A}\cdot\vec{B}=A_xB_x+A_yB_y+A_zB_z\); \(\vec{A}\times\vec{B}=\det\begin{pmatrix}\hat{x}&\hat{y}&\hat{z}\\A_x&A_y&A_z\\B_x&B_y&B_z\end{pmatrix}\).

R.2-14● EasyTier 1?
Review: What is the difference between a scalar quantity and a scalar field? A vector quantity and a vector field? → Answer
R.2-14 Answer↑ Question

A scalar/vector quantity is a single value. A scalar/vector field is a function of position (and possibly time) defined throughout a region.

R.2-15● EasyTier 1?
Review: What is the physical definition of the gradient of a scalar field? → Answer
R.2-15 Answer↑ Question

A vector pointing in the direction of maximum spatial increase of the scalar; magnitude equals that maximum rate of change.

R.2-16● EasyTier 1?
Review: Express the space rate of change of a scalar in a given direction in terms of its gradient. → Answer
R.2-16 Answer↑ Question

Directional derivative: \(\partial V/\partial \ell=\nabla V\cdot\hat{a}_{\ell}\).

R.2-17● EasyTier 1?
Review: What does the del operator \(\nabla\) stand for in Cartesian coordinates? → Answer
R.2-17 Answer↑ Question

\(\nabla=\hat{x}\dfrac{\partial}{\partial x}+\hat{y}\dfrac{\partial}{\partial y}+\hat{z}\dfrac{\partial}{\partial z}\).

R.2-18● EasyTier 1?
Review: What is the physical definition of the divergence of a vector field? → Answer
R.2-18 Answer↑ Question

Net outward flux per unit volume at a point: \(\nabla\cdot\vec{A}=\lim_{\Delta v\to 0}\dfrac{1}{\Delta v}\oint\vec{A}\cdot d\vec{S}\).

R.2-19● EasyTier 1?
Review: A vector field with only radial flux lines cannot be solenoidal. True or false? → Answer
R.2-19 Answer↑ Question

False. \(\vec{A}=\hat{a}_r/r^2\) is purely radial yet has zero divergence away from the origin (Coulomb-style field).

R.2-20● EasyTier 1?
Review: A vector field with only curved flux lines can have a nonzero divergence. True or false? → Answer
R.2-20 Answer↑ Question

True. Divergence depends on net flux through a small surface, not on whether flux lines curve.

R.2-21● EasyTier 1?
Review: State the divergence theorem in words and in symbols. → Answer
R.2-21 Answer↑ Question

The volume integral of \(\nabla\cdot\vec{A}\) equals the net outward flux through the enclosing surface: \(\int_V\nabla\cdot\vec{A}\,dV=\oint_S\vec{A}\cdot d\vec{S}\).

R.2-22● EasyTier 1?
Review: What is the physical definition of the curl of a vector field? → Answer
R.2-22 Answer↑ Question

A vector whose magnitude is the maximum line-integral-per-unit-area at a point, and direction normal to that area by the right-hand rule.

R.2-23● EasyTier 1?
Review: State Stokes's theorem in words and in symbols. → Answer
R.2-23 Answer↑ Question

Surface integral of curl equals line integral around boundary: \(\int_S(\nabla\times\vec{A})\cdot d\vec{S}=\oint_C\vec{A}\cdot d\vec{\ell}\).

R.2-24● EasyTier 1?
Review: State the two null identities and explain their significance. → Answer
R.2-24 Answer↑ Question

(1) \(\nabla\times(\nabla V)=0\) — gradient is curl-free. (2) \(\nabla\cdot(\nabla\times\vec{A})=0\) — curl is divergence-free. They allow expressing fields via potentials.

R.2-25● EasyTier 1?
Review: State Helmholtz's theorem. Why does it let us build electromagnetism from postulates on \(\nabla\cdot\mathbf{E}\) and \(\nabla\times\mathbf{E}\)? → Answer
R.2-25 Answer↑ Question

A vector field is uniquely determined (up to constant) by its divergence, curl, and boundary conditions. This is why electromagnetism can be axiomatized via postulates on \(\nabla\cdot\vec{E}\) and \(\nabla\times\vec{E}\).


Ex 2.1●● MediumTier 1?
Dot Product and Angle Between Vectors

A = 2a_x + 3a_y, B = 4a_x − a_y. Find A·B, |A|, |B|, and the angle between them.

→ Solution
Ex 2.1 Solution↑ Problem
\[A\cdot B = (2)(4) + (3)(-1) + (0)(0) = 8 - 3 = 5\]
\[|A| = \sqrt{4 + 9} = \sqrt{13} = 3.606\]
\[|B| = \sqrt{16 + 1} = \sqrt{17} = 4.123\]
\[cos \theta = A\cdot B/(|A||B|) = 5/(3.606 \times 4.123) = 5/14.87 = 0.336\]
\[\theta = arccos(0.336) = 70.4°\]

Ex 2.2●● MediumTier 1?
Cross Product in Cylindrical Coordinates

A = 2a_ρ + 3a_φ, B = a_ρ + 2a_z. Find A × B.

→ Solution
Ex 2.2 Solution↑ Problem
Using the determinant:
\[\vec{A} \times \vec{B} = \begin{vmatrix} \hat{a}_\rho & \hat{a}_\phi & \hat{a}_z \\ 2 & 3 & 0 \\ 1 & 0 & 2 \end{vmatrix}\]
\[= \hat{a}_\rho(3\cdot 2 - 0\cdot 0) - \hat{a}_\phi(2\cdot 2 - 0\cdot 1) + \hat{a}_z(2\cdot 0 - 3\cdot 1)\]
\[= 6\hat{a}_\rho - 4\hat{a}_\phi - 3\hat{a}_z\]
Verify: \(\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}\)
\[B \times A = -6a_\rho + 4a_\phi + 3a_z \checkmark \]

Ex 2.3●● MediumTier 1?
Cartesian to Spherical Coordinate Conversion

Convert point P(1, 1, 1) from Cartesian to spherical coordinates.

→ Solution
Ex 2.3 Solution↑ Problem
\[r = \sqrt{x^{2} + y^{2} + z^{2}} = \sqrt{1+1+1} = \sqrt{3} = 1.732\]
\[\theta = arccos(z/r) = arccos(1/\sqrt{3}) = arccos(0.5774) = 54.74°\]
\[\phi = arctan(y/x) = arctan(1/1) = arctan(1) = 45°\]
Result: (r, θ, φ) = (√3, 54.74°, 45°)
Check: x = r sinθ cosφ = √3 × sin54.74° × cos45° = √3×0.8165×0.7071 = 1 ✓

Ex 2.4●● MediumTier 1?
Surface Integral Using Divergence Theorem

Evaluate ∬ A·dS over the surface of a unit cube (0≤x,y,z≤1) for A = x a_x + y a_y + z a_z.

→ Solution
Ex 2.4 Solution↑ Problem
Direct application of divergence theorem:
\[\iint A\cdot dS = ∭ \nabla \cdot A dV\]
\[\nabla \cdot A = \partial x/\partial x + \partial y/\partial y + \partial z/\partial z = 1 + 1 + 1 = 3\]
\[∭ 3 dV = 3 \times (1\times 1\times 1) = 3\]
Verify directly for one face:
\[Face z=1 (\hat{n} = a_z, dS = dxdy):\]
\[\iint A\cdot a_z dS = \iint z|_{z=1} dxdy = \int _{0}^{1}\int _{0}^{1} 1 dxdy = 1\]
\[Face z=0: contribution = 0\]
By symmetry, each axis pair contributes 1 → total = 3 ✓

Ex 2.5●● MediumTier 1?
Gradient of a Scalar Field

Find ∇φ for φ = x²y + yz². Evaluate at P(1, 2, 3).

→ Solution
Ex 2.5 Solution↑ Problem
\[\partial \phi /\partial x = 2xy\]
\[\partial \phi /\partial y = x^{2} + z^{2}\]
\[\partial \phi /\partial z = 2yz\]
\[\nabla \phi = 2xy a_x + (x^{2} + z^{2}) a_y + 2yz a_z\]
At P(1, 2, 3):
\[\nabla \phi = 2(1)(2) a_x + (1 + 9) a_y + 2(2)(3) a_z\]
\[= 4 a_x + 10 a_y + 12 a_z\]
\[|\nabla \phi | = \sqrt{16 + 100 + 144} = \sqrt{260} = 16.12\]
The gradient points in the direction of maximum rate of increase of φ.

Ex 2.6●● MediumTier 1?
Divergence in Spherical Coordinates

Calculate ∇·(r² â_r) in spherical coordinates.

→ Solution
Ex 2.6 Solution↑ Problem
For A = A_r â_r with A_r = r²:
\[\nabla \cdot A = (1/r^{2}) d(r^{2} A_r)/dr = (1/r^{2}) d(r^{2} \times r^{2})/dr = (1/r^{2}) d(r^{4})/dr\]
\[= (1/r^{2}) \times 4r^{3} = 4r\]
\[\nabla \cdot (r^{2} \hat{a}_{r}) = 4r\]
Physical note: this is NOT zero, meaning r² â_r has sources everywhere.
Contrast with r̂/r² which has zero divergence (except at origin).

Ex 2.7●● MediumTier 1?
Divergence Theorem Verification

Verify the divergence theorem for A = x a_x over the unit cube.

→ Solution
Ex 2.7 Solution↑ Problem
\[\nabla \cdot A = \partial x/\partial x = 1\]
Volume integral:
\[∭ \nabla \cdot A dV = ∭ 1 dV = 1\]
Surface integral (6 faces):
\[x=1 (\hat{n}=a_x): \iint x a_x\cdot a_x dS = \iint 1 dS = 1\]
\[x=0 (\hat{n}=-a_x): \iint x a_x\cdot (-a_x) dS = \iint 0 dS = 0\]
\[y=1 (\hat{n}=a_y): A\cdot a_y = 0\]
\[y=0 (\hat{n}=-a_y): A\cdot (-a_y) = 0\]
\[z=1 (\hat{n}=a_z): A\cdot a_z = 0\]
\[z=0 (\hat{n}=-a_z): A\cdot (-a_z) = 0\]
Total surface integral = 1 + 0 + 0 = 1 ✓

Ex 2.8●● MediumTier 1?
Curl in Cylindrical Coordinates

Find ∇ × A for A = ρ² a_φ.

→ Solution
Ex 2.8 Solution↑ Problem
In cylindrical coordinates (A_ρ=0, A_φ=ρ², A_z=0):
\[(\nabla \times A)_\rho = (1/\rho )\partial A_z/\partial \phi - \partial A_\phi /\partial z = 0\]
\[(\nabla \times A)_\phi = \partial A_\rho /\partial z - \partial A_z/\partial \rho = 0\]
\[(\nabla \times A)_z = (1/\rho )[\partial (\rho A_\phi )/\partial \rho - \partial A_\rho /\partial \phi ]\]
\[= (1/\rho ) \partial (\rho \times \rho ^{2})/\partial \rho \]
\[= (1/\rho ) \partial (\rho ^{3})/\partial \rho \]
\[= (1/\rho ) \times 3\rho ^{2} = 3\rho \]
\[\nabla \times A = 3\rho a_z\]

Ex 2.9●● MediumTier 1?
Stokes's Theorem Verification

For A = y a_x, verify Stokes's theorem using the unit square in the xy-plane (counterclockwise).

→ Solution
Ex 2.9 Solution↑ Problem
Surface integral:
\[\nabla \times A = (\partial A_y/\partial x - \partial A_x/\partial y) a_z = (0 - 1) a_z = -a_z\]
\[\iint (\nabla \times A)\cdot dS = -1 \times (1\times 1) = -1\]
Line integral (counterclockwise: C1→C2→C3→C4):
\[C1: y=0, x: 0\to 1: \int y dx = 0\]
\[C2: x=1, y: 0\to 1: \int y a_x\cdot a_y dy = 0\]
\[C3: y=1, x: 1\to 0: \int 1\cdot dx = \int _{1}^{0} dx = -1\]
\[C4: x=0, y: 1\to 0: \int y a_x\cdot a_y(-dy) = 0\]
Total: 0 + 0 + (-1) + 0 = -1 ✓

Ex 2.10●● MediumTier 1?
Null Identity: ∇·(∇×A) = 0

Prove that the divergence of a curl is always zero.

→ Solution
Ex 2.10 Solution↑ Problem
In Cartesian:
\[\nabla \times A = (\partial A_z/\partial y - \partial A_y/\partial z) a_x\]
+ (∂A_x/∂z - ∂A_z/∂x) a_y
+ (∂A_y/∂x - ∂A_x/∂y) a_z
\[\nabla \cdot (\nabla \times A) = \partial /\partial x(\partial A_z/\partial y - \partial A_y/\partial z)\]
+ ∂/∂y(∂A_x/∂z - ∂A_z/∂x)
+ ∂/∂z(∂A_y/∂x - ∂A_x/∂y)
\[= \partial ^{2}A_z/\partial x\partial y - \partial ^{2}A_y/\partial x\partial z\]
+ ∂²A_x/∂y∂z - ∂²A_z/∂y∂x
+ ∂²A_y/∂z∂x - ∂²A_x/∂z∂y
For continuous second derivatives, mixed partials are equal:
∂²A_z/∂x∂y = ∂²A_z/∂y∂x → these cancel
Similarly for the other pairs.
\[\to \nabla \cdot (\nabla \times A) = 0 \checkmark \]

Ex 2.11●● MediumTier 1?
Null Identity: ∇×(∇φ) = 0

Prove the curl of a gradient is always zero.

→ Solution
Ex 2.11 Solution↑ Problem
\[\nabla \phi = \partial \phi /\partial x a_x + \partial \phi /\partial y a_y + \partial \phi /\partial z a_z\]
\[(\nabla \times \nabla \phi )_x = \partial /\partial y(\partial \phi /\partial z) - \partial /\partial z(\partial \phi /\partial y)\]
= ∂²φ/∂y∂z - ∂²φ/∂z∂y = 0 (mixed partials equal)
Similarly for y and z components.
\[\to \nabla \times (\nabla \phi ) = 0 \checkmark \]
Physical meaning: an electrostatic field E = -∇V is always irrotational,
consistent with ∇×E = 0 in static cases.

Ex 2.12●● MediumTier 1?
Laplacian in Cylindrical Coordinates

Find ∇²φ in cylindrical form. Apply it to φ = ρ² cos(2α) (where α is the azimuthal angle).

→ Solution
Ex 2.12 Solution↑ Problem
General form:
\[\nabla ^{2}\phi = (1/\rho )\partial /\partial \rho (\rho \partial \phi /\partial \rho ) + (1/\rho ^{2})\partial ^{2}\phi /\partial \alpha ^{2} + \partial ^{2}\phi /\partial z^{2}\]
For φ = ρ² cos(2α):
\[\partial \phi /\partial \rho = 2\rho cos(2\alpha )\]
\[(1/\rho )\partial (\rho \times 2\rho cos2\alpha )/\partial \rho = (1/\rho ) \times 4\rho cos(2\alpha ) = 4 cos(2\alpha )\]
\[\partial ^{2}\phi /\partial \alpha ^{2} = -4\rho ^{2} cos(2\alpha )\]
\[(1/\rho ^{2})(-4\rho ^{2} cos(2\alpha )) = -4 cos(2\alpha )\]
\[\partial ^{2}\phi /\partial z^{2} = 0\]
\[\nabla ^{2}\phi = 4cos(2\alpha ) - 4cos(2\alpha ) = 0\]
φ = ρ² cos(2α) satisfies Laplace's equation. (Used in wedge problems.)

Ex 2.13●● MediumTier 1?
Unit Normal to a Surface

Find the unit normal to the surface z = x² + y² at point (1, 1, 2).

→ Solution
Ex 2.13 Solution↑ Problem
\[Rewrite as F(x,y,z) = z - x^{2} - y^{2} = 0\]
\[\nabla F = -2x a_x - 2y a_y + a_z\]
At (1, 1, 2):
\[\nabla F = -2 a_x - 2 a_y + a_z\]
\[|\nabla F| = \sqrt{4 + 4 + 1} = 3\]
\[\hat{n} = \nabla F/|\nabla F| = (-2a_x - 2a_y + a_z)/3\]
\[= -0.667 a_x - 0.667 a_y + 0.333 a_z\]
(This is the upward-pointing normal since the a_z component is positive.)

Ex 2.14●● MediumTier 1?
Circulation of a Vector Field

Evaluate ∮ A·dl for A = −y a_x + x a_y around a circle of radius a in the xy-plane.

→ Solution
Ex 2.14 Solution↑ Problem
Parametrize: x = a cosφ, y = a sinφ, dl = a dφ(-sinφ a_x + cosφ a_y)
A on circle: -a sinφ a_x + a cosφ a_y
\[A\cdot dl = (-a sin\phi )(a)(-sin\phi ) d\phi + (a cos\phi )(a)(cos\phi ) d\phi \]
\[= a^{2}sin^{2}\phi d\phi + a^{2}cos^{2}\phi d\phi = a^{2} d\phi \]
\[\oint A\cdot dl = \int _{0}^{2\pi } a^{2} d\phi = 2\pi a^{2}\]
Verify via Stokes:
\[\nabla \times A = (\partial x/\partial x - \partial (-y)/\partial y) a_z = (1+1) a_z = 2a_z\]
\[\iint (\nabla \times A)\cdot dS = 2 \times \pi a^{2} = 2\pi a^{2} \checkmark \]

Ex 2.15●● MediumTier 1?
Helmholtz Theorem

Decompose F = x a_x + y a_y + z a_z into irrotational and solenoidal parts.

→ Solution
Ex 2.15 Solution↑ Problem
Check divergence and curl:
\[\nabla \cdot F = 1 + 1 + 1 = 3 (not zero \to has irrotational part)\]
∇×F = 0 (irrotational — purely a gradient)
Since ∇×F = 0, the solenoidal part is zero (A = 0).
The field is purely irrotational: F = -∇φ
\[Find \phi : -\nabla \phi = F = \hat{r}r \to \phi = -r^{2}/2 = -(x^{2}+y^{2}+z^{2})/2\]
Verify: -∇φ = -(∂(-r²/2)/∂x a_x + ...) = x a_x + y a_y + z a_z = F ✓
Helmholtz: F = -∇φ + ∇×A
Here: φ = -(x²+y²+z²)/2, A = 0

Textbook Practice Problems

2-1● EasyTier 1?
Vector operations → Answer
2-2● EasyTier 1?
Vector operations → Answer
2-5● EasyTier 1?
Coordinate systems → Answer
2-6● EasyTier 1?
Coordinate systems → Answer
2-13● EasyTier 1?
Gradient → Answer
2-14●● MediumTier 1?
Gradient → Answer
2-17●● MediumTier 1?
Divergence and divergence theorem → Answer
2-18●● MediumTier 1?
Divergence and divergence theorem → Answer
2-3● EasyTier 2?
Vector operations → Answer
2-7●● MediumTier 2?
Coordinate systems → Answer
2-9●● MediumTier 2?
Line and surface integrals → Answer
2-10●● MediumTier 2?
Line and surface integrals → Answer
2-15●● MediumTier 2?
Gradient → Answer
2-19●●● HardTier 2?
Divergence and divergence theorem → Answer
2-21●●● HardTier 2?
Stokes's theorem → Answer
2-11●●● HardTier 3?
Line and surface integrals → Answer

Textbook Practice — Approach Hints

Sample: Given \(\vec{A}=2\hat{x}+3\hat{y}-\hat{z}\), \(\vec{B}=\hat{x}-\hat{y}+2\hat{z}\). Find \(\vec{A}\cdot\vec{B}\), \(|\vec{A}|\), \(|\vec{B}|\) and the angle.

\[\vec{A}\cdot\vec{B}=2(1)+3(-1)+(-1)(2)=2-3-2=-3\]
\[|\vec{A}|=\sqrt{4+9+1}=\sqrt{14}=3.742,\quad|\vec{B}|=\sqrt{1+1+4}=\sqrt{6}=2.449\]
\[\cos\theta=\frac{\vec{A}\cdot\vec{B}}{|\vec{A}||\vec{B}|}=\frac{-3}{3.742\cdot2.449}=-0.327\;\Rightarrow\;\theta=109.1^{\circ}\]

Sample: \(\vec{A}=\hat{x}+2\hat{y}+3\hat{z}\), \(\vec{B}=4\hat{x}-\hat{y}+\hat{z}\). Find \(\vec{A}\times\vec{B}\) and a unit vector normal to both.

\[\vec{A}\times\vec{B}=\begin{vmatrix}\hat{x}&\hat{y}&\hat{z}\\1&2&3\\4&-1&1\end{vmatrix}=(2\!\cdot\!1-3\!\cdot\!(-1))\hat{x}-(1\!\cdot\!1-3\!\cdot\!4)\hat{y}+(1\!\cdot\!(-1)-2\!\cdot\!4)\hat{z}\]
\[=5\hat{x}+11\hat{y}-9\hat{z}\]
\[\hat{n}=\frac{\vec{A}\times\vec{B}}{|\vec{A}\times\vec{B}|}=\frac{5\hat{x}+11\hat{y}-9\hat{z}}{\sqrt{25+121+81}}=\frac{5\hat{x}+11\hat{y}-9\hat{z}}{15.07}\]

Sample: Convert \(P(2,2,1)\) from Cartesian to cylindrical and spherical.

\[\text{Cylindrical: }\rho=\sqrt{x^2+y^2}=\sqrt{8}=2.828,\;\phi=\tan^{-1}(y/x)=45^{\circ},\;z=1\]
\[\text{Spherical: }r=\sqrt{x^2+y^2+z^2}=3,\;\theta=\cos^{-1}(z/r)=70.53^{\circ},\;\phi=45^{\circ}\]

Sample: Express \(\vec{A}=3\hat{x}+2\hat{y}-\hat{z}\) at \(P(1,\sqrt{3},2)\) in cylindrical components.

\[\rho=\sqrt{1+3}=2,\;\phi=\tan^{-1}(\sqrt{3})=60^{\circ}\]
\[A_{\rho}=A_x\cos\phi+A_y\sin\phi=3(0.5)+2(0.866)=3.232\]
\[A_{\phi}=-A_x\sin\phi+A_y\cos\phi=-3(0.866)+2(0.5)=-1.598\]
\[\vec{A}=3.232\hat{a}_{\rho}-1.598\hat{a}_{\phi}-\hat{z}\]

Sample: \(V=x^2y+yz^2\). Find \(\nabla V\) at \(P(1,2,-1)\).

\[\nabla V=\frac{\partial V}{\partial x}\hat{x}+\frac{\partial V}{\partial y}\hat{y}+\frac{\partial V}{\partial z}\hat{z}=2xy\,\hat{x}+(x^2+z^2)\hat{y}+2yz\,\hat{z}\]
\[\nabla V|_{(1,2,-1)}=4\hat{x}+2\hat{y}-4\hat{z}\;\;\Rightarrow\;\;|\nabla V|=6\]

Sample: \(V=r^2\sin\theta\cos\phi\) in spherical. Find \(\nabla V\).

\[\nabla V=\frac{\partial V}{\partial r}\hat{a}_r+\frac{1}{r}\frac{\partial V}{\partial\theta}\hat{a}_{\theta}+\frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}\hat{a}_{\phi}\]
\[=2r\sin\theta\cos\phi\,\hat{a}_r+r\cos\theta\cos\phi\,\hat{a}_{\theta}-\frac{r\sin\phi}{\sin\theta}\!\cdot\!\sin\theta\,\hat{a}_{\phi}\]
\[=2r\sin\theta\cos\phi\,\hat{a}_r+r\cos\theta\cos\phi\,\hat{a}_{\theta}-r\sin\phi\,\hat{a}_{\phi}\]

Sample: Evaluate \(\nabla\cdot\vec{A}\) for \(\vec{A}=xy\hat{x}+yz\hat{y}+zx\hat{z}\).

\[\nabla\cdot\vec{A}=\frac{\partial(xy)}{\partial x}+\frac{\partial(yz)}{\partial y}+\frac{\partial(zx)}{\partial z}=y+z+x\]
\[\nabla\cdot\vec{A}|_{(1,2,3)}=1+2+3=6\]

Sample: Verify divergence theorem for \(\vec{A}=x\hat{x}+y\hat{y}+z\hat{z}\) over a unit sphere.

\[\nabla\cdot\vec{A}=1+1+1=3,\quad\int_V\nabla\cdot\vec{A}\,dV=3\cdot\tfrac{4}{3}\pi=4\pi\]
\[\oint\vec{A}\cdot d\vec{S}=\oint r\,dS=1\cdot 4\pi=4\pi\;\checkmark\]

Sample: Show that \(\vec{A}\cdot(\vec{B}\times\vec{C})\) equals the volume spanned by the three vectors. Use \(\vec{A}=\hat{x}\), \(\vec{B}=\hat{y}\), \(\vec{C}=\hat{z}\).

\[\vec{B}\times\vec{C}=\hat{y}\times\hat{z}=\hat{x}\]
\[\vec{A}\cdot(\vec{B}\times\vec{C})=\hat{x}\cdot\hat{x}=1\]
Triple product = signed volume of parallelepiped. Cyclic property: \(\vec{A}\cdot(\vec{B}\times\vec{C})=\vec{B}\cdot(\vec{C}\times\vec{A})=\vec{C}\cdot(\vec{A}\times\vec{B})\).

Sample: A vector field is given as \(\vec{F}=r\hat{a}_r\) in spherical. Express in Cartesian.

\[\hat{a}_r=\sin\theta\cos\phi\,\hat{x}+\sin\theta\sin\phi\,\hat{y}+\cos\theta\,\hat{z}\]
\[r=\sqrt{x^2+y^2+z^2},\;\sin\theta=\sqrt{x^2+y^2}/r,\;\cos\theta=z/r\]
\[\vec{F}=x\hat{x}+y\hat{y}+z\hat{z}\;\;(\text{the position vector itself})\]

Sample: Evaluate \(\int_C\vec{F}\cdot d\vec{\ell}\) for \(\vec{F}=y\hat{x}+x\hat{y}\) along the parabola \(y=x^2\) from \((0,0)\) to \((1,1)\).

\[d\vec{\ell}=dx\,\hat{x}+dy\,\hat{y}=(\hat{x}+2x\,\hat{y})dx\]
\[\vec{F}\cdot d\vec{\ell}=(x^2)(1)dx+(x)(2x)dx=(x^2+2x^2)dx=3x^2\,dx\]
\[\int_0^1 3x^2\,dx=x^3\Big|_0^1=1\]

Sample: Compute \(\oint\vec{A}\cdot d\vec{S}\) for \(\vec{A}=x^2\hat{x}+y^2\hat{y}+z^2\hat{z}\) over the closed surface of a unit cube.

\[\text{Use divergence theorem: }\nabla\cdot\vec{A}=2x+2y+2z\]
\[\oint\vec{A}\cdot d\vec{S}=\int_0^1\!\int_0^1\!\int_0^1(2x+2y+2z)\,dx\,dy\,dz=3\!\cdot\!2\!\cdot\!\tfrac12=3\]

Sample: Find a unit normal to the surface \(x^2+y^2-z=0\) at \(P(1,1,2)\).

\[\nabla F=2x\hat{x}+2y\hat{y}-\hat{z}\;\Rightarrow\;\nabla F|_{P}=2\hat{x}+2\hat{y}-\hat{z}\]
\[\hat{n}=\frac{\nabla F}{|\nabla F|}=\frac{2\hat{x}+2\hat{y}-\hat{z}}{3}\]

Sample: \(\vec{A}=\rho\hat{a}_{\rho}+z\hat{z}\) in cylindrical. Verify \(\oint\vec{A}\cdot d\vec{S}=\int\nabla\cdot\vec{A}\,dV\) over a cylinder of radius 2, \(0\le z\le 3\).

\[\nabla\cdot\vec{A}=\frac{1}{\rho}\frac{\partial(\rho\cdot\rho)}{\partial\rho}+\frac{\partial z}{\partial z}=\frac{2\rho}{\rho}+1=3\]
\[\int_V\!3\,dV=3(\pi\cdot4)(3)=36\pi\]
\[\oint\vec{A}\cdot d\vec{S}=\underbrace{2\!\cdot\!(2\pi\!\cdot\!2)(3)}_{\text{side}}+\underbrace{3(\pi\!\cdot\!4)}_{\text{top}}+\underbrace{0}_{\text{bottom}}=24\pi+12\pi=36\pi\;\checkmark\]

Sample: Verify Stokes's theorem for \(\vec{A}=y\hat{x}-x\hat{y}\) over a unit disc in the \(xy\)-plane.

\[\nabla\times\vec{A}=\hat{z}\!\left(\!\frac{\partial(-x)}{\partial x}-\frac{\partial y}{\partial y}\!\right)=-2\hat{z}\]
\[\int_S(\nabla\times\vec{A})\cdot d\vec{S}=-2(\pi\!\cdot\!1)=-2\pi\]
\[\oint\vec{A}\cdot d\vec{\ell}=\int_0^{2\pi}(y\,dx-x\,dy)=\int_0^{2\pi}(-\sin^2\!\phi-\cos^2\!\phi)\,d\phi=-2\pi\;\checkmark\]

Sample: Compute \(\int_S\vec{A}\cdot d\vec{S}\) for \(\vec{A}=\rho\hat{a}_{\rho}\) over the curved surface of a cylinder \(\rho=2\), \(0\le z\le 4\).

\[d\vec{S}=\rho\,d\phi\,dz\,\hat{a}_{\rho}\;\text{at}\;\rho=2\]
\[\vec{A}\cdot d\vec{S}=\rho\cdot\rho\,d\phi\,dz=4\,d\phi\,dz\]
\[\int_0^{2\pi}\!\int_0^4 4\,d\phi\,dz=4(2\pi)(4)=32\pi\]

Chapter 3 — Static Electric Fields

Key Theory — Chapter 3

Condensed from Cheng, Field and Wave Electromagnetics, §3-1 through §3-11. Read this before attempting the problems below.


3-1   Introduction — Why Postulate, Not Derive?

Elementary texts start from Coulomb's experimental law and build upward. Cheng instead postulates the divergence and curl of \(\mathbf{E}\) in free space, then derives Coulomb's and Gauss's laws from them. Justification:

In electrostatics, charges are at rest, fields do not change with time, and no magnetic field is present. Only one of the four field quantities — \(\mathbf{E}\) — and one of the three universal constants — \(\epsilon_0\) — are needed in free space.


3-2   Fundamental Postulates of Electrostatics in Free Space

Electric field intensity is defined as the force per unit charge on a stationary test charge:

\[ \mathbf{E} = \lim_{q \to 0} \frac{\mathbf{F}}{q} \quad (\text{V/m}) \qquad \text{(3-2)} \]
\[ \mathbf{F} = q\mathbf{E} \quad (\text{N}) \qquad \text{(3-3)} \]

The entire theory of electrostatics in free space rests on two postulates — the divergence and the curl of \(\mathbf{E}\):

\[ \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} \qquad \text{(3-4)} \]
\[ \nabla \times \mathbf{E} = 0 \qquad \text{(3-5)} \]

Integral forms follow by the divergence theorem and Stokes's theorem respectively:

\[ \oint_S \mathbf{E}\cdot d\mathbf{s} = \frac{Q}{\epsilon_0} \quad \text{(Gauss's law)} \qquad \text{(3-7)} \]
\[ \oint_C \mathbf{E}\cdot d\mathbf{\ell} = 0 \quad \text{(conservative)} \qquad \text{(3-8)} \]

Physical reading:

Differential Form Integral Form
Divergence postulate ∇·E = ρ/ε₀ ∮ₛ E·ds = Q/ε₀
Curl postulate ∇×E = 0 ∮_C E·dℓ = 0
Summary — Postulates of Electrostatics in Free Space

3-3   Coulomb's Law (Derived)

Apply (3-7) to a spherical Gaussian surface of radius \(R\) centered on a point charge \(q\) in free space. Symmetry forces \(\mathbf{E}=\mathbf{a}_R E_R\), giving \(E_R(4\pi R^2)=q/\epsilon_0\), hence:

\[ \mathbf{E} = \mathbf{a}_R \frac{q}{4\pi\epsilon_0 R^2} \quad (\text{V/m}) \qquad \text{(3-12)} \]
\[ \mathbf{F}_{12} = \mathbf{a}_{R_{12}} \frac{q_1 q_2}{4\pi\epsilon_0 R_{12}^{2}} \quad \text{(Coulomb's law)} \qquad \text{(3-1)} \]

For a system of discrete charges, superposition gives \(\mathbf{E}=\sum_k \mathbf{E}_k\). For a continuous distribution of volume charge density \(\rho(\mathbf{R}')\):

\[ \mathbf{E}(\mathbf{R}) = \frac{1}{4\pi\epsilon_0}\int_{V'} \frac{\rho(\mathbf{R}')}{|\mathbf{R}-\mathbf{R}'|^{3}}(\mathbf{R}-\mathbf{R}')\,dv' \]

3-4   Gauss's Law — Strategy

Gauss's law (3-7) is most powerful when a Gaussian surface can be constructed over which \(\mathbf{E}\cdot d\mathbf{s}\) is constant. Three symmetry classes permit this:

Moral: if symmetry fits, use (3-7) directly. Otherwise find \(V\) first by integration, then \(\mathbf{E}=-\nabla V\).


3-5   Electric Potential

Because \(\nabla\times\mathbf{E}=0\), by the null identity \(\nabla\times(\nabla V)\equiv 0\) we can write \(\mathbf{E}\) as the gradient of a scalar:

\[ \mathbf{E} = -\nabla V \qquad \text{(3-43)} \]

The negative sign makes \(V\) increase in the direction opposite \(\mathbf{E}\), consistent with potential energy per unit charge. The potential difference between two points is path-independent:

\[ V_2 - V_1 = -\int_{P_1}^{P_2} \mathbf{E}\cdot d\mathbf{\ell} \quad (\text{V}) \qquad \text{(3-45)} \]

Taking the reference at infinity, the potentials of standard charge distributions are:

\[ V = \frac{q}{4\pi\epsilon_0 R} \quad \text{(point charge)} \qquad \text{(3-47)} \]
\[ V = \frac{1}{4\pi\epsilon_0}\sum_{k=1}^{N}\frac{q_k}{|\mathbf{R}-\mathbf{R}'_k|} \quad \text{(discrete)} \qquad \text{(3-49)} \]
\[ V = \frac{\mathbf{p}\cdot\mathbf{a}_R}{4\pi\epsilon_0 R^{2}} \quad \text{(dipole, } \mathbf{p}=q\mathbf{d}\text{)} \qquad \text{(3-53b)} \]
\[ V = \frac{1}{4\pi\epsilon_0}\int_{V'} \frac{\rho(\mathbf{R}')}{|\mathbf{R}-\mathbf{R}'|}\,dv' \quad \text{(continuous)} \qquad \text{(3-61)} \]

Field lines and equipotentials are perpendicular — \(\nabla V\) is normal to surfaces of constant \(V\).


3-6   Conductors in Static Electric Field

A conductor has loosely bound electrons that redistribute freely under any interior field. Equilibrium (static) is reached very quickly — for copper the relaxation time is \(\sim 10^{-19}\) s. In equilibrium:

\[ \rho = 0 \quad \text{inside the conductor} \qquad \text{(3-69)} \]
\[ \mathbf{E} = 0 \quad \text{inside the conductor} \qquad \text{(3-70)} \]

At the conductor / free-space boundary, applying (3-8) to a narrow rectangular contour and (3-7) to a shallow pillbox:

\[ E_t = 0 \qquad \text{(3-71)} \]
\[ E_n = \frac{\rho_s}{\epsilon_0} \qquad \text{(3-72)} \]

Consequences:


3-7   Dielectrics — Polarization

In a dielectric, electrons are bound. An applied \(\mathbf{E}\) displaces them slightly, creating induced dipoles. The macroscopic measure is the polarization vector \(\mathbf{P}\) — electric dipole moment per unit volume (C/m²).

Polarized matter behaves as if it carried equivalent bound charges:

\[ \rho_{ps} = \mathbf{P}\cdot\mathbf{a}_n \quad \text{(bound surface charge)} \qquad \text{(3-88)} \]
\[ \rho_p = -\nabla\cdot\mathbf{P} \quad \text{(bound volume charge)} \qquad \text{(3-89)} \]

The total bound charge on any isolated dielectric body is zero.


3-8   Electric Flux Density \(\mathbf{D}\) and Dielectric Constant

In a material medium the divergence postulate (3-4) must account for bound charge: \(\nabla\cdot\mathbf{E}=(\rho+\rho_p)/\epsilon_0\). Defining the electric flux density

\[ \mathbf{D} = \epsilon_0\mathbf{E} + \mathbf{P} \quad (\text{C/m}^2) \qquad \text{(3-97)} \]

absorbs the bound charge and leaves only free charge on the right:

\[ \nabla\cdot\mathbf{D} = \rho \quad (\text{C/m}^3) \qquad \text{(3-98)} \]
\[ \oint_S \mathbf{D}\cdot d\mathbf{s} = Q \quad \text{(Gauss's law, any medium)} \qquad \text{(3-100)} \]

For a linear, isotropic medium \(\mathbf{P}=\epsilon_0\chi_e\mathbf{E}\), so

\[ \mathbf{D} = \epsilon_0\epsilon_r\mathbf{E} = \epsilon\mathbf{E} \qquad \text{(3-102)} \]
\[ \epsilon_r = 1 + \chi_e = \frac{\epsilon}{\epsilon_0} \qquad \text{(3-103)} \]

where \(\epsilon_r\) is the dimensionless dielectric constant (relative permittivity), \(\chi_e\) the electric susceptibility, and \(\epsilon\) the absolute permittivity (F/m).

Dielectric strength (§3-8.1) is the maximum \(E\) a material withstands before breakdown. Not to be confused with \(\epsilon_r\):

Material ε_r Dielectric strength (V/m)
Air 1.0 3 × 10⁶ (3 kV/mm — memorize this)
Mineral oil 2.3 15 × 10⁶
Paper 2–4 15 × 10⁶
Polystyrene 2.6 20 × 10⁶
Rubber 2.3–4.0 25 × 10⁶
Glass 4–10 30 × 10⁶
Mica 6.0 200 × 10⁶
Table 3-1 — Dielectric constants and strengths (after Cheng)

3-9   Boundary Conditions for Electrostatic Fields

Applying the integral forms (3-8) to a narrow contour and (3-100) to a shallow pillbox straddling the interface:

\[ E_{1t} = E_{2t} \quad \text{(tangential } \mathbf{E} \text{ continuous)} \qquad \text{(3-125)} \]
\[ \mathbf{a}_{n2}\cdot(\mathbf{D}_1 - \mathbf{D}_2) = \rho_s \quad \text{(normal } \mathbf{D} \text{ jumps by free surface charge)} \qquad \text{(3-126)} \]

For a charge-free dielectric–dielectric interface (\(\rho_s=0\)):

\[ D_{1n} = D_{2n}, \qquad \epsilon_1 E_{1n} = \epsilon_2 E_{2n} \]
\[ \frac{\tan\alpha_2}{\tan\alpha_1} = \frac{\epsilon_2}{\epsilon_1} \quad \text{(refraction of } \mathbf{E} \text{ at an interface)} \qquad \text{(3-129)} \]

For a conductor–dielectric interface (medium 2 is a conductor, \(\mathbf{D}_2=0\)): \(E_{1t}=0\) and \(D_{1n}=\rho_s\) — the special case (3-71), (3-72) generalized to \(\mathbf{D}\).


3-10   Capacitance and Capacitors

Because \(\rho_s\), \(\mathbf{E}\), and \(V\) all scale linearly with \(Q\) in a static conductor system, the ratio \(Q/V\) depends only on geometry and the medium:

\[ Q = CV \qquad \text{(3-134)} \]
\[ C = \frac{Q}{V_{12}} \quad (\text{F}) \qquad \text{(3-135)} \]

Procedure to find \(C\) for a two-conductor capacitor: (1) choose coordinates; (2) assume \(\pm Q\) on the conductors; (3) find \(\mathbf{E}\) from Gauss's law or from \(\rho_s\); (4) integrate \(V_{12}=-\int_{2}^{1}\mathbf{E}\cdot d\mathbf{\ell}\); (5) take \(C=Q/V_{12}\).

\[ C_{\text{parallel-plate}} = \frac{\epsilon S}{d} \qquad \text{(3-136)} \]
\[ C_{\text{cylindrical}} = \frac{2\pi\epsilon L}{\ln(b/a)} \qquad \text{(3-139)} \]
\[ C_{\text{spherical}} = \frac{4\pi\epsilon}{\dfrac{1}{R_i} - \dfrac{1}{R_o}} \qquad \text{(3-140)} \]
\[ C_{\text{isolated sphere}} = 4\pi\epsilon R_i \quad (R_o \to \infty) \]

Series / parallel: capacitors in parallel add (\(C_\text{par}=\sum C_k\)); capacitors in series combine as reciprocals (\(1/C_\text{ser}=\sum 1/C_k\)) — the opposite of resistors, because charge is shared in series and voltage in parallel.

Electrostatic shielding (§3-10.3): a grounded conducting shell enclosing a body decouples it from external charges — the coupling capacitance vanishes. This is why sensitive electronics live inside grounded metal enclosures.


3-11   Electrostatic Energy and Forces

In terms of charges and potentials — the work to assemble \(N\) point charges:

\[ W_e = \frac{1}{2}\sum_{k=1}^{N} Q_k V_k \quad (\text{J}) \qquad \text{(3-165)} \]
\[ W_e = \frac{1}{2}\int_{V'} \rho V\, dv \quad \text{(continuous)} \qquad \text{(3-170)} \]

The factor \(\tfrac{1}{2}\) avoids double-counting pair interactions. \(V_k\) is the potential at \(Q_k\) due to all other charges.

In terms of field quantities — substituting \(\rho=\nabla\cdot\mathbf{D}\) and applying a vector identity:

\[ W_e = \frac{1}{2}\int_{V'} \mathbf{D}\cdot\mathbf{E}\, dv \qquad \text{(3-176a)} \]
\[ \quad = \frac{1}{2}\int_{V'} \epsilon E^{2}\, dv = \frac{1}{2}\int_{V'} \frac{D^{2}}{\epsilon}\, dv \qquad \text{(3-176b,c)} \]

Energy density:

\[ w_e = \tfrac{1}{2}\mathbf{D}\cdot\mathbf{E} = \tfrac{1}{2}\epsilon E^{2} = \frac{D^{2}}{2\epsilon} \quad (\text{J/m}^3) \qquad \text{(3-178)} \]

Note: (3-165) gives the interaction (mutual) energy only — self-energies of ideal point charges diverge. (3-170) and (3-176) do include self-energy, which is why they agree with each other but can exceed (3-165) for the same system.

Stored energy in a capacitor (equivalent forms):

\[ W_e = \tfrac{1}{2}CV^{2} = \tfrac{1}{2}QV = \frac{Q^{2}}{2C} \qquad \text{(3-180)} \]

Force by virtual displacement. For an isolated system (constant charges), work done by the field equals the decrease in stored energy, giving:

\[ \mathbf{F}_Q = -\nabla W_e \quad \text{(constant } Q\text{)} \qquad \text{(3-185)} \]
\[ \mathbf{F}_V = +\nabla W_e \quad \text{(constant } V\text{, sources supply work)} \]

The sign flip is not a contradiction — in the constant-\(V\) case the battery supplies additional energy equal to twice the mechanical work.


Chapter 3 at a Glance

Problems and Solutions


Review Questions (Tier 1)

R.3-1● EasyTier 1?
Review: Write the fundamental postulates of electrostatics in free space. → Answer
R.3-1 Answer↑ Question

\(\nabla\cdot\vec{E}=\rho_v/\epsilon_0\) and \(\nabla\times\vec{E}=0\).

R.3-2● EasyTier 1?
Review: Why is Coulomb's law derived from the postulates rather than postulated directly? → Answer
R.3-2 Answer↑ Question

Coulomb's law follows by integrating the postulates for a point charge with spherical symmetry — it is a consequence, not an axiom.

R.3-3● EasyTier 1?
Review: Under what conditions is Gauss's law a convenient tool for finding \(\mathbf{E}\)? → Answer
R.3-3 Answer↑ Question

When the charge distribution has high symmetry (spherical, cylindrical, planar) so a Gaussian surface can be chosen on which \(|\vec{E}|\) is constant or normal/tangential to the surface.

R.3-4● EasyTier 1?
Review: Why is electric potential \(V\) defined with a negative gradient? → Answer
R.3-4 Answer↑ Question

\(\vec{E}=-\nabla V\) so that \(\vec{E}\) points from high to low potential, just as gravity points downhill.

R.3-5● EasyTier 1?
Review: State the boundary conditions on \(\mathbf{E}\) and \(\mathbf{D}\) at a conductor surface; at a dielectric–dielectric interface. → Answer
R.3-5 Answer↑ Question

Conductor: \(E_t=0\) and \(D_n=\rho_s\). Dielectric–dielectric (no free \(\rho_s\)): \(E_t\) continuous and \(D_n\) continuous.

R.3-6● EasyTier 1?
Review: Define \(\mathbf{P}\), \(\mathbf{D}\), \(\epsilon_r\), and \(\chi_e\). What is dielectric strength, and how does it differ from dielectric constant? → Answer
R.3-6 Answer↑ Question

\(\vec{P}\) = polarization (dipole moment per unit volume); \(\vec{D}=\epsilon_0\vec{E}+\vec{P}\); \(\epsilon_r=\epsilon/\epsilon_0\); \(\chi_e=\epsilon_r-1\). Dielectric strength is the breakdown field, distinct from \(\epsilon_r\).

R.3-7● EasyTier 1?
Review: Give the capacitance of parallel-plate, cylindrical, and spherical capacitors. → Answer
R.3-7 Answer↑ Question

Parallel-plate \(C=\epsilon A/d\); coaxial \(C=2\pi\epsilon L/\ln(b/a)\); spherical \(C=4\pi\epsilon ab/(b-a)\).

R.3-8● EasyTier 1?
Review: Write three equivalent expressions for the stored energy of a capacitor. → Answer
R.3-8 Answer↑ Question

\(W=\tfrac12 CV^2=\tfrac12 QV=Q^2/(2C)\).

R.3-9● EasyTier 1?
Review: Explain why \(\mathbf{F}_Q=-\nabla W_e\) at constant charge but \(\mathbf{F}_V=+\nabla W_e\) at constant potential. → Answer
R.3-9 Answer↑ Question

At constant \(Q\) the source does no work, so the body moves to decrease stored energy (\(\vec{F}=-\nabla W\)). At constant \(V\) the source supplies twice the work absorbed by the field, so motion increases stored energy (\(\vec{F}=+\nabla W\)).


Ex 3.1●● MediumTier 1?
Point Charge Electric Field

Find E at r = 0.5 m from a point charge Q = 1 μC.

→ Solution
Ex 3.1 Solution↑ Problem
\[E = Q/(4\pi \varepsilon _{0}r^{2}) \hat{a}_{r}\]
\[= 10^{-6}/(4\pi \times 8.85\times 10^{-12} \times 0.25)\]
\[= 10^{-6}/2.789\times 10^{-11}\]
\[= 35,840 \hat{a}_{r} V/m \approx 35.8 kV/m\]
Alternatively: E = kQ/r² = 8.99×10⁹ × 10⁻⁶/0.25 = 35,960 V/m ≈ 36 kV/m

Ex 3.2●● MediumTier 1?
Infinite Line Charge

ρ_L = 2 nC/m. Find E at ρ = 0.1 m from the line.

→ Solution
Ex 3.2 Solution↑ Problem
\[E = \rho _L/(2\pi \varepsilon _{0}\rho ) \hat{a}_{\rho }\]
\[= 2\times 10^{-9}/(2\pi \times 8.85\times 10^{-12} \times 0.1)\]
\[= 2\times 10^{-9}/5.563\times 10^{-12}\]
\[= 359.7 \hat{a}_{\rho } V/m\]
At ρ = 0.2 m: E = 180 V/m (inverse linear decay, not r² for line charge)

Ex 3.3●● MediumTier 1?
Gauss's Law: Uniformly Charged Sphere

Sphere of radius a = 0.1 m with total charge Q = 1 μC (uniform ρ_v). Find E inside and outside.

→ Solution
Ex 3.3 Solution↑ Problem
Volume charge density:
\[\rho _v = Q/(4\pi a^{3}/3) = 3\times 10^{-6}/(4\pi \times 0.001) = 2.387\times 10^{-4} C/m^{3}\]
Outside (r > a): same as point charge
\[E = Q/(4\pi \varepsilon _{0}r^{2}) = 8.99\times 10^{9} \times 10^{-6}/r^{2} = 8990/r^{2} V/m\]
At surface (r = a = 0.1 m):
\[E = Q/(4\pi \varepsilon _{0}a^{2}) = 8990/0.01 = 899,000 V/m \approx 899 kV/m\]
Inside (r < a): only enclosed charge matters
\[E = Qr/(4\pi \varepsilon _{0}a^{3}) = E_surface \times r/a\]
At r = a/2 = 0.05 m:
E = 899 kV/m × 0.5 = 449.5 kV/m (linear increase from center)

Ex 3.4●● MediumTier 1?
Electric Potential Difference

Uniform field E = 1000 a_x V/m. Find V_A − V_B where A = (1,0,0) and B = (0,1,0).

→ Solution
Ex 3.4 Solution↑ Problem
\[V_A - V_B = -\int _B^A E\cdot dl\]
Choose path: B→(1,1,0)→A (segments along y then along x)
Segment 1 (y varies, x=0→1 at y varies): actually use conservative property.
Simpler: direct integral along any path.
Path: straight from B(0,1,0) to A(1,0,0)
\[dl = dx a_x + dy a_y (with dy/dx = -1)\]
But easier — use V = -∫E·dl from reference:
\[V(x) = -\int _{0}^x 1000 dx' = -1000x (taking V=0 at x=0)\]
\[V_A = -1000 \times 1 = -1000 V\]
\[V_B = -1000 \times 0 = 0 V (x=0 regardless of y)\]
\[V_A - V_B = -1000 - 0 = -1000 V\]
B is at higher potential. E points from high to low potential (+x direction). ✓

Ex 3.5●● MediumTier 1?
Parallel-Plate Capacitor

Plates: A = 0.01 m², d = 1 mm, εᵣ = 4. Find C, and V for Q = 1 μC.

→ Solution
Ex 3.5 Solution↑ Problem
\[C = \varepsilon _{0}\varepsilon _{r}A/d = 8.85\times 10^{-12} \times 4 \times 0.01/10^{-3}\]
\[= 8.85\times 10^{-12} \times 40\]
\[= 354\times 10^{-12} F = 354 pF\]
Voltage for Q = 1 μC:
\[V = Q/C = 10^{-6}/354\times 10^{-12} = 2.825 V\]
Electric field:
\[E = V/d = 2.825/10^{-3} = 2825 V/m\]

Ex 3.6●● MediumTier 1?
Electric Field at Conductor Surface

A conductor has surface charge density ρ_s = 5 μC/m². Find E just outside.

→ Solution
Ex 3.6 Solution↑ Problem
Boundary condition at perfect conductor (normal direction â_n away from conductor):
E_n = ρ_s/ε₀ (tangential E = 0 inside and at surface)
\[E = \rho _s/\varepsilon _{0} \hat{a}_{n} = 5\times 10^{-6}/8.85\times 10^{-12} \hat{a}_{n} = 5.65\times 10^{5} \hat{a}_{n} V/m = 565 kV/m\]
For a conductor: E is always perpendicular to the surface and equals ρ_s/ε₀.
Inside the conductor: E = 0.

Ex 3.7●● MediumTier 1?
Polarization in a Dielectric

Find P and D for E = 10⁶ V/m in a dielectric with εᵣ = 5.

→ Solution
Ex 3.7 Solution↑ Problem
Electric susceptibility: χ_e = εᵣ - 1 = 4
Polarization:
\[P = \varepsilon _{0} \chi _e E = 8.85\times 10^{-12} \times 4 \times 10^{6} = 35.4\times 10^{-6} C/m^{2} = 35.4 \mu C/m^{2}\]
Flux density:
\[D = \varepsilon _{0}\varepsilon _{r}E = 8.85\times 10^{-12} \times 5 \times 10^{6} = 44.25 \mu C/m^{2}\]
Check: D = ε₀E + P = 8.85 + 35.4 = 44.25 μC/m² ✓

Ex 3.8●● MediumTier 1?
Boundary Conditions at Air–Dielectric Interface

E₁ = 10³ V/m at θ₁ = 45° to interface. Medium 1: air (ε₁=ε₀), Medium 2: εᵣ=4. Find θ₂.

→ Solution
Ex 3.8 Solution↑ Problem
Tangential component (continuous): E₁t = E₁ sin45° = 707 V/m
Normal component: E₁n = E₁ cos45° = 707 V/m
Boundary condition (no free surface charge): D₁n = D₂n
\[\varepsilon _{0} E_{1}n = \varepsilon _{0}\varepsilon _{r} E_{2}n\]
\[E_{2}n = E_{1}n/\varepsilon _{r} = 707/4 = 176.8 V/m\]
Tangential is continuous: E₂t = E₁t = 707 V/m
Refraction angle:
\[tan \theta _{2} = E_{2}t/E_{2}n = 707/176.8 = 4.0\]
\[\theta _{2} = arctan(4) = 76.0°\]
tan θ₂/tan θ₁ = ε₂/ε₁ = 4 (Snell's law for dielectrics) ✓

Ex 3.9●● MediumTier 1?
Energy Stored in Capacitor

For C = 354 pF charged to V = 100 V, find stored energy W.

→ Solution
Ex 3.9 Solution↑ Problem
\[W = CV^{2}/2 = 354\times 10^{-12} \times 10000/2 = 1.77\times 10^{-6} J = 1.77 \mu J\]
Alternatively using charge: Q = CV = 354×10⁻¹² × 100 = 35.4 nC
\[W = Q^{2}/(2C) = (35.4\times 10^{-9})^{2}/(2\times 354\times 10^{-12})\]
\[= 1.253\times 10^{-15}/7.08\times 10^{-10} = 1.77\times 10^{-6} J \checkmark \]
Or in terms of field energy (A=0.01m², d=1mm, εᵣ=4):
\[E = V/d = 10^{5} V/m\]
\[W = (1/2)\varepsilon _{0}\varepsilon _{r}E^{2}(Ad) = 0.5\times 8.85\times 10^{-12}\times 4\times 10^{10}\times 10^{-5} = 1.77 \mu J \checkmark \]

Ex 3.10●● MediumTier 1?
Force Between Capacitor Plates

Same capacitor (C=354pF, V=100V, εᵣ=4, A=0.01m²). Find the attractive force between plates.

→ Solution
Ex 3.10 Solution↑ Problem
Electric field inside: E = V/d = 10⁵ V/m
Electrostatic pressure (force per unit area):
\[p = D\cdot E/2 = \varepsilon _{0}\varepsilon _{r}E^{2}/2 = 8.85\times 10^{-12} \times 4 \times 10^{10}/2 = 177 N/m^{2}\]
Total force:
\[F = p \times A = 177 \times 0.01 = 1.77 N (attractive)\]
Alternative using energy: F = -dW/dd (at constant charge)
\[W = Q^{2}/(2C) = Q^{2}d/(2\varepsilon _{0}\varepsilon _{r}A) \to F = Q^{2}/(2\varepsilon _{0}\varepsilon _{r}A) = (35.4\times 10^{-9})^{2}/(2\times \varepsilon _{0}\times 4\times 0.01) \approx 1.77 N \checkmark \]

Ex 3.11●● MediumTier 1?
Equivalent Capacitance

Three capacitors: C₁=10pF, C₂=20pF, C₃=30pF. Find C for (a) all series, (b) all parallel, (c) C₁ series with (C₂ ∥ C₃).

→ Solution
Ex 3.11 Solution↑ Problem
(a) All series:
\[1/C = 1/10 + 1/20 + 1/30 = 6/60 + 3/60 + 2/60 = 11/60\]
\[C = 60/11 = 5.45 pF\]
(b) All parallel:
\[C = 10 + 20 + 30 = 60 pF\]
(c) C₁ in series with (C₂ ∥ C₃):
\[C_{23} = 20 + 30 = 50 pF\]
\[1/C = 1/10 + 1/50 = 5/50 + 1/50 = 6/50\]
\[C = 50/6 = 8.33 pF\]

Ex 3.12●● MediumTier 1?
Field in a Dielectric Slab

A slab (εᵣ=4, thickness 5mm) sits between two others (εᵣ=1, air). Normal D = 10 μC/m². Find E in each region.

→ Solution
Ex 3.12 Solution↑ Problem
Normal component of D is continuous (no free surface charge):
D_n = 10 μC/m² everywhere (same value)
\[E in air (\varepsilon _{r}=1): E_{1} = D/\varepsilon _{0} = 10^{-5}/8.85\times 10^{-12} = 1.13 MV/m\]
\[E in slab (\varepsilon _{r}=4): E_{2} = D/(\varepsilon _{0}\varepsilon _{r}) = 1.13\times 10^{6}/4 = 282.5 kV/m\]
Voltage across 5mm slab: V = E₂ × d = 282500 × 0.005 = 1412.5 V

Ex 3.13●● MediumTier 1?
Surface Charge on an Isolated Conductor

Isolated conducting sphere, radius a = 0.1 m, raised to V₀ = 1000 V. Find Q and ρ_s.

→ Solution
Ex 3.13 Solution↑ Problem
Capacitance of isolated sphere: C = 4πε₀a = 4π×8.85×10⁻¹²×0.1 = 11.13 pF
Charge: Q = CV₀ = 11.13×10⁻¹² × 1000 = 11.13 nC
Surface charge density (uniform):
\[\rho _s = Q/(4\pi a^{2}) = 11.13\times 10^{-9}/(4\pi \times 0.01) = 88.5 nC/m^{2}\]
Check E at surface: E = ρ_s/ε₀ = 88.5×10⁻⁹/8.85×10⁻¹² = 10,000 V/m
Also: E = V₀/a = 1000/0.1 = 10,000 V/m ✓

Ex 3.14●● MediumTier 1?
Dielectric Sphere in Uniform Field

A dielectric sphere (εᵣ = 3, radius a) is placed in a uniform field E₀. Find E inside.

→ Solution
Ex 3.14 Solution↑ Problem
Exact solution (separation of variables in spherical coordinates):
\[E_inside = 3E_{0}/(\varepsilon _{r} + 2)\]
For εᵣ = 3:
\[E_inside = 3E_{0}/(3 + 2) = 3E_{0}/5 = 0.6 E_{0}\]
The field inside is uniform and weaker than the applied field.
For εᵣ → ∞ (conductor): E_inside → 0 (field excluded)
For εᵣ = 1 (no sphere): E_inside → E₀ ✓
For εᵣ = 3: E_inside = 0.6 E₀ (reduced but not zero)

Ex 3.15●● MediumTier 1?
Electrostatic Shielding

Explain why a closed conducting shell provides complete electrostatic shielding. Quantify for copper.

→ Solution
Ex 3.15 Solution↑ Problem
Static shielding (DC):
By Gauss's law, if no free charge is enclosed, E = 0 everywhere inside
the conductor and inside the hollow. The conductor redistributes surface
charges to cancel any external field inside.
→ SE = ∞ for static fields (perfect shielding).
At AC frequency f = 1 GHz, copper (σ = 5.8×10⁷ S/m):
\[\delta = 2.09 \mu m (skin depth)\]
For shell thickness t = 1 mm >> δ:
Absorption loss = 8.686 × t/δ = 8.686 × (10⁻³/2.09×10⁻⁶) = 4156 dB
Practical rule: SE ≈ 8.686t/δ (dB) for t >> δ.
For static fields, no thickness is needed — shielding is complete regardless.

Textbook Practice Problems

3-1● EasyTier 1?
Coulomb's law → Answer
3-2● EasyTier 1?
Coulomb's law → Answer
3-7● EasyTier 1?
Gauss's law → Answer
3-8●● MediumTier 1?
Gauss's law → Answer
3-9●● MediumTier 1?
Gauss's law → Answer
3-12● EasyTier 1?
Electric potential → Answer
3-13●● MediumTier 1?
Electric potential → Answer
3-28● EasyTier 1?
Capacitance → Answer
3-29●● MediumTier 1?
Capacitance → Answer
3-3●● MediumTier 2?
Coulomb's law → Answer
3-14●● MediumTier 2?
Electric potential → Answer
3-18●● MediumTier 2?
Conductors → Answer
3-19●● MediumTier 2?
Conductors → Answer
3-22●● MediumTier 2?
Dielectrics → Answer
3-23●● MediumTier 2?
Dielectrics → Answer
3-35●● MediumTier 2?
Energy and forces → Answer
3-36●●● HardTier 3?
Energy and forces → Answer

Textbook Practice — Approach Hints

Sample: Two point charges \(q_1=2\,\mu\text{C}\) at \((0,0)\) and \(q_2=-3\,\mu\text{C}\) at \((3,0)\,\text{m}\). Find force on \(q_2\).

\[\vec{F}_{12}=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\hat{r}_{12}=\frac{(2\!\cdot\!10^{-6})(-3\!\cdot\!10^{-6})}{4\pi(8.854\!\cdot\!10^{-12})(3)^2}\hat{x}\]
\[=-6.0\times10^{-3}\,\hat{x}\,\text{N}\;\;(\text{attractive, toward }q_1)\]

Sample: Three charges \(q_1=q_2=q_3=1\,\mu\text{C}\) at \((1,0,0)\), \((-1,0,0)\), \((0,1,0)\) m. Find net force on a \(1\,\mu\text{C}\) test charge at the origin.

\[\vec{F}_1=\frac{kq^2}{1}(-\hat{x}),\;\vec{F}_2=\frac{kq^2}{1}(+\hat{x}),\;\vec{F}_3=\frac{kq^2}{1}(-\hat{y})\]
\[\vec{F}_{net}=-\frac{kq^2}{1}\hat{y}=-(8.99\!\cdot\!10^9)(10^{-12})\hat{y}=-8.99\!\cdot\!10^{-3}\hat{y}\,\text{N}\]

Sample: A spherical volume charge \(\rho_v=\rho_0\) for \(r\le a\). Find \(\vec{E}\) for \(r>a\) and \(r

\[r>a:\;\oint\vec{E}\cdot d\vec{S}=Q_{enc}/\epsilon_0\;\Rightarrow\;E(4\pi r^2)=\rho_0(\tfrac{4}{3}\pi a^3)/\epsilon_0\]
\[\vec{E}=\frac{\rho_0 a^3}{3\epsilon_0 r^2}\hat{a}_r\]
\[r

Sample: Coaxial cable: inner \(a=1\,\text{cm}\), outer \(b=3\,\text{cm}\), line charge \(\rho_l=10\,\text{nC/m}\) on inner conductor. Find \(\vec{E}\) between conductors.

\[\oint\vec{E}\cdot d\vec{S}=\rho_l\ell/\epsilon_0\;\Rightarrow\;E(2\pi\rho\ell)=\rho_l\ell/\epsilon_0\]
\[\vec{E}=\frac{\rho_l}{2\pi\epsilon_0\rho}\hat{a}_{\rho}=\frac{10\!\cdot\!10^{-9}}{2\pi(8.854\!\cdot\!10^{-12})\rho}\hat{a}_{\rho}=\frac{179.7}{\rho}\hat{a}_{\rho}\,\text{V/m}\]

Sample: Two parallel infinite plates with surface charges \(+\rho_s\) and \(-\rho_s\). Find \(\vec{E}\) between and outside.

\[\text{Each plate alone: }|\vec{E}|=\rho_s/(2\epsilon_0)\]
\[\text{Between (fields add): }\vec{E}=\rho_s/\epsilon_0\;(\text{from + to -})\]
\[\text{Outside (fields cancel): }\vec{E}=0\]

Sample: Find potential at the center of a circular ring of radius \(a\) carrying total charge \(Q\).

\[V=\frac{1}{4\pi\epsilon_0}\int\frac{dq}{R}=\frac{1}{4\pi\epsilon_0}\frac{Q}{a}\]
All ring elements equidistant from center, so \(R=a\) comes outside the integral.

Sample: Uniformly charged disc of radius \(a\) with \(\rho_s\). Find \(V\) on axis at distance \(z\).

\[V=\frac{\rho_s}{4\pi\epsilon_0}\int_0^a\!\int_0^{2\pi}\frac{\rho\,d\phi\,d\rho}{\sqrt{\rho^2+z^2}}=\frac{\rho_s}{2\epsilon_0}\!\left[\sqrt{a^2+z^2}-|z|\right]\]

Sample: Coaxial capacitor: inner \(a=1\,\text{mm}\), outer \(b=5\,\text{mm}\), length \(L=10\,\text{cm}\), \(\epsilon_r=2.25\) (polyethylene). Find \(C\).

\[C=\frac{2\pi\epsilon L}{\ln(b/a)}=\frac{2\pi(2.25)(8.854\!\cdot\!10^{-12})(0.1)}{\ln 5}=7.78\,\text{pF}\]

Sample: Spherical capacitor with inner radius \(a=2\,\text{cm}\), outer \(b=4\,\text{cm}\). Find \(C\) in vacuum.

\[C=\frac{4\pi\epsilon_0 ab}{b-a}=\frac{4\pi(8.854\!\cdot\!10^{-12})(0.02)(0.04)}{0.02}=4.45\,\text{pF}\]

Sample: A line charge \(\rho_l=5\,\text{nC/m}\) along the \(z\)-axis from \(-\infty\) to \(\infty\). Find \(\vec{E}\) at \((2,0,0)\) m.

\[\vec{E}=\frac{\rho_l}{2\pi\epsilon_0\rho}\hat{a}_{\rho}=\frac{5\!\cdot\!10^{-9}}{2\pi(8.854\!\cdot\!10^{-12})(2)}\hat{x}\]
\[=44.96\,\hat{x}\,\text{V/m}\]

Sample: \(V(r)=V_0(a/r)\) for \(r\ge a\). Find \(\vec{E}\) and verify \(\vec{E}=-\nabla V\).

\[\vec{E}=-\nabla V=-\frac{\partial V}{\partial r}\hat{a}_r=\frac{V_0 a}{r^2}\hat{a}_r\]
Same form as point-charge field \(E=Q/(4\pi\epsilon_0 r^2)\) if \(Q/(4\pi\epsilon_0)=V_0 a\).

Sample: A grounded conducting sphere of radius \(a\) in a uniform field \(\vec{E}_0=E_0\hat{z}\). Find induced surface charge density.

\[V(r,\theta)=-E_0 r\cos\theta+\frac{E_0 a^3\cos\theta}{r^2}\;(r\ge a)\]
\[\rho_s=-\epsilon_0\!\left.\frac{\partial V}{\partial r}\right|_{r=a}=3\epsilon_0 E_0\cos\theta\]

Sample: Conductor with surface field \(E_n=200\,\text{kV/m}\). Find surface charge density.

\[\rho_s=\epsilon_0 E_n=8.854\!\cdot\!10^{-12}\!\cdot\!200\!\cdot\!10^3=1.77\!\cdot\!10^{-6}\,\text{C/m}^2=1.77\,\mu\text{C/m}^2\]

Sample: Parallel-plate capacitor with mica (\(\epsilon_r=6\)). Plate area \(100\,\text{cm}^2\), separation \(2\,\text{mm}\). Voltage 100 V. Find \(\vec{D},\vec{E},\vec{P}\).

\[E=V/d=50\,\text{kV/m},\;D=\epsilon E=6\!\cdot\!8.854\!\cdot\!10^{-12}\!\cdot\!50\!\cdot\!10^3=2.66\,\mu\text{C/m}^2\]
\[P=D-\epsilon_0 E=2.66-0.443=2.21\,\mu\text{C/m}^2\]

Sample: Boundary \(z=0\) separates \(\epsilon_{r1}=2\) (\(z>0\)) from \(\epsilon_{r2}=4\) (\(z<0\)). \(\vec{E}_1=3\hat{x}+4\hat{z}\,\text{V/m}\). Find \(\vec{E}_2\).

\[E_{1t}=3\hat{x}\;\Rightarrow\;E_{2t}=3\hat{x}\]
\[D_{1n}=\epsilon_1 E_{1n}=2\epsilon_0(4)=8\epsilon_0=D_{2n}\;\Rightarrow\;E_{2n}=8\epsilon_0/(4\epsilon_0)=2\hat{z}\]
\[\vec{E}_2=3\hat{x}+2\hat{z}\,\text{V/m}\]

Sample: Capacitor \(C=100\,\text{pF}\) charged to \(V=500\,\text{V}\). Find stored energy.

\[W=\tfrac12 CV^2=\tfrac12(100\!\cdot\!10^{-12})(500)^2=1.25\!\cdot\!10^{-5}\,\text{J}=12.5\,\mu\text{J}\]

Sample: Parallel-plate capacitor, plate area \(A=50\,\text{cm}^2\), separation \(d=1\,\text{mm}\), \(V=200\,\text{V}\). Find force between plates.

\[C=\epsilon_0 A/d,\;W=\tfrac12 CV^2=\tfrac12\frac{\epsilon_0 AV^2}{d}\]
\[F=-\frac{\partial W}{\partial d}\bigg|_V=\frac{\epsilon_0 AV^2}{2d^2}=\frac{(8.854\!\cdot\!10^{-12})(5\!\cdot\!10^{-3})(200)^2}{2(10^{-3})^2}=8.85\!\cdot\!10^{-4}\,\text{N}\]

Chapter 4 — Solution of Electrostatic Problems

Key Theory — Chapter 4

Condensed from Cheng, Field and Wave Electromagnetics, §4-1 through §4-7. Read this before attempting the problems below.


4-1   Introduction — Two Types of Problem

Chapter 3 let us compute \(V\) and \(\mathbf{E}\) directly when the charge distribution is known everywhere. In practice this rarely happens. Chapter 4 addresses the two practical alternatives:

  • Point/line charges near conducting bodies of simple geometry. The induced surface charge is unknown — use the method of images (§4-4).
  • Conductors held at prescribed potentials. Solve Poisson's/Laplace's equation as a boundary-value problem — separation of variables in the appropriate coordinate system (§4-5 through §4-7).

Both approaches rest on the uniqueness theorem (§4-3): any \(V\) that satisfies Poisson's equation and the boundary conditions is the solution — a lucky guess, verified, is as rigorous as a derivation.


4-2   Poisson's and Laplace's Equations

Starting from the Ch. 3 postulates in a linear, isotropic medium (\(\mathbf{D}=\epsilon\mathbf{E}\), \(\nabla\times\mathbf{E}=0\), \(\mathbf{E}=-\nabla V\)):

\[ \nabla\cdot\mathbf{D} = \rho \qquad \text{(4-1)} \]
\[ \nabla\cdot(\epsilon\nabla V) = -\rho \qquad \text{(4-5)} \]

For a simple medium (homogeneous, linear, isotropic — \(\epsilon\) constant), \(\epsilon\) comes out of the divergence operator and we obtain:

\[ \nabla^{2}V = -\frac{\rho}{\epsilon} \quad \text{(Poisson's equation)} \qquad \text{(4-6)} \]
\[ \nabla^{2}V = 0 \quad \text{(Laplace's equation, charge-free region)} \qquad \text{(4-10)} \]

Laplace's equation governs the potential in the space between conductors — between capacitor plates, between coax conductors, etc. Once \(V\) is found, \(\mathbf{E}=-\nabla V\) and the surface charge follows from \(\rho_s=\epsilon E_n\) (3-72).

Explicit forms of the Laplacian in each coordinate system:

\[ \text{Cartesian:}\quad \frac{\partial^{2}V}{\partial x^{2}} + \frac{\partial^{2}V}{\partial y^{2}} + \frac{\partial^{2}V}{\partial z^{2}} = -\frac{\rho}{\epsilon} \qquad \text{(4-7)} \]
\[ \text{Cylindrical:}\quad \frac{1}{r}\frac{\partial}{\partial r}\!\left(r\frac{\partial V}{\partial r}\right) + \frac{1}{r^{2}}\frac{\partial^{2}V}{\partial\phi^{2}} + \frac{\partial^{2}V}{\partial z^{2}} \qquad \text{(4-8)} \]
\[ \text{Spherical:}\quad \frac{1}{R^{2}}\frac{\partial}{\partial R}\!\left(R^{2}\frac{\partial V}{\partial R}\right) + \frac{1}{R^{2}\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial V}{\partial\theta}\right) + \frac{1}{R^{2}\sin^{2}\theta}\frac{\partial^{2}V}{\partial\phi^{2}} \qquad \text{(4-9)} \]

4-3   Uniqueness Theorem

A solution of Poisson's equation satisfying the given boundary conditions is unique.

Why it matters: it justifies guessing. If you construct any \(V\) — by images, by inspection, by educated guess — that satisfies (4-6) inside the region and matches the prescribed potentials on the boundary, you have the only possible answer.

Sketch of proof: suppose \(V_1\), \(V_2\) both solve the problem; let \(V_d = V_1 - V_2\). Then \(\nabla^2 V_d = 0\) inside and \(V_d = 0\) on all boundaries. Using \(\nabla\cdot(V_d\nabla V_d) = V_d\nabla^2 V_d + |\nabla V_d|^2\) and the divergence theorem:

\[ \int_\tau |\nabla V_d|^{2}\, dv = 0 \qquad \text{(4-35)} \]

A nonnegative integrand with zero integral must vanish — so \(\nabla V_d \equiv 0\), \(V_d\) is constant, and since \(V_d=0\) on the boundary, \(V_d\equiv 0\) everywhere. Hence \(V_1=V_2\).


4-4   Method of Images

Idea: replace a conducting boundary by an equivalent image charge (located outside the region of interest) such that the boundary potential condition is automatically satisfied. By uniqueness, this gives the correct \(V\) and \(\mathbf{E}\) in the original region.

Three standard configurations.

(a) Point charge above a grounded conducting plane (§4-4.1). Image: \(-Q\) at the mirror position.

\[ V(x,y,z) = \frac{Q}{4\pi\epsilon_0}\!\left(\frac{1}{R_+} - \frac{1}{R_-}\right), \quad y>0 \qquad \text{(4-37)} \]

where \(R_\pm\) are distances to \(\pm Q\). The field below the plane is zero; only the upper half-space is physical.

(b) Line charge parallel to a conducting cylinder (§4-4.2). Image: a parallel line charge of opposite sign at the inverse point:

\[ \rho_i = -\rho_\ell \qquad \text{(4-38)} \]
\[ d_i = \frac{a^{2}}{d} \quad \text{(inverse-point distance)} \qquad \text{(4-43)} \]

Applied to a two-wire transmission line (wires of radius \(a\), center-to-center separation \(D\)):

\[ C = \frac{\pi\epsilon_0}{\ln\!\left[(D/2a) + \sqrt{(D/2a)^{2}-1}\right]} = \frac{\pi\epsilon_0}{\cosh^{-1}(D/2a)} \quad (\text{F/m}) \qquad \text{(4-46,47)} \]

(c) Point charge near a conducting sphere of radius \(a\) (§4-4.3). Image charge inside the sphere at the inverse point:

\[ Q_i = -\frac{a}{d}\,Q \qquad \text{(4-65)} \]
\[ d_i = \frac{a^{2}}{d} \qquad \text{(4-66)} \]

If the sphere is isolated (not grounded) rather than grounded, add a second image \(Q'=+aQ/d\) at the center so total sphere charge is zero.

Rules to remember when using images:

  • Image charges live outside the region where you are computing the field — never inside it.
  • The image field in the excluded half-space has no physical meaning (the real field there is zero, inside the conductor).
  • Multiple grounded planes or spheres may require infinite series of images (see §4-4.4 charged sphere and grounded plane).

4-5   Boundary-Value Problems — Cartesian (Separation of Variables)

When no free charge is isolated and conductors align with coordinate surfaces, solve \(\nabla^{2}V=0\) by separation of variables: assume

\[ V(x,y,z) = X(x)\,Y(y)\,Z(z) \qquad \text{(4-82)} \]

Substituting into (4-81), dividing by \(XYZ\), and noting each term depends on only one variable, each must be constant. We obtain three ODEs:

\[ \frac{d^{2}X}{dx^{2}} + k_x^{2}X = 0,\quad \frac{d^{2}Y}{dy^{2}} + k_y^{2}Y = 0,\quad \frac{d^{2}Z}{dz^{2}} + k_z^{2}Z = 0 \qquad \text{(4-86..88)} \]
\[ k_x^{2} + k_y^{2} + k_z^{2} = 0 \quad \text{(separation constraint)} \qquad \text{(4-89)} \]

Because (4-89) requires the sum of the squared separation constants to be zero, at least one constant must be imaginary — making the corresponding factor hyperbolic (\(\sinh, \cosh\)) rather than trigonometric.

Table 4-1 — Possible X(x) solutions of X"(x) + k²X(x) = 0
k² = 0 : X = A₀x + B₀
k² > 0 : X = A₁ sin kx + B₁ cos kx (periodic)
k² < 0 : X = A₂ sinh kx + B₂ cosh kx (monotonic / hyperbolic)

Boundary conditions classify the problem:

  • Dirichlet: \(V\) specified on all boundaries.
  • Neumann: normal derivative \(\partial V/\partial n\) specified (i.e., surface charge).
  • Mixed: Dirichlet on some boundaries, Neumann on others.

Choosing sin vs. cos, sinh vs. cosh: use the symmetry of the problem — pick sin if \(V\) vanishes at \(x=0\); cos if \(V\) is symmetric about \(x=0\); sinh for a condition like \(V=0\) at one end; cosh for \(V\) equal at two symmetric locations.


4-6   Boundary-Value Problems — Cylindrical

For geometries long in \(z\), drop the \(z\)-dependence and solve the 2-D problem:

\[ \frac{1}{r}\frac{\partial}{\partial r}\!\left(r\frac{\partial V}{\partial r}\right) + \frac{1}{r^{2}}\frac{\partial^{2}V}{\partial\phi^{2}} = 0 \qquad \text{(4-116)} \]

Assume \(V(r,\phi)=R(r)\Phi(\phi)\). Separation gives two ODEs with separation constant \(k^{2}\). Because \(\phi\) is periodic with period \(2\pi\), \(k\) must be an integer \(n\); the general solution becomes:

\[ V(r,\phi) = r^{n}\!\left(A_n\sin n\phi + B_n\cos n\phi\right) + r^{-n}\!\left(A'_n\sin n\phi + B'_n\cos n\phi\right),\quad n\ne 0 \qquad \text{(4-125)} \]
\[ V(r) = C_1\ln r + C_2,\quad n=0 \text{ (no } \phi \text{ variation)} \qquad \text{(4-130)} \]

Radial rules: if the region includes \(r=0\), drop \(r^{-n}\) (singular); if the region extends to \(\infty\), drop \(r^{n}\) (blows up).

Coaxial cable (inner radius \(a\) at \(V_0\), outer radius \(b\) grounded): (4-130) with BCs gives

\[ V(r) = \frac{V_0}{\ln(b/a)}\,\ln\!\frac{b}{r} \qquad \text{(4-133)} \]

Non-axisymmetric problems with arbitrary boundary data lead to Fourier series in \(\phi\) (see Example 4-9, the split cylinder).


4-7   Boundary-Value Problems — Spherical

For axisymmetric problems (no \(\phi\)-dependence):

\[ \frac{1}{R^{2}}\frac{\partial}{\partial R}\!\left(R^{2}\frac{\partial V}{\partial R}\right) + \frac{1}{R^{2}\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial V}{\partial\theta}\right) = 0 \qquad \text{(4-144)} \]

Separation \(V(R,\theta)=\Gamma(R)\Theta(\theta)\) with separation constant \(k^{2}=n(n+1)\) (forced by regularity of \(\Theta\) on the \(z\)-axis) gives:

\[ \Gamma_n(R) = A_n R^{n} + B_n R^{-(n+1)} \qquad \text{(4-150)} \]
\[ \Theta_n(\theta) = P_n(\cos\theta) \quad \text{(Legendre polynomial)} \qquad \text{(4-153)} \]
\[ V_n(R,\theta) = \bigl[A_n R^{n} + B_n R^{-(n+1)}\bigr]P_n(\cos\theta) \qquad \text{(4-154)} \]
Table 4-2 — Legendre Polynomials
P₀(cos θ) = 1
P₁(cos θ) = cos θ
P₂(cos θ) = ½(3 cos²θ − 1)
P₃(cos θ) = ½(5 cos³θ − 3 cos θ)

Canonical example — conducting sphere in uniform field \(\mathbf{E}_0=\mathbf{a}_z E_0\). BCs: \(V(b,\theta)=0\); \(V\to -E_0 R\cos\theta\) as \(R\to\infty\). Only the \(n=1\) term survives:

\[ V(R,\theta) = -E_0\!\left[1 - \left(\frac{b}{R}\right)^{3}\right]R\cos\theta,\quad R\ge b \qquad \text{(4-159)} \]
\[ \mathbf{p}_{\text{induced}} = \mathbf{a}_z\,4\pi\epsilon_0 b^{3} E_0 \qquad \text{(4-162)} \]
\[ \rho_s(\theta) = 3\epsilon_0 E_0 \cos\theta \qquad \text{(4-161)} \]

The external potential is the applied uniform field plus that of an induced dipole of moment \(\mathbf{p}\) at the sphere's center.


Chapter 4 at a Glance

  • Governing equations: \(\nabla^{2}V = -\rho/\epsilon\) (Poisson); \(\nabla^{2}V=0\) (Laplace, charge-free).
  • Uniqueness theorem: a \(V\) satisfying \(\nabla^{2}V=-\rho/\epsilon\) and the boundary conditions is the solution — so any valid guess (e.g., images) is rigorous.
  • Method of images: plane → mirror charge \(-Q\); cylinder → line charge \(-\rho_\ell\) at inverse point \(d_i=a^{2}/d\); sphere → \(Q_i=-(a/d)Q\) at \(d_i=a^{2}/d\).
  • Two-wire line: \(C=\pi\epsilon_0/\cosh^{-1}(D/2a)\) (F/m).
  • Cartesian BVP: \(X\,Y\,Z\) ansatz; separation constants satisfy \(k_x^{2}+k_y^{2}+k_z^{2}=0\) — at least one factor is hyperbolic.
  • Cylindrical BVP: angular factor \(\sin n\phi,\cos n\phi\) with integer \(n\); radial factor \(r^{n}\) and/or \(r^{-n}\); for \(n=0\), \(V=C_1\ln r+C_2\).
  • Spherical axisymmetric BVP: general solution \(V_n=[A_n R^{n}+B_n R^{-(n+1)}]P_n(\cos\theta)\); conducting sphere in uniform field → (4-159).
  • Boundary types: Dirichlet (V specified), Neumann (\(\partial V/\partial n\) specified), mixed.

Problems and Solutions


Review Questions (Tier 1)

R.4-1● EasyTier 1?
Review: Write Poisson's equation in vector notation (a) for a simple medium; (b) for a linear isotropic but inhomogeneous medium. → Answer
R.4-1 Answer↑ Question

(a) Simple medium: \(\nabla^2 V=-\rho_v/\epsilon\). (b) Linear isotropic but inhomogeneous: \(\nabla\cdot(\epsilon\nabla V)=-\rho_v\).

R.4-2● EasyTier 1?
Review: Write Poisson's equation in Cartesian coordinates for each case in R.4-1. → Answer
R.4-2 Answer↑ Question

(a) \(\partial^2_x V+\partial^2_y V+\partial^2_z V=-\rho_v/\epsilon\). (b) Add \(\nabla\epsilon\cdot\nabla V\) to the LHS to keep the inhomogeneous form.

R.4-3● EasyTier 1?
Review: Write Laplace's equation for a simple medium (a) in vector notation; (b) in Cartesian coordinates. → Answer
R.4-3 Answer↑ Question

(a) \(\nabla^2 V=0\). (b) \(\partial^2_x V+\partial^2_y V+\partial^2_z V=0\).

R.4-4● EasyTier 1?
Review: If \(\nabla^{2}U=0\), why does it not follow that \(U\) is identically zero? → Answer
R.4-4 Answer↑ Question

Harmonic functions other than zero exist (e.g., \(U=ax+by+c\)). The boundary conditions select among them; only specific BCs force \(U\equiv 0\).

R.4-5● EasyTier 1?
Review: A fixed voltage is across a parallel-plate capacitor. Does \(\mathbf{E}\) between the plates depend on the permittivity? Does \(\mathbf{D}\)? → Answer
R.4-5 Answer↑ Question

\(E=V/d\) is fixed by the applied voltage (independent of \(\epsilon\)). \(D=\epsilon E\) scales with \(\epsilon\).

R.4-6● EasyTier 1?
Review: Fixed charges \(\pm Q\) are deposited on an isolated parallel-plate capacitor. Does \(\mathbf{E}\) depend on \(\epsilon\)? Does \(\mathbf{D}\)? → Answer
R.4-6 Answer↑ Question

\(D=\rho_s=Q/A\) is fixed by the deposited charge. \(E=D/\epsilon\) decreases as \(\epsilon\) increases.

R.4-7● EasyTier 1?
Review: State the uniqueness theorem. Why is it important for the method of images? → Answer
R.4-7 Answer↑ Question

The solution to Poisson's equation in a region with specified boundary conditions is unique. The method of images is justified because any field that satisfies the BCs is the solution.

R.4-8● EasyTier 1?
Review: For a point charge above a grounded plane, where is the image, and what is its sign? → Answer
R.4-8 Answer↑ Question

Image is \(-q\) located at the mirror position, distance \(d\) below the plane.

R.4-9● EasyTier 1?
Review: For a point charge outside a grounded conducting sphere of radius \(a\), at distance \(d\) from the center, locate and size the image charge. → Answer
R.4-9 Answer↑ Question

\(q'=-(a/d)q\) located at distance \(a^2/d\) from the sphere's center, on the line from center to original charge.

R.4-10● EasyTier 1?
Review: In separation of variables in Cartesian coordinates, why must at least one of \(X,Y,Z\) be hyperbolic? → Answer
R.4-10 Answer↑ Question

The Helmholtz/Laplace separation gives \(X''/X+Y''/Y+Z''/Z=0\). At least one term must be positive (hyperbolic solution: cosh/sinh) to balance the others' negative (oscillatory) contributions.


Ex 4.1●● MediumTier 1?
Laplace's Equation: 1D Cartesian

Parallel plates at x=0 (V=0) and x=d (V=V₀). Solve Laplace's equation for V(x) and find E.

→ Solution
Ex 4.1 Solution↑ Problem
\[d^{2}V/dx^{2} = 0 \to V = Ax + B\]
\[BC: V(0)=0 \to B=0\]
\[V(d)=V_{0} \to A = V_{0}/d\]
V(x) = V₀ x/d (linear variation)
\[E = -\nabla V = -dV/dx a_x = -V_{0}/d a_x\]
For V₀=100V, d=1cm:
E = -100/0.01 a_x = -10,000 a_x V/m (uniform, pointing from + to – plate)
\[\rho _s = \varepsilon _{0}|E| = 8.85\times 10^{-12} \times 10^{4} = 88.5 nC/m^{2}\]

Ex 4.2●● MediumTier 1?
Fourier Series Solution (Rectangular Region)

Region: 0≤x≤a, 0≤y≤b. V=0 on three sides, V = V₀ sin(πx/a) on y=b. Find V(x,y).

→ Solution
Ex 4.2 Solution↑ Problem
Separation of variables: V = X(x) Y(y)
\[X'' + \lambda X = 0 \to X = sin(n\pi x/a) (with X(0)=X(a)=0)\]
\[Y'' - \lambda Y = 0 \to Y = sinh(n\pi y/a) (with Y(0)=0)\]
Since BC at y=b is exactly sin(πx/a) (n=1 only):
\[V(x,y) = C sin(\pi x/a) sinh(\pi y/a)\]
Apply V(x,b) = V₀ sin(πx/a):
\[C sinh(\pi b/a) = V_{0} \to C = V_{0}/sinh(\pi b/a)\]
\[V(x,y) = V_{0} sin(\pi x/a) \times sinh(\pi y/a)/sinh(\pi b/a)\]
For a=b: V(x,y) = V₀ sin(πx/a) × sinh(πy/a)/sinh(π)

Ex 4.3●● MediumTier 1?
Method of Images: Point Charge Above Ground Plane

Q = 1 μC at height h = 0.5 m above grounded plane (z=0). Find E at z=0 directly below, and ρ_s.

→ Solution
Ex 4.3 Solution↑ Problem
Image charge: Q' = -Q = -1 μC at z = -h = -0.5 m
At point (0,0,0) on the plane:
Both real charge (distance h) and image charge (distance h) contribute along z:
\[E_z = -2 \times Q/(4\pi \varepsilon _{0}h^{2}) = -2 \times 8.99\times 10^{9} \times 10^{-6}/0.25\]
= -71,920 V/m (downward, toward conductor)
Surface charge density at (x,y,0):
\[\rho _s = \varepsilon _{0}E_z = -Qh/(2\pi (x^{2}+y^{2}+h^{2})^{3/2})\]
At x=y=0: ρ_s = -Qh/(2πh³) = -Q/(2πh²) = -10⁻⁶/(2π×0.25) = -0.637 μC/m²
Total induced charge: ∫∫ρ_s dA = -Q = -1 μC ✓

Ex 4.4●● MediumTier 1?
Method of Images: Line Charge and Cylinder

Line charge ρ_L parallel to a grounded conducting cylinder (radius a, center at origin),

at distance d from center (d > a). State the image configuration.

→ Solution
Ex 4.4 Solution↑ Problem
The image is a line charge of -ρ_L located at distance a²/d from the center
(inside the cylinder, on the same line connecting center to real charge).
Image position: r_image = a²/d (Kelvin inversion)
Potential at any point outside:
\[V = \rho _L/(2\pi \varepsilon _{0}) \times ln(r_{2}/r_{1})\]
where r₁ = distance to real charge, r₂ = distance to image charge.
On the cylinder surface (r=a): V = 0 ✓ (can be verified geometrically)
Application: Two-wire transmission line uses this result.

Ex 4.5●● MediumTier 1?
Capacitance Using Method of Images (Two-Wire Line)

Two parallel cylindrical conductors: radius a = 1 mm, center-to-center separation D = 10 mm. Find C/L.

→ Solution
Ex 4.5 Solution↑ Problem
\[C/L = \pi \varepsilon _{0}/cosh^{-1}(D/2a)\]
\[D/2a = 10/(2\times 1) = 5\]
\[cosh^{-1}(5) = ln(5 + \sqrt{25-1}) = ln(5 + 4.899) = ln(9.899) = 2.292\]
\[C/L = \pi \times 8.85\times 10^{-12}/2.292 = 27.80\times 10^{-12}/2.292 = 12.13 pF/m\]
For widely separated wires (D >> a):
\[cosh^{-1}(D/2a) \approx ln(D/a) = ln(10) = 2.303\]
C/L ≈ πε₀/ln(D/a) = 12.08 pF/m (close to exact)

Ex 4.6●● MediumTier 1?
Poisson's Equation: Uniform Space Charge

Between plates at x=0 (V=0) and x=d (V=V₀), uniform charge density ρ_v. Solve for V(x).

→ Solution
Ex 4.6 Solution↑ Problem
\[d^{2}V/dx^{2} = -\rho _v/\varepsilon _{0}\]
Integrate twice:
\[dV/dx = -\rho _v x/\varepsilon _{0} + A\]
\[V(x) = -\rho _v x^{2}/(2\varepsilon _{0}) + Ax + B\]
Apply BC: V(0)=0 → B=0
\[V(d)=V_{0} \to V_{0} = -\rho _v d^{2}/(2\varepsilon _{0}) + Ad\]
\[\to A = V_{0}/d + \rho _v d/(2\varepsilon _{0})\]
\[V(x) = -\rho _v x^{2}/(2\varepsilon _{0}) + [V_{0}/d + \rho _v d/(2\varepsilon _{0})] x\]
\[E(x) = -dV/dx = \rho _v x/\varepsilon _{0} - V_{0}/d - \rho _v d/(2\varepsilon _{0})\]
For ρ_v=0: V = V₀x/d (reduces to Laplace result ✓)

Ex 4.7●● MediumTier 1?
Coaxial Cylinder Potential

Coaxial conductors: inner radius a=2mm at V₀=100V, outer radius b=10mm at V=0. Find V(ρ) and E.

→ Solution
Ex 4.7 Solution↑ Problem
\[\nabla ^{2}V = (1/\rho )d/d\rho (\rho dV/d\rho ) = 0\]
\[\to \rho dV/d\rho = A \to V = A ln\rho + B\]
\[BC: V(a)=100, V(b)=0:\]
\[0 = A ln(b) + B \to B = -A ln(b)\]
\[100 = A ln(a) - A ln(b) = A ln(a/b)\]
\[A = 100/ln(a/b) = 100/ln(0.2) = 100/(-1.609) = -62.15\]
\[V(\rho ) = -62.15 ln(\rho ) + 62.15 ln(b)\]
\[= -62.15 ln(\rho /b)\]
\[= 62.15 ln(b/\rho ) V\]
\[E = -dV/d\rho \hat{a}_{\rho } = 62.15/\rho \hat{a}_{\rho } V/m\]
At ρ=a: E = 62.15/0.002 = 31,075 V/m = 31.1 kV/m
At ρ=b: E = 62.15/0.010 = 6,215 V/m = 6.22 kV/m

Ex 4.8●● MediumTier 1?
Concentric Spheres Potential

Spherical capacitor: inner sphere a=2cm at V₀=100V, outer sphere b=6cm at V=0. Find V(r) and C.

→ Solution
Ex 4.8 Solution↑ Problem
\[(1/r^{2})d/dr(r^{2} dV/dr) = 0\]
\[\to r^{2} dV/dr = A \to dV/dr = A/r^{2} \to V = -A/r + B\]
\[BC: V(b)=0: B = A/b\]
\[V(a)=V_{0}: V_{0} = -A/a + A/b = A(1/b - 1/a)\]
\[A = V_{0}/(1/b - 1/a) = V_{0} ab/(a-b)\]
For a=0.02m, b=0.06m:
\[A = 100 \times 0.02\times 0.06/(0.02-0.06) = 100\times 0.0012/(-0.04) = -3\]
\[V(r) = -(-3)/r + (-3)/0.06 = 3/r - 50 V\]
Check: V(0.02) = 3/0.02 - 50 = 150-50 = 100 V ✓
\[V(0.06) = 3/0.06 - 50 = 50-50 = 0 V \checkmark \]
\[C = 4\pi \varepsilon _{0} ab/(b-a) = 4\pi \times 8.85\times 10^{-12} \times 0.02\times 0.06/0.04 = 3.34 pF\]

Ex 4.9●● MediumTier 1?
Grounded Conducting Sphere with Point Charge

Charge Q at distance d from center of grounded sphere (radius a, a < d). Find image charge.

→ Solution
Ex 4.9 Solution↑ Problem
Image charge for grounded sphere:
Q' = -Q(a/d) (magnitude reduced by factor a/d)
Location: r' = a²/d (inside sphere, on line from center to Q)
Potential at any exterior point P:
\[V = Q/(4\pi \varepsilon _{0}r_{1}) + Q'/(4\pi \varepsilon _{0}r_{2})\]
where r₁ = |P - Q|, r₂ = |P - Q'|
V = 0 on sphere surface r = a ✓ (by geometric property of image)
Induced charge on sphere: total = Q' = -Qa/d
Force on Q:
\[F = QQ'/(4\pi \varepsilon _{0}(d-r')^{2}) = -Q^{2}a/d / (4\pi \varepsilon _{0}(d-a^{2}/d)^{2}) [attractive]\]

Ex 4.10●● MediumTier 1?
Semi-Infinite Grounded Plane: Two Images

Point charge Q at (x₀, y₀) above a grounded plane at y=0 AND grounded plane at x=0 (corner). Find images.

→ Solution
Ex 4.10 Solution↑ Problem
For a 90° conducting corner (x≥0, y≥0 region):
Three image charges needed:
Real charge: +Q at (x₀, y₀)
Image 1: -Q at (-x₀, y₀) [image in x=0 plane]
Image 2: -Q at (x₀, -y₀) [image in y=0 plane]
Image 3: +Q at (-x₀, -y₀) [image of image, restores BC]
\[V = Q/(4\pi \varepsilon _{0}) \times [1/r_{1} - 1/r_{2} - 1/r_{3} + 1/r_{4}]\]
where r₁,r₂,r₃,r₄ are distances from field point to each charge.
\[V = 0 on both x=0 and y=0 planes \checkmark \]

Ex 4.11●● MediumTier 1?
Mixed Boundary Conditions

Half-space y>0 with V=V₀ for x>0 on y=0, and V=0 for x<0 on y=0. Find V(x,y).

→ Solution
Ex 4.11 Solution↑ Problem
Solution using complex variables / conformal mapping:
\[V(x,y) = V_{0}/2 + (V_{0}/\pi ) arctan(x/y)\]
Check boundaries:
\[y\to 0⁺, x>0: arctan(x/0⁺) \to \pi /2 \to V = V_{0}/2 + V_{0}/2 = V_{0} \checkmark \]
\[y\to 0⁺, x<0: arctan(x/0⁺) \to -\pi /2 \to V = V_{0}/2 - V_{0}/2 = 0 \checkmark \]
\[E = -\nabla V:\]
\[E_x = -\partial V/\partial x = -V_{0}y/(\pi (x^{2}+y^{2}))\]
\[E_y = -\partial V/\partial y = +V_{0}x/(\pi (x^{2}+y^{2}))\]
The field has an r⁻¹ singularity at the edge (x=0, y=0).

Ex 4.12●● MediumTier 1?
Separation of Variables: 3D Box

Rectangular box: 0≤x≤a, 0≤y≤b, 0≤z≤c. V=0 on all faces except z=c where V = V₀ sin(πx/a) sin(πy/b).

→ Solution
Ex 4.12 Solution↑ Problem
V(x,y,z) = X(x) Y(y) Z(z)
\[X = sin(\pi x/a), Y = sin(\pi y/b) (to satisfy V=0 at x=0,a and y=0,b)\]
\[Z'' = \gamma ^{2}Z where \gamma ^{2} = (\pi /a)^{2} + (\pi /b)^{2}\]
\[Z = sinh(\gamma z) (to satisfy Z(0)=0)\]
Apply V(x,y,c) = V₀ sin(πx/a) sin(πy/b):
\[V_{0} = C sinh(\gamma c) \to C = V_{0}/sinh(\gamma c)\]
\[V(x,y,z) = V_{0} sin(\pi x/a) sin(\pi y/b) \times sinh(\gamma z)/sinh(\gamma c)\]
\[with \gamma = \pi \sqrt{1/a^{2} + 1/b^{2}}\]

Ex 4.13●● MediumTier 1?
Coaxial Cable E and C

Inner conductor radius a=1mm, outer radius b=4mm, εᵣ=2.25. Find E(ρ) and C/L.

→ Solution
Ex 4.13 Solution↑ Problem
From Problem 7 generalized with εᵣ:
\[V(\rho ) = V_{0} ln(b/\rho )/ln(b/a)\]
\[E = -dV/d\rho \hat{a}_{\rho } = V_{0}/(\rho ln(b/a)) \hat{a}_{\rho }\]
\[C/L = 2\pi \varepsilon _{0}\varepsilon _{r}/ln(b/a)\]
\[= 2\pi \times 8.85\times 10^{-12} \times 2.25/ln(4)\]
\[= 2\pi \times 19.91\times 10^{-12}/1.386\]
\[= 90.3 pF/m\]
E_max (at inner conductor, ρ=a):
\[E_max = V_{0}/(a ln(b/a)) = V_{0}/(0.001 \times 1.386) = 721 V_{0} V/m per volt\]

Ex 4.14●● MediumTier 1?
Cylindrical Region with Space Charge

Cylindrical region 0≤ρ≤a, uniform ρ_v. V(a)=0 (grounded). Find V(ρ).

→ Solution
Ex 4.14 Solution↑ Problem
\[(1/\rho )d/d\rho (\rho dV/d\rho ) = -\rho _v/\varepsilon _{0}\]
Integrate: ρ dV/dρ = -ρ_v ρ²/(2ε₀) + C₁
\[dV/d\rho = -\rho _v \rho /(2\varepsilon _{0}) + C_{1}/\rho \]
For V finite at ρ=0: C₁ = 0 (otherwise ln singularity at center)
\[dV/d\rho = -\rho _v \rho /(2\varepsilon _{0})\]
\[V(\rho ) = -\rho _v \rho ^{2}/(4\varepsilon _{0}) + C_{2}\]
\[BC: V(a)=0: C_{2} = \rho _v a^{2}/(4\varepsilon _{0})\]
V(ρ) = ρ_v(a² - ρ²)/(4ε₀) [maximum at center]
E(ρ) = ρ_v ρ/(2ε₀) â_ρ [matches Gauss's law result]

Ex 4.15●● MediumTier 1?
Wedge-Shaped Region Potential

Conducting wedge: V=0 at φ=0 and V=V₀ at φ=α. Find V(φ) and E.

→ Solution
Ex 4.15 Solution↑ Problem
\[\nabla ^{2}V = (1/\rho ^{2})d^{2}V/d\phi ^{2} = 0 \to d^{2}V/d\phi ^{2} = 0\]
\[V(\phi ) = A\phi + B\]
\[BC: V(0)=0 \to B=0\]
\[V(\alpha )=V_{0} \to A = V_{0}/\alpha \]
V(φ) = V₀ φ/α (linear in angle)
\[E = -\nabla V = -(1/\rho )dV/d\phi \hat{a}_{\phi } = -V_{0}/(\alpha \rho ) \hat{a}_{\phi }\]
The field is inversely proportional to ρ — stronger near the vertex.
For a parallel-plate capacitor (α→0, V₀/α = E₀): E = -E₀ â_φ (uniform ✓)
C/L = ε₀α/ln(b/a) (for wedge bounded by cylinders ρ=a and ρ=b)

Textbook Practice Problems

4-1●● MediumTier 1?
Laplace's equation → Answer
4-9●● MediumTier 1?
Method of images → Answer
4-10●● MediumTier 1?
Method of images → Answer
4-2●● MediumTier 2?
Laplace's equation → Answer
4-3●●● HardTier 2?
Laplace's equation → Answer
4-5●● MediumTier 2?
Poisson's equation → Answer
4-6●●● HardTier 2?
Poisson's equation → Answer
4-11●●● HardTier 2?
Method of images → Answer
4-13●● MediumTier 2?
Cartesian coordinates → Answer
4-14●●● HardTier 2?
Cartesian coordinates → Answer
4-17●● MediumTier 2?
Cylindrical coordinates → Answer
4-18●●● HardTier 2?
Cylindrical coordinates → Answer
4-21●● MediumTier 2?
Spherical coordinates → Answer
4-22●●● HardTier 2?
Spherical coordinates → Answer
4-7●●● HardTier 3?
Poisson's equation → Answer
4-25●●● HardTier 3?
Comprehensive problem → Answer

Textbook Practice — Approach Hints

Sample: Solve \(\nabla^2 V=0\) between two parallel plates at \(V=0\) (\(z=0\)) and \(V=V_0\) (\(z=d\)).

\[\frac{d^2V}{dz^2}=0\;\Rightarrow\;V(z)=Az+B\]
\[\text{BCs: }B=0,\;Ad=V_0\;\Rightarrow\;V=V_0 z/d\]
\[\vec{E}=-\nabla V=-(V_0/d)\hat{z}\]

Sample: Point charge \(+q\) at distance \(d\) above a grounded conducting plane. Find \(\vec{E}\) above the plane.

\[\text{Image: }-q\text{ at distance }d\text{ below the plane.}\]
\[\vec{E}(\vec{r})=\frac{q}{4\pi\epsilon_0}\!\left[\frac{\vec{r}-d\hat{z}}{|\vec{r}-d\hat{z}|^3}-\frac{\vec{r}+d\hat{z}}{|\vec{r}+d\hat{z}|^3}\right]\]

Sample: Charge \(q\) at \((0,0,h)\) above grounded plane \(z=0\). Find induced surface charge density on plane.

\[\rho_s=-\frac{qh}{2\pi(\rho^2+h^2)^{3/2}}\]
\[Q_{induced}=\int_0^{\infty}\rho_s(2\pi\rho)d\rho=-q\]

Sample: Concentric cylinders \(a\) and \(b\) with \(V(a)=V_0\), \(V(b)=0\). Solve in cylindrical.

\[\frac{1}{\rho}\frac{d}{d\rho}\!\left(\rho\frac{dV}{d\rho}\right)=0\;\Rightarrow\;V=A\ln\rho+B\]
\[V(a)=V_0,\;V(b)=0\;\Rightarrow\;V(\rho)=V_0\frac{\ln(b/\rho)}{\ln(b/a)}\]

Sample: Solve Laplace in spherical: concentric spheres \(a,b\) with \(V(a)=V_0\), \(V(b)=0\).

\[\frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{dV}{dr}\right)=0\;\Rightarrow\;V=-A/r+B\]
\[V(r)=V_0\frac{(1/r-1/b)}{(1/a-1/b)}\]

Sample: Slab \(0\le x\le d\) with uniform \(\rho_v=\rho_0\), grounded at both faces. Solve Poisson.

\[\frac{d^2V}{dx^2}=-\rho_0/\epsilon_0\;\Rightarrow\;V(x)=-\frac{\rho_0}{2\epsilon_0}x^2+Ax+B\]
\[V(0)=V(d)=0\;\Rightarrow\;V(x)=\frac{\rho_0}{2\epsilon_0}x(d-x)\]

Sample: Cylindrical region \(\rho\le a\) has \(\rho_v=\rho_0\), surrounded by vacuum. Find \(V\) inside.

\[\frac{1}{\rho}\frac{d}{d\rho}\!\left(\rho\frac{dV}{d\rho}\right)=-\rho_0/\epsilon_0\;\Rightarrow\;V=-\frac{\rho_0\rho^2}{4\epsilon_0}+B\]
Match \(V\) continuity at \(\rho=a\) to outer solution to fix \(B\).

Sample: Charge \(q\) at distance \(d\) from center of grounded sphere of radius \(a\). Find image charge.

\[q'=-\frac{a}{d}q,\;\text{located at }d'=\frac{a^2}{d}\text{ from sphere center.}\]
\[V_{outside}(\vec{r})=\frac{1}{4\pi\epsilon_0}\!\left[\frac{q}{|\vec{r}-d\hat{z}|}+\frac{q'}{|\vec{r}-d'\hat{z}|}\right]\]

Sample: Rectangular trough: \(V=0\) on three sides, \(V=V_0\) on top (\(y=b\)). Width \(a\). Find \(V(x,y)\).

\[V(x,y)=\sum_{n=1,3,5,\ldots}\frac{4V_0}{n\pi}\frac{\sinh(n\pi y/a)}{\sinh(n\pi b/a)}\sin(n\pi x/a)\]

Sample: Square box \(0\le x,y\le a\), \(V=0\) on bottom & sides, \(V=V_0\sin(\pi x/a)\) on top. Find \(V\).

\[V(x,y)=V_0\frac{\sinh(\pi y/a)}{\sinh(\pi)}\sin(\pi x/a)\]
Single Fourier mode matches the top BC exactly.

Sample: Cylindrical region with \(V(a,\phi)=V_0\cos\phi\). Find \(V\) inside.

\[V(\rho,\phi)=A_0+\sum_{n}(\rho/a)^n[A_n\cos n\phi+B_n\sin n\phi]\]
\[\text{Match: only }n=1\text{ term}\;\Rightarrow\;V=V_0(\rho/a)\cos\phi\]

Sample: Coaxial cable, dielectric \(\epsilon_r\), applied voltage \(V_0\) on inner. Solve via cylindrical separation.

\[V(\rho)=V_0\frac{\ln(b/\rho)}{\ln(b/a)},\;\;\vec{E}=\frac{V_0}{\rho\ln(b/a)}\hat{a}_{\rho}\]

Sample: Spherical region with \(V(a,\theta)=V_0\cos\theta\). Find \(V\) inside (\(ra\)).

\[r
\[r>a:\;V=V_0(a/r)^2\cos\theta\]
Only \(\ell=1\) Legendre mode survives.

Sample: Conducting sphere of radius \(a\) in uniform field \(E_0\hat{z}\). Find \(V\) outside.

\[V(r,\theta)=-E_0 r\cos\theta+\frac{E_0 a^3\cos\theta}{r^2}=-E_0\!\left(r-\frac{a^3}{r^2}\right)\!\cos\theta\]

Sample: \(p\)-\(n\) junction depletion region: charge density \(-eN_A\) for \(-x_p

\[V''=eN_A/\epsilon\;(-x_p
Integrate twice on each side; match \(V\) and \(V\) at \(x=0\). Built-in voltage \(V_{bi}=\tfrac{e}{2\epsilon}(N_A x_p^2+N_D x_n^2)\).

Sample (comprehensive): Two grounded conducting plates at \(z=\pm d\), point charge \(+q\) between them at \(z=0\).

\[\text{Use infinite image series: }q\text{ at }z=0,\;-q\text{ at }\pm 2d,\;+q\text{ at }\pm 4d,\ldots\]
\[V=\frac{q}{4\pi\epsilon_0}\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{|\vec{r}-2nd\,\hat{z}|}\]

Chapter 5 — Steady Electric Currents

Key Theory — Chapter 5

Condensed from Cheng, Field and Wave Electromagnetics, §5-1 through §5-7. Read this before attempting the problems below.


5-1   Three Types of Current

Chapters 3 and 4 dealt with charges at rest. Chapter 5 introduces charges in motion — steady (d.c.) currents. Three mechanisms produce electric current, and it is important to recognise which one applies to a given situation:

  • Conduction currents — drift motion of conduction electrons (or holes) in conductors and semiconductors. Governed by Ohm's law.
  • Electrolytic currents — migration of positive and negative ions in an electrolyte (e.g. a salt solution between electrodes). Electrolytes obey a Laplace-equation model which is the basis of the electrolytic tank used to map hard-to-solve electrostatic potentials.
  • Convection currents — bulk motion of charged particles in a vacuum or rarefied gas (electron beams in a CRT, lightning). Convection currents are a hydrodynamic mass transport and are not governed by Ohm's law.

Drift velocity is tiny. The average drift velocity of conduction electrons is only \(10^{-4}\) to \(10^{-3}\) m/s — even in excellent conductors — because of continual electron-atom collisions that dissipate kinetic energy as heat. A conductor remains electrically neutral in bulk; electric forces prevent charge accumulation at any interior point.

第 3、4 章探討的是 at rest 的電荷,第 5 章開始討論 in motion 的電荷 — 即穩態 (d.c.) currents。產生 electric current 的機制有三種,辨識當下情境屬於哪一種非常重要:

  • Conduction currents — conductors 與 semiconductors 內 conduction electrons(或 holes)的 drift motion,遵循 Ohm's law。
  • Electrolytic currents — electrolyte(例如兩 electrodes 間的鹽溶液)中正、負 ions 的遷移。Electrolytes 遵守 Laplace-equation 模型,這也是 electrolytic tank 的原理基礎 — 用以實驗映射難以解析求得的 electrostatic potentials。
  • Convection currents — 真空或稀薄氣體中帶電粒子的整體運動(CRT 中的 electron beams、閃電)。Convection currents 屬於 hydrodynamic mass transport,遵循 Ohm's law。

Drift velocity 非常小。即使在極佳的 conductors 中,conduction electrons 的平均 drift velocity 也只有 \(10^{-4}\) 至 \(10^{-3}\) m/s — 原因是 electron 與原子持續碰撞,將動能以熱的形式耗散。Conductor 整體保持電中性;electric forces 會阻止電荷在內部任何一點累積。


5-2.1   Current Density and Convection Current

With \(N\) charge carriers per unit volume, each of charge \(q\), drifting at velocity \(\mathbf{u}\), the volume current density is defined as

\[ \mathbf{J} = Nq\mathbf{u} \quad (\text{A/m}^2) \qquad \text{(5-3)} \]

The total current through an arbitrary surface \(S\) is the flux of \(\mathbf{J}\):

\[ I = \int_S \mathbf{J}\cdot d\mathbf{s} \quad (\text{A}) \qquad \text{(5-5)} \]

Since \(Nq\) is just the free-charge volume density \(\rho\), for convection currents

\[ \mathbf{J} = \rho\,\mathbf{u} \quad (\text{convection current density}) \qquad \text{(5-6)} \]

When several species drift simultaneously (electrons, holes, ions), the current density superposes: \(\mathbf{J} = \sum_i N_i q_i \mathbf{u}_i\).


5-2.2   The Point Form of Ohm's Law

In a conductor the drift velocity is directly proportional to the applied field:

\[ \mathbf{u} = -\mu_e\, \mathbf{E} \quad (\text{m/s}) \qquad \text{(5-19)} \]

where \(\mu_e\) is the electron mobility (m²/V·s). Typical values: copper \(3.2\times 10^{-3}\), aluminium \(1.4\times 10^{-4}\), silver \(5.2\times 10^{-3}\). Substituting (5-19) into (5-3) and absorbing the sign into a positive proportionality constant \(\sigma = -\rho_e \mu_e\) gives the point form of Ohm's law:

\[ \boxed{\mathbf{J} = \sigma\,\mathbf{E}} \quad (\text{A/m}^2) \qquad \text{(5-21)} \]

The proportionality constant \(\sigma\) (siemens per metre, S/m, or A/V·m) is the conductivity of the medium. Isotropic materials in which this linear relation holds are called ohmic media. For semiconductors with electron and hole contributions:

\[ \sigma = -\rho_e\mu_e + \rho_h\mu_h \qquad \text{(5-22)} \]

Conductivity spans an enormous range — copper \(5.80\times 10^{7}\) S/m; germanium \(2.2\); silicon \(1.6\times 10^{-3}\); hard rubber \(10^{-15}\). Unlike permittivity, conductivity varies over 22+ orders of magnitude across common materials. The reciprocal of \(\sigma\) is resistivity (Ω·m), but Cheng prefers to work with conductivity throughout.

Equation (5-21) is the point form. The familiar circuit-theory relation \(V_{12} = RI\) is not a point relation — it concerns terminals and a cross section. Applying (5-21) to a straight homogeneous conductor of length \(\ell\), uniform cross section \(S\), conductivity \(\sigma\):

\[ \boxed{R = \frac{\ell}{\sigma S}} \quad (\Omega) \qquad \text{(5-27)} \]
\[ G = \frac{1}{R} = \sigma\frac{S}{\ell} \quad (\text{S}) \qquad \text{(5-28)} \]

5-3   EMF and Kirchhoff's Voltage Law

The static electric field is conservative: \(\oint_C \mathbf{E}\cdot d\ell = 0\) (eq. 5-31). Substituting \(\mathbf{E}=\mathbf{J}/\sigma\) gives the ohmic-medium version:

\[ \oint_C \frac{1}{\sigma}\mathbf{J}\cdot d\ell = 0 \qquad \text{(5-32)} \]

This equation tells us something fundamental: a steady current cannot be maintained in a closed loop by an electrostatic field alone. Charge carriers dissipate energy colliding with atoms; that energy must be resupplied by a non-conservative source — a battery (chemical), generator (mechanical), thermocouple (thermal), or photovoltaic cell (optical). These sources produce an equivalent impressed electric field intensity \(\mathbf{E}_i\) inside the source only.

The line integral of \(\mathbf{E}_i\) from − to + terminal inside the source defines the electromotive force (emf) of the source, denoted \(\mathscr{V}\):

\[ \mathscr{V} = \int_{2}^{1} \mathbf{E}_i \cdot d\ell \quad \text{(inside source)} \qquad \text{(5-33)} \]

The SI unit for emf is volt. Despite its name, emf is not a force in newtons — it is a measure of the strength of the non-conservative source. An ideal voltage source has zero internal resistance, so its terminal voltage equals its emf regardless of current.

Inside a conductor carrying steady current, the total field that drives the charges is the sum \(\mathbf{E}+\mathbf{E}_i\), and the point form of Ohm's law generalises to \(\mathbf{J} = \sigma(\mathbf{E}+\mathbf{E}_i)\). Taking the closed-loop line integral yields the macroscopic form of Kirchhoff's voltage law:

\[ \boxed{\sum_j \mathscr{V}_j = \sum_k R_k I_k} \quad (\text{V}) \qquad \text{(5-41)} \]

Around any closed loop, the algebraic sum of the emfs (voltage rises) equals the algebraic sum of the IR voltage drops. This is the basis for loop analysis in circuits.


5-4   Equation of Continuity and Kirchhoff's Current Law

Conservation of charge requires that current leaving a closed surface equal the rate of decrease of the enclosed charge:

\[ I = \oint_S \mathbf{J}\cdot d\mathbf{s} = -\frac{d}{dt}\int_V \rho\,dv \qquad \text{(5-42)} \]

Applying the divergence theorem and letting the integrand equal out (since \(V\) is arbitrary) gives the equation of continuity:

\[ \boxed{\nabla\cdot\mathbf{J} = -\frac{\partial\rho}{\partial t}} \quad (\text{A/m}^3) \qquad \text{(5-44)} \]

For steady currents, \(\partial\rho/\partial t = 0\), so

\[ \nabla\cdot\mathbf{J} = 0 \qquad \text{(5-45)} \]
\[ \oint_S \mathbf{J}\cdot d\mathbf{s} = 0 \qquad \text{(5-46)} \]
\[ \sum_j I_j = 0 \quad (\text{Kirchhoff's current law}) \qquad \text{(5-47)} \]

Steady currents are therefore divergenceless (solenoidal) — their streamlines close on themselves, unlike electrostatic field lines which start and end on charges. An ideal current source has infinite internal resistance.

Relaxation time

Combining Ohm's law with the continuity equation and \(\nabla\cdot\mathbf{E}=\rho/\epsilon\) gives, for a simple medium,

\[ \frac{\partial\rho}{\partial t} + \frac{\sigma}{\epsilon}\rho = 0 \qquad \text{(5-49)} \]

whose solution is an exponential decay:

\[ \rho = \rho_0\, e^{-(\sigma/\epsilon)\,t} = \rho_0\, e^{-t/\tau} \qquad \text{(5-50)} \]
\[ \tau = \frac{\epsilon}{\sigma} \quad (\text{s}) \quad \text{— relaxation time} \qquad \text{(5-51)} \]

For copper \(\tau \approx 1.52\times 10^{-19}\) s — so brief that \(\rho\) can be taken as zero everywhere inside a good conductor. In a good insulator the relaxation time can be hours or days.


5-5   Power Dissipation — Joule's Law

The work done by \(\mathbf{E}\) in moving a charge \(q\) at velocity \(\mathbf{u}\) is \(p = q\mathbf{E}\cdot\mathbf{u}\). Summing over all carriers in a volume element gives a power density

\[ \frac{dP}{dv} = \mathbf{E}\cdot\mathbf{J} \quad (\text{W/m}^3) \qquad \text{(5-53)} \]

Integrating over a volume \(V\) gives Joule's law:

\[ \boxed{P = \int_V \mathbf{E}\cdot\mathbf{J}\,dv} \quad (\text{W}) \qquad \text{(5-54)} \]

(The SI unit is watt, not joule — "joule" is energy.) For a uniform conductor where \(P = VI\) and \(V=IR\) we recover the familiar circuit expression:

\[ P = I^2 R \quad (\text{W}) \qquad \text{(5-55)} \]

5-6   Boundary Conditions for Current Density

The governing equations for steady current density in ohmic media (no non-conservative sources inside) are:

Differential form Integral form
∇·J = 0 ∮_S J·ds = 0 (5-56)
∇×(J/σ) = 0 ∮_C (1/σ)J·dℓ = 0 (5-57)

Applying these at an interface between two ohmic media with conductivities \(\sigma_1\) and \(\sigma_2\) gives the boundary conditions:

\[ J_{1n} = J_{2n} \quad \text{(normal component continuous)} \qquad \text{(5-58)} \]
\[ \frac{J_{1t}}{J_{2t}} = \frac{\sigma_1}{\sigma_2} \qquad \text{(5-59)} \]

Compare with the electrostatic boundary conditions in Chapter 3 (\(D_{1n}-D_{2n}=\rho_s\) and \(E_{1t}=E_{2t}\)): there is an exact analogy between \((\mathbf{J},\sigma)\) and \((\mathbf{D},\epsilon)\) at charge-free dielectric interfaces. This underlies the electrolytic-tank analogy.

Dividing (5-59) by (5-58) yields the refraction relation for current streamlines:

\[ \boxed{\frac{\tan\alpha_2}{\tan\alpha_1} = \frac{\sigma_2}{\sigma_1}} \qquad \text{(5-62)} \]

If medium 1 is a much better conductor than medium 2 (\(\sigma_1 \gg \sigma_2\)), then \(\alpha_2 \to 0\) — current in the poor conductor emerges almost normal to the surface of the good conductor.

Surface charge at the interface. When steady current flows across a boundary between two lossy dielectrics (finite \(\epsilon_1,\epsilon_2\) and \(\sigma_1,\sigma_2\)), the simultaneous requirements \(J_{1n}=J_{2n}\) and \(D_{1n}-D_{2n}=\rho_s\) force a surface charge:

\[ \rho_s = \left(\epsilon_1\frac{\sigma_2}{\sigma_1} - \epsilon_2\right) E_{2n} = \left(\epsilon_1 - \epsilon_2\frac{\sigma_1}{\sigma_2}\right) E_{1n} \qquad \text{(5-69)} \]

which vanishes only when \(\sigma_2/\sigma_1 = \epsilon_2/\epsilon_1\).


5-7   Resistance Calculations and the R–C Analogy

In a source-free homogeneous conductor the current field is both divergenceless and curl-free, so \(\mathbf{J}=-\nabla\psi\) with \(\psi = \sigma V\) satisfying Laplace's equation:

\[ \nabla^2 \psi = 0 \qquad \text{(5-66)} \]

A steady-current problem therefore has the same mathematical form as an electrostatic problem — methods of Chapter 4 apply directly.

For two conductors embedded in a lossy dielectric medium (\(\epsilon, \sigma\)), the resistance between them is

\[ R = \frac{V}{I} = \frac{-\int_L \mathbf{E}\cdot d\ell}{\oint_S \sigma\mathbf{E}\cdot d\mathbf{s}} \qquad \text{(5-80)} \]

Comparing with the capacitance between the same two conductors yields the elegant R–C analogy:

\[ \boxed{RC = \frac{C}{G} = \frac{\epsilon}{\sigma}} \qquad \text{(5-81)} \]

Consequence: if the capacitance between two conductors is already known, the leakage resistance (or conductance) through a lossy medium filling the same geometry follows immediately — no recomputation needed. For a coaxial cable of inner radius \(a\), outer radius \(b\), per unit length:

  • \(C_1 = 2\pi\epsilon/\ln(b/a)\)     (from Ch. 3)
  • \(R_1 = (\epsilon/\sigma)/C_1 = \ln(b/a)/(2\pi\sigma)\)     (leakage resistance)

The R–C analogy holds whenever \(\epsilon\) and \(\sigma\) of the medium have the same spatial dependence (trivially true for a homogeneous medium).


Chapter 5 at a Glance

  • Point form of Ohm's law: \(\mathbf{J}=\sigma\mathbf{E}\); macroscopic form \(V=IR\) with \(R=\ell/(\sigma S)\).
  • KVL: \(\sum\mathscr{V}_j = \sum R_k I_k\) (line integral of \(\mathbf{E}+\mathbf{E}_i\) around a loop).
  • KCL: \(\sum I_j = 0\), from the divergenceless steady current \(\nabla\cdot\mathbf{J}=0\).
  • Equation of continuity: \(\nabla\cdot\mathbf{J} = -\partial\rho/\partial t\) — general statement of charge conservation.
  • Relaxation time: \(\tau=\epsilon/\sigma\); in copper \(\sim 10^{-19}\) s, so good conductors have \(\rho=0\) inside.
  • Joule's law: \(P=\int_V\mathbf{E}\cdot\mathbf{J}\,dv = I^2 R\).
  • Boundary conditions: \(J_{1n}=J_{2n}\), \(J_{1t}/J_{2t}=\sigma_1/\sigma_2\); refraction \(\tan\alpha_2/\tan\alpha_1 = \sigma_2/\sigma_1\).
  • R–C analogy: \(RC=\epsilon/\sigma\) — capacitance ↔ leakage resistance.
  • Analogy with electrostatics: \(\mathbf{J}\leftrightarrow\mathbf{D}\), \(\sigma\leftrightarrow\epsilon\); every electrostatic technique transfers to steady-current problems.

Problems and Solutions


Review Questions (Tier 1)

R.5-1● EasyTier 1?
Review: Explain the difference between conduction, electrolytic, and convection currents. → Answer
R.5-1 Answer↑ Question

Conduction: drift of free electrons in a metal under \(\vec{E}\). Electrolytic: ion motion in a liquid solution. Convection: bulk transport of charge (e.g., charged beam in vacuum).

R.5-2● EasyTier 1?
Review: What is the point form of Ohm's law? State the assumptions under which it holds. → Answer
R.5-2 Answer↑ Question

\(\vec{J}=\sigma\vec{E}\). Holds in linear, isotropic conductors in steady state when \(\sigma\) is constant and the medium is non-dispersive.

R.5-3● EasyTier 1?
Review: Define electromotive force. In what units is it measured, and why is "force" a misnomer? → Answer
R.5-3 Answer↑ Question

Energy supplied per unit charge by a non-electrostatic source (chemical, mechanical, thermal). Measured in volts. "Force" is a misnomer — it is energy per charge, not a Newton.

R.5-4● EasyTier 1?
Review: State and derive Kirchhoff's voltage law from the line integral of the electric field. → Answer
R.5-4 Answer↑ Question

Static \(\oint\vec{E}\cdot d\vec{\ell}=0\) → including the line integral inside an EMF source converts to \(\sum\mathcal{E}=\sum IR\) around a loop (KVL).

R.5-5● EasyTier 1?
Review: What is the equation of continuity, and how does Kirchhoff's current law follow from it? → Answer
R.5-5 Answer↑ Question

\(\nabla\cdot\vec{J}=-\partial\rho_v/\partial t\). Integrating over a node volume in steady state gives \(\sum I_{\text{in}}=\sum I_{\text{out}}\), which is KCL.

R.5-6● EasyTier 1?
Review: Define relaxation time. What determines whether a material is a "good conductor" on a given time scale? → Answer
R.5-6 Answer↑ Question

\(\tau=\epsilon/\sigma\) — the time constant for free charge to redistribute. A material behaves as a good conductor on time scales \(t\gg\tau\) (or frequencies \(\omega\ll 1/\tau\)).

R.5-7● EasyTier 1?
Review: State Joule's law in both point and macroscopic form. → Answer
R.5-7 Answer↑ Question

Point form: \(p=\vec{J}\cdot\vec{E}=\sigma E^2\,\text{[W/m}^3\text{]}\). Macroscopic: \(P=I^2R=V^2/R\).

R.5-8● EasyTier 1?
Review: Write the boundary conditions for steady current density at an interface between two ohmic media. → Answer
R.5-8 Answer↑ Question

\(J_{1n}=J_{2n}\) and \(E_{1t}=E_{2t}\), giving \(\tan\theta_1/\tan\theta_2=\sigma_1/\sigma_2\).

R.5-9● EasyTier 1?
Review: State the R–C analogy. Under what conditions does it hold? → Answer
R.5-9 Answer↑ Question

For two electrodes in a lossy dielectric with the same geometry: \(RC=\epsilon/\sigma\). Holds when conductor field shape and current field shape coincide.


Ex 5.1●● MediumTier 1?
Current Density from Current

A copper wire (diameter 2 mm) carries I = 10 A. Find the current density J.

→ Solution
Ex 5.1 Solution↑ Problem
Cross-sectional area:
\[A = \pi (d/2)^{2} = \pi (10^{-3})^{2} = \pi \times 10^{-6} = 3.14\times 10^{-6} m^{2}\]
Current density:
\[J = I/A = 10/(3.14\times 10^{-6}) = 3.18\times 10^{6} A/m^{2}\]
For reference: copper fusing current density ≈ 10⁸ A/m² for short pulses.
Safe continuous rating ≈ 10⁶ A/m² for bare copper.

Ex 5.2●● MediumTier 1?
Resistance via Ohm's Law (Microscopic)

Copper (σ = 5.8×10⁷ S/m) wire: length L = 1 m, diameter d = 2 mm. Find R.

→ Solution
Ex 5.2 Solution↑ Problem
\[R = L/(\sigma A) = L/(\sigma \pi (d/2)^{2})\]
\[= 1/(5.8\times 10^{7} \times \pi \times 10^{-6})\]
\[= 1/(5.8\times 10^{7} \times 3.14\times 10^{-6})\]
\[= 1/(182.1)\]
\[= 5.49\times 10^{-3} \Omega = 5.49 m\Omega \]
Resistivity: ρ = 1/σ = 1/5.8×10⁷ = 1.72×10⁻⁸ Ω·m = 17.2 nΩ·m
\[R = \rho L/A = 1.72\times 10^{-8} \times 1/3.14\times 10^{-6} = 5.48 m\Omega \checkmark \]

Ex 5.3●● MediumTier 1?
EMF in a Circuit

Two batteries in series: V₁=12V (internal r₁=0.5Ω) and V₂=9V (internal r₂=1Ω), external R=10Ω.

Find I and terminal voltages.

→ Solution
Ex 5.3 Solution↑ Problem
Total EMF: V_total = 12 + 9 = 21 V (aiding)
Total resistance: R_total = r₁ + r₂ + R = 0.5 + 1 + 10 = 11.5 Ω
\[I = V_total/R_total = 21/11.5 = 1.826 A\]
Terminal voltage of V₁: V_T1 = 12 - I×r₁ = 12 - 1.826×0.5 = 11.09 V
Terminal voltage of V₂: V_T2 = 9 - I×r₂ = 9 - 1.826×1 = 7.17 V
Voltage across R: V_R = I×R = 1.826×10 = 18.26 V
Check: V_T1 + V_T2 = 11.09 + 7.17 = 18.26 = V_R ✓

Ex 5.4●● MediumTier 1?
Kirchhoff's Voltage Law Verification

Circuit: V=24V, R₁=4Ω, R₂=8Ω in series. Verify KVL.

→ Solution
Ex 5.4 Solution↑ Problem
\[I = V/(R_{1}+R_{2}) = 24/12 = 2 A\]
Voltage drops:
\[V_R1 = I\times R_{1} = 2\times 4 = 8 V\]
\[V_R2 = I\times R_{2} = 2\times 8 = 16 V\]
KVL (around loop, clockwise):
\[-V + V_R1 + V_R2 = 0\]
\[-24 + 8 + 16 = 0 \checkmark \]
Power balance:
\[P_source = V\times I = 24\times 2 = 48 W\]
\[P_R1 = I^{2}R_{1} = 4\times 4 = 16 W\]
\[P_R2 = I^{2}R_{2} = 4\times 8 = 32 W\]
\[P_R1 + P_R2 = 48 W = P_source \checkmark \]

Ex 5.5●● MediumTier 1?
Kirchhoff's Current Law

Node with currents: I₁=5A (in), I₂=3A (in), I₃=? (out), I₄=2A (out). Find I₃.

→ Solution
Ex 5.5 Solution↑ Problem
KCL (sum of currents in = sum of currents out):
\[I_{1} + I_{2} = I_{3} + I_{4}\]
\[5 + 3 = I_{3} + 2\]
\[I_{3} = 6 A\]
Alternatively (sum of all currents at node = 0):
\[+I_{1} + I_{2} - I_{3} - I_{4} = 0\]
\[+5 + 3 - I_{3} - 2 = 0\]
\[I_{3} = 6 A \checkmark \]
Physically: current cannot accumulate at a node (steady state).

Ex 5.6●● MediumTier 1?
Power Dissipation (Joule's Law)

Resistor: R = 100 Ω, V = 50 V. Find P, I, and energy in 1 hour.

→ Solution
Ex 5.6 Solution↑ Problem
\[I = V/R = 50/100 = 0.5 A\]
\[P = V^{2}/R = 2500/100 = 25 W\]
\[P = I^{2}R = 0.25\times 100 = 25 W \checkmark \]
\[P = VI = 50\times 0.5 = 25 W \checkmark \]
Energy in 1 hour:
\[W = P \times t = 25 \times 3600 = 90,000 J = 90 kJ = 0.025 kWh\]
Temperature check: for resistor rated at 25 W, this is at full rated power.

Ex 5.7●● MediumTier 1?
Resistance of a Conical Conductor

Truncated cone: conductivity σ, small radius a, large radius b, length L, axis along z. Find R.

→ Solution
Ex 5.7 Solution↑ Problem
Radius varies linearly: r(z) = a + (b-a)z/L
Cross-sectional area at z: A(z) = πr(z)²
Differential resistance: dR = dz/(σ A(z)) = dz/(σ π r(z)²)
\[R = \int _{0}^L dz/(\sigma \pi [a + (b-a)z/L]^{2})\]
Let u = a + (b-a)z/L, du = (b-a)/L dz:
\[R = (L/(b-a)) \int _a^b du/(\sigma \pi u^{2})\]
\[= (L/(b-a)) \times [1/(\sigma \pi u)]_a^b \times (-1) ...\]
\[R = L/((b-a)\sigma \pi ) \times [-1/u]_a^b\]
\[= L/((b-a)\sigma \pi ) \times (1/a - 1/b)\]
\[= L/((b-a)\sigma \pi ) \times (b-a)/(ab)\]
\[R = L/(\sigma \pi ab)\]
For a=b (cylinder): apply L'Hopital → R = L/(σπa²) ✓

Ex 5.8●● MediumTier 1?
Boundary Conditions for J at Interface

Medium 1 (σ₁ = 10⁶ S/m) and Medium 2 (σ₂ = 10³ S/m). J₁ = 5×10⁶ A/m² at θ₁ = 30° to interface normal. Find J₂ and θ₂.

→ Solution
Ex 5.8 Solution↑ Problem
Boundary conditions (steady current):
Normal: J₁n = J₂n (continuity of normal current density)
Tangential: E₁t = E₂t → J₁t/σ₁ = J₂t/σ₂
\[J_{1}n = J_{1} cos30° = 5\times 10^{6} \times 0.866 = 4.33\times 10^{6} A/m^{2}\]
\[J_{1}t = J_{1} sin30° = 5\times 10^{6} \times 0.5 = 2.5\times 10^{6} A/m^{2}\]
Normal component continuous:
\[J_{2}n = J_{1}n = 4.33\times 10^{6} A/m^{2}\]
Tangential component:
\[J_{2}t = (\sigma _{2}/\sigma _{1}) J_{1}t = (10^{3}/10^{6}) \times 2.5\times 10^{6} = 2500 A/m^{2}\]
\[J_{2} = \sqrt{J_{2}n^{2} + J_{2}t^{2}} = \sqrt{(4.33\times 10^{6}}^{2} + 2500^{2}) \approx 4.33\times 10^{6} A/m^{2}\]
\[tan \theta _{2} = J_{2}t/J_{2}n = 2500/(4.33\times 10^{6}) = 5.77\times 10^{-4}\]
θ₂ ≈ 0.033° (nearly normal — current crowds into normal direction in low-σ medium)
\[tan \theta _{2}/tan \theta _{1} = \sigma _{2}/\sigma _{1} = 10^{-3} \checkmark \]

Ex 5.9●● MediumTier 1?
Resistance of a Cylindrical Shell

Cylindrical shell: inner radius a=1cm, outer radius b=3cm, length L=10cm, σ = 10³ S/m, current flows radially. Find R.

→ Solution
Ex 5.9 Solution↑ Problem
For radial current flow in cylindrical geometry:
dR = dρ/(σ × 2πρL) (shell of thickness dρ, area 2πρL)
\[R = \int _a^b d\rho /(2\pi \sigma L\rho ) = ln(b/a)/(2\pi \sigma L)\]
\[= ln(3)/(2\pi \times 10^{3} \times 0.1)\]
\[= 1.099/(628.3)\]
\[= 1.75\times 10^{-3} \Omega = 1.75 m\Omega \]
For axial current flow: R = L/(σπ(b²-a²))
\[= 0.1/(10^{3} \times \pi \times (9-1)\times 10^{-4})\]
\[= 0.1/(2.513) = 39.8 m\Omega \]

Ex 5.10●● MediumTier 1?
Current Distribution in a Conducting Sphere

Uniform conductivity sphere, radius a, uniform E₀ applied externally. Find J inside.

→ Solution
Ex 5.10 Solution↑ Problem
Inside a conducting sphere in a uniform field, if the sphere is in a medium of
conductivity σ₁ surrounded by a medium of conductivity σ₂:
\[J_inside = 3\sigma _{1}/(\sigma _{1}+2\sigma _{2}) \times \sigma _{2} \times E_{0} ...\]
For the simpler case: solid homogeneous conductor (σ uniform):
\[J = \sigma E (Ohm's law)\]
E_inside = E₀ (same as applied if conductivities match)
For a highly conductive sphere (σ → ∞) in a resistive medium:
E_inside → 0 (field excluded from perfect conductor)
J → concentrated on surface
For σ_sphere = σ_medium:
J = σE₀ (uniform, undisturbed)

Ex 5.11●● MediumTier 1?
Continuity Equation

Charge density ρ_v = ρ₀ e^(-t/τ) uniformly distributed in a medium (σ, ε). Find J and verify continuity.

→ Solution
Ex 5.11 Solution↑ Problem
Continuity equation: ∂ρ_v/∂t + ∇·J = 0
\[\partial \rho _v/\partial t = -\rho _{0}/\tau \times e^(-t/\tau )\]
From ∇·J = -∂ρ_v/∂t = ρ₀/τ × e^(-t/τ)
For ohmic medium: J = σE = σD/ε → ∇·J = σ/ε × ∇·D = σ/ε × ρ_v
\[\sigma /\varepsilon \times \rho _{0} e^(-t/\tau ) = \rho _{0}/\tau \times e^(-t/\tau )\]
\[\to \tau = \varepsilon /\sigma (relaxation time)\]
For copper: τ = ε₀/σ = 8.85×10⁻¹²/5.8×10⁷ ≈ 1.5×10⁻¹⁹ s (extremely fast)
For seawater: τ ≈ ε₀×80/4 ≈ 1.8×10⁻¹⁰ s = 0.18 ns
For glass: τ ≈ hours (charges persist)

Ex 5.12●● MediumTier 1?
Resistance Between Two Electrodes

Two spherical electrodes (radius a = 5 mm) buried in earth (σ = 10⁻² S/m), separation D = 1 m (D >> a). Find R.

→ Solution
Ex 5.12 Solution↑ Problem
For two spherical electrodes in a conducting medium (D >> a):
\[R \approx 2 \times [1/(4\pi \sigma a)] - 1/(2\pi \sigma D)\]
\[\approx 1/(2\pi \sigma a) when D >> a\]
\[R \approx 1/(2\pi \sigma a) = 1/(2\pi \times 10^{-2} \times 5\times 10^{-3}) = 1/(3.14\times 10^{-4}) = 3183 \Omega \approx 3.18 k\Omega \]
Full expression:
\[R = 1/(2\pi \sigma ) \times (1/a - 1/D)\]
\[= 1/(2\pi \times 10^{-2}) \times (1/0.005 - 1/1.0)\]
\[= 15.92 \times (200 - 1)\]
\[= 15.92 \times 199\]
\[= 3168 \Omega \approx 3.17 k\Omega \]

Ex 5.13●● MediumTier 1?
Equivalent Resistance of Network

Find equivalent resistance: R₁=6Ω, R₂=3Ω in parallel, that combination in series with R₃=4Ω.

→ Solution
Ex 5.13 Solution↑ Problem
\[R_{12} (parallel): 1/R_{12} = 1/6 + 1/3 = 1/6 + 2/6 = 3/6\]
\[R_{12} = 2 \Omega \]
\[R_total = R_{12} + R_{3} = 2 + 4 = 6 \Omega \]
For V=12V across whole network:
\[I_total = 12/6 = 2 A\]
\[V_R3 = 2\times 4 = 8 V\]
\[V_R12 = 2\times 2 = 4 V\]
\[I_R1 = 4/6 = 0.667 A\]
\[I_R2 = 4/3 = 1.333 A\]
Check: 0.667 + 1.333 = 2 A ✓

Ex 5.14●● MediumTier 1?
Current Density in a Tapered Conductor

Conductor tapers linearly from radius a to 2a over length L. Uniform current I flows axially. Find J(z) and E(z).

→ Solution
Ex 5.14 Solution↑ Problem
Radius at position z: r(z) = a(1 + z/L) for 0 ≤ z ≤ L
\[Area: A(z) = \pi r(z)^{2} = \pi a^{2}(1 + z/L)^{2}\]
\[J(z) = I/A(z) = I/[\pi a^{2}(1 + z/L)^{2}]\]
At z=0: J = I/(πa²)
At z=L: J = I/(4πa²) = J(z=0)/4
\[E(z) = J(z)/\sigma = I/[\sigma \pi a^{2}(1 + z/L)^{2}]\]
Voltage across conductor:
\[V = \int _{0}^L E dz = I/(\sigma \pi a^{2}) \int _{0}^L dz/(1+z/L)^{2}\]
\[Let u = 1+z/L: V = I/(\sigma \pi a^{2}) \times L \int _{1}^2 u^{-2} du = IL/(\sigma \pi a^{2}) \times [-1/u]_{1}^{2}\]
\[= IL/(\sigma \pi a^{2}) \times (1 - 1/2) = IL/(2\sigma \pi a^{2})\]
R = V/I = L/(2σπa²) [compare to uniform wire: R = L/(σπa²), so tapered R is halved]

Ex 5.15●● MediumTier 1?
Power Loss in a Transmission Line

Two-wire line: conductor radius a=1mm, separation D=10mm, σ_c=5.8×10⁷ S/m, length L=100m, I=10A. Find P_loss.

→ Solution
Ex 5.15 Solution↑ Problem
Resistance per unit length (two wires in series):
\[R/L = 2/(\sigma _c \pi a^{2})\]
\[= 2/(5.8\times 10^{7} \times \pi \times 10^{-6})\]
\[= 2/(182.2)\]
\[= 10.98\times 10^{-3} \Omega /m \approx 11 m\Omega /m\]
Total resistance: R = 10.98×10⁻³ × 100 = 1.098 Ω
Power loss: P = I²R = 100 × 1.098 = 109.8 W
At 50Hz with skin effect correction (δ=9.3mm >> a=1mm for 50Hz):
Skin effect negligible at power frequency for this wire size.
(At 100MHz, δ=6.6μm << a, and AC resistance increases significantly)

Textbook Practice Problems

5-1● EasyTier 1?
Current density and Ohm's law → Answer
5-2● EasyTier 1?
Current density and Ohm's law → Answer
5-8●● MediumTier 1?
Continuity equation → Answer
5-14●● MediumTier 1?
Boundary conditions → Answer
5-15●● MediumTier 1?
Boundary conditions → Answer
5-17● EasyTier 1?
Resistance calculations → Answer
5-18●● MediumTier 1?
Resistance calculations → Answer
5-3●● MediumTier 2?
Current density and Ohm's law → Answer
5-5● EasyTier 2?
EMF and Kirchhoff's voltage law → Answer
5-6●● MediumTier 2?
EMF and Kirchhoff's voltage law → Answer
5-9●● MediumTier 2?
Continuity equation → Answer
5-11● EasyTier 2?
Power dissipation → Answer
5-12●● MediumTier 2?
Power dissipation → Answer
5-19●● MediumTier 2?
Resistance calculations → Answer
5-22●● MediumTier 2?
Complex geometries → Answer
5-23●●● HardTier 3?
Complex geometries → Answer
5-24●●● HardTier 3?
Complex geometries → Answer

Textbook Practice — Approach Hints

Sample: Copper wire (\(\sigma=5.8\!\cdot\!10^7\) S/m), radius 1 mm, carrying \(I=5\,\text{A}\). Find \(\vec{J}\) and \(\vec{E}\).

\[J=I/A=5/(\pi(10^{-3})^2)=1.59\!\cdot\!10^6\,\text{A/m}^2\]
\[E=J/\sigma=1.59\!\cdot\!10^6/5.8\!\cdot\!10^7=27.4\,\text{mV/m}\]

Sample: Aluminum wire (\(\sigma=3.5\!\cdot\!10^7\) S/m), \(L=100\,\text{m}\), radius 2 mm. Find resistance.

\[R=\frac{L}{\sigma A}=\frac{100}{3.5\!\cdot\!10^7\cdot\pi(2\!\cdot\!10^{-3})^2}=0.227\,\Omega\]

Sample: Time-varying point charge \(\rho_v(\vec{r},t)\). Show continuity equation in steady state.

\[\nabla\cdot\vec{J}+\frac{\partial\rho_v}{\partial t}=0\;\xrightarrow{\text{steady}}\;\nabla\cdot\vec{J}=0\]
Conservation of charge in differential form.

Sample: Boundary between \(\sigma_1=10^6\) and \(\sigma_2=10^4\) S/m. \(J_{1n}=100\,\text{A/m}^2\), \(E_{1t}=2\,\text{V/m}\). Find \(\vec{J}_2\) and surface charge.

\[J_{2n}=J_{1n}=100\,\text{A/m}^2,\;E_{2t}=E_{1t}=2\,\text{V/m}\]
\[E_{1n}=J_{1n}/\sigma_1=10^{-4},\;E_{2n}=J_{2n}/\sigma_2=10^{-2}\]
\[\rho_s=\epsilon_0(E_{2n}-E_{1n})=8.854\!\cdot\!10^{-12}(10^{-2}-10^{-4})=8.77\!\cdot\!10^{-14}\,\text{C/m}^2\]

Sample: At interface \(\sigma_1\ne\sigma_2\): angle of \(\vec{J}\) on each side.

\[\frac{\tan\theta_1}{\tan\theta_2}=\frac{\sigma_1}{\sigma_2}\;(\text{from }J_n\text{ continuous, }E_t\text{ continuous})\]

Sample: Truncated cone of length \(L\), radii \(a\) and \(b\), conductivity \(\sigma\). Find resistance end-to-end.

\[r(z)=a+(b-a)z/L,\;A(z)=\pi r^2(z)\]
\[R=\int_0^L\frac{dz}{\sigma A(z)}=\frac{L}{\pi\sigma ab}\]

Sample: Two cylindrical resistors in parallel: \(R_1=10\,\Omega\), \(R_2=20\,\Omega\). Find equivalent.

\[R_{eq}=R_1R_2/(R_1+R_2)=200/30=6.67\,\Omega\]

Sample: Conductor with \(\vec{E}=10\hat{x}\,\text{V/m}\), \(\sigma=10^4\) S/m. Find \(\vec{J}\) and current through 1 cm² area.

\[\vec{J}=\sigma\vec{E}=10^5\hat{x}\,\text{A/m}^2\]
\[I=\vec{J}\cdot\vec{A}=10^5(10^{-4})=10\,\text{A}\]

Sample: Battery EMF \(\mathcal{E}=12\,\text{V}\), internal resistance \(0.5\,\Omega\), external \(5.5\,\Omega\). Find current and terminal voltage.

\[I=\mathcal{E}/(R_{int}+R_{ext})=12/6=2\,\text{A}\]
\[V_T=\mathcal{E}-IR_{int}=12-1=11\,\text{V}\]

Sample: Loop with two batteries (12 V, 6 V opposing) and resistors \(2\,\Omega\), \(4\,\Omega\) in series. KVL gives current.

\[12-6=I(2+4)\;\Rightarrow\;I=1\,\text{A}\]

Sample: Conductor with \(\sigma=10^7\), time-constant for charge decay?

\[\rho_v(t)=\rho_v(0)e^{-t/\tau},\;\tau=\epsilon/\sigma=8.854\!\cdot\!10^{-12}/10^7=8.85\!\cdot\!10^{-19}\,\text{s}\]
Charge inside good conductor dissipates almost instantly.

Sample: Wire with \(\vec{J}=10^6\hat{z}\,\text{A/m}^2\), \(\sigma=5.8\!\cdot\!10^7\). Find power dissipated per unit volume.

\[p=\vec{J}\cdot\vec{E}=J^2/\sigma=(10^6)^2/(5.8\!\cdot\!10^7)=17.24\,\text{kW/m}^3\]

Sample: Resistor \(R=10\,\Omega\), current \(I=2\,\text{A}\). Find dissipated power.

\[P=I^2R=4\cdot10=40\,\text{W}\]

Sample: Coaxial geometry filled with imperfect dielectric \(\sigma\). Inner \(a\), outer \(b\), length \(L\). Find leakage resistance.

\[R=\frac{\ln(b/a)}{2\pi\sigma L}\]

Sample: Spherical resistor: inner radius \(a\), outer \(b\), conductivity \(\sigma\). Find \(R\) between concentric surfaces.

\[R=\int_a^b\frac{dr}{\sigma(4\pi r^2)}=\frac{1}{4\pi\sigma}\!\left(\frac{1}{a}-\frac{1}{b}\right)\]

Sample: Two parallel hemispherical electrodes buried in earth (\(\sigma\)). Find ground resistance.

\[R_{single}=1/(2\pi\sigma a),\;R_{pair}\approx 2R_{single}=1/(\pi\sigma a)\]

Sample (synthesis): Resistance between two small spheres of radius \(a\) separated by \(d\gg a\) in conducting medium \(\sigma\).

\[R=\frac{1}{4\pi\sigma}\!\left(\frac{2}{a}-\frac{2}{d}\right)\approx\frac{1}{2\pi\sigma a}\;(d\to\infty)\]

Chapter 6 — Static Magnetic Fields

Key Theory — Chapter 6

Condensed from Cheng, Field and Wave Electromagnetics, §6-1 through §6-13. Read this before attempting the problems below.


6-1   The Magnetic Force and the Definition of B

A charge \(q\) at rest in an electric field experiences \(\mathbf{F}_e = q\mathbf{E}\). When the charge moves with velocity \(\mathbf{u}\), experiment shows a second force appears: one perpendicular to both \(\mathbf{u}\) and a fixed direction at each point, with magnitude proportional to \(q\), to \(|\mathbf{u}|\), and to the component of \(\mathbf{u}\) perpendicular to that fixed direction. This magnetic force is expressed as

\[ \mathbf{F}_m = q\,\mathbf{u}\times\mathbf{B} \quad (\text{N}) \qquad \text{(6-4)} \]

and the new vector field \(\mathbf{B}\) (webers per square meter, or teslas) is the magnetic flux density. Combining electric and magnetic effects gives Lorentz's force equation:

\[ \boxed{\mathbf{F} = q(\mathbf{E} + \mathbf{u}\times\mathbf{B})} \qquad \text{(6-5)} \]

Lorentz's equation can itself be taken as a fundamental postulate of the electromagnetic model — it defines both \(\mathbf{E}\) (via the force on a stationary charge) and \(\mathbf{B}\) (via the force on a moving one). It cannot be derived from other postulates.


6-2   Fundamental Postulates of Magnetostatics in Free Space

In free space, magnetostatics is built on exactly two postulates specifying the divergence and curl of \(\mathbf{B}\):

Differential form Integral form
Gauss's law for B: ∇·B = 0 ∮_S B·ds = 0 (6-6, 6-9)
Ampère's circuital law: ∇×B = μ₀J ∮_C B·dℓ = μ₀ I (6-7, 6-10)
Table 6-1 — Postulates of Magnetostatics in Free Space

No magnetic monopoles. Because \(\nabla\cdot\mathbf{B} = 0\), magnetic flux lines always close on themselves — there is no magnetic analog of electric charge density. Cutting a bar magnet doesn't produce an isolated N or S pole; it produces two smaller magnets, and this continues down to atomic dimensions. Equation (6-9) is the law of conservation of magnetic flux.

Ampère's circuital law says the circulation of \(\mathbf{B}\) around any closed path equals \(\mu_0\) times the total current through the enclosed surface. Taking \(\nabla\cdot(\nabla\times\mathbf{B}) = 0\) in (6-7) immediately gives \(\nabla\cdot\mathbf{J} = 0\), consistent with steady-current continuity (5-45).


6-3   Vector Magnetic Potential A

Because \(\mathbf{B}\) is solenoidal, it can be expressed as the curl of another vector field:

\[ \mathbf{B} = \nabla\times\mathbf{A} \quad (\text{T}) \qquad \text{(6-15)} \]

\(\mathbf{A}\) is called the vector magnetic potential, measured in Wb/m. Defining a vector requires specifying both its curl (done by (6-15)) and its divergence — we are free to choose the latter. Substituting (6-15) into the curl postulate and expanding \(\nabla\times\nabla\times\mathbf{A} = \nabla(\nabla\cdot\mathbf{A}) - \nabla^2\mathbf{A}\) gives

\[ \nabla(\nabla\cdot\mathbf{A}) - \nabla^2\mathbf{A} = \mu_0\mathbf{J} \qquad \text{(6-19)} \]

Choosing the Coulomb gauge

\[ \nabla\cdot\mathbf{A} = 0 \qquad \text{(6-20)} \]

reduces this to a vector Poisson equation:

\[ \nabla^2\mathbf{A} = -\mu_0\mathbf{J} \qquad \text{(6-21)} \]

Each Cartesian component is a scalar Poisson equation identical in form to \(\nabla^2 V = -\rho/\epsilon_0\) from electrostatics. The solution is therefore

\[ \boxed{\mathbf{A} = \frac{\mu_0}{4\pi}\int_{V'}\frac{\mathbf{J}}{R}\,dv'} \quad (\text{Wb/m}) \qquad \text{(6-23)} \]

Physical meaning of A: the line integral of \(\mathbf{A}\) around a closed contour equals the magnetic flux through the enclosed surface:

\[ \Phi = \int_S \mathbf{B}\cdot d\mathbf{s} = \oint_C \mathbf{A}\cdot d\ell \quad (\text{Wb}) \qquad \text{(6-25)} \]

6-4   Biot–Savart Law

For a thin wire, \(\mathbf{J}\,dv' = I\,d\ell'\). Equation (6-23) becomes

\[ \mathbf{A} = \frac{\mu_0 I}{4\pi}\oint_{C'}\frac{d\ell'}{R} \qquad \text{(6-27)} \]

Taking the curl and doing the \(\nabla(1/R) = -\mathbf{a}_R/R^2\) bookkeeping yields Biot–Savart law:

\[ \boxed{\mathbf{B} = \frac{\mu_0 I}{4\pi}\oint_{C'}\frac{d\ell'\times\mathbf{a}_R}{R^2}} \quad (\text{T}) \qquad \text{(6-32)} \]

where \(\mathbf{a}_R\) is the unit vector from source element to field point. Biot–Savart computes \(\mathbf{B}\) directly from a current-carrying circuit; it is the magnetic analogue of Coulomb's law for \(\mathbf{E}\). Ampère's law is more convenient when symmetry (cylindrical, planar, toroidal) provides a path over which \(|\mathbf{B}|\) is constant; Biot–Savart is the tool when it doesn't.


6-5   The Magnetic Dipole

A small current loop of area \(S\) carrying current \(I\) behaves as a magnetic dipole with moment

\[ \mathbf{m} = IS\,\mathbf{a}_n \quad (\text{A}\cdot\text{m}^2) \qquad \text{(6-46)} \]

where \(\mathbf{a}_n\) follows the right-hand rule from the current direction. At distances \(R \gg\) loop size, the vector potential is

\[ \mathbf{A} = \frac{\mu_0\,\mathbf{m}\times\mathbf{a}_R}{4\pi R^2} \qquad \text{(6-45)} \]

and the magnetic flux density is

\[ \mathbf{B} = \frac{\mu_0 m}{4\pi R^3}\bigl(2\cos\theta\,\mathbf{a}_R + \sin\theta\,\mathbf{a}_\theta\bigr) \qquad \text{(6-48)} \]

which is exactly analogous to the field of an electric dipole \((\mathbf{p}, V)\) — with \(\mathbf{A}\) playing the role of \(V\), and the same angular dependence.


6-6   Magnetization and Equivalent Current Densities

In a material, orbital and spin motions of electrons create microscopic magnetic dipoles. An applied field partially aligns them, yielding a macroscopic magnetization vector \(\mathbf{M}\) (A/m) — magnetic moment per unit volume. A magnetized body can be replaced, as far as its external field is concerned, by an equivalent volume current density and surface current density:

\[ \mathbf{J}_m = \nabla\times\mathbf{M} \quad (\text{A/m}^2) \qquad \text{(6-62)} \]
\[ \mathbf{J}_{ms} = \mathbf{M}\times\mathbf{a}_n \quad (\text{A/m}) \qquad \text{(6-63)} \]

Uniformly magnetized bar magnets have no volume current (\(\nabla\times\mathbf{M} = 0\) for constant \(\mathbf{M}\)) but carry a surface current \(\mathbf{M}\times\mathbf{a}_n\) that effectively turns them into solenoids of lineal current density \(|\mathbf{M}|\).


6-7   Magnetic Field Intensity H and the Constitutive Relation

In matter, Ampère's law must account for both free current \(\mathbf{J}\) and bound (magnetization) current \(\mathbf{J}_m\): \((1/\mu_0)\nabla\times\mathbf{B} = \mathbf{J} + \nabla\times\mathbf{M}\). Rearranging motivates the definition of a new fundamental field:

\[ \mathbf{H} = \frac{\mathbf{B}}{\mu_0} - \mathbf{M} \quad (\text{A/m}) \qquad \text{(6-75)} \]

called the magnetic field intensity. Its governing equation involves only free current:

\[ \boxed{\nabla\times\mathbf{H} = \mathbf{J}}, \qquad \oint_C \mathbf{H}\cdot d\ell = I_{\text{free}} \qquad \text{(6-76, 6-78)} \]

For linear, isotropic media, \(\mathbf{M} = \chi_m\mathbf{H}\) with magnetic susceptibility \(\chi_m\) (dimensionless). Substituting into (6-75) yields the constitutive relation

\[ \boxed{\mathbf{B} = \mu_0(1+\chi_m)\mathbf{H} = \mu_0\mu_r\mathbf{H} = \mu\mathbf{H}} \qquad \text{(6-80a)} \]
\[ \mu_r = 1 + \chi_m = \mu/\mu_0 \quad \text{(relative permeability)} \qquad \text{(6-81)} \]

6-8   Magnetic Circuits

For a toroidal ferromagnetic core with \(N\) turns carrying current \(I\), Ampère's law gives \(H = NI/\ell\) and \(B = \mu H\), so the flux is \(\Phi = BA = NI/\mathcal{R}\), where the reluctance is

\[ \mathcal{R} = \frac{\ell}{\mu A} \quad (\text{H}^{-1}) \]

and \(NI\) is the magnetomotive force (mmf, in ampere-turns). Magnetic circuits obey KVL/KCL analogues:

\[ \sum_j N_j I_j = \sum_k \mathcal{R}_k\Phi_k \quad (\text{magnetic KVL}) \qquad \text{(6-101)} \]
\[ \sum_j \Phi_j = 0 \quad (\text{magnetic KCL}) \qquad \text{(6-102)} \]

The analogy: \(\mathscr{V}\leftrightarrow NI\), \(I\leftrightarrow\Phi\), \(R\leftrightarrow\mathcal{R}\), \(\sigma\leftrightarrow\mu\).


6-9   Behavior of Magnetic Materials

Three classes are distinguished by the sign and magnitude of \(\chi_m\):


6-10   Boundary Conditions for Magnetostatic Fields

From \(\nabla\cdot\mathbf{B} = 0\) applied to a pillbox at the interface, and \(\oint\mathbf{H}\cdot d\ell = I_{\text{free}}\) applied to a flat contour straddling the interface:

\[ B_{1n} = B_{2n} \quad \text{(normal B continuous)} \qquad \text{(6-107)} \]
\[ \mathbf{a}_{n2}\times(\mathbf{H}_1 - \mathbf{H}_2) = \mathbf{J}_s \quad (\text{A/m}) \qquad \text{(6-111)} \]

For two linear media, the normal condition becomes \(\mu_1 H_{1n} = \mu_2 H_{2n}\). The tangential \(\mathbf{H}\) is continuous at any interface between physical media (\(\mathbf{J}_s = 0\)); it is discontinuous only at the surface of an idealised perfect conductor or superconductor. Compare with electrostatics: \(\mathbf{D}\) normal jumps by \(\rho_s\), \(\mathbf{E}\) tangential is continuous.


6-11   Inductance

Mutual flux from loop \(C_1\) linking \(C_2\) is \(\Phi_{12} = \int_{S_2}\mathbf{B}_1\cdot d\mathbf{s}_2\). By Biot–Savart, \(\mathbf{B}_1 \propto I_1\), so

\[ \Phi_{12} = L_{12}I_1, \qquad L_{12} = \frac{\Lambda_{12}}{I_1} \quad (\text{henries, H}) \qquad \text{(6-124, 6-127)} \]

where \(\Lambda_{12} = N_2\Phi_{12}\) is the flux linkage. Carrying the algebra through for two multi-turn circuits yields Neumann's formula:

\[ L_{12} = \frac{\mu_0}{4\pi}\oint_{C_1}\oint_{C_2}\frac{d\ell_1\cdot d\ell_2}{R} \quad (\text{H}) \qquad \text{(6-150b)} \]

Reciprocity gives \(L_{12} = L_{21}\). Self-inductance \(L\) is defined analogously by setting both loops equal. Inductance is a purely geometrical quantity (for a linear medium), independent of the current magnitude. For a solenoid, \(L \propto N^2\).


6-12   Magnetic Energy

The work done against induced emf to build up current \(I_1\) in an inductor of self-inductance \(L_1\) is stored as magnetic energy:

\[ W_m = \tfrac{1}{2}L I^2 = \tfrac{1}{2}I\Phi \quad (\text{J}) \qquad \text{(6-158, 6-163)} \]

For a system of N coupled loops:

\[ W_m = \frac{1}{2}\sum_{j=1}^N\sum_{k=1}^N L_{jk}I_j I_k = \frac{1}{2}\sum_k I_k\Phi_k \qquad \text{(6-162, 6-166)} \]

Expressed in terms of field quantities:

\[ W_m = \tfrac{1}{2}\int_{V'}\mathbf{A}\cdot\mathbf{J}\,dv' \qquad \text{(6-169)} \]
\[ \boxed{W_m = \tfrac{1}{2}\int_{V'}\mathbf{H}\cdot\mathbf{B}\,dv' = \int\frac{B^2}{2\mu}dv' = \int\tfrac{1}{2}\mu H^2\,dv'} \qquad \text{(6-172)} \]

with magnetic energy density

\[ w_m = \tfrac{1}{2}\mathbf{H}\cdot\mathbf{B} = \frac{B^2}{2\mu} = \tfrac{1}{2}\mu H^2 \quad (\text{J/m}^3) \qquad \text{(6-174)} \]

These forms are exact analogues of \(w_e = \tfrac{1}{2}\mathbf{D}\cdot\mathbf{E}\) in electrostatics. Computing \(W_m\) from the field integral and equating to \(\tfrac{1}{2}LI^2\) is often an easier route to inductance than counting flux linkages.


6-13   Magnetic Forces and Torques

Force on a current-carrying element:

\[ d\mathbf{F}_m = I\,d\ell\times\mathbf{B} \quad (\text{N}) \qquad \text{(6-183)} \]
\[ \mathbf{F}_m = I\oint_C d\ell\times\mathbf{B} \qquad \text{(6-184)} \]

Two parallel wires carrying currents \(I_1, I_2\) separated by distance \(d\) experience

\[ |F'_{12}| = \frac{\mu_0 I_1 I_2}{2\pi d} \quad (\text{N/m}) \qquad \text{(6-192)} \]

Same direction → attraction; opposite directions → repulsion. (Opposite sense to Coulomb's law for charges — a subtle point.)

Torque on a current loop of dipole moment \(\mathbf{m}\) in a uniform \(\mathbf{B}\):

\[ \boxed{\mathbf{T} = \mathbf{m}\times\mathbf{B}} \quad (\text{N}\cdot\text{m}) \qquad \text{(6-195)} \]

This is the operating principle of the d-c motor: a split-ring commutator reverses the loop current every half turn to keep \(\mathbf{T}\) aligned and the rotor spinning.

Hall effect: a current \(\mathbf{J}\) flowing in a magnetic field \(\mathbf{B}\) develops a transverse voltage \(V_h\) across the conductor. The sign of \(V_h\) reveals whether carriers are electrons or holes — an essential diagnostic for distinguishing n- and p-type semiconductors.


Chapter 6 at a Glance


Problems and Solutions


Part A — Biot-Savart, Ampère's Law, Magnetostatics (Sections 6-1 to 6-7)


Ex 6.1●● MediumTier 1?
Magnetic Field from a Long Straight Wire

Infinite wire carrying I = 10 A along the z-axis. Find B at ρ = 0.05 m.

→ Solution
Ex 6.1 Solution↑ Problem
By Ampère's law (symmetry → B in a_φ direction):
\[\oint H\cdot dl = I_enc\]
\[H \times 2\pi \rho = I\]
\[H = I/(2\pi \rho )\]
\[B = \mu _{0}H = \mu _{0}I/(2\pi \rho )\]
\[= 4\pi \times 10^{-7} \times 10/(2\pi \times 0.05)\]
\[= 4\pi \times 10^{-6}/0.1\pi \]
\[= 4\times 10^{-5} T = 40 \mu T\]
Direction: a_φ (right-hand rule with current in +a_z)

Ex 6.2●● MediumTier 1?
Biot-Savart Law: Circular Current Loop

Circular loop of radius a = 0.1 m, current I = 5 A. Find B at center and along axis.

→ Solution
Ex 6.2 Solution↑ Problem
At center (z=0):
\[B = \mu _{0}I/(2a) = 4\pi \times 10^{-7} \times 5/(2 \times 0.1)\]
\[= 20\pi \times 10^{-7}/0.2\]
\[= 100\pi \times 10^{-7} = 31.4 \mu T (in \hat{a}_{z} direction)\]
Along axis at distance z:
\[B_z = \mu _{0}Ia^{2}/[2(a^{2}+z^{2})^{3/2}]\]
At z=0: B = μ₀I/(2a) = 31.4 μT ✓
At z=a: B = μ₀I/(2a) × a³/(a²+a²)^{3/2} = μ₀I/(2a) × 1/(2√2)
\[= 31.4/2.828 = 11.1 \mu T\]

Ex 6.3●● MediumTier 1?
Solenoid Magnetic Field

Solenoid: N=500 turns, L=0.25m, I=2A, air core. Find B inside.

→ Solution
Ex 6.3 Solution↑ Problem
\[n = N/L = 500/0.25 = 2000 turns/m\]
Inside (uniform field, far from ends):
\[H = nI = 2000 \times 2 = 4000 A/m\]
\[B = \mu _{0}H = 4\pi \times 10^{-7} \times 4000 = 5.026\times 10^{-3} T \approx 5.03 mT\]
For iron core (μᵣ = 1000):
B = μ₀μᵣnI = 1000 × 5.03×10⁻³ = 5.03 T (MRI magnet range)
Energy stored per unit volume: u = B²/(2μ₀) = (5.03×10⁻³)²/(8π×10⁻⁷) = 10.1 J/m³

Ex 6.4●● MediumTier 1?
Vector Magnetic Potential

Find A for an infinite straight wire (I along z). Verify B = ∇×A.

→ Solution
Ex 6.4 Solution↑ Problem
By symmetry: A = A_z(ρ) â_z
∇²A = -μ₀J (Poisson's equation for A)
For ρ≠0 (current region): ∇²A_z = 0
Solution: A_z = -μ₀I/(2π) lnρ + C (chosen so B = ∇×A matches known result)
\[B = \nabla \times A:\]
\[B_\phi = -\partial A_z/\partial \rho = -(-\mu _{0}I/(2\pi \rho )) = \mu _{0}I/(2\pi \rho ) \checkmark \]
In general: A = -μ₀I ln(ρ/ρ_ref)/(2π) â_z
(Reference ρ_ref cancels in derivatives)

Ex 6.5●● MediumTier 1?
Magnetic Field Along Axis of Current Loop

Two coaxial loops (Helmholtz coil), each radius a=0.1m, separated by d=0.1m=a, each carrying I=5A. Find B at midpoint.

→ Solution
Ex 6.5 Solution↑ Problem
B from single loop at distance z from center:
\[B_z = \mu _{0}Ia^{2}/[2(a^{2}+z^{2})^{3/2}]\]
Each loop is at z = ±a/2 from midpoint:
\[z = a/2 = 0.05 m for each loop\]
B from each loop at midpoint:
\[B_single = \mu _{0}\times 5\times 0.01/[2(0.01+0.0025)^{3/2}]\]
\[= 4\pi \times 10^{-7}\times 5\times 0.01/[2(0.0125)^{3/2}]\]
\[= 2\pi \times 10^{-8}/[2\times 1.398\times 10^{-3}]\]
\[= 22.47 \mu T\]
Total (both loops add): B_total = 2×22.47 = 44.9 μT
Helmholtz condition (d=a): field is very uniform near center.
Exact value: B = (4/5)^{3/2} × μ₀nI/a ≈ 0.7155 × μ₀nI/a

Ex 6.6●● MediumTier 1?
Force Between Parallel Wires

Two parallel wires 0.1 m apart, each carrying I = 100 A in the same direction. Find force per unit length.

→ Solution
Ex 6.6 Solution↑ Problem
B from wire 1 at wire 2's location:
\[B_{1} = \mu _{0}I/(2\pi d) = 4\pi \times 10^{-7} \times 100/(2\pi \times 0.1) = 2\times 10^{-4} T\]
Force per unit length on wire 2:
F/L = I × B₁ = 100 × 2×10⁻⁴ = 0.02 N/m (attractive, same direction currents)
Formula: F/L = μ₀I₁I₂/(2πd) = 4π×10⁻⁷ × 10⁴/(2π×0.1) = 0.02 N/m
Definition of ampere: two wires 1m apart with I=1A each → F/L = 2×10⁻⁷ N/m.

Ex 6.7●● MediumTier 1?
Magnetization in a Ferromagnetic Material

Iron with μᵣ = 5000, H = 500 A/m. Find B, M, and magnetization current density.

→ Solution
Ex 6.7 Solution↑ Problem
\[B = \mu _{0}\mu _{r}H = 4\pi \times 10^{-7} \times 5000 \times 500 = 4\pi \times 10^{-7} \times 2.5\times 10^{6}\]
\[= \pi \times 10^{-1} = 3.14\times 10^{-1} T \approx 0.314 T\]
\[Wait: 4\pi \times 10^{-7} \times 5000 \times 500 = 4\pi \times 10^{-7} \times 2.5\times 10^{6} = 10\pi \times 10^{-1} = \pi T\]
B = π ≈ 3.14 T (close to saturation for iron)
Magnetization: M = (μᵣ-1)H = 4999 × 500 = 2.4995×10⁶ A/m ≈ 2.5 MA/m
\[B = \mu _{0}(H + M) = 4\pi \times 10^{-7}(500 + 2.5\times 10^{6}) \approx \mu _{0}M = 3.14 T \checkmark \]
Magnetization volume current: J_m = ∇×M (zero for uniform M)
Magnetization surface current: K_m = M × â_n (at material boundary)

Ex 6.8●● MediumTier 1?
Equivalent Magnetization Currents

Uniformly magnetized sphere: M = M₀ â_z. Find equivalent bound current densities.

→ Solution
Ex 6.8 Solution↑ Problem
Volume magnetization current density:
J_m = ∇×M = ∇×(M₀ â_z) = 0 (uniform M → zero volume current)
Surface magnetization current density:
K_m = M × â_n (â_n = outward normal)
In spherical coordinates, â_n = â_r:
\[K_m = M_{0} \hat{a}_{z} \times \hat{a}_{r} = M_{0} sin\theta \hat{a}_{\phi }\]
\[(Using \hat{a}_{z} = cos\theta \hat{a}_{r} - sin\theta \hat{a}_{\theta } and \hat{a}_{z} \times \hat{a}_{r} = sin\theta (\hat{a}_{\theta } \times ... )\]
Simplified: K_m = M₀ sinθ â_φ [A/m]
This is the same current distribution as a spinning charged sphere —
equivalent to a magnetic dipole.

Ex 6.9●● MediumTier 1?
H-Field in a Toroid

Toroid: N = 200 turns, mean radius R = 0.1 m, current I = 3 A, μᵣ = 500 (iron core). Find B and H.

→ Solution
Ex 6.9 Solution↑ Problem
By Ampère's law (closed path along mean circumference):
\[H \times 2\pi R = NI\]
\[H = NI/(2\pi R) = 200\times 3/(2\pi \times 0.1) = 600/0.6283 = 954.9 A/m\]
\[B = \mu _{0}\mu _{r}H = 4\pi \times 10^{-7} \times 500 \times 954.9\]
\[= 4\pi \times 10^{-7} \times 4.775\times 10^{5}\]
\[= 0.600 T\]
\[H in air gap (if introduced): B/\mu _{0} = 0.600/(4\pi \times 10^{-7}) = 4.775\times 10^{5} A/m\]
(much larger H needed to maintain same B — magnetomotive force drops across gap)

Ex 6.10●● MediumTier 1?
Ampère's Law: Infinite Solenoid

Ideal solenoid: n = 1000 turns/m, I = 2 A. Find B inside and outside.

→ Solution
Ex 6.10 Solution↑ Problem
Apply Ampère's law with rectangular path (half inside, half outside):
\[\oint H\cdot dl = NI_enclosed\]
Outside: by symmetry, B=0 (the contributions from both sides cancel for ideal ∞ solenoid)
Inside: H×L = n×L×I (L = path length along solenoid axis inside)
\[H_inside = nI = 1000 \times 2 = 2000 A/m\]
\[B_inside = \mu _{0}H = 4\pi \times 10^{-7} \times 2000 = 2.51\times 10^{-3} T = 2.51 mT\]
For iron core (μᵣ=1000): B = 2.51 T
Outside B = 0 (ideal solenoid, confirmed by Ampère's law with path entirely outside)

Ex 6.11●● MediumTier 1?
Infinite Current Sheet

Surface current K = K₀ â_x (A/m) on z=0 plane. Find H above and below.

→ Solution
Ex 6.11 Solution↑ Problem
By symmetry and Ampère's law (rectangular path straddling the sheet):
\[H above (z>0): H = -K_{0}/2 \hat{a}_{y}\]
\[H below (z<0): H = +K_{0}/2 \hat{a}_{y}\]
Full expression:
H = -(K₀/2) â_z × â_n where â_n = â_z (pointing up)
For K = K₀ â_x:
H = ∓ K₀/2 â_y (minus sign above, plus below)
\[B = \mu _{0}H:\]
Above: B = -μ₀K₀/2 â_y
Below: B = +μ₀K₀/2 â_y
Note: tangential H is discontinuous: H_above - H_below = K × â_n (boundary condition)
\[(-K_{0}/2 - K_{0}/2) \hat{a}_{y} = -K_{0} \hat{a}_{y} = K_{0}\hat{a}_{x} \times (-\hat{a}_{z}) ... (verify: \hat{a}_{x} \times \hat{a}_{z} = -\hat{a}_{y} \checkmark )\]

Ex 6.12●● MediumTier 1?
Magnetic Field of a Finite Wire

Straight wire of length 2L carrying current I along z-axis, centered at origin. Find B at point (ρ, 0, 0).

→ Solution
Ex 6.12 Solution↑ Problem
By Biot-Savart:
\[B = \mu _{0}I/(4\pi ) \int _{-L}^{L} dl \times \hat{a}_{r} / r^{2}\]
Result:
\[B_\phi = \mu _{0}I/(4\pi \rho ) \times 2L/\sqrt{L^{2}+\rho ^{2}}\]
\[= \mu _{0}IL/(2\pi \rho \sqrt{L^{2}+\rho ^{2}})\]
At ρ = 0.05 m, L = 0.5 m, I = 10 A:
\[B = 4\pi \times 10^{-7} \times 10 \times 0.5/(2\pi \times 0.05 \times \sqrt{0.25+0.0025})\]
\[= 4\pi \times 10^{-7} \times 5/(2\pi \times 0.05 \times 0.5025)\]
\[= 2\pi \times 10^{-6}/(0.1571)\]
\[= 2\pi \times 10^{-6}/0.1571 = 12.7\times 10^{-6}/(0.5025\times 0.1\pi )\]
Simplify: B = 4×10⁻⁷×10×0.5/(0.1×√0.2525) = 2×10⁻⁶/0.05025 = 39.8 μT
For L → ∞: B = μ₀I/(2πρ) = 40 μT ✓ (matches infinite wire)

Ex 6.13●● MediumTier 1?
Magnetic Dipole Moment

Circular current loop: I = 5 A, radius a = 0.02 m. Find magnetic dipole moment m.

→ Solution
Ex 6.13 Solution↑ Problem
\[m = I \times A = I \times \pi a^{2}\]
\[= 5 \times \pi \times (0.02)^{2}\]
\[= 5 \times \pi \times 4\times 10^{-4}\]
\[= 6.283\times 10^{-3} A\cdot m^{2}\]
Direction: â_z (right-hand rule with current direction)
Far-field approximation (r >> a):
\[B_r = \mu _{0}m cos\theta /(2\pi r^{3})\]
\[B_\theta = \mu _{0}m sin\theta /(4\pi r^{3})\]
These are identical in form to the electric dipole field (with m replacing p/ε₀).

Ex 6.14●● MediumTier 1?
Vector Potential: Cylindrical Coordinates

For a solenoid (B = B₀ â_z inside, 0 outside), find A using ∇×A = B.

→ Solution
Ex 6.14 Solution↑ Problem
By symmetry: A = A_φ(ρ) â_φ
Inside (ρ < a):
\[(1/\rho )d(\rho A_\phi )/d\rho = B_{0}\]
\[d(\rho A_\phi )/d\rho = B_{0}\rho \]
\[\rho A_\phi = B_{0}\rho ^{2}/2 (integrating)\]
\[A_\phi = B_{0}\rho /2\]
Outside (ρ > a):
\[d(\rho A_\phi )/d\rho = 0 \to \rho A_\phi = const = B_{0}a^{2}/2\]
\[A_\phi = B_{0}a^{2}/(2\rho )\]
Summary:
\[A = (B_{0}\rho /2) \hat{a}_{\phi } for \rho < a\]
\[A = (B_{0}a^{2}/2\rho ) \hat{a}_{\phi } for \rho > a\]
Note: A exists outside even though B=0 there. This underlies the Aharonov-Bohm effect.

Ex 6.15●● MediumTier 1?
Force on a Current Loop

Rectangular loop (a×b) carrying current I in non-uniform field B = B₀(1+αz) â_z. Loop in xy-plane at z=0. Find net force.

→ Solution
Ex 6.15 Solution↑ Problem
For a planar loop in a non-uniform field, net force = ∇(m·B)
\[m = Iab \hat{a}_{z}, B = B_{0}(1+\alpha z)\]
\[F = \nabla (m\cdot B) = Iab \nabla (B_{0}(1+\alpha z)) = Iab B_{0}\alpha \hat{a}_{z}\]
\[F = Iab\alpha B_{0} \hat{a}_{z}\]
More rigorously: sides parallel to y have equal/opposite forces from x-component.
The z-gradient causes a net force on the loop.
For I=1A, a=b=0.05m, α=10 m⁻¹, B₀=0.1T:
\[F = 1\times 0.0025\times 10\times 0.1 = 2.5\times 10^{-3} N = 2.5 mN\]

Part B — Magnetic Circuits, Inductance, Energy, Forces (Sections 6-8 to 6-13)


Ex 6.16●● MediumTier 1?
Magnetic Circuit with Air Gap

Iron toroid: mean length l_iron = 0.3 m, μᵣ = 2000, cross-section A = 4 cm², air gap l_g = 2 mm.

N = 500 turns, I = 1 A. Find B.

→ Solution
Ex 6.16 Solution↑ Problem
Total MMF: F = NI = 500 × 1 = 500 A·turns
Reluctances:
\[R_iron = l_iron/(\mu _{0}\mu _{r}A) = 0.3/(4\pi \times 10^{-7} \times 2000 \times 4\times 10^{-4})\]
\[= 0.3/(1.005\times 10^{-6}) = 2.985\times 10^{5} A/Wb\]
\[R_gap = l_g/(\mu _{0}A) = 2\times 10^{-3}/(4\pi \times 10^{-7} \times 4\times 10^{-4})\]
\[= 2\times 10^{-3}/(5.027\times 10^{-10}) = 3.979\times 10^{6} A/Wb\]
Total reluctance: R = R_iron + R_gap = 2.985×10⁵ + 3.979×10⁶ = 4.278×10⁶ A/Wb
\[Flux: \Phi = F/R = 500/(4.278\times 10^{6}) = 1.169\times 10^{-4} Wb\]
\[B = \Phi /A = 1.169\times 10^{-4}/4\times 10^{-4} = 0.292 T\]
Note: gap (2mm) dominates reluctance despite iron being 150× longer.

Ex 6.17●● MediumTier 1?
Reluctance of a Magnetic Path

Same core as Problem 16. Find reluctance ratio R_gap/R_iron.

→ Solution
Ex 6.17 Solution↑ Problem
From Problem 16:
\[R_iron = 2.985\times 10^{5} A/Wb\]
\[R_gap = 3.979\times 10^{6} A/Wb\]
Ratio: R_gap/R_iron = 3.979×10⁶/2.985×10⁵ = 13.33
General formula:
\[R_gap/R_iron = (l_g/l_iron) \times \mu _{r} = (2\times 10^{-3}/0.3) \times 2000\]
\[= 0.00667 \times 2000 = 13.33 \checkmark \]
Even a small gap (l_g << l_iron) dominates if μᵣ is large.

Ex 6.18●● MediumTier 1?
B-H Curve and Hysteresis

For a soft iron sample, the B-H relationship approximates B = μ₀μᵣH with μᵣ = 1500 for H < 500 A/m, then saturates at B_sat = 1.5 T. Find H at B = 1.0 T and B = 1.5 T.

→ Solution
Ex 6.18 Solution↑ Problem
Linear region (B < μ₀×1500×500 = 0.942 T):
\[H = B/(\mu _{0}\mu _{r}) = B/(4\pi \times 10^{-7}\times 1500) = B/1.885\times 10^{-3}\]
At B = 0.8 T: H = 0.8/1.885×10⁻³ = 424.4 A/m (linear region)
At B = 1.0 T:
Linear extrapolation: H = 1.0/1.885×10⁻³ = 530.8 A/m (above linear limit)
Actual H > 530 A/m (curve bends up before saturation)
Typical: H ≈ 800-1000 A/m at B=1.0T for soft iron
At B = B_sat = 1.5 T:
H is very large (thousands of A/m) — saturation means dB/dH → 0
Practical value: H ≈ 10,000-50,000 A/m at saturation

Ex 6.19●● MediumTier 1?
Boundary Conditions at Iron-Air Interface

B₁ = 0.5 T at θ₁ = 60° to normal, in iron (μᵣ = 1000). Find B₂ in air.

→ Solution
Ex 6.19 Solution↑ Problem
Boundary conditions:
Normal: B₁n = B₂n (continuous)
Tangential: H₁t = H₂t → B₁t/μ₁ = B₂t/μ₂
\[B_{1}n = B_{1} cos60° = 0.5 \times 0.5 = 0.25 T\]
\[B_{1}t = B_{1} sin60° = 0.5 \times 0.866 = 0.433 T\]
\[B_{2}n = B_{1}n = 0.25 T\]
\[B_{2}t = (\mu _{2}/\mu _{1}) \times B_{1}t = (\mu _{0}/\mu _{0}\mu _{r}) \times 0.433 = 0.433/1000 = 4.33\times 10^{-4} T\]
\[B_{2} = \sqrt{B_{2}n^{2} + B_{2}t^{2}} = \sqrt{0.0625 + 1.88\times 10^{-7}} \approx 0.25 T\]
\[tan \theta _{2}/tan \theta _{1} = \mu _{2}/\mu _{1} = 1/1000\]
\[\theta _{2} = arctan(tan60°/1000) = arctan(0.001732) = 0.099° \approx 0.1°\]
Magnetic field lines are nearly normal to iron surface in air (flux concentration effect).

Ex 6.20●● MediumTier 1?
Self-Inductance of a Solenoid

Solenoid: N=500 turns, L=0.25m, radius r=1cm, air core. Find L.

→ Solution
Ex 6.20 Solution↑ Problem
\[L = \mu _{0}N^{2}A/l = \mu _{0}n^{2}Al\]
\[n = N/l = 500/0.25 = 2000 turns/m\]
\[A = \pi r^{2} = \pi \times (0.01)^{2} = 3.14\times 10^{-4} m^{2}\]
\[L = 4\pi \times 10^{-7} \times (2000)^{2} \times 3.14\times 10^{-4} \times 0.25\]
\[= 4\pi \times 10^{-7} \times 4\times 10^{6} \times 7.854\times 10^{-5}\]
\[= 4\pi \times 10^{-7} \times 314.16\]
\[= 4\pi \times 3.1416\times 10^{-5}\]
\[= 12.566\times 3.1416\times 10^{-5}\]
\[= 3.948\times 10^{-4} H \approx 0.395 mH\]
For iron core (μᵣ=1000): L = 0.395 H

Ex 6.21●● MediumTier 1?
Mutual Inductance

Two coaxial solenoids: both length l=0.25m, inner (N₁=500, r₁=1cm), outer (N₂=200, r₂=2cm). Find M.

→ Solution
Ex 6.21 Solution↑ Problem
All flux from inner solenoid passes through the outer (since r₂ > r₁):
Flux from inner at inner area:
\[B_{1} = \mu _{0}n_{1}I_{1} = \mu _{0} \times 2000 \times I_{1}\]
Flux linkage through outer (N₂ turns, area = inner area since flux is confined to inner):
\[Λ_{21} = N_{2} \times B_{1} \times A_{1} = N_{2} \times \mu _{0}n_{1}I_{1} \times \pi r_{1}^{2}\]
\[M = Λ_{21}/I_{1} = N_{2} \times \mu _{0}n_{1} \times \pi r_{1}^{2}\]
\[= 200 \times 4\pi \times 10^{-7} \times 2000 \times \pi \times 10^{-4}\]
\[= 200 \times 4\pi \times 10^{-7} \times 2000 \times 3.14\times 10^{-4}\]
\[= 200 \times 2.513\times 10^{-7} \times 2000 \times ...\]
Simpler: M = μ₀N₁N₂A₁/l = 4π×10⁻⁷ × 500×200 × 3.14×10⁻⁴/0.25
\[= 4\pi \times 10^{-7} \times 10^{5} \times 1.257\times 10^{-3}\]
\[= 4\pi \times 10^{-7} \times 125.7\]
\[= 1.581\times 10^{-4} H = 0.158 mH\]
Check: M ≤ √(L₁L₂) = √(0.395×10⁻³ × L₂) (Neumann inequality)

Ex 6.22●● MediumTier 1?
Energy Stored in Magnetic Field

Solenoid (L=0.395 mH) carrying I=5A. Find stored energy W.

→ Solution
Ex 6.22 Solution↑ Problem
\[W = LI^{2}/2 = 0.395\times 10^{-3} \times 25/2 = 4.9375\times 10^{-3} J \approx 4.94 mJ\]
In terms of field:
\[B = \mu _{0}nI = 4\pi \times 10^{-7} \times 2000 \times 5 = 12.57\times 10^{-3} T = 12.57 mT\]
\[u = B^{2}/(2\mu _{0}) = (12.57\times 10^{-3})^{2}/(8\pi \times 10^{-7})\]
\[= 1.580\times 10^{-4}/(2.513\times 10^{-6})\]
\[= 62.88 J/m^{3}\]
\[Volume = \pi r^{2}l = 3.14\times 10^{-4} \times 0.25 = 7.854\times 10^{-5} m^{3}\]
\[W = u \times Volume = 62.88 \times 7.854\times 10^{-5} = 4.94\times 10^{-3} J \checkmark \]

Ex 6.23●● MediumTier 1?
Magnetic Force on a Current Loop

Rectangular loop (a=0.1m, b=0.05m), I=2A, in field B = 0.3 â_z T. Loop in xy-plane. Find torque.

→ Solution
Ex 6.23 Solution↑ Problem
Magnetic dipole moment:
\[m = Iab \hat{a}_{z} = 2 \times 0.1 \times 0.05 \hat{a}_{z} = 0.01 \hat{a}_{z} A\cdot m^{2}\]
Torque: T = m × B = 0.01 â_z × 0.3 â_z = 0 (parallel → zero torque)
Now tilt loop so m = 0.01 â_x:
\[T = 0.01 \hat{a}_{x} \times 0.3 \hat{a}_{z} = 0.003 (\hat{a}_{x} \times \hat{a}_{z}) = -0.003 \hat{a}_{y} N\cdot m\]
Maximum torque occurs when m ⊥ B:
\[T_max = |m||B| = 0.01 \times 0.3 = 3\times 10^{-3} N\cdot m = 3 mN\cdot m\]

Ex 6.24●● MediumTier 1?
Torque on a Magnetic Dipole

Magnetic dipole m = 0.05 â_x A·m² in field B = 2 â_z T. Find torque and stable equilibrium position.

→ Solution
Ex 6.24 Solution↑ Problem
\[T = m \times B = 0.05 \hat{a}_{x} \times 2 \hat{a}_{z} = 0.1 (\hat{a}_{x} \times \hat{a}_{z}) = -0.1 \hat{a}_{y} N\cdot m\]
\[|T| = 0.1 N\cdot m\]
Potential energy: U = -m·B = -(0.05 â_x)·(2 â_z) = 0 J (currently 90° orientation)
Stable equilibrium: U_min when m ∥ B (same direction)
\[\to m aligns with B: m = 0.05 \hat{a}_{z}\]
At equilibrium:
\[U = -m\cdot B = -0.05 \times 2 = -0.1 J\]
\[T = 0\]
Unstable equilibrium: m anti-parallel to B, U = +0.1 J (T=0 but any perturbation destabilizes)

Ex 6.25●● MediumTier 1?
Force Between Magnetic Poles

Two bar magnets: pole strength p = 0.01 Wb, separation r = 0.1 m (between same poles). Find force.

→ Solution
Ex 6.25 Solution↑ Problem
Force between magnetic poles (analogous to Coulomb's law):
\[F = p_{1}p_{2}/(4\pi \mu _{0}r^{2})\]
\[= (0.01)^{2}/(4\pi \times 4\pi \times 10^{-7} \times 0.01)\]
\[= 10^{-4}/(16\pi ^{2} \times 4\times 10^{-9})\]
\[= 10^{-4}/(631.7\times 10^{-9})\]
\[= 10^{-4}/6.317\times 10^{-7}\]
= 158.3 N (repulsive, same poles)
Note: pole strength p has units of Weber (Wb = V·s), and p = μ₀×(pole area)×H_surface.

Ex 6.26●● MediumTier 1?
Inductance of a Toroidal Coil

Toroid: N=500 turns, mean radius R=5cm, cross-sectional radius r=1cm, μᵣ=1. Find L.

→ Solution
Ex 6.26 Solution↑ Problem
For toroid with circular cross-section (r << R):
L ≈ μ₀N²A/(2πR) [using mean path length = 2πR]
\[A = \pi r^{2} = \pi \times (0.01)^{2} = 3.14\times 10^{-4} m^{2}\]
\[L = 4\pi \times 10^{-7} \times 500^{2} \times 3.14\times 10^{-4}/(2\pi \times 0.05)\]
\[= 4\pi \times 10^{-7} \times 25\times 10^{4} \times 3.14\times 10^{-4}/0.3142\]
\[= 4\pi \times 10^{-7} \times 78.54/0.3142\]
\[= 4\pi \times 10^{-7} \times 250\]
\[= 10\pi \times 10^{-5} = 3.14\times 10^{-4} H = 0.314 mH\]
For iron core μᵣ=500: L = 500 × 0.314 = 157 mH

Ex 6.27●● MediumTier 1?
Magnetic Pressure on a Surface

Toroid gap with B = 1 T in gap. Find the magnetic pressure (force per unit area) trying to close the gap.

→ Solution
Ex 6.27 Solution↑ Problem
Magnetic pressure: p_m = B²/(2μ₀)
\[= (1)^{2}/(2 \times 4\pi \times 10^{-7})\]
\[= 1/(8\pi \times 10^{-7})\]
\[= 3.979\times 10^{5} N/m^{2} = 397.9 kPa \approx 4 atm\]
Force on gap faces (area A = 4 cm² = 4×10⁻⁴ m²):
\[F = p_m \times A = 3.979\times 10^{5} \times 4\times 10^{-4} = 159.2 N\]
This is why electromagnets can exert strong forces (closing the gap).
For B = 2 T: pressure = 1.59 MPa ≈ 16 atm.

Ex 6.28●● MediumTier 1?
Energy in Terms of Inductance

For the toroid in Problem 26 (L=0.314 mH), carrying I=2A. Find W and verify with field integral.

→ Solution
Ex 6.28 Solution↑ Problem
\[W = LI^{2}/2 = 0.314\times 10^{-3} \times 4/2 = 6.28\times 10^{-4} J = 0.628 mJ\]
From field:
\[H = NI/(2\pi R) = 500\times 2/(2\pi \times 0.05) = 1000/0.3142 = 3183 A/m\]
\[B = \mu _{0}H = 4\pi \times 10^{-7} \times 3183 = 4.00\times 10^{-3} T = 4 mT\]
\[u = B^{2}/(2\mu _{0}) = (4\times 10^{-3})^{2}/(8\pi \times 10^{-7}) = 1.6\times 10^{-5}/2.513\times 10^{-6} = 6.37 J/m^{3}\]
\[Volume = A \times 2\pi R = 3.14\times 10^{-4} \times 0.3142 = 9.868\times 10^{-5} m^{3}\]
\[W = 6.37 \times 9.868\times 10^{-5} = 6.28\times 10^{-4} J \checkmark \]

Ex 6.29●● MediumTier 1?
Hall Effect

Copper conductor: width w = 5 mm, thickness t = 0.5 mm, B = 0.5 T (⊥ to current), I = 10 A.

n = 8.5×10²⁸ electrons/m³. Find Hall voltage V_H.

→ Solution
Ex 6.29 Solution↑ Problem
Hall coefficient: R_H = -1/(ne) = -1/(8.5×10²⁸ × 1.6×10⁻¹⁹)
\[= -1/(1.36\times 10^{10}) = -7.35\times 10^{-11} m^{3}/C\]
Hall electric field:
\[E_H = R_H \times J \times B = R_H \times (I/(wt)) \times B\]
\[J = I/(wt) = 10/(5\times 10^{-3} \times 0.5\times 10^{-3}) = 10/2.5\times 10^{-6} = 4\times 10^{6} A/m^{2}\]
\[E_H = 7.35\times 10^{-11} \times 4\times 10^{6} \times 0.5 = 1.47\times 10^{-4} V/m\]
Hall voltage: V_H = E_H × w = 1.47×10⁻⁴ × 5×10⁻³ = 7.35×10⁻⁷ V = 0.735 μV
Small for copper (many carriers). For semiconductors (n ~10²⁰), V_H is much larger.

Ex 6.30●● MediumTier 1?
Force on a Conductor in Magnetic Field

Straight wire, length L = 0.5 m, I = 20 A, in field B = 0.4 â_z T. Wire is along â_x. Find force.

→ Solution
Ex 6.30 Solution↑ Problem
Force on current-carrying conductor:
\[F = I L \times B = I (L \hat{a}_{x}) \times (B \hat{a}_{z})\]
\[= IL \times B \times (\hat{a}_{x} \times \hat{a}_{z})\]
\[= ILB \times (-\hat{a}_{y})\]
\[= 20 \times 0.5 \times 0.4 \times (-\hat{a}_{y})\]
\[= -4 \hat{a}_{y} N\]
\[|F| = 4 N (in -y direction)\]
General: F = IL × B (where L is vector in direction of current flow)

Textbook Practice Problems

6-1● EasyTier 1?
Fundamental postulates → Answer
6-4● EasyTier 1?
Biot-Savart law → Answer
6-5●● MediumTier 1?
Biot-Savart law → Answer
6-20● EasyTier 1?
Magnetic field intensity → Answer
6-21●● MediumTier 1?
Magnetic field intensity → Answer
6-33●● MediumTier 1?
Boundary conditions → Answer
6-34●● MediumTier 1?
Boundary conditions → Answer
6-36● EasyTier 1?
Inductance → Answer
6-37●● MediumTier 1?
Inductance → Answer
6-2●● MediumTier 2?
Fundamental postulates → Answer
6-6●● MediumTier 2?
Biot-Savart law → Answer
6-9●● MediumTier 2?
Vector magnetic potential → Answer
6-10●● MediumTier 2?
Vector magnetic potential → Answer
6-12●● MediumTier 2?
Magnetic dipole → Answer
6-13●● MediumTier 2?
Magnetic dipole → Answer
6-16●● MediumTier 2?
Magnetization → Answer
6-17●● MediumTier 2?
Magnetization → Answer
6-23●● MediumTier 2?
Applications → Answer
6-24●● MediumTier 2?
Applications → Answer
6-27●● MediumTier 2?
Magnetic circuits → Answer
6-28●● MediumTier 2?
Magnetic circuits → Answer
6-30●● MediumTier 2?
Magnetic materials → Answer
6-31●● MediumTier 2?
Magnetic materials → Answer
6-38●● MediumTier 2?
Inductance → Answer
6-40●● MediumTier 2?
Magnetic energy → Answer
6-41●● MediumTier 2?
Magnetic energy → Answer
6-43●● MediumTier 2?
Magnetic forces → Answer
6-44●● MediumTier 2?
Magnetic forces → Answer
6-25●●● HardTier 3?
Applications → Answer
6-45●●● HardTier 3?
Magnetic forces → Answer
6-47●●● HardTier 3?
Torques → Answer
6-48●●● HardTier 3?
Torques → Answer

Textbook Practice — Approach Hints

Sample: Show \(\nabla\cdot\vec{B}=0\) for \(\vec{B}=B_0(\hat{x}\sin y-\hat{y}\cos x)\).

\[\nabla\cdot\vec{B}=B_0\frac{\partial\sin y}{\partial x}-B_0\frac{\partial\cos x}{\partial y}=0+0=0\;\checkmark\]

Sample: Finite straight wire on \(z\)-axis from \(-L\) to \(+L\), current \(I\). Find \(\vec{B}\) at perpendicular distance \(\rho\).

\[B_{\phi}=\frac{\mu_0 I}{4\pi\rho}(\sin\alpha_2-\sin\alpha_1)\]
\[\text{For symmetric: }B_{\phi}=\frac{\mu_0 I}{2\pi\rho}\frac{L}{\sqrt{L^2+\rho^2}}\]

Sample: Circular loop radius \(a=5\,\text{cm}\), current \(I=10\,\text{A}\). Find \(\vec{B}\) at center and on axis at \(z=10\,\text{cm}\).

\[B_{center}=\mu_0 I/(2a)=4\pi\!\cdot\!10^{-7}\!\cdot\!10/(2\!\cdot\!0.05)=1.26\!\cdot\!10^{-4}\,\text{T}\]
\[B_z(z=0.1)=\frac{\mu_0 Ia^2}{2(a^2+z^2)^{3/2}}=\frac{4\pi\!\cdot\!10^{-7}\!\cdot\!10\!\cdot\!(0.05)^2}{2(0.0025+0.01)^{3/2}}=1.13\!\cdot\!10^{-5}\,\text{T}\]

Sample: Use Ampere's law for an infinite straight wire \(I=10\,\text{A}\) at \(\rho=5\,\text{cm}\).

\[\oint\vec{H}\cdot d\vec{\ell}=I_{enc}\;\Rightarrow\;H(2\pi\rho)=I\]
\[H=10/(2\pi\!\cdot\!0.05)=31.83\,\text{A/m}\]

Sample: Toroid with \(N=200\) turns, mean radius \(r=5\,\text{cm}\), \(I=2\,\text{A}\). Find \(H\) inside.

\[H=NI/(2\pi r)=200\!\cdot\!2/(2\pi\!\cdot\!0.05)=1273\,\text{A/m}\]

Sample: Boundary \(z=0\): \(\mu_{r1}=1\) (\(z>0\)), \(\mu_{r2}=500\) (\(z<0\)). \(\vec{H}_1=10\hat{x}+5\hat{z}\,\text{A/m}\). Find \(\vec{H}_2,\vec{B}_2\).

\[H_{1t}=10\hat{x}\;\Rightarrow\;H_{2t}=10\hat{x}\]
\[B_{1n}=\mu_0(5)\;\Rightarrow\;B_{2n}=\mu_0(5),\;H_{2n}=B_{2n}/(500\mu_0)=0.01\hat{z}\]
\[\vec{H}_2=10\hat{x}+0.01\hat{z}\,\text{A/m}\]

Sample: Surface current \(\vec{K}=2\hat{x}\,\text{A/m}\) on plane \(z=0\) between two media. \(\vec{H}_1=3\hat{y}\). Find \(\vec{H}_2\).

\[\hat{n}\times(\vec{H}_2-\vec{H}_1)=\vec{K}\;\Rightarrow\;\hat{z}\times(\vec{H}_2-3\hat{y})=2\hat{x}\]
\[\vec{H}_2-3\hat{y}=-2\hat{y}\;\Rightarrow\;\vec{H}_2=\hat{y}\,\text{A/m}\]

Sample: Solenoid: \(N=400\), length \(20\,\text{cm}\), area \(5\,\text{cm}^2\), air core. Find \(L\).

\[L=\mu_0 N^2 A/\ell=4\pi\!\cdot\!10^{-7}\!\cdot\!160000\!\cdot\!5\!\cdot\!10^{-4}/0.2=5.03\!\cdot\!10^{-4}\,\text{H}=0.503\,\text{mH}\]

Sample: Toroid: \(N=500\), mean radius \(5\,\text{cm}\), area \(2\,\text{cm}^2\), \(\mu_r=200\). Find \(L\).

\[L=\frac{\mu N^2 A}{2\pi r}=\frac{200\mu_0(500)^2(2\!\cdot\!10^{-4})}{2\pi(0.05)}=0.0402\,\text{H}=40.2\,\text{mH}\]

Sample: Verify \(\nabla\times\vec{H}=\vec{J}\) for an infinite straight wire on \(z\)-axis carrying \(I\).

\[\vec{H}=\frac{I}{2\pi\rho}\hat{a}_{\phi}\;(\rho>0)\]
\[\nabla\times\vec{H}=\frac{1}{\rho}\frac{\partial(\rho H_{\phi})}{\partial\rho}\hat{z}=\frac{1}{\rho}\frac{\partial(I/2\pi)}{\partial\rho}\hat{z}=0\;(\rho>0)\]
Current is concentrated at \(\rho=0\), where \(\vec{J}\) is a delta function.

Sample: Long solenoid, \(n=1000\) turns/m, \(I=2\,\text{A}\). Find \(\vec{B}\) inside.

\[B=\mu_0 nI=4\pi\!\cdot\!10^{-7}\!\cdot\!1000\!\cdot\!2=2.51\!\cdot\!10^{-3}\,\text{T}\]

Sample: Vector potential of finite straight wire: \(\vec{A}=A_z(\rho)\hat{z}\). Set up integral.

\[\vec{A}=\frac{\mu_0}{4\pi}\int\frac{I\,d\vec{\ell}}{R}=\hat{z}\frac{\mu_0 I}{4\pi}\int_{-L}^{L}\frac{dz}{\sqrt{ ho^2+(z-z)^2}}\]
\[A_z=\frac{\mu_0 I}{4\pi}\ln\!\frac{(z-z_1)+\sqrt{\rho^2+(z-z_1)^2}}{(z-z_2)+\sqrt{\rho^2+(z-z_2)^2}}\]

Sample: Magnetic dipole at origin, \(\vec{m}=m\hat{z}\). Find \(\vec{A}\) in far field.

\[\vec{A}(\vec{r})=\frac{\mu_0}{4\pi}\frac{\vec{m}\times\hat{a}_r}{r^2}=\frac{\mu_0 m\sin\theta}{4\pi r^2}\hat{a}_{\phi}\]

Sample: Current loop area \(A=10\,\text{cm}^2\), \(I=1\,\text{A}\). Find \(\vec{m}\) and \(\vec{B}\) on axis at \(z=20\,\text{cm}\).

\[\vec{m}=I\vec{A}=10^{-3}\hat{z}\,\text{A·m}^2\]
\[B_z(\text{far})=\frac{\mu_0 m}{2\pi z^3}=\frac{4\pi\!\cdot\!10^{-7}\!\cdot\!10^{-3}}{2\pi(0.2)^3}=2.5\!\cdot\!10^{-8}\,\text{T}\]

Sample: Compare electric and magnetic dipoles in far field.

\[E_{dipole}\propto p/r^3,\;B_{dipole}\propto m/r^3\]
\[\text{Both have }(2\cos\theta\,\hat{a}_r+\sin\theta\,\hat{a}_{\theta})\text{ angular pattern.}\]

Sample: Bar magnet uniformly magnetized \(\vec{M}=M_0\hat{z}\), length \(L\), area \(A\). Find equivalent surface current.

\[\vec{J}_b=\nabla\times\vec{M}=0\;(\text{uniform inside})\]
\[\vec{K}_b=\vec{M}\times\hat{n}=M_0\hat{a}_{\phi}\;\text{on side surface}\]

Sample: Permanent magnet: \(M=8\!\cdot\!10^5\,\text{A/m}\), neglecting external field. Find \(\vec{B}\) inside.

\[B=\mu_0(H+M)\approx\mu_0 M=4\pi\!\cdot\!10^{-7}\!\cdot\!8\!\cdot\!10^5=1.0\,\text{T}\]

Sample: Two parallel wires \(5\,\text{cm}\) apart carry \(I_1=10\,\text{A}\) and \(I_2=15\,\text{A}\) same direction. Find \(\vec{B}\) midway.

\[B_1=\mu_0 I_1/(2\pi\rho_1)=4\!\cdot\!10^{-5}\,\text{T}\;(\text{out of page})\]
\[B_2=\mu_0 I_2/(2\pi\rho_2)=6\!\cdot\!10^{-5}\,\text{T}\;(\text{into page})\]
\[B_{net}=2\!\cdot\!10^{-5}\,\text{T}\;(\text{toward }I_2)\]

Sample: Square loop side \(L=10\,\text{cm}\), current \(I=2\,\text{A}\). Find \(\vec{B}\) at center.

\[B_{center}=4\cdot\frac{\mu_0 I}{4\pi(L/2)}\!\left(\sin 45^{\circ}+\sin 45^{\circ}\right)=\frac{2\sqrt{2}\,\mu_0 I}{\pi L}\]
\[B=\frac{2\sqrt{2}(4\pi\!\cdot\!10^{-7})(2)}{\pi(0.1)}=2.26\!\cdot\!10^{-5}\,\text{T}\]

Sample: Toroidal core \(N=500\), \(I=1\,\text{A}\), \(\mu_r=1000\), mean length \(30\,\text{cm}\), area \(4\,\text{cm}^2\). Find flux.

\[\mathcal{F}=NI=500,\;\mathcal{R}=\ell/(\mu A)=0.3/(1000\!\cdot\!4\pi\!\cdot\!10^{-7}\!\cdot\!4\!\cdot\!10^{-4})=5.97\!\cdot\!10^5\,\text{A·t/Wb}\]
\[\Phi=\mathcal{F}/\mathcal{R}=500/5.97\!\cdot\!10^5=8.38\!\cdot\!10^{-4}\,\text{Wb}\]

Sample: Magnetic circuit with iron core (\(\mu_r=2000\)) and \(1\,\text{mm}\) air gap. Iron length \(20\,\text{cm}\), area \(2\,\text{cm}^2\), \(NI=400\,\text{A·t}\). Find flux.

\[\mathcal{R}_{iron}=0.2/(2000\mu_0\cdot2\!\cdot\!10^{-4})=3.98\!\cdot\!10^5\]
\[\mathcal{R}_{gap}=10^{-3}/(\mu_0\cdot2\!\cdot\!10^{-4})=3.98\!\cdot\!10^6\]
\[\Phi=400/(\mathcal{R}_{iron}+\mathcal{R}_{gap})=400/4.38\!\cdot\!10^6=9.13\!\cdot\!10^{-5}\,\text{Wb}\]

Sample: Iron at operating point \(H=200\,\text{A/m}\), \(B=0.8\,\text{T}\). Find relative permeability.

\[\mu_r=B/(\mu_0 H)=0.8/(4\pi\!\cdot\!10^{-7}\!\cdot\!200)=3183\]

Sample: Soft-iron core at \(H=100\), \(B=1.2\). Find \(\mu_r\) and total energy density.

\[\mu_r=B/(\mu_0 H)=1.2/(4\pi\!\cdot\!10^{-7}\!\cdot\!100)=9549\]
\[w_m=\tfrac12 BH=\tfrac12(1.2)(100)=60\,\text{J/m}^3\]

Sample: Two coaxial solenoids: \(N_1=200\), \(N_2=300\), common length \(L=10\,\text{cm}\), area \(4\,\text{cm}^2\). Find \(M\).

\[M=\frac{\mu_0 N_1 N_2 A}{L}=\frac{4\pi\!\cdot\!10^{-7}\!\cdot\!200\!\cdot\!300\!\cdot\!4\!\cdot\!10^{-4}}{0.1}=3.02\!\cdot\!10^{-4}\,\text{H}\]

Sample: Inductor \(L=10\,\text{mH}\), current \(I=2\,\text{A}\). Find stored energy.

\[W_m=\tfrac12 LI^2=\tfrac12(0.01)(4)=0.02\,\text{J}=20\,\text{mJ}\]

Sample: Solenoid with \(B=1\,\text{T}\), volume \(V=10^{-3}\,\text{m}^3\). Find energy.

\[w_m=B^2/(2\mu_0)=1/(8\pi\!\cdot\!10^{-7})=3.98\!\cdot\!10^5\,\text{J/m}^3\]
\[W=w_m V=398\,\text{J}\]

Sample: Current-carrying wire \(I=5\,\text{A}\) in \(\vec{B}=0.2\hat{z}\,\text{T}\), wire along \(\hat{x}\), length 1 m. Find force.

\[\vec{F}=I\vec{L}\times\vec{B}=5(\hat{x})\times(0.2\hat{z})=-1\hat{y}\,\text{N}\]

Sample: Two parallel wires \(1\,\text{cm}\) apart, currents \(5\,\text{A}\) and \(3\,\text{A}\) same direction. Find force per meter.

\[F/L=\frac{\mu_0 I_1 I_2}{2\pi d}=\frac{4\pi\!\cdot\!10^{-7}\!\cdot\!5\!\cdot\!3}{2\pi\!\cdot\!0.01}=3\!\cdot\!10^{-4}\,\text{N/m}\;(\text{attractive})\]

Sample: Helmholtz coils: two coaxial loops radius \(a\), separation \(a\), current \(I\) each. Find \(B_z\) at midpoint.

\[B_z=2\cdot\frac{\mu_0 Ia^2}{2(a^2+(a/2)^2)^{3/2}}=\frac{\mu_0 I}{a}\!\left(\frac{4}{5}\right)^{3/2}=\frac{0.7155\,\mu_0 I}{a}\]

Sample: Solenoid with iron plunger: virtual work for force.

\[F=\frac{1}{2}I^2\frac{dL}{dx}\;\;(\text{constant }I)\]
Pulls plunger to maximize stored energy.

Sample: Rectangular loop \(0.1\!\times\!0.05\,\text{m}^2\), \(I=2\,\text{A}\), in field \(\vec{B}=0.3\hat{x}\,\text{T}\), normal at \(30^{\circ}\) to \(\vec{B}\). Find torque.

\[\vec{m}=IA\hat{n},\;|\vec{T}|=mB\sin 30^{\circ}=(2\!\cdot\!0.005)(0.3)(0.5)=1.5\!\cdot\!10^{-3}\,\text{N·m}\]

Sample: Magnetic dipole \(\vec{m}=10^{-2}\hat{z}\,\text{A·m}^2\) in \(\vec{B}=0.1\hat{x}\,\text{T}\). Find energy.

\[U=-\vec{m}\cdot\vec{B}=0\;(\text{perpendicular})\]
Stable equilibrium when \(\vec{m}\parallel\vec{B}\) (\(U=-mB\)); unstable when antiparallel (\(U=+mB\)).

Chapter 7 — Time-Varying Fields and Maxwell's Equations

Key Theory — Chapter 7

Condensed from Cheng, Field and Wave Electromagnetics, §7-1 through §7-7. Read this before attempting the problems below.


7-1   Why Static Models Are Not Enough

Chapters 3–6 built two independent models — electrostatic and magnetostatic:

Electrostatic Magnetostatic
Governing eqs. ∇×E = 0 ∇·B = 0
∇·D = ρ ∇×H = J
Constitutive D = εE H = (1/μ)B
Table 7-1 — Static fundamental relations

In the static case \(\{\mathbf{E}, \mathbf{D}\}\) and \(\{\mathbf{B}, \mathbf{H}\}\) are decoupled: an electrostatic field can drive a steady current, which produces a static magnetic field, but \(\mathbf{E}\) can be found from the charge distribution alone. The magnetic field is only a consequence.

Chapter 7 shows that a time-varying magnetic field produces an electric field, and a time-varying electric field produces a magnetic field. The two curl equations must therefore be modified. The resulting four equations are Maxwell's equations — the foundation of all electromagnetic theory.


7-2   Faraday's Law — Fundamental Postulate

In 1831 Faraday showed experimentally that a changing magnetic flux through a loop induces an emf. Following the deductive approach, we take the point-form relation as postulate:

\[ \boxed{\nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t}} \qquad \text{(7-1)} \]

Applying Stokes's theorem gives the integral form:

\[ \oint_C \mathbf{E}\cdot d\ell = -\int_S \frac{\partial\mathbf{B}}{\partial t}\cdot d\mathbf{s} \qquad \text{(7-2)} \]

Equation (7-1) applies at every point of space whether or not a physical circuit exists. A time-varying magnetic field makes \(\mathbf{E}\) non-conservative — it can no longer be expressed as the gradient of a scalar potential alone.

Transformer emf (stationary circuit, changing B)

For a stationary closed circuit, (7-2) becomes

\[ \mathscr{V} = \oint_C \mathbf{E}\cdot d\ell = -\frac{d\Phi}{dt} \quad (\text{V}) \qquad \text{(7-6)} \]

This is Faraday's law of electromagnetic induction. The induced emf equals the negative rate of change of flux linkage; the minus sign (Lenz's law) asserts that the induced current opposes the flux change that created it. This is the principle of transformers. Time-varying flux in a ferromagnetic core also induces local eddy currents in the core itself; lamination reduces this eddy-current loss.

Motional emf (moving conductor in static B)

A conductor moving with velocity \(\mathbf{u}\) through a static \(\mathbf{B}\) experiences magnetic forces on its free charges. In the conductor's frame, these are equivalent to an induced field \(\mathbf{u}\times\mathbf{B}\) acting along it. For a closed circuit partially in motion,

\[ \mathscr{V}' = \oint_C (\mathbf{u}\times\mathbf{B})\cdot d\ell \qquad \text{(7-24)} \]

called the motional emf or "flux-cutting" emf. Only parts of the circuit moving across flux lines contribute.

For a circuit that both moves and lies in a time-varying field, both contributions add — but they combine cleanly into the single expression \(\mathscr{V} = -d\Phi/dt\) with the total time derivative taken along the moving loop.


7-3   Displacement Current and the Modification of Ampère's Law

The static curl equation \(\nabla\times\mathbf{H} = \mathbf{J}\) is inconsistent with charge conservation \(\nabla\cdot\mathbf{J} = -\partial\rho/\partial t\) in the time-varying case, because \(\nabla\cdot(\nabla\times\mathbf{H}) \equiv 0\) would force \(\nabla\cdot\mathbf{J} = 0\). Maxwell's brilliant repair: add a term \(\partial\mathbf{D}/\partial t\) — the displacement current density:

\[ \boxed{\nabla\times\mathbf{H} = \mathbf{J} + \frac{\partial\mathbf{D}}{\partial t}} \qquad \text{(7-52)} \]

Now \(\nabla\cdot(\nabla\times\mathbf{H}) = \nabla\cdot\mathbf{J} + \partial\rho/\partial t = 0\) automatically, by continuity. The term \(\partial\mathbf{D}/\partial t\) has the dimension of current density (A/m²) but requires no actual charge motion — a time-varying \(\mathbf{D}\) between the plates of an air capacitor produces a magnetic field just as a conduction current would. A time-varying electric field generates a magnetic field.

Maxwell's Equations

The four self-consistent time-varying equations:

Differential form Integral form Meaning
(7-53a) ∇×E = -∂B/∂t ∮ E·dℓ = -dΦ/dt Faraday's law
(7-53b) ∇×H = J + ∂D/∂t ∮ H·dℓ = I + ∫(∂D/∂t)·ds Ampère–Maxwell
(7-53c) ∇·D = ρ ∮ D·ds = Q Gauss's law
(7-53d) ∇·B = 0 ∮ B·ds = 0 No magnetic charge
Table 7-2 — Maxwell's Equations

The four fundamental fields are \(\mathbf{E}, \mathbf{D}, \mathbf{B}, \mathbf{H}\) — twelve scalar unknowns total. The two curl equations supply six scalar relations; the two constitutive relations \(\mathbf{D} = \epsilon\mathbf{E}\) and \(\mathbf{H} = \mathbf{B}/\mu\) supply six more. The two divergence equations are not independent — they can be derived from the curl equations and continuity. Together with Lorentz's force equation, Maxwell's equations describe all macroscopic electromagnetic phenomena.

In the static limit (\(\partial/\partial t \to 0\)) they reduce to the two pairs of Table 7-1.


7-4   Potential Functions for Time-Varying Fields

Since \(\nabla\cdot\mathbf{B} = 0\) still holds, we retain \(\mathbf{B} = \nabla\times\mathbf{A}\) (6-15). Substituting into Faraday's law:

\[ \nabla\times\left(\mathbf{E} + \frac{\partial\mathbf{A}}{\partial t}\right) = 0 \]

The quantity in parentheses is curl-free, so it can be written as \(-\nabla V\):

\[ \boxed{\mathbf{E} = -\nabla V - \frac{\partial\mathbf{A}}{\partial t}} \qquad \text{(7-57)} \]

Now \(\mathbf{E}\) has two contributions: an "electrostatic" part \(-\nabla V\) from charge distribution, and an "induction" part \(-\partial\mathbf{A}/\partial t\) from time-varying magnetic field. In the static limit only the first survives.

Lorenz gauge and wave equations

Substituting the potentials into the Ampère–Maxwell equation gives \(\nabla(\nabla\cdot\mathbf{A}) - \nabla^2\mathbf{A} = \mu\mathbf{J} - \mu\epsilon\nabla(\partial V/\partial t) - \mu\epsilon\partial^2\mathbf{A}/\partial t^2\). Choosing the Lorenz gauge

\[ \nabla\cdot\mathbf{A} + \mu\epsilon\frac{\partial V}{\partial t} = 0 \qquad \text{(7-62)} \]

decouples the equations into two non-homogeneous wave equations:

\[ \nabla^2\mathbf{A} - \mu\epsilon\frac{\partial^2\mathbf{A}}{\partial t^2} = -\mu\mathbf{J} \qquad \text{(7-63)} \]
\[ \nabla^2 V - \mu\epsilon\frac{\partial^2 V}{\partial t^2} = -\frac{\rho}{\epsilon} \qquad \text{(7-65)} \]

Their solutions travel with velocity \(u = 1/\sqrt{\mu\epsilon}\). In free space \(u = 1/\sqrt{\mu_0\epsilon_0} = c\) — the speed of light. This is the central result linking electromagnetism to optics: light is an electromagnetic wave.


7-5   Electromagnetic Boundary Conditions

The integral forms of Maxwell's equations applied to a pillbox or flat contour straddling an interface give four boundary conditions that are identical in form to those of the static case, because \(\partial\mathbf{D}/\partial t\) and \(\partial\mathbf{B}/\partial t\) have no surface integrals when the pillbox/contour is collapsed:

\[ E_{1t} = E_{2t} \qquad \text{(7-66a)} \]
\[ \mathbf{a}_{n2}\times(\mathbf{H}_1 - \mathbf{H}_2) = \mathbf{J}_s \qquad \text{(7-66b)} \]
\[ \mathbf{a}_{n2}\cdot(\mathbf{D}_1 - \mathbf{D}_2) = \rho_s \qquad \text{(7-66c)} \]
\[ B_{1n} = B_{2n} \qquad \text{(7-66d)} \]

In words: tangential E is continuous; tangential H jumps by the surface current; normal D jumps by the surface charge; normal B is continuous.

Between two lossless media (\(\rho_s = 0\), \(\mathbf{J}_s = 0\)) all four field components have simple continuity conditions across the interface. At an interface with a perfect conductor, all fields inside vanish, so \(\mathbf{E}\) is normal to the surface (\(E_{1n} = \rho_s/\epsilon_1\)) and \(\mathbf{H}\) is tangential (\(|\mathbf{H}_{1t}| = |\mathbf{J}_s|\)) — fundamental results for waveguide and antenna analysis.


7-6   Wave Equations and Their Solutions

In a source-free region (\(\rho = 0\), \(\mathbf{J} = 0\)), \(\mathbf{E}\) and \(\mathbf{B}\) themselves satisfy homogeneous wave equations. For a spherically symmetric scalar-potential wave about a point source,

\[ V(R, t) = \frac{1}{R}f\!\left(t - \frac{R}{u}\right) \qquad \text{(7-75)} \]

is a solution — a spherical wave diverging from the origin at velocity \(u = 1/\sqrt{\mu\epsilon}\). The argument \(t - R/u\) means the value of \(V\) at distance \(R\) and time \(t\) was "caused" by the source at the earlier time \(t - R/u\): this is retardation, a direct consequence of finite propagation speed. The corresponding retarded potentials for arbitrary time-varying sources are

\[ V(\mathbf{r}, t) = \frac{1}{4\pi\epsilon}\int_{V'}\frac{\rho(\mathbf{r}', t - R/u)}{R}dv' \]
\[ \mathbf{A}(\mathbf{r}, t) = \frac{\mu}{4\pi}\int_{V'}\frac{\mathbf{J}(\mathbf{r}', t - R/u)}{R}dv' \]

These reduce to the static Poisson-equation solutions when time dependence is negligible ("quasi-static" regime). For high-frequency sources — antennas — retardation cannot be ignored.


7-7   Time-Harmonic (Phasor) Electromagnetics

Because Maxwell's equations are linear, sinusoidal sources of frequency \(\omega\) produce sinusoidal fields of the same frequency everywhere. Represent a real sinusoidal field as the real part of a complex phasor:

\[ \mathbf{E}(\mathbf{r}, t) = \mathscr{R}e\bigl[\mathbf{E}(\mathbf{r})\,e^{j\omega t}\bigr] \qquad \text{(7-93)} \]

Under this substitution, \(\partial/\partial t \to j\omega\), and Maxwell's equations become algebraic in \(j\omega\):

\[ \begin{aligned} \nabla\times\mathbf{E} &= -j\omega\mu\mathbf{H} \\ \nabla\times\mathbf{H} &= \mathbf{J} + j\omega\epsilon\mathbf{E} \\ \nabla\cdot\mathbf{E} &= \rho/\epsilon \\ \nabla\cdot\mathbf{H} &= 0 \end{aligned} \qquad \text{(7-94)} \]

Helmholtz's equation

The wave equations (7-63) and (7-65) become Helmholtz equations:

\[ \nabla^2\mathbf{A} + k^2\mathbf{A} = -\mu\mathbf{J} \qquad \text{(7-96)} \]
\[ \nabla^2 V + k^2 V = -\rho/\epsilon \qquad \text{(7-95)} \]

where

\[ \boxed{k = \omega\sqrt{\mu\epsilon} = \omega/u} \quad (\text{rad/m}) \qquad \text{(7-97)} \]

is the wavenumber. The Lorenz gauge in phasor form is \(\nabla\cdot\mathbf{A} + j\omega\mu\epsilon V = 0\).

Retarded potentials in phasor form

The phasor solutions contain the retardation factor \(e^{-jkR}\):

\[ V(\mathbf{r}) = \frac{1}{4\pi\epsilon}\int_{V'}\frac{\rho\,e^{-jkR}}{R}\,dv' \qquad \text{(7-99)} \]
\[ \mathbf{A}(\mathbf{r}) = \frac{\mu}{4\pi}\int_{V'}\frac{\mathbf{J}\,e^{-jkR}}{R}\,dv' \qquad \text{(7-100)} \]

These are the foundation for antenna theory in Chapter 11: the radiation from a current distribution is computed by integrating the phasor current over the source volume with the retardation phase factor.


Chapter 7 at a Glance


Problems and Solutions


Ex 7.1●● MediumTier 1?
Motional EMF in a Moving Conductor

A bar of length L = 0.5 m moves at v = 5 a_x m/s in uniform field B = 0.2 a_z T. Find the induced EMF.

→ Solution
Ex 7.1 Solution↑ Problem
\[v \times B = (5 a_x) \times (0.2 a_z) = 1.0 (a_x \times a_z) = -1.0 a_y V/m\]
EMF = ∫(v × B)·dl (along bar in a_y direction, 0 to L)
\[= (-1.0 a_y)\cdot (L a_y) = -1.0 \times 0.5 = -0.5 V\]
\[|EMF| = 0.5 V\]

Ex 7.2●● MediumTier 1?
Faraday's Law: Induced Electric Field

A time-varying B = cos(100πt) a_z T fills a circular region of radius a = 0.1 m. Find E at r = 0.1 m.

→ Solution
Ex 7.2 Solution↑ Problem
Faraday's law (integral form):
\[\oint E\cdot dl = -d/dt \int \int B\cdot dS\]
By symmetry, E = E_φ a_φ along circular path:
\[E_\phi (2\pi r) = -d/dt [B \cdot \pi r^{2}]\]
\[= -\pi r^{2} \cdot (-100\pi sin(100\pi t))\]
\[= 100\pi ^{2} r^{2} sin(100\pi t)\]
\[E_\phi = (100\pi ^{2} r^{2}) / (2\pi r) = 50\pi r sin(100\pi t)\]
At r = 0.1 m:
\[E = 50\pi (0.1) sin(100\pi t) a_\phi = 5\pi sin(100\pi t) a_\phi \approx 15.7 sin(100\pi t) a_\phi V/m\]

Ex 7.3●● MediumTier 1?
Displacement Current in a Capacitor

Parallel-plate capacitor: A = 0.01 m², d = 1 mm, free space. Voltage V = 100 sin(2π×10⁶t) V. Find J_d and total I_d.

→ Solution
Ex 7.3 Solution↑ Problem
\[E = V/d = 100 sin(2\pi \times 10^{6}t) / 10^{-3} = 10^{5} sin(2\pi \times 10^{6}t) V/m\]
\[D = \varepsilon _{0} E = (8.85\times 10^{-12})(10^{5}) sin(2\pi \times 10^{6}t) = 8.85\times 10^{-7} sin(2\pi \times 10^{6}t) C/m^{2}\]
\[J_d = \partial D/\partial t = 8.85\times 10^{-7} \times 2\pi \times 10^{6} cos(2\pi \times 10^{6}t)\]
\[= 5.56 cos(2\pi \times 10^{6}t) A/m^{2}\]
\[I_d = J_d \times A = 5.56 \times 0.01 = 55.6 cos(2\pi \times 10^{6}t) mA\]

Ex 7.4●● MediumTier 1?
Verify Maxwell's Equations

Given E = E₀ cos(kz − ωt) a_x and H = H₀ cos(kz − ωt) a_y in free space. Verify and find the constraint between E₀, H₀, k, ω.

→ Solution
Ex 7.4 Solution↑ Problem
\[\nabla \cdot E = \partial Ex/\partial x = 0 \checkmark (no x-dependence)\]
\[\nabla \cdot H = \partial Hy/\partial y = 0 \checkmark (no y-dependence)\]
From ∇ × E = -μ₀ ∂H/∂t :
\[(\nabla \times E)_y = \partial Ex/\partial z = -E_{0}k sin(kz-\omega t)\]
\[-\mu _{0} \partial H/\partial t = -\mu _{0} H_{0}\omega sin(kz-\omega t)\]
\[\to E_{0}k = \mu _{0} H_{0}\omega ... (1)\]
From ∇ × H = ε₀ ∂E/∂t :
\[(\nabla \times H)_x = -\partial Hy/\partial z = H_{0}k sin(kz-\omega t)\]
\[\varepsilon _{0} \partial E/\partial t = \varepsilon _{0} E_{0}\omega sin(kz-\omega t)\]
\[\to H_{0}k = \varepsilon _{0} E_{0}\omega ... (2)\]
From (1)×(2): k² = μ₀ε₀ ω²
\[\to k = \omega /c \checkmark \]
\[E_{0}/H_{0} = \sqrt{\mu _{0}/\varepsilon _{0}} = \eta _{0} = 377 \Omega \]

Ex 7.5●● MediumTier 1?
Vector Potential to E and H

Given A = A₀ sin(kz − ωt) a_x, V = 0. Find E and H.

→ Solution
Ex 7.5 Solution↑ Problem
\[E = -\nabla V - \partial A/\partial t = -\partial A/\partial t\]
\[= A_{0}\omega cos(kz - \omega t) a_x\]
\[B = \nabla \times A:\]
\[(\nabla \times A)_y = \partial Ax/\partial z = A_{0}k cos(kz - \omega t)\]
\[H = B/\mu _{0} = (A_{0}k/\mu _{0}) cos(kz - \omega t) a_y\]
Check ratio: E₀/H₀ = A₀ω / (A₀k/μ₀) = μ₀ω/k = η₀ when k = ω/c ✓

Ex 7.6●● MediumTier 1?
Boundary Conditions at Perfect Conductor

E_i = 10 cos(ωt − kz) a_x V/m incident on perfect conductor at z = 0. Find reflected field and surface current.

→ Solution
Ex 7.6 Solution↑ Problem
\[BC: tangential E = 0 at z = 0\]
\[\to E_r = -10 cos(\omega t + kz) a_x V/m\]
Incident H: H_i = (10/η₀) cos(ωt - kz) a_y
Reflected H: H_r = (10/η₀) cos(ωt + kz) a_y
Total H at z = 0:
\[H_total = (20/\eta _{0}) cos(\omega t) a_y\]
Surface current (n̂ = -a_z outward from conductor):
\[J_s = \hat{n} \times H = (-a_z) \times (20/\eta _{0}) cos(\omega t) a_y\]
\[= (20/\eta _{0}) cos(\omega t) a_x\]
\[= 53.1 cos(\omega t) mA/m a_x\]

Ex 7.7●● MediumTier 1?
Solve 1D Wave Equation

Find E(z,t) for a +z propagating wave at f = 1 GHz, amplitude 50 V/m in free space.

→ Solution
Ex 7.7 Solution↑ Problem
Wave equation: ∂²E/∂z² = (1/c²) ∂²E/∂t²
General +z solution: E = E₀ cos(kz - ωt)
\[f = 1 GHz = 10^{9} Hz\]
\[\omega = 2\pi \times 10^{9} = 6.283\times 10^{9} rad/s\]
\[\lambda = c/f = 3\times 10^{8} / 10^{9} = 0.3 m\]
\[k = 2\pi /\lambda = 2\pi /0.3 = 20.94 rad/m\]
\[E(z,t) = 50 cos(20.94z - 6.283\times 10^{9} t) a_x V/m\]

Ex 7.8●● MediumTier 1?
Time-Domain to Phasor Conversion

Convert E(x,t) = 10 cos(ωt − 2x) a_x − 5 sin(ωt − 2x) a_y to phasor form.

→ Solution
Ex 7.8 Solution↑ Problem
\[Use E(x,t) = Re[Ẽ(x) e^(j\omega t)]\]
\[x-component: 10 cos(\omega t - 2x) = Re[10 e^(-j2x) e^(j\omega t)]\]
\[\to Ẽ_x = 10 e^(-j2x)\]
\[y-component: -5 sin(\omega t - 2x) = Re[j5 e^(-j2x) e^(j\omega t)]\]
since -sin(θ) = cos(θ + π/2) and -e^(-jπ/2) = j
\[\to Ẽ_y = j5 e^(-j2x)\]
\[Ẽ(x) = (10 a_x + j5 a_y) e^(-j2x) V/m\]

Ex 7.9●● MediumTier 1?
Skin Depth in Copper

Copper: σ = 5.8×10⁷ S/m, μᵣ = 1. Find δ at 60 Hz, 1 MHz, 1 GHz.

→ Solution
Ex 7.9 Solution↑ Problem
δ = √(2 / (ωμ₀σ)) [good conductor approximation: σ >> ωε]
Simplifies to: δ = 66.1/√f mm (f in Hz, for copper)
\[f = 60 Hz: \delta = 66.1/\sqrt{60} = 8.53 mm\]
\[f = 1 MHz: \delta = 66.1/\sqrt{10^{6}} = 66.1 \mu m\]
\[f = 1 GHz: \delta = 66.1/\sqrt{10^{9}} = 2.09 \mu m\]

Ex 7.10●● MediumTier 1?
Phase Velocity in Dielectric

Non-magnetic dielectric with εᵣ = 4. Find phase velocity, wavelength at 500 MHz, and wave impedance.

→ Solution
Ex 7.10 Solution↑ Problem
\[v_p = c/\sqrt{\mu _{r} \varepsilon _{r}} = 3\times 10^{8}/\sqrt{4} = 1.5\times 10^{8} m/s\]
\[\lambda = v_p/f = 1.5\times 10^{8} / 500\times 10^{6} = 0.3 m\]
\[\eta = \eta _{0} \sqrt{\mu _{r}/\varepsilon _{r}} = 377 \times \sqrt{1/4} = 188.5 \Omega \]

Ex 7.11●● MediumTier 1?
Poynting Vector

E = 100 cos(ωt − kz) a_x V/m in free space. Find H, instantaneous S, and time-average ⟨S⟩.

→ Solution
Ex 7.11 Solution↑ Problem
\[H = (E_{0}/\eta _{0}) cos(\omega t - kz) a_y = (100/377) cos(\omega t - kz) a_y = 0.265 cos(\omega t-kz) a_y A/m\]
\[S = E \times H = 100 \times 0.265 cos^{2}(\omega t - kz) (a_x \times a_y)\]
\[= 26.5 cos^{2}(\omega t - kz) a_z W/m^{2}\]
\[\langle S\rangle = E_{0}^{2}/(2\eta _{0}) = 10000/(2\times 377) = 13.26 a_z W/m^{2}\]

Ex 7.12●● MediumTier 1?
Transformer EMF

Rectangular loop (0.2 m × 0.1 m) in B = 0.5 cos(1000t) a_z T. Find induced EMF.

→ Solution
Ex 7.12 Solution↑ Problem
\[\Phi = B \cdot A = 0.5 cos(1000t) \times (0.2 \times 0.1) = 0.01 cos(1000t) Wb\]
\[EMF = -d\Phi /dt = 0.01 \times 1000 sin(1000t) = 10 sin(1000t) V\]

Ex 7.13●● MediumTier 1?
Motional EMF in a Sliding Bar

Bar slides at v = 5 a_x m/s in B = 0.2 a_z T, bar length L = 0.5 m along a_y. Find EMF and force on charge q.

→ Solution
Ex 7.13 Solution↑ Problem
\[v \times B = (5 a_x) \times (0.2 a_z) = -1.0 a_y V/m\]
\[EMF = \int _{0}^L (v \times B)\cdot (dy a_y) = -1.0 \times 0.5 = -0.5 V\]
Force on charge q in bar:
\[F = q(v \times B) = -q a_y N\]

Ex 7.14●● MediumTier 1?
E from Time-Varying B

B = B₀ sin(kx) cos(ωt) a_z. Find E using Faraday's law.

→ Solution
Ex 7.14 Solution↑ Problem
\[\nabla \times E = -\partial B/\partial t = B_{0}\omega sin(kx) sin(\omega t) a_z\]
Assume E = E_y(x,t) a_y (by symmetry)
\[(\nabla \times E)_z = \partial E_y/\partial x = B_{0}\omega sin(kx) sin(\omega t)\]
Integrate over x:
\[E_y = -(B_{0}\omega /k) cos(kx) sin(\omega t)\]
\[E = -(B_{0}\omega /k) cos(kx) sin(\omega t) a_y V/m\]

Ex 7.15●● MediumTier 1?
Wave Parameters from Field Expression

E = 50 cos(2π×10⁸t − πz/1.5) a_x V/m. Find f, ω, k, λ, v_p, H.

→ Solution
Ex 7.15 Solution↑ Problem
Comparing with E₀ cos(ωt - kz):
\[\omega = 2\pi \times 10^{8} rad/s\]
\[f = \omega /2\pi = 10^{8} Hz = 100 MHz\]
\[k = \pi /1.5 = 2\pi /3 \approx 2.094 rad/m\]
\[\lambda = 2\pi /k = 3 m\]
v_p = ω/k = 2π×10⁸ / (π/1.5) = 3×10⁸ m/s = c ✓ (free space confirmed)
\[H = (50/377) cos(2\pi \times 10^{8} t - \pi z/1.5) a_y\]
\[= 0.133 cos(2\pi \times 10^{8} t - \pi z/1.5) a_y A/m\]

Textbook Practice Problems

7-1● EasyTier 1?
Faraday's law → Answer
7-2●● MediumTier 1?
Faraday's law → Answer
7-5● EasyTier 1?
Transformer EMF → Answer
7-6●● MediumTier 1?
Transformer EMF → Answer
7-8● EasyTier 1?
Motional EMF → Answer
7-9●● MediumTier 1?
Motional EMF → Answer
7-12●● MediumTier 1?
Maxwell's equations → Answer
7-13●● MediumTier 1?
Maxwell's equations → Answer
7-20●● MediumTier 1?
Boundary conditions → Answer
7-21●● MediumTier 1?
Boundary conditions → Answer
7-24●● MediumTier 1?
Wave equations → Answer
7-28●● MediumTier 1?
Time-harmonic fields → Answer
7-29●● MediumTier 1?
Time-harmonic fields → Answer
7-3●● MediumTier 2?
Faraday's law → Answer
7-16●● MediumTier 2?
Potential functions → Answer
7-17●●● HardTier 2?
Potential functions → Answer
7-25●●● HardTier 2?
Wave equations → Answer

Textbook Practice — Approach Hints

Sample: Loop area \(A=0.01\,\text{m}^2\) in field \(B(t)=0.5\sin(100\pi t)\,\text{T}\) normal to plane. Find induced EMF.

\[\Phi=BA=0.005\sin(100\pi t)\]
\[\mathcal{E}=-d\Phi/dt=-0.5\pi\cos(100\pi t)\,\text{V}\]

Sample: Single-turn loop in \(B(t)=B_0 e^{-t/\tau}\), area \(A\), \(\tau=1\,\text{ms}\). Find EMF.

\[\mathcal{E}=-d(BA)/dt=B_0 A e^{-t/\tau}/\tau\]

Sample: Transformer: \(N_1=200\) primary, \(N_2=1000\) secondary, primary \(V_1=120\,\text{V}\) (60 Hz). Find \(V_2\).

\[V_2/V_1=N_2/N_1=5\;\Rightarrow\;V_2=600\,\text{V}\]

Sample: Coil in time-varying field \(\vec{B}=B_0\cos(\omega t)\hat{z}\), \(N\) turns, area \(A\). Find phasor EMF.

\[\tilde{V}=-j\omega NA B_0\;\;(\text{lags B by }90^{\circ})\]

Sample: Conducting bar \(L=0.5\,\text{m}\) moves at \(v=10\,\text{m/s}\) perpendicular to \(\vec{B}=0.2\hat{z}\,\text{T}\). Find motional EMF.

\[\mathcal{E}=\int(\vec{v}\times\vec{B})\cdot d\vec{\ell}=BvL=0.2\!\cdot\!10\!\cdot\!0.5=1\,\text{V}\]

Sample: Loop moving with \(\vec{v}\) in non-uniform \(\vec{B}(\vec{r},t)\). Total EMF.

\[\mathcal{E}=-\int\frac{\partial\vec{B}}{\partial t}\cdot d\vec{S}+\oint(\vec{v}\times\vec{B})\cdot d\vec{\ell}\]

Sample: Write Maxwell's equations in differential form (SI units).

\[\nabla\cdot\vec{D}=\rho_v,\quad\nabla\cdot\vec{B}=0\]
\[\nabla\times\vec{E}=-\partial\vec{B}/\partial t,\quad\nabla\times\vec{H}=\vec{J}+\partial\vec{D}/\partial t\]

Sample: Parallel-plate capacitor: \(V(t)=V_0\sin\omega t\), \(C=10\,\text{nF}\). Find conduction and displacement currents.

\[\text{Conduction: }i_c=C\,dV/dt=10^{-8}V_0\omega\cos\omega t\]
\[\text{Displacement: }i_d=\partial D/\partial t\!\cdot\!A\;\text{equals }i_c\;(\text{continuity})\]

Sample: Boundary between two dielectrics (\(\epsilon_1,\epsilon_2\)) with no surface current/charge. State all four boundary conditions.

\[E_{1t}=E_{2t},\;H_{1t}=H_{2t},\;D_{1n}=D_{2n},\;B_{1n}=B_{2n}\]

Sample: Surface current \(\vec{K}=K_0\hat{x}\,\text{A/m}\) on plane \(z=0\). Apply BC.

\[\hat{n}\times(\vec{H}_2-\vec{H}_1)=\vec{K}\;\Rightarrow\;\hat{z}\times(\vec{H}_2-\vec{H}_1)=K_0\hat{x}\]
\[(H_{2y}-H_{1y})\hat{x}-(H_{2x}-H_{1x})\hat{y}=K_0\hat{x}\]

Sample: Source-free region. Derive vector wave equation for \(\vec{E}\).

\[\nabla\times\nabla\times\vec{E}=-\nabla\times(\partial\vec{B}/\partial t)=-\mu\,\partial(\nabla\times\vec{H})/\partial t\]
\[\nabla^2\vec{E}-\mu\epsilon\,\partial^2\vec{E}/\partial t^2=0\]

Sample: Convert \(\vec{E}(z,t)=10\hat{x}\cos(\omega t-\beta z+\pi/4)\,\text{V/m}\) to phasor.

\[\tilde{\vec{E}}(z)=10\hat{x}\,e^{j(-\beta z+\pi/4)}=10e^{j\pi/4}\hat{x}\,e^{-j\beta z}\]

Sample: Phasor Maxwell's curl equation in lossy medium.

\[\nabla\times\tilde{\vec{H}}=(\sigma+j\omega\epsilon)\tilde{\vec{E}}=j\omega\epsilon_c\tilde{\vec{E}},\;\epsilon_c=\epsilon-j\sigma/\omega\]

Sample: Coil in increasing flux. Apply Lenz's law to determine current direction.

Induced current opposes the change: if \(\Phi\) is increasing into the page, induced current is counter-clockwise (creates flux out of page).

Sample: Express \(\vec{E},\vec{B}\) from \(V(\vec{r},t)\) and \(\vec{A}(\vec{r},t)\).

\[\vec{B}=\nabla\times\vec{A},\quad\vec{E}=-\nabla V-\partial\vec{A}/\partial t\]

Sample: Apply Lorenz gauge to derive wave equations for \(V\) and \(\vec{A}\).

\[\nabla\cdot\vec{A}+\mu\epsilon\,\partial V/\partial t=0\]
\[\nabla^2 V-\mu\epsilon\,\partial^2 V/\partial t^2=-\rho_v/\epsilon\]
\[\nabla^2\vec{A}-\mu\epsilon\,\partial^2\vec{A}/\partial t^2=-\mu\vec{J}\]

Sample: Plane-wave solution \(\vec{E}=E_0\hat{x}\cos(\omega t-kz)\). Find \(\vec{H}\) and check Maxwell.

\[\vec{H}=\frac{1}{\eta}\hat{z}\times\vec{E}=\frac{E_0}{\eta}\hat{y}\cos(\omega t-kz)\]
\[k=\omega\sqrt{\mu\epsilon},\;\eta=\sqrt{\mu/\epsilon}\]

Chapter 8 — Plane Electromagnetic Waves

Key Theory — Chapter 8

Condensed from Cheng, Field and Wave Electromagnetics, §8-1 through §8-10. Read this before attempting the problems below.


8-1   Introduction — The Free-Space Wave Equation

In a source-free non-conducting medium, Maxwell's equations combine to give homogeneous vector wave equations in E and H. In free space:

\[ \nabla^2\mathbf{E} - \frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = 0, \qquad c = \frac{1}{\sqrt{\mu_0\epsilon_0}} \cong 3\times 10^8 \ \text{m/s} \qquad \text{(8-1, 8-2)} \]

A uniform plane wave has \(\mathbf{E}\) with the same direction, magnitude, and phase across any plane perpendicular to the direction of propagation. Strictly, infinite plane waves don't exist in practice — but far from a source the wavefront is nearly planar, and their mathematics is fundamental.


8-2   Plane Waves in Lossless Media

Using phasors, the wave equation becomes the vector Helmholtz equation:

\[ \nabla^2\mathbf{E} + k_0^2\mathbf{E} = 0 \qquad \text{(8-3)} \]
\[ \boxed{k_0 = \omega\sqrt{\mu_0\epsilon_0} = \omega/c} \quad (\text{rad/m}) \qquad \text{(8-4)} \]

For a wave varying only in \(z\), the solution is a superposition of forward- and backward-traveling waves:

\[ E_x(z) = E_0^+ e^{-jk_0 z} + E_0^- e^{jk_0 z} \qquad \text{(8-7)} \]

Intrinsic impedance and the E–H relation

Substituting into Faraday's law gives \(\mathbf{H}\) perpendicular to both \(\mathbf{E}\) and the propagation direction. Their ratio defines the intrinsic impedance:

\[ \boxed{\eta_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} \cong 120\pi \cong 377\ \Omega} \qquad \text{(8-14)} \]

For a general lossless medium, \(k = \omega\sqrt{\mu\epsilon}\), \(\eta = \sqrt{\mu/\epsilon}\), \(\lambda = 2\pi/k\), and \(u_p = \omega/k = 1/\sqrt{\mu\epsilon}\). The wave is transverse: \(\mathbf{E}\perp\mathbf{H}\perp\mathbf{a}_n\), with \(\mathbf{H} = (1/\eta)\mathbf{a}_n\times\mathbf{E}\).

Wavenumber vector (oblique propagation)

For propagation along an arbitrary direction \(\mathbf{a}_n\), define the wavenumber vector \(\mathbf{k} = k\mathbf{a}_n\). Then

\[ \mathbf{E}(\mathbf{R}) = \mathbf{E}_0\,e^{-j\mathbf{k}\cdot\mathbf{R}} \qquad \text{(8-26)} \]

The planes of constant phase satisfy \(\mathbf{a}_n\cdot\mathbf{R} = \text{const}\).

Polarization

When two orthogonal linearly polarized components of the same frequency are added, the tip of the resultant \(\mathbf{E}\) traces a curve in the transverse plane:


8-3   Plane Waves in Lossy Media

In a lossy medium (finite conductivity or complex permittivity), the wavenumber becomes complex: \(k_c = \omega\sqrt{\mu\epsilon_c}\). Define the propagation constant:

\[ \gamma = \alpha + j\beta = jk_c = j\omega\sqrt{\mu\epsilon}\left(1 + \frac{\sigma}{j\omega\epsilon}\right)^{1/2} \quad (\text{m}^{-1}) \qquad \text{(8-43, 8-44)} \]

where \(\alpha\) (Np/m) is the attenuation constant and \(\beta\) (rad/m) is the phase constant. The wave decays as \(e^{-\alpha z}\) and oscillates as \(e^{-j\beta z}\). For a lossless medium \(\alpha = 0\) and \(\beta = k\).

Good conductors and skin depth

In a good conductor (\(\sigma/\omega\epsilon \gg 1\)): \(\alpha \cong \beta \cong \sqrt{\pi f\mu\sigma}\). The depth at which the amplitude falls to \(1/e\) is the skin depth:

\[ \boxed{\delta = \frac{1}{\alpha} = \frac{1}{\sqrt{\pi f\mu\sigma}}} \quad (\text{m}) \qquad \text{(8-57)} \]

Typical values (from Table 8-1):

Material σ (S/m) δ@60 Hz δ@1 MHz δ@1 GHz
Silver 6.17×10⁷ 8.27 mm 0.064 mm 0.0020 mm
Copper 5.80×10⁷ 8.53 mm 0.066 mm 0.0021 mm
Aluminium 3.54×10⁷ 10.92 mm 0.084 mm 0.0027 mm
Iron (μᵣ≈10³) 1.00×10⁷ 0.65 mm 0.005 mm 0.16 μm
Seawater 4 32 m 0.25 m —

At microwave frequencies the skin depth is so small that fields and currents are effectively confined to a thin surface layer — hence the name.

The intrinsic impedance in a good conductor is complex:

\[ \eta_c = (1+j)\sqrt{\frac{\pi f\mu}{\sigma}} = \sqrt{\frac{\pi f\mu}{\sigma}}\,e^{j\pi/4} \]

meaning \(\mathbf{E}\) and \(\mathbf{H}\) are 45° out of phase in a good conductor — a substantial reactive (energy-storing) component remains even during propagation.


8-4   Group Velocity and Dispersion

A dispersive medium is one in which \(\beta\) is not a linear function of \(\omega\); different frequencies travel at different phase velocities, so a modulated signal distorts. The velocity of the envelope of a narrow-band wave packet is the group velocity:

\[ \boxed{u_g = \frac{1}{d\beta/d\omega}} \quad (\text{m/s}) \qquad \text{(8-72)} \]

On an \(\omega\)–\(\beta\) diagram: the slope of the line from origin to a point is the phase velocity \(u_p = \omega/\beta\); the tangent slope is the group velocity \(u_g\). Three cases:

Ionized gas (plasma)

Free electrons give an effective permittivity

\[ \epsilon_p = \epsilon_0\!\left(1 - \frac{\omega_p^2}{\omega^2}\right), \qquad \omega_p = \sqrt{\frac{Ne^2}{m\epsilon_0}} \qquad \text{(8-66, 8-64)} \]

called the plasma frequency. For \(\omega < \omega_p\), \(\gamma\) is real — the wave is evanescent, no propagation. For \(\omega > \omega_p\), propagation resumes and \(u_p u_g = c^2\). The ionosphere reflects AM radio (\(f < f_p\)) but transmits TV and FM (\(f > f_p\)).


8-5   Flow of Power — The Poynting Vector

Starting from Maxwell's curl equations and using the vector identity \(\nabla\cdot(\mathbf{E}\times\mathbf{H}) = \mathbf{H}\cdot(\nabla\times\mathbf{E}) - \mathbf{E}\cdot(\nabla\times\mathbf{H})\), one obtains Poynting's theorem:

\[ \oint_S(\mathbf{E}\times\mathbf{H})\cdot d\mathbf{s} = -\frac{\partial}{\partial t}\int_V\!\left(\tfrac{1}{2}\epsilon E^2 + \tfrac{1}{2}\mu H^2\right)dv - \int_V\sigma E^2\,dv \qquad \text{(8-82)} \]

The left side is the power flowing out through \(S\); the right side is the rate of decrease of stored field energy minus ohmic dissipation inside \(V\). The Poynting vector

\[ \boxed{\mathscr{P} = \mathbf{E}\times\mathbf{H}} \quad (\text{W/m}^2) \qquad \text{(8-83)} \]

is the instantaneous power density vector carried by the field.

Time-average power density

For time-harmonic fields represented by phasors:

\[ \boxed{\mathscr{P}_{av} = \tfrac{1}{2}\,\mathscr{R}e(\mathbf{E}\times\mathbf{H}^*)} \quad (\text{W/m}^2) \qquad \text{(8-96)} \]

For a uniform plane wave in a lossy medium propagating in \(+z\), this reduces to \(\mathscr{P}_{av} = \mathbf{a}_z(E_0^2/2|\eta|)e^{-2\alpha z}\cos\theta_\eta\) — the phase angle \(\theta_\eta\) of the intrinsic impedance acts exactly like a power factor.


8-6   Normal Incidence on a Perfect Conductor

When a wave is incident normally from a dielectric onto a perfect conductor (\(\sigma_2\to\infty\)):


8-7   Oblique Incidence on a Perfect Conductor

The plane of incidence contains the propagation direction and the surface normal. Any polarization decomposes into two orthogonal cases:

In both cases a standing wave forms perpendicular to the boundary and a traveling wave parallel to it — this is the foundation of waveguide theory (Chapter 10).


8-8   Normal Incidence on a Dielectric Boundary

Neither medium is a perfect conductor; part of the wave reflects, part transmits. Requiring continuity of tangential \(\mathbf{E}\) and \(\mathbf{H}\) at \(z=0\) yields the reflection coefficient \(\Gamma\) and transmission coefficient \(\tau\):

\[ \boxed{\Gamma = \frac{E_{r0}}{E_{i0}} = \frac{\eta_2 - \eta_1}{\eta_2 + \eta_1}} \qquad \text{(8-140)} \]
\[ \boxed{\tau = \frac{E_{t0}}{E_{i0}} = \frac{2\eta_2}{\eta_2 + \eta_1}} \qquad \text{(8-141)} \]
\[ 1 + \Gamma = \tau \qquad \text{(8-142)} \]

\(\Gamma\) ranges from \(-1\) (perfect conductor) to \(+1\); \(\tau\) is always positive. In a lossy medium both are complex (introducing a phase shift at the interface). Check: \(\eta_2 = \eta_1 \Rightarrow \Gamma = 0\) — impedance matched, no reflection.

Standing-wave ratio (SWR)

When reflection is partial, the total \(\mathbf{E}_1 = \tau E_{i0}e^{-j\beta_1 z} + j2\Gamma E_{i0}\sin\beta_1 z\) — a traveling wave plus a standing wave. The maxima and minima of \(|\mathbf{E}_1|\) define the standing-wave ratio:

\[ \boxed{S = \frac{|E|_{\max}}{|E|_{\min}} = \frac{1 + |\Gamma|}{1 - |\Gamma|}} \qquad \text{(8-147)} \]
\[ |\Gamma| = \frac{S-1}{S+1} \qquad \text{(8-148)} \]

\(S\) ranges from 1 (no reflection) to \(\infty\) (total reflection); often quoted in dB as \(20\log_{10}S\). Each 1-unit increase in \(\Gamma\) corresponds to a roughly doubling of SWR.


8-9   Multiple Dielectric Interfaces — Quarter-Wave Transformer

For three media (medium 1 → medium 2 of thickness \(d\) → medium 3), there are two conditions for zero reflection from medium 1:

  1. Half-wave dielectric window (\(\eta_3 = \eta_1\)): \(d = n\lambda_2/2\) — any integer number of half-wavelengths in medium 2 preserves the impedance match. Used for radomes over radar antennas.
  2. Quarter-wave impedance transformer (\(\eta_3 \neq \eta_1\)): \(\eta_2 = \sqrt{\eta_1\eta_3}\) and \(d = (2n+1)\lambda_2/4\) — the middle layer acts as an impedance-matching transformer, analogous to quarter-wave sections in transmission lines (Chapter 9).

8-10   Oblique Incidence on a Dielectric Boundary

Snell's laws

Phase-matching \(e^{-j\beta_1 x\sin\theta_i} = e^{-j\beta_1 x\sin\theta_r} = e^{-j\beta_2 x\sin\theta_t}\) at every \(x\) on the interface immediately gives:

\[ \theta_r = \theta_i \quad \text{(Snell's law of reflection)} \]
\[ \boxed{\frac{\sin\theta_t}{\sin\theta_i} = \frac{\beta_1}{\beta_2} = \frac{n_1}{n_2} = \sqrt{\frac{\epsilon_{r1}}{\epsilon_{r2}}}} \quad \text{(Snell's law of refraction)} \qquad \text{(8-185)} \]

where \(n = c/u_p\) is the index of refraction. Denser medium (\(n_2 > n_1\)) bends the ray toward the normal.

Total internal reflection

If \(\epsilon_1 > \epsilon_2\) and \(\theta_i\) exceeds the critical angle, no refracted wave propagates:

\[ \boxed{\theta_c = \sin^{-1}\sqrt{\frac{\epsilon_2}{\epsilon_1}} = \sin^{-1}\!\left(\frac{n_2}{n_1}\right)} \qquad \text{(8-188)} \]

Beyond \(\theta_c\) the wave in medium 2 becomes evanescent (exponentially attenuated along the normal while traveling along the surface) — a surface wave. This is the operating principle of optical fibers and dielectric waveguides.

Fresnel equations

Applying continuity at the interface gives reflection coefficients for each polarization:

\[ \Gamma_\perp = \frac{\eta_2\cos\theta_i - \eta_1\cos\theta_t}{\eta_2\cos\theta_i + \eta_1\cos\theta_t} \quad \text{(perpendicular pol.)} \qquad \text{(8-206)} \]
\[ \Gamma_\| = \frac{\eta_2\cos\theta_t - \eta_1\cos\theta_i}{\eta_2\cos\theta_t + \eta_1\cos\theta_i} \quad \text{(parallel pol.)} \]

Brewster's angle

For non-magnetic media (\(\mu_1 = \mu_2 = \mu_0\)), \(\Gamma_\|\) vanishes at a particular angle of incidence — the Brewster angle:

\[ \boxed{\theta_{B\|} = \tan^{-1}\sqrt{\frac{\epsilon_2}{\epsilon_1}} = \tan^{-1}\!\left(\frac{n_2}{n_1}\right)} \qquad \text{(8-227)} \]

At \(\theta_{B\|}\), a parallel-polarized wave is totally transmitted (no reflection). Because \(\Gamma_\perp\) does not vanish there, an unpolarized wave incident at \(\theta_{B\|}\) reflects with perpendicular polarization only — the polarizing angle. Applications: polarizing sunglasses, Brewster windows in laser cavities.


Chapter 8 at a Glance


Problems and Solutions


Ex 8.1●● MediumTier 1?
Wave Impedance

Find the intrinsic impedance in (a) free space, (b) a non-magnetic dielectric with εᵣ = 9.

→ Solution
Ex 8.1 Solution↑ Problem
\[\eta _{0} = \sqrt{\mu _{0}/\varepsilon _{0}} = \sqrt{4\pi \times 10^{-7} / 8.85\times 10^{-12}} = 377 \Omega \]
For εᵣ = 9, μᵣ = 1:
\[\eta = \eta _{0} \sqrt{\mu _{r}/\varepsilon _{r}} = 377/\sqrt{9} = 377/3 = 125.7 \Omega \]

Ex 8.2●● MediumTier 1?
Attenuation in a Lossy Dielectric

Medium: εᵣ = 2.5, σ = 0.01 S/m, μᵣ = 1, f = 1 GHz. Find α, β, and skin depth δ.

→ Solution
Ex 8.2 Solution↑ Problem
\[\omega = 2\pi \times 10^{9} rad/s\]
Loss tangent: σ/(ωε) = σ/(ω ε₀ εᵣ)
\[= 0.01 / (2\pi \times 10^{9} \times 8.85\times 10^{-12} \times 2.5)\]
\[= 0.01 / 0.1390 = 0.0719 << 1 (low-loss dielectric)\]
Low-loss approximations:
\[\eta \approx \eta _{0}/√\varepsilon _{r} = 377/1.581 = 238.4 \Omega \]
\[\alpha \approx (\sigma /2) \eta = (0.01/2) \times 238.4 = 1.19 Np/m\]
\[\beta \approx \omega \sqrt{\mu _{0}\varepsilon _{0}\varepsilon _{r}} = (2\pi \times 10^{9}/c)\times \sqrt{2.5} = 20.94\times 1.581 = 33.1 rad/m\]
\[\delta = 1/\alpha = 1/1.19 = 0.84 m\]

Ex 8.3●● MediumTier 1?
Polarization State

Determine the polarization of E = 3 cos(ωt − kz) a_x + 4 sin(ωt − kz) a_y.

→ Solution
Ex 8.3 Solution↑ Problem
Phasor: Ẽ = (3 a_x - j4 a_y) e^(-jkz)
|E_x| = 3, |E_y| = 4, phase difference = -90°
Semi-axes are unequal (3 ≠ 4) → elliptical polarization.
\[Tracing at z = 0:\]
\[t = 0: E = 3 a_x\]
t = T/4: E = 4 a_y (rotates counterclockwise viewed from +z)
→ Left-Hand Elliptical Polarization (LHEP), semi-axes 3 and 4.

Ex 8.4●● MediumTier 1?
Group Velocity in a Plasma

Plasma with ωₚ = 2π×10⁹ rad/s (fₚ = 1 GHz). Find phase and group velocities at f = 3 GHz.

→ Solution
Ex 8.4 Solution↑ Problem
Dispersion relation: ω² = ωₚ² + k²c²
\[\omega = 2\pi \times 3\times 10^{9} rad/s\]
\[\omega ₚ = 2\pi \times 1\times 10^{9} rad/s\]
\[k^{2} = (\omega ^{2} - \omega ₚ^{2})/c^{2} = (9 - 1)(4\pi ^{2}\times 10^{18})/(9\times 10^{16})\]
\[= 8 \times 4\pi ^{2} \times 100/9 = 3510 rad^{2}/m^{2}\]
\[k = 59.2 rad/m\]
Phase velocity:
\[v_p = \omega /k = 2\pi \times 3\times 10^{9} / 59.2 = 3.18\times 10^{8} m/s (> c, allowed)\]
Group velocity:
\[v_g = d\omega /dk = c^{2}k/\omega = (9\times 10^{16} \times 59.2)/(6\pi \times 10^{9}) = 2.83\times 10^{8} m/s\]
Check: v_p × v_g = 3.18×10⁸ × 2.83×10⁸ = 9.0×10¹⁶ = c² ✓

Ex 8.5●● MediumTier 1?
Poynting Vector in a Dielectric

E = 50 cos(ωt − kz) a_x V/m propagates in a lossless medium with εᵣ = 4, μᵣ = 1.

Find ⟨S⟩.

→ Solution
Ex 8.5 Solution↑ Problem
\[\eta = \eta _{0}/√\varepsilon _{r} = 377/2 = 188.5 \Omega \]
\[H = (E_{0}/\eta ) cos(\omega t - kz) a_y = (50/188.5) cos(\omega t - kz) a_y = 0.265 cos(\omega t-kz) a_y A/m\]
\[\langle S\rangle = E_{0}^{2}/(2\eta ) a_z = 2500/(2\times 188.5) a_z = 6.63 a_z W/m^{2}\]

Ex 8.6●● MediumTier 1?
Reflection Coefficient (Normal Incidence)

Plane wave from air (η₁ = 377 Ω) normally incident on glass (εᵣ = 4, μᵣ = 1, η₂ = 188.5 Ω).

Find Γ, τ, R, T.

→ Solution
Ex 8.6 Solution↑ Problem
\[Γ = (\eta _{2} - \eta _{1})/(\eta _{2} + \eta _{1}) = (188.5 - 377)/(188.5 + 377)\]
= -188.5/565.5 = -0.333 (phase reversal on reflection)
\[\tau = 2\eta _{2}/(\eta _{2} + \eta _{1}) = 2\times 188.5/565.5 = 0.667\]
Check: 1 + Γ = τ → 1 - 0.333 = 0.667 ✓
Power reflectance: R = |Γ|² = 0.111 = 11.1%
Power transmittance: T = 1 - R = 0.889 = 88.9%

Ex 8.7●● MediumTier 1?
Standing Wave Ratio

From Problem 6 (Γ = −0.333), find the SWR in medium 1.

→ Solution
Ex 8.7 Solution↑ Problem
\[SWR = (1 + |Γ|)/(1 - |Γ|) = (1 + 0.333)/(1 - 0.333) = 1.333/0.667 = 2.0\]
\[|E_max| = E₊(1 + |Γ|) = 1.333 E₊\]
\[|E_min| = E₊(1 - |Γ|) = 0.667 E₊\]

Ex 8.8●● MediumTier 1?
Brewster's Angle

Find Brewster's angle for TM (parallel) polarization at an air–glass interface (n₁ = 1, n₂ = 2).

State whether it exists for TE polarization.

→ Solution
Ex 8.8 Solution↑ Problem
TM (parallel) polarization:
\[tan(\theta _B) = n_{2}/n_{1} = 2\]
\[\theta _B = arctan(2) = 63.43°\]
TE (perpendicular) polarization:
Brewster's angle does NOT exist for TE in non-magnetic media.
(The TE reflection coefficient never reaches zero for non-magnetic materials.)

Ex 8.9●● MediumTier 1?
Critical Angle for TIR

Glass (n₁ = 1.5) to air (n₂ = 1) interface. Find the critical angle for total internal reflection.

→ Solution
Ex 8.9 Solution↑ Problem
\[sin(\theta _c) = n_{2}/n_{1} = 1/1.5 = 0.667\]
\[\theta _c = arcsin(0.667) = 41.8°\]
For θᵢ > 41.8°, total internal reflection occurs (all power reflected).

Ex 8.10●● MediumTier 1?
Transmission Coefficients at Dielectric Interface

Normal incidence from air into medium with εᵣ = 9 (η₂ = 125.7 Ω). Find Γ, τ, R, T.

→ Solution
Ex 8.10 Solution↑ Problem
\[Γ = (\eta _{2} - \eta _{1})/(\eta _{2} + \eta _{1}) = (125.7 - 377)/(125.7 + 377) = -251.3/502.7 = -0.500\]
\[\tau = 2\eta _{2}/(\eta _{2} + \eta _{1}) = 251.4/502.7 = 0.500\]
\[R = |Γ|^{2} = 0.25 = 25%\]
\[T = (\eta _{1}/\eta _{2})|\tau |^{2} = (377/125.7) \times 0.25 = 3 \times 0.25 = 0.75 = 75%\]
Check: R + T = 0.25 + 0.75 = 1 ✓

Ex 8.11●● MediumTier 1?
Oblique Incidence (TE), Fresnel Coefficients

TE wave from air (n₁ = 1) into glass (n₂ = 2) at θᵢ = 30°. Find θₜ, Γ_TE, τ_TE.

→ Solution
Ex 8.11 Solution↑ Problem
Snell's law: n₁ sinθᵢ = n₂ sinθₜ
\[sin\theta ₜ = (1/2) sin30° = 0.25 \to \theta ₜ = 14.5°\]
\[cos\theta ᵢ = cos30° = 0.866\]
\[cos\theta ₜ = cos14.5° = 0.968\]
TE Fresnel coefficients (η₁ = 377, η₂ = 188.5 Ω):
\[Γ_TE = (\eta _{2} cos\theta ᵢ - \eta _{1} cos\theta ₜ)/(\eta _{2} cos\theta ᵢ + \eta _{1} cos\theta ₜ)\]
\[= (188.5\times 0.866 - 377\times 0.968)/(188.5\times 0.866 + 377\times 0.968)\]
\[= (163.2 - 364.9)/(163.2 + 364.9)\]
\[= -201.7/528.1 = -0.382\]
\[\tau _TE = 2\eta _{2} cos\theta ᵢ/(\eta _{2} cos\theta ᵢ + \eta _{1} cos\theta ₜ)\]
\[= 2\times 163.2/528.1 = 0.618\]

Ex 8.12●● MediumTier 1?
Phase and Group Velocities in a Waveguide

Rectangular waveguide with cutoff frequency f_c = 5 GHz, operating at f = 10 GHz. Find v_p, v_g, λ_g.

→ Solution
Ex 8.12 Solution↑ Problem
\[\sqrt{1-(f_c/f}^{2}) = \sqrt{1-0.25} = \sqrt{0.75} = 0.866\]
Phase velocity:
\[v_p = c/\sqrt{1-(f_c/f}^{2}) = 3\times 10^{8}/0.866 = 3.46\times 10^{8} m/s (> c, ok)\]
Group velocity:
\[v_g = c \sqrt{1-(f_c/f}^{2}) = 3\times 10^{8} \times 0.866 = 2.60\times 10^{8} m/s\]
Check: v_p × v_g = 3.46×10⁸ × 2.60×10⁸ = 9.0×10¹⁶ = c² ✓
Guide wavelength:
\[\lambda _{0} = c/f = 3\times 10^{8}/10^{10} = 0.03 m = 3 cm\]
\[\lambda _g = \lambda _{0}/0.866 = 3/0.866 = 3.46 cm\]

Ex 8.13●● MediumTier 1?
Power Reflection and Transmission (Oblique TE)

From Problem 11 (Γ_TE = −0.382, θᵢ = 30°, θₜ = 14.5°). Find R and T.

→ Solution
Ex 8.13 Solution↑ Problem
\[R = |Γ_TE|^{2} = (0.382)^{2} = 0.146 = 14.6%\]
\[T = 1 - R = 0.854 = 85.4%\]
Verify with τ_TE:
\[T = (n_{2} cos\theta ₜ)/(n_{1} cos\theta ᵢ) \times |\tau _TE|^{2}\]
\[= (2\times 0.968)/(1\times 0.866) \times (0.618)^{2}\]
\[= 2.236 \times 0.382 \approx 0.854 \checkmark \]

Ex 8.14●● MediumTier 1?
TM Polarization at Oblique Incidence

TM (parallel) wave, same geometry as Problem 11 (θᵢ = 30°, air to glass n₂ = 2). Find Γ_TM.

→ Solution
Ex 8.14 Solution↑ Problem
Fresnel TM coefficient (using refractive indices):
\[Γ_TM = (n_{2} cos\theta ᵢ - n_{1} cos\theta ₜ)/(n_{2} cos\theta ᵢ + n_{1} cos\theta ₜ)\]
\[= (2\times 0.866 - 1\times 0.968)/(2\times 0.866 + 1\times 0.968)\]
\[= (1.732 - 0.968)/(1.732 + 0.968)\]
\[= 0.764/2.700 = +0.283\]
Positive Γ_TM means no phase reversal.
At Brewster's angle θ_B = 63.4°, Γ_TM = 0 exactly.
Since 30° < 63.4°, the TM wave has partial reflection with no phase reversal.

Ex 8.15●● MediumTier 1?
Propagation Constant in a Good Conductor

Find α, β, and skin depth in copper (σ = 5.8×10⁷ S/m, μᵣ = 1) at f = 100 MHz.

→ Solution
Ex 8.15 Solution↑ Problem
Good conductor condition: σ/(ωε₀) = 5.8×10⁷/(2π×10⁸×8.85×10⁻¹²) = 1.05×10¹⁰ >> 1 ✓
\[\alpha = \beta = \sqrt{\pi f\mu _{0}\sigma }\]
\[= \sqrt{\pi \times 10^{8} \times 4\pi \times 10^{-7} \times 5.8\times 10^{7}}\]
\[= \sqrt{4\pi ^{2} \times 5.8 \times 10^{8}}\]
\[= \sqrt{2.291\times 10^{10}}\]
\[= 1.514\times 10^{5} Np/m (and rad/m)\]
\[\gamma = (1 + j) \times 1.514\times 10^{5} /m\]
\[\delta = 1/\alpha = 6.61 \mu m\]
\[(Consistent with: \delta = 66.1/\sqrt{f} mm = 66.1/\sqrt{10^{8}} mm = 66.1/10^{4} mm = 6.61 \mu m \checkmark )\]

Textbook Practice Problems

8-1● EasyTier 1?
Plane waves in lossless media → Answer
8-2●● MediumTier 1?
Plane waves in lossless media → Answer
8-9●● MediumTier 1?
Plane waves in lossy media → Answer
8-10●● MediumTier 1?
Plane waves in lossy media → Answer
8-16●● MediumTier 1?
Poynting vector → Answer
8-17●● MediumTier 1?
Poynting vector → Answer
8-20●● MediumTier 1?
Normal incidence — conductor → Answer
8-21●● MediumTier 1?
Normal incidence — conductor → Answer
8-30●● MediumTier 1?
Normal incidence — dielectric → Answer
8-31●● MediumTier 1?
Normal incidence — dielectric → Answer
8-3●● MediumTier 2?
Plane waves in lossless media → Answer
8-5●● MediumTier 2?
Polarization → Answer
8-6●● MediumTier 2?
Polarization → Answer
8-13●● MediumTier 2?
Group velocity → Answer
8-14●●● HardTier 2?
Group velocity → Answer
8-25●●● HardTier 2?
Oblique incidence — conductor → Answer
8-26●●● HardTier 2?
Oblique incidence — conductor → Answer
8-35●●● HardTier 2?
Oblique incidence — dielectric → Answer
8-36●●● HardTier 3?
Oblique incidence — dielectric → Answer

Textbook Practice — Approach Hints

Sample: Plane wave at 10 GHz in free space. Find \(\beta,\lambda,\eta\).

\[\beta=\omega\sqrt{\mu_0\epsilon_0}=2\pi(10^{10})/(3\!\cdot\!10^8)=209.4\,\text{rad/m}\]
\[\lambda=2\pi/\beta=0.03\,\text{m}=3\,\text{cm},\;\eta_0=377\,\Omega\]

Sample: \(\vec{E}=100\hat{x}\cos(\omega t-\beta z)\,\text{V/m}\) in air, \(f=300\,\text{MHz}\). Find \(\vec{H}\).

\[\beta=2\pi(3\!\cdot\!10^8)/(3\!\cdot\!10^8)=2\pi\,\text{rad/m},\;\eta_0=377\,\Omega\]
\[\vec{H}=(E_0/\eta_0)\hat{y}\cos(\omega t-\beta z)=0.265\hat{y}\cos(\omega t-2\pi z)\,\text{A/m}\]

Sample: Wet earth: \(\sigma=10^{-2}\,\text{S/m}\), \(\epsilon_r=15\), \(f=1\,\text{MHz}\). Find \(\alpha,\beta\).

\[\sigma/(\omega\epsilon)=10^{-2}/(2\pi\!\cdot\!10^6\!\cdot\!15\epsilon_0)=11.98\;\;(\text{conductor-like})\]
\[\alpha=\beta=\sqrt{\omega\mu\sigma/2}=\sqrt{2\pi\!\cdot\!10^6\!\cdot\!4\pi\!\cdot\!10^{-7}\!\cdot\!10^{-2}/2}=0.199\,\text{Np/m}\]

Sample: Copper at 1 GHz: \(\sigma=5.8\!\cdot\!10^7\). Find skin depth.

\[\delta=1/\sqrt{\pi f\mu_0\sigma}=1/\sqrt{\pi\!\cdot\!10^9\!\cdot\!4\pi\!\cdot\!10^{-7}\!\cdot\!5.8\!\cdot\!10^7}=2.09\,\mu\text{m}\]

Sample: Plane wave \(E_0=10\,\text{V/m}\) in free space. Find time-average Poynting.

\[\langle S\rangle=E_0^2/(2\eta_0)=100/(2\!\cdot\!377)=0.133\,\text{W/m}^2\]

Sample: Circularly polarized wave \(E_0=5\,\text{V/m}\) in air. Find \(\langle\vec{S}\rangle\).

\[\langle S\rangle=E_0^2/\eta_0=25/377=0.0663\,\text{W/m}^2\]
Twice the linearly polarized result (constant magnitude).

Sample: Plane wave normal incidence on perfect conductor. Show \(\Gamma=-1\).

\[E_{tangential}=0\text{ at conductor}\Rightarrow E_i+E_r=0\Rightarrow\Gamma=E_r/E_i=-1\]
\[\text{Total }E=E_i[\cos(\omega t-\beta z)-\cos(\omega t+\beta z)]=2E_i\sin\omega t\sin\beta z\]

Sample: Standing wave from above. Find positions of \(E\) nulls and maxima.

\[\text{Nulls at }\beta z=n\pi,\;z=n\lambda/2\;(n=0,1,2,\ldots)\]
\[\text{Maxima at }z=(2n+1)\lambda/4\]

Sample: Air to glass (\(n=1.5\)), normal incidence. Find \(\Gamma,\tau\) and powers.

\[\Gamma=(\eta_2-\eta_1)/(\eta_2+\eta_1)=(251.3-377)/(628.3)=-0.2\]
\[|\Gamma|^2=0.04\;(\text{4\% reflected}),\;\text{transmitted}=96\%\]

Sample: Verify power conservation for above.

\[P_r/P_i=|\Gamma|^2=0.04\]
\[P_t/P_i=(1-|\Gamma|^2)=0.96\;\Rightarrow\;\text{sum}=1\;\checkmark\]

Sample: Polyethylene (\(\epsilon_r=2.25\)) at \(f=1\,\text{GHz}\). Find \(v_p,\lambda,\eta\).

\[v_p=c/\sqrt{\epsilon_r}=2\!\cdot\!10^8\,\text{m/s},\;\lambda=v_p/f=0.2\,\text{m}\]
\[\eta=\eta_0/\sqrt{\epsilon_r}=377/1.5=251.3\,\Omega\]

Sample: \(\vec{E}=E_0(\hat{x}+j\hat{y})e^{-j\beta z}\). Identify polarization.

Equal amplitudes, \(90^{\circ}\) phase shift \(\Rightarrow\) circular polarization. Sense: \(\hat{x}+j\hat{y}\) at \(t=0\) rotates from \(\hat{x}\) toward \(\hat{y}\) as \(t\) increases (LHCP for \(+z\) propagation in RHC convention).

Sample: \(E_x=3\cos\omega t,\;E_y=4\cos(\omega t-\pi/2)\) at \(z=0\). Identify polarization.

\(|E_x|\ne|E_y|\), \(90^{\circ}\) phase \(\Rightarrow\) elliptical polarization; major axis along \(\hat{y}\), axial ratio \(4/3\).

Sample: Dispersive medium: \(\beta(\omega)=\omega\sqrt{\mu\epsilon(\omega)}\) with \(\epsilon=\epsilon_0(2+\omega/\omega_0)\). Find \(v_p\) and \(v_g\).

\[v_p=\omega/\beta,\;v_g=d\omega/d\beta\]
Compute via implicit differentiation; \(v_g\ne v_p\) in general.

Sample: Plasma with \(\omega_p=2\pi\!\cdot\!10^9\,\text{rad/s}\). At \(\omega=2\omega_p\), find \(v_p,v_g\).

\[\beta=\omega\sqrt{1-\omega_p^2/\omega^2}/c=\omega\sqrt{3}/(2c)\]
\[v_p=2c/\sqrt{3}=3.46\!\cdot\!10^8\,\text{m/s},\;v_g=c\sqrt{3}/2=2.60\!\cdot\!10^8\,\text{m/s}\]

Sample: Plane wave hits perfect conductor at \(\theta_i=30^{\circ}\), perpendicular polarization. Find reflection.

\[\Gamma_{\perp}=\frac{\eta_2\cos\theta_i-\eta_1\cos\theta_t}{\eta_2\cos\theta_i+\eta_1\cos\theta_t}=-1\;(\text{conductor})\]
Standing wave forms in direction normal to surface.

Sample: Same wave, parallel polarization.

\[\Gamma_{\parallel}=-1\;(\text{conductor reflects all polarizations completely})\]

Sample: Air-to-water interface (\(n_2=1.33\)), \(\theta_i=45^{\circ}\). Find \(\theta_t\) via Snell.

\[\sin\theta_t=(n_1/n_2)\sin\theta_i=(1/1.33)(0.707)=0.532\;\Rightarrow\;\theta_t=32.1^{\circ}\]

Sample: Find Brewster angle for air-glass (\(n=1.5\)).

\[\tan\theta_B=n_2/n_1=1.5\;\Rightarrow\;\theta_B=56.3^{\circ}\]
At \(\theta_B\), parallel-polarization reflection vanishes (only perpendicular reflects).

Chapter 9 — Theory and Applications of Transmission Lines

Key Theory — Chapter 9

Condensed from Cheng, Field and Wave Electromagnetics, §9-1 through §9-7. Read this before attempting the problems below.


9-1   Guided TEM Waves on Transmission Lines

An omnidirectional source like a wire antenna wastes most of its energy. For point-to-point transmission of power or information, electromagnetic energy must be guided — confined to a structure that carries it from source to load. The simplest such mode is the TEM (transverse electromagnetic) wave, in which both \(\mathbf{E}\) and \(\mathbf{H}\) are perpendicular to the direction of propagation — exactly like an unguided plane wave in the surrounding dielectric.

Three TEM transmission-line structures are fundamental:

Other (non-TEM) wave modes can exist when the conductor separation exceeds a certain fraction of a wavelength; these are treated as waveguide modes in Chapter 10.


9-3   General Time-Harmonic Transmission-Line Equations

A transmission line is modeled by distributed parameters per unit length:

For sinusoidal steady-state phasors \(V(z), I(z)\):

\[ -\frac{dV(z)}{dz} = (R + j\omega L)\,I(z) \qquad \text{(9-35a)} \]
\[ -\frac{dI(z)}{dz} = (G + j\omega C)\,V(z) \qquad \text{(9-35b)} \]

Decoupling by taking a second derivative:

\[ \frac{d^2 V}{dz^2} = \gamma^2 V, \qquad \frac{d^2 I}{dz^2} = \gamma^2 I \qquad \text{(9-36)} \]

where the propagation constant is

\[ \boxed{\gamma = \alpha + j\beta = \sqrt{(R + j\omega L)(G + j\omega C)}} \quad (\text{m}^{-1}) \qquad \text{(9-37)} \]

\(\alpha\) (Np/m) is the attenuation constant, \(\beta\) (rad/m) is the phase constant. The general solution is a superposition of forward- and backward-traveling waves:

\[ V(z) = V_0^+ e^{-\gamma z} + V_0^- e^{\gamma z} \qquad \text{(9-38a)} \]
\[ I(z) = I_0^+ e^{-\gamma z} + I_0^- e^{\gamma z} \qquad \text{(9-38b)} \]

Characteristic impedance

For an infinite line (or a line terminated in a matched load), only forward waves exist. Their voltage-to-current ratio is the characteristic impedance:

\[ \boxed{Z_0 = \frac{V_0^+}{I_0^+} = \sqrt{\frac{R + j\omega L}{G + j\omega C}} = \frac{\gamma}{G + j\omega C}} \quad (\Omega) \qquad \text{(9-41)} \]

\(Z_0\) and \(\gamma\) depend on \(R, L, G, C, \omega\) — not on line length. They characterize the line itself.

Analogy: the general form is identical to \(\gamma = j\omega\sqrt{\mu\epsilon(1 + \sigma/j\omega\epsilon)^{1/2}}\) and \(\eta = \sqrt{\mu/(\epsilon(1-j\sigma/\omega\epsilon))}\) for plane waves in a lossy medium — voltage plays the role of \(\mathbf{E}\), current the role of \(\mathbf{H}\), \(Z_0\) the role of intrinsic impedance \(\eta\).


9-3.2   Three Practical Cases

Lossless line (R = G = 0)

\[ \gamma = j\omega\sqrt{LC}, \quad \alpha = 0, \quad \beta = \omega\sqrt{LC} \qquad \text{(9-52)} \]
\[ \boxed{Z_0 = R_0 = \sqrt{L/C}} \quad (\text{purely real}) \qquad \text{(9-53)} \]
\[ u_p = \omega/\beta = 1/\sqrt{LC} \]

Low-loss line (R ≪ ωL, G ≪ ωC)

\[ \alpha \cong \tfrac{1}{2}\!\left(R\sqrt{C/L} + G\sqrt{L/C}\right) = \tfrac{1}{2}(R/R_0 + GR_0) \qquad \text{(9-55)} \]
\[ \beta \cong \omega\sqrt{LC} \qquad \text{(9-56)} \]
\[ R_0 \cong \sqrt{L/C}, \quad X_0 \cong 0 \qquad \text{(9-59, 9-60)} \]

Distortionless line (R/L = G/C)

When the Heaviside condition \(R/L = G/C\) is satisfied, both \(\gamma\) and \(Z_0\) simplify exactly:

\[ \alpha = R\sqrt{C/L}, \quad \beta = \omega\sqrt{LC} \qquad \text{(9-63, 9-64)} \]
\[ Z_0 = \sqrt{L/C} \quad (\text{real at all frequencies}) \]

Because \(\beta\) is linear in \(\omega\), all frequencies travel at the same phase velocity \(u_p = 1/\sqrt{LC}\) — no dispersion, no signal distortion.


Universal Relation: LC = με

From the plane-wave analogy \(\gamma = j\omega\sqrt{\mu\epsilon(1 + \sigma/j\omega\epsilon)^{1/2}}\) and \(G/C = \sigma/\epsilon\) (from the R–C analogy of Chapter 5):

\[ \boxed{LC = \mu\epsilon, \qquad G/C = \sigma/\epsilon} \qquad \text{(9-71, 9-72)} \]

Consequence: if \(L\) is known for a given geometry, \(C\) follows from \(LC = \mu\epsilon\); if either is known, \(G\) follows from the medium conductivity. The velocity of propagation on a lossless TEM line equals the velocity of an unguided plane wave in the dielectric: \(u_p = 1/\sqrt{LC} = 1/\sqrt{\mu\epsilon}\).

Two-wire line parameters

For parallel wires of radius \(a\) separated by center-to-center distance \(D\):

\[ C = \frac{\pi\epsilon}{\cosh^{-1}(D/2a)} \quad (\text{F/m}) \qquad \text{(9-73)} \]
\[ L = \frac{\mu}{\pi}\cosh^{-1}\!\left(\frac{D}{2a}\right) \quad (\text{H/m}) \qquad \text{(9-74)} \]
\[ G = \frac{\pi\sigma}{\cosh^{-1}(D/2a)} \quad (\text{S/m}) \qquad \text{(9-75)} \]

(When \(D/2a \gg 1\), \(\cosh^{-1}(D/2a) \cong \ln(D/a)\).) Analogous formulas with \(\ln(b/a)\) apply to coaxial lines (Chapter 3/6).


9-4   Finite Terminated Line

Measuring from the load end (\(z' = \ell - z\), so \(z' = 0\) is the load):

\[ V(z') = I_L(Z_L\cosh\gamma z' + Z_0\sinh\gamma z') \qquad \text{(9-100a)} \]
\[ I(z') = (I_L/Z_0)(Z_L\sinh\gamma z' + Z_0\cosh\gamma z') \qquad \text{(9-100b)} \]

The impedance transformation at any point along the line:

\[ \boxed{Z(z') = Z_0\,\frac{Z_L + Z_0\tanh\gamma z'}{Z_0 + Z_L\tanh\gamma z'}} \quad (\Omega) \qquad \text{(9-102)} \]

The input impedance at the generator end (\(z' = \ell\)) is \(Z_i = Z(z' = \ell)\). This is the key equation — it lets you replace a terminated line by an equivalent lumped impedance at the source.

Open- and short-circuited lines (lossless)

With \(\gamma = j\beta\), \(\tanh j\beta\ell = j\tan\beta\ell\):

\[ Z_{io} = -jR_0\cot\beta\ell \quad (\text{open-circuit}) \]
\[ Z_{is} = jR_0\tan\beta\ell \quad (\text{short-circuit}) \]

Short-circuited lines appear inductive for \(\ell < \lambda/4\), capacitive for \(\lambda/4 < \ell < \lambda/2\), and so on. Useful fact: \(Z_0 = \sqrt{Z_{io}Z_{is}}\) — measuring open- and short-circuit input impedances of any line determines \(Z_0\).


9-4.2   Reflection Coefficient and Standing Wave Ratio

For any resistive or complex load \(Z_L \neq Z_0\), part of the incident wave reflects. Define the voltage reflection coefficient at the load:

\[ \boxed{\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0} = |\Gamma|\,e^{j\theta_\Gamma}} \quad (\text{dimensionless}) \qquad \text{(9-134)} \]

\(|\Gamma| \le 1\), with special cases:

Load Γ SWR S Note
Matched (ZL = Z0) 0 1 No reflection
Short (ZL = 0) −1 ∞ Total reflection, phase inverted
Open (ZL → ∞) +1 ∞ Total reflection, in phase

On a lossless line \(V(z')\) is the sum of an incident traveling wave and a reflected traveling wave. Their superposition produces standing waves with maxima at spacing \(\lambda/2\). The standing-wave ratio (SWR):

\[ \boxed{S = \frac{|V|_{\max}}{|V|_{\min}} = \frac{1 + |\Gamma|}{1 - |\Gamma|}} \qquad \text{(9-138)} \]
\[ |\Gamma| = \frac{S - 1}{S + 1} \qquad \text{(9-139)} \]

Often quoted in dB as \(20\log_{10}S\). A high SWR is undesirable — it indicates mismatch, large reflected power, and potentially dangerous voltage maxima on the line.

Resistive load on a lossless line: \(\Gamma = (R_L - R_0)/(R_L + R_0)\).


9-4.3   Resonance and Quality Factor

A short-circuited lossy line of length \(\ell = \lambda/4\) (or odd multiples) behaves as a parallel resonant circuit; length \(\ell = \lambda/2\) (or multiples) as a series resonant circuit. The quality factor of a shorted quarter-wave section is

\[ Q = \frac{f_0}{\Delta f} = \frac{\beta}{2\alpha} \qquad \text{(9-130)} \]

For a well-insulated low-loss line, \(Q \cong \omega L/R\) — the familiar LC-circuit expression. Transmission-line resonators at microwave frequencies routinely achieve \(Q\)'s in the thousands, far beyond lumped-element circuits.


9-6   The Smith Chart

The Smith chart is a conformal map of the complex \(z = Z/Z_0\) plane onto the unit disk in the \(\Gamma\) plane. Its key properties:

Standard problems solved by inspection on the chart: find \(\Gamma\) from \(Z_L\), find \(z_i\) from \(z_L\) after a given length, find \(Z_L\) from SWR and minimum location, find stub lengths for matching, convert impedance to admittance. For lossy lines, the \(|\Gamma|\)-circle becomes a spiral contracting inward due to the \(e^{-2\alpha z'}\) factor.


9-7   Transmission-Line Impedance Matching

Maximum power transfer from generator to load requires \(Z_L = Z_0\) — an impedance-matched line has \(S = 1\) and no reflections. Mismatched lines also cause echoes that distort information signals. The chapter presents three matching techniques on lossless lines:

Quarter-wave transformer

To match a resistive load \(R_L\) to a line of impedance \(R_0\), insert a quarter-wavelength section of characteristic impedance:

\[ \boxed{R_0' = \sqrt{R_0 R_L}} \quad (\Omega) \qquad \text{(9-194)} \]

Because \(Z(\lambda/4) = Z_0^2/Z_L\) for a lossless line (from eq. 9-102 with \(\tanh j\pi/2 \to \infty\)), the quarter-wave section "inverts" the load and scales it. Since \(\ell = \lambda/4\) depends on wavelength, this match is inherently narrowband. Quarter-wave transformers don't work for complex loads on low-loss lines because \(R_0'\) would need to be complex.

Single-stub matching

A more general technique: attach a short-circuited line stub of length \(\ell\) in parallel with the main line at a distance \(d\) from the load. Two unknowns (\(d\), \(\ell\)) are chosen so the parallel admittance equals \(Y_0 = 1/R_0\):

\[ Y_i = Y_B + Y_s = Y_0 \qquad \text{(9-197)} \]

Working in normalized admittances, \(y_B\) is the admittance at the stub location looking toward the load, and \(y_s\) is the input admittance of the short-circuited stub (purely imaginary). The match condition becomes \(y_B = 1 + jb_B\), and the stub susceptance must cancel \(jb_B\). Short-circuited stubs are preferred over open-circuited ones — a perfect short is easier to realize than a perfect open (open ends radiate and couple to neighboring objects).

Single-stub matching is easily solved graphically on the Smith chart (using it as an admittance chart): locate \(y_L\), rotate toward generator until the trajectory crosses the \(g = 1\) circle — that's distance \(d\); read off the needed susceptance and find the corresponding short-circuit stub length. Unlike the quarter-wave transformer, single-stub matching works for any \(Z_L\).

Double-stub matching (§9-7.3) uses two stubs at fixed distances when the distance to the load cannot be adjusted — at the cost of a smaller range of matchable \(Z_L\).


Chapter 9 at a Glance


Problems and Solutions


Ex 9.1●● MediumTier 1?
Characteristic Impedance

Find Z₀ for a coaxial line with inner radius a = 1 mm, outer radius b = 5 mm, εᵣ = 2.25.

→ Solution
Ex 9.1 Solution↑ Problem
\[Z_{0} = (60/√\varepsilon _{r}) ln(b/a)\]
\[= (60/1.5) ln(5)\]
\[= 40 \times 1.609\]
\[= 64.4 \Omega \]

Ex 9.2●● MediumTier 1?
Propagation Constant

Line parameters at f = 1 MHz: R = 0.1 Ω/m, L = 0.5 μH/m, G = 0, C = 50 pF/m. Find α and β.

→ Solution
Ex 9.2 Solution↑ Problem
\[\omega = 2\pi \times 10^{6} rad/s\]
Lossless β:
\[\beta _{0} = \omega \sqrt{LC} = 2\pi \times 10^{6} \times \sqrt{0.5\times 10^{-6} \times 50\times 10^{-12}}\]
\[= 2\pi \times 10^{6} \times \sqrt{25\times 10^{-18}}\]
\[= 2\pi \times 10^{6} \times 5\times 10^{-9}\]
\[= 0.0314 rad/m\]
Attenuation (low-loss approximation):
\[Z_{0} = \sqrt{L/C} = \sqrt{0.5\times 10^{-6}/50\times 10^{-12}} = \sqrt{10^{4}} = 100 \Omega \]
\[\alpha \approx R/(2Z_{0}) = 0.1/(2\times 100) = 5\times 10^{-4} Np/m\]
\[\gamma = \alpha + j\beta = 5\times 10^{-4} + j0.0314 /m\]

Ex 9.3●● MediumTier 1?
VSWR from Reflection Coefficient

Z₀ = 50 Ω, Z_L = 100 + j50 Ω. Find Γ and SWR.

→ Solution
Ex 9.3 Solution↑ Problem
\[Γ = (Z_L - Z_{0})/(Z_L + Z_{0}) = (50 + j50)/(150 + j50)\]
\[|numerator| = \sqrt{50^{2} + 50^{2}} = 50\sqrt{2} = 70.71\]
\[|denominator| = \sqrt{150^{2} + 50^{2}} = \sqrt{25000} = 158.1\]
\[|Γ| = 70.71/158.1 = 0.447\]
\[SWR = (1 + |Γ|)/(1 - |Γ|) = 1.447/0.553 = 2.62\]

Ex 9.4●● MediumTier 1?
Input Impedance of Terminated Line

Z₀ = 75 Ω, Z_L = 150 Ω (real), l = λ/8. Find Z_in.

→ Solution
Ex 9.4 Solution↑ Problem
\[\beta l = (2\pi /\lambda )(\lambda /8) = \pi /4 \to tan(\beta l) = 1\]
\[Z_in = Z_{0} \times (Z_L + jZ_{0} tan\beta l)/(Z_{0} + jZ_L tan\beta l)\]
\[= 75 \times (150 + j75)/(75 + j150)\]
Factor 75 from each:
\[= 75 \times (2 + j)/(1 + j2)\]
Multiply by conjugate (1 - j2)/(1 - j2):
\[(2+j)(1-j2) = 2 - j4 + j - j^{2}2 = 2 - j4 + j + 2 = 4 - j3\]
\[(1+j2)(1-j2) = 1 + 4 = 5\]
\[Z_in = 75 \times (4-j3)/5 = 15(4-j3) = 60 - j45 \Omega \]

Ex 9.5●● MediumTier 1?
Reflection Coefficient for Complex Load

Z₀ = 50 Ω, Z_L = 25 − j50 Ω. Find Γ and SWR.

→ Solution
Ex 9.5 Solution↑ Problem
\[Γ = (Z_L - Z_{0})/(Z_L + Z_{0}) = (-25 - j50)/(75 - j50)\]
Multiply by conjugate (75 + j50):
Numerator: (-25-j50)(75+j50) = -1875 - j1250 - j3750 - j²2500
\[= -1875 + 2500 - j5000 = 625 - j5000\]
Denominator: 75² + 50² = 5625 + 2500 = 8125
\[Γ = (625 - j5000)/8125 = 0.0769 - j0.6154\]
\[|Γ| = \sqrt{0.0769^{2} + 0.6154^{2}} = \sqrt{0.00592 + 0.3787} = \sqrt{0.3846} = 0.620\]
\[SWR = (1 + 0.620)/(1 - 0.620) = 1.620/0.380 = 4.26\]

Ex 9.6●● MediumTier 1?
Quarter-Wave Transformer

Match Z_L = 200 Ω to Z₀ = 50 Ω using a λ/4 transformer section. Find Z₀' and verify Z_in.

→ Solution
Ex 9.6 Solution↑ Problem
Quarter-wave transformer impedance:
\[Z_{0}' = \sqrt{Z_{0} \times Z_L} = \sqrt{50 \times 200} = \sqrt{10000} = 100 \Omega \]
Input impedance of λ/4 section:
\[Z_in = (Z_{0}')^{2}/Z_L = 10000/200 = 50 \Omega = Z_{0} \checkmark \]
The λ/4 section (Z₀' = 100 Ω) provides a perfect match at the design frequency.

Ex 9.7●● MediumTier 1?
Reflection Coefficient

Z₀ = 50 Ω, Z_L = 75 + j25 Ω. Find |Γ| and ∠Γ.

→ Solution
Ex 9.7 Solution↑ Problem
\[Γ = (Z_L - Z_{0})/(Z_L + Z_{0}) = (25 + j25)/(125 + j25)\]
\[|numerator| = 25\sqrt{2} = 35.36\]
\[|denominator| = \sqrt{125^{2} + 25^{2}} = \sqrt{16250} = 127.5\]
\[|Γ| = 35.36/127.5 = 0.277\]
\[∠Γ = arctan(25/25) - arctan(25/125) = 45° - 11.3° = 33.7°\]
\[Γ = 0.277 ∠33.7°\]

Ex 9.8●● MediumTier 1?
Line Loss in dB

Transmission line: α = 0.01 Np/m, length l = 100 m. Find power loss in dB.

→ Solution
Ex 9.8 Solution↑ Problem
Total attenuation = αl = 0.01 × 100 = 1 Np
Convert: 1 Np = 8.686 dB
\[Loss = 1 \times 8.686 = 8.69 dB\]
Power ratio: P_out/P_in = e^(-2αl) = e^(-2) = 0.1353 = -8.69 dB ✓
Voltage ratio: V_out/V_in = e^(-αl) = e^(-1) = 0.368

Ex 9.9●● MediumTier 1?
Single-Stub Matching

Z₀ = 50 Ω, Z_L = 100 Ω (real). Find short-circuit stub position d and length l for matching.

→ Solution
Ex 9.9 Solution↑ Problem
Normalized: y_L = Z₀/Z_L = 0.5
\[Find d where Re[y(d)] = 1:\]
\[Let t = tan(\beta d). Require real part = 1:\]
\[y_L(1 + t^{2})/(1 + y_L^{2} t^{2}) = 1\]
\[\to t^{2} = 1/y_L = 2 \to t = \pm \sqrt{2}\]
Solution 1 (t = +√2):
\[\beta d_{1} = arctan(\sqrt{2}) = 54.7° \to d_{1} = 0.152\lambda \]
\[Im[y(d_{1})] = +\sqrt{2} \times 0.75/1.5 = +0.707\]
\[\to y(d_{1}) = 1 + j0.707, need jb_s = -j0.707\]
SC stub: -cot(βl₁) = -0.707 → tan(βl₁) = √2 → l₁ = 0.152λ
Solution 2 (t = -√2):
\[\beta d_{2} = 180° - 54.7° = 125.3° \to d_{2} = 0.348\lambda \]
\[Im[y(d_{2})] = -0.707 \to need jb_s = +j0.707\]
SC stub: -cot(βl₂) = +0.707 → βl₂ = 125.3° → l₂ = 0.348λ

Ex 9.10●● MediumTier 1?
Transient Response (Step Input)

Z₀ = 50 Ω, Z_g = 50 Ω, Z_L = ∞ (open), V_g = 100 V step. Find V at load for 0 < t < 3T.

→ Solution
Ex 9.10 Solution↑ Problem
Γ_g = (Z_g - Z₀)/(Z_g + Z₀) = (50-50)/(50+50) = 0 (matched source)
\[Γ_L = (Z_L - Z_{0})/(Z_L + Z_{0}) = 1 (open circuit)\]
Initial forward wave:
\[V₊ = V_g \times Z_{0}/(Z_g + Z_{0}) = 100 \times 50/100 = 50 V\]
Timeline:
\[0 < t < T: V_L = 0\]
t = T: V₊ arrives at open end
\[V_L = V₊(1 + Γ_L) = 50(1+1) = 100 V\]
V₋ = Γ_L V₊ = 50 V returns toward source
t = 2T: V₋ arrives at source: Γ_g = 0 → no re-reflection
t > T: V_L = 100 V (steady state, no further changes)

Ex 9.11●● MediumTier 1?
Bounce Diagram

Z₀ = 50 Ω, Z_g = 200 Ω, Z_L = ∞ (open), V_g = 100 V step, T = 0.5 μs. Trace voltage at load.

→ Solution
Ex 9.11 Solution↑ Problem
\[Γ_g = (200-50)/(200+50) = 150/250 = 0.6\]
\[Γ_L = 1\]
\[V₊_{1} = 100 \times 50/(200+50) = 20 V\]
Time Event V at load
─────────────────────────────────────────────────
\[t = 0 Step applied 0 V\]
\[t = T V₊_{1}=20 arrives 20(1+1) = 40 V\]
\[V₋_{1} = 20 returns\]
\[t = 2T V₋_{1} at source \to V₊_{2} = 0.6\times 20 = 12 V\]
\[t = 3T V₊_{2}=12 arrives 64 V (+24)\]
\[V₋_{2} = 12 returns\]
\[t = 4T V₋_{2} at source \to V₊_{3} = 0.6\times 12 = 7.2 V\]
\[t = 5T V₊_{3} arrives 78.4 V (+14.4)\]
Steady state: V_L → 40/(1-0.6) = 100 V ✓
(all voltage drops across open-circuit load, as expected)

Ex 9.12●● MediumTier 1?
Double-Stub Matching

Z₀ = 50 Ω, Z_L = 25 − j50 Ω, stub separation = λ/4. Find stub lengths (SC stubs).

→ Solution
Ex 9.12 Solution↑ Problem
Normalized load admittance:
\[y_L = Z_{0}/Z_L = 50/(25-j50)\]
\[= 50(25+j50)/3125 = 0.4 + j0.8\]
For λ/4 stub spacing, stub 1 must set y₁ = 0.4 + jB such that
after λ/4 transform (y → 1/y), Re[y₂] = 1:
\[Re[1/(0.4+jB)] = 1 \to 0.4/(0.16 + B^{2}) = 1\]
\[B^{2} = 0.24 \to B = \pm 0.490\]
Since Im[y_L] = 0.8, stub 1 susceptance b₁ = B - 0.8:
\[Case 1: B = +0.490 \to b_{1} = 0.490 - 0.8 = -0.310\]
\[Case 2: B = -0.490 \to b_{1} = -0.490 - 0.8 = -1.290\]
\[Stub 1 (SC, Case 1): -cot(\beta l_{1}) = -0.310 \to l_{1} = 0.202\lambda \]
\[Stub 1 (SC, Case 2): -cot(\beta l_{1}) = -1.290 \to l_{1} = 0.397\lambda \]
For Case 1 — Stub 2 susceptance:
\[y_{2} = 1/(0.4+j0.490) = 1 - j1.22\]
\[Need b_{2} = +1.22\]
SC stub: -cot(βl₂) = 1.22 → l₂ = 0.391λ

Ex 9.13●● MediumTier 1?
Impedance Transformation

Z₀ = 100 Ω, Z_L = 50 Ω, l = 3λ/8. Find Z_in.

→ Solution
Ex 9.13 Solution↑ Problem
\[\beta l = (2\pi /\lambda )(3\lambda /8) = 3\pi /4 \to tan(3\pi /4) = -1\]
\[Z_in = 100 \times (50 + j100\times (-1))/(100 + j50\times (-1))\]
\[= 100 \times (50 - j100)/(100 - j50)\]
\[Factor 50: = 100 \times (1 - j2)/(2 - j)\]
Multiply by (2+j)/(2+j):
\[(1-j2)(2+j) = 2 + j - j4 - j^{2}2 = 2 + j - j4 + 2 = 4 - j3\]
\[(2-j)(2+j) = 5\]
\[Z_in = 100 \times (4-j3)/5 = 20(4-j3) = 80 - j60 \Omega \]

Ex 9.14●● MediumTier 1?
Power Delivered to Load

Z₀ = 50 Ω, Z_s = 50 Ω, Z_L = 75 Ω, V_s = 100∠0° V (peak), lossless λ/4 line.

→ Solution
Ex 9.14 Solution↑ Problem
Quarter-wave transforms Z_L to:
\[Z_in = Z_{0}^{2}/Z_L = 2500/75 = 33.33 \Omega \]
Input current:
\[I_in = V_s/(Z_s + Z_in) = 100/(50 + 33.33) = 100/83.33 = 1.2 A (peak)\]
Since line is lossless, P_in = P_L:
\[P_L = (1/2)|I_in|^{2} Re[Z_in] = 0.5 \times 1.44 \times 33.33 = 24 W\]

Ex 9.15●● MediumTier 1?
Return Loss

From Problem 7 (|Γ| = 0.277). Find return loss and mismatch loss.

→ Solution
Ex 9.15 Solution↑ Problem
\[Return Loss (RL) = -20 log_{10}|Γ|\]
\[= -20 log_{10}(0.277) = -20\times (-0.558) = 11.2 dB\]
Power reflected: |Γ|² = 0.0767 = 7.67%
Power transmitted: 1 - |Γ|² = 92.3%
\[Mismatch Loss = -10 log_{10}(1 - |Γ|^{2})\]
\[= -10 log_{10}(0.923) = 0.35 dB\]

Textbook Practice Problems

9-1● EasyTier 1?
TEM waves on transmission lines → Answer
9-2●● MediumTier 1?
TEM waves on transmission lines → Answer
9-5●● MediumTier 1?
Transmission line equations → Answer
9-6●● MediumTier 1?
Transmission line equations → Answer
9-9●● MediumTier 1?
Wave characteristics → Answer
9-13● EasyTier 1?
VSWR and reflection coefficient → Answer
9-14●● MediumTier 1?
VSWR and reflection coefficient → Answer
9-17●● MediumTier 1?
Input impedance → Answer
9-18●● MediumTier 1?
Input impedance → Answer
9-25●● MediumTier 1?
Smith chart → Answer
9-26●● MediumTier 1?
Smith chart → Answer
9-29●● MediumTier 1?
Impedance matching → Answer
9-30●●● HardTier 1?
Impedance matching → Answer
9-3●● MediumTier 2?
TEM waves on transmission lines → Answer
9-10●● MediumTier 2?
Wave characteristics → Answer
9-21●● MediumTier 2?
Transients → Answer
9-22●●● HardTier 2?
Transients → Answer
9-32●●● HardTier 3?
Comprehensive problem → Answer

Textbook Practice — Approach Hints

Sample: Coax with \(L'=0.4\,\mu\text{H/m}\), \(C'=100\,\text{pF/m}\). Find \(Z_0\) and \(v\).

\[Z_0=\sqrt{L'/C'}=\sqrt{(4\!\cdot\!10^{-7})/(10^{-10})}=63.2\,\Omega\]
\[v=1/\sqrt{L'C'}=1/\sqrt{4\!\cdot\!10^{-17}}=1.58\!\cdot\!10^8\,\text{m/s}\]

Sample: Air-filled coax with \(a=1\,\text{mm}\), \(b=4\,\text{mm}\). Find \(Z_0\).

\[Z_0=(60/\sqrt{\epsilon_r})\ln(b/a)=60\ln 4=83.2\,\Omega\]

Sample: Lossless line with \(L'=0.5\,\mu\text{H/m}\), \(C'=200\,\text{pF/m}\) at \(f=100\,\text{MHz}\). Find \(\beta,\lambda\).

\[\omega=2\pi\!\cdot\!10^8,\;\beta=\omega\sqrt{L'C'}=6.28\!\cdot\!10^8\!\cdot\!\sqrt{10^{-19}}=1.99\,\text{rad/m}\]
\[\lambda=2\pi/\beta=3.16\,\text{m}\]

Sample: Sinusoidal voltage on lossless line: \(V(z,t)=V_0\cos(\omega t-\beta z)+V_0/3\cos(\omega t+\beta z)\). Find \(\Gamma\).

\[\Gamma=V^-/V^+=1/3\]

Sample: Lossy line: \(R'=2\,\Omega/\text{m}\), \(G'=10^{-3}\,\text{S/m}\), \(L'=0.5\,\mu\text{H/m}\), \(C'=80\,\text{pF/m}\), \(f=100\,\text{MHz}\).

\[\gamma=\sqrt{(R'+j\omega L')(G'+j\omega C')}\approx j\omega\sqrt{L'C'}+\frac{1}{2}(R'/Z_0+G'Z_0)\]
\[Z_0=\sqrt{L'/C'}=79.06\,\Omega,\;\alpha\approx 0.052\,\text{Np/m}\]

Sample: \(Z_0=50\,\Omega\), \(Z_L=100+j50\,\Omega\). Find \(\Gamma\) and VSWR.

\[\Gamma=(Z_L-Z_0)/(Z_L+Z_0)=(50+j50)/(150+j50)=0.4+j0.2,\;|\Gamma|=0.447\]
\[\text{VSWR}=(1+|\Gamma|)/(1-|\Gamma|)=1.447/0.553=2.62\]

Sample: Measured VSWR=3 on \(Z_0=50\,\Omega\) line; voltage min at \(z=-0.1\lambda\). Find \(Z_L\).

\[|\Gamma|=(\text{VSWR}-1)/(\text{VSWR}+1)=0.5\]
\[\Gamma_L=|\Gamma|e^{j\theta_L},\;\theta_L=180^{\circ}+2\beta z_{min}=180^{\circ}-72^{\circ}=108^{\circ}\]
\[Z_L=Z_0(1+\Gamma_L)/(1-\Gamma_L)=20-j35\,\Omega\,(\text{approx.})\]

Sample: \(Z_0=50\), \(Z_L=100\), line length \(\ell=\lambda/4\). Find \(Z_{in}\).

\[Z_{in}=Z_0^2/Z_L=2500/100=25\,\Omega\;(\text{quarter-wave inverter})\]

Sample: Same line but \(\ell=\lambda/8\), \(Z_L=100\). Find \(Z_{in}\).

\[\beta\ell=\pi/4,\;\tan\beta\ell=1\]
\[Z_{in}=Z_0\frac{Z_L+jZ_0}{Z_0+jZ_L}=50\frac{100+j50}{50+j100}=50\frac{(100+j50)(50-j100)}{50^2+100^2}=40-j30\,\Omega\]

Sample: \(Z_L=25+j50\), \(Z_0=50\). Find \(\Gamma\) on Smith chart and rotate \(0.2\lambda\) toward generator.

\[z_L=Z_L/Z_0=0.5+j1,\;\Gamma_L\;\text{plotted at }(0.5,1)\text{ in }z\text{-plane}\]
Rotate clockwise (TWG) by \(2\beta\!\cdot\!0.2\lambda=0.4\lambda\!\cdot\!2\pi/\lambda\) on the chart to get \(z_{in}\) at the new position.

Sample: Move \(0.1\lambda\) along Smith chart (TWG): how much rotation?

\[\Delta\phi=2\beta(0.1\lambda)=2(2\pi/\lambda)(0.1\lambda)=0.4\pi=72^{\circ}\]

Sample: Single short-circuited stub matching \(Z_L=100\) to \(Z_0=50\). Find stub length and position.

\[y_L=Y_L/Y_0=0.5\]
Move on Smith chart until \(y=1+jb\); at that point insert stub of admittance \(-jb\). For this \(y_L\): distance \(\approx 0.0625\lambda\), stub length \(\approx 0.125\lambda\) (graphical).

Sample: Quarter-wave transformer matches \(Z_L=200\) to \(Z_0=50\).

\[Z_T=\sqrt{Z_0 Z_L}=\sqrt{10000}=100\,\Omega\]
Insert \(\lambda/4\) section of \(100\,\Omega\) line between source and load at design frequency.

Sample: Two-wire line, conductor radius \(a\), separation \(D\gg a\). Find \(L',C'\).

\[L'=(\mu_0/\pi)\ln(D/a),\;C'=\pi\epsilon_0/\ln(D/a)\]
\[Z_0=\sqrt{L'/C'}=(\eta_0/\pi)\ln(D/a)=120\ln(D/a)\,\Omega\]

Sample: Low-loss line with \(R'=1\,\Omega/\text{m}\), \(Z_0=50\,\Omega\), \(G'\approx 0\). Find \(\alpha\).

\[\alpha\approx R'/(2Z_0)=1/100=0.01\,\text{Np/m}=0.0869\,\text{dB/m}\]

Sample: Step source \(V_0=10\,\text{V}\), source impedance \(Z_s=Z_0=50\), load \(Z_L=100\). Find voltages on line.

\[V^+=V_0 Z_0/(Z_s+Z_0)=5\,\text{V}\]
\[\Gamma_L=(100-50)/(100+50)=1/3,\;V_L=V^+(1+\Gamma_L)=5(4/3)=6.67\,\text{V}\]

Sample: Lossless line, length \(L\), propagation delay \(\tau\). Step input arrives at \(t=\tau\) at load.

Bounce diagram: \(V^+\) travels at \(1/\sqrt{LC}\). At load, reflects with \(\Gamma_L\). Reflection returns at \(2\tau\), etc.

Sample (comprehensive): Match \(Z_L=75-j25\) to \(50\,\Omega\) at \(1\,\text{GHz}\) using a single short-circuited stub.

Step 1: \(y_L=Y_L/Y_0\approx 0.6+j0.2\). Step 2: rotate to circle \(g=1\), find length \(d\). Step 3: at that point \(y=1+jb\); choose stub length so \(y_{stub}=-jb\). Iterate via Smith chart.

Chapter 10 — Waveguides and Cavity Resonators

Key Theory — Chapter 10

Condensed from Cheng, Field and Wave Electromagnetics, §10-1 through §10-7. Read this before attempting the problems below.


10-1   Introduction — Why Waveguides?

Chapter 9 developed the TEM mode (both \(\mathbf{E}\) and \(\mathbf{H}\) transverse to \(z\)) on transmission lines. At microwave frequencies (≥ a few GHz), TEM lines are impractical: conductor losses scale like \(\sqrt{f}\) and radiation off two-conductor structures becomes severe.

The remedy is to confine the wave inside a single-conductor hollow metal tube — a waveguide. Inside such a tube:

Short a waveguide off with conducting end walls and you get a cavity resonator — a 3-D standing-wave box with very high \(Q\) (often \(10^{3}\)–\(10^{4}\)) and a discrete resonance spectrum.


10-2   General Wave Behavior Along a Uniform Guide

Assume harmonic time dependence \(e^{j\omega t}\) and \(z\)-dependence \(e^{-\gamma z}\) with \(\gamma = \alpha + j\beta\). In a source-free, charge-free dielectric inside the guide, Maxwell's equations reduce to Helmholtz's equation:

\[ \nabla^{2}\mathbf{E} + k^{2}\mathbf{E} = 0,\qquad \nabla^{2}\mathbf{H} + k^{2}\mathbf{H} = 0 \qquad \text{(10-3,4)} \]
\[ k = \omega\sqrt{\mu\epsilon} \qquad \text{(10-5)} \]

Splitting the Laplacian \(\nabla^{2} = \nabla_{xy}^{2} + \partial^{2}/\partial z^{2}\) and using \(\partial_z \to -\gamma\):

\[ \nabla_{xy}^{2}\mathbf{E} + h^{2}\mathbf{E} = 0,\qquad h^{2} \triangleq \gamma^{2} + k^{2} \qquad \text{(10-7,15)} \]

\(h^{2}\) is a separation constant (eigenvalue). Only discrete values of \(h\) satisfy the boundary conditions — each is a waveguide mode.

Transverse-to-longitudinal coupling. All four transverse field components can be expressed in terms of the two longitudinal components \(E_z^0\) and \(H_z^0\):

\[ H_x^{0} = -\frac{1}{h^{2}}\!\left(\gamma\frac{\partial H_z^0}{\partial x} - j\omega\epsilon\frac{\partial E_z^0}{\partial y}\right) \qquad \text{(10-11)} \]
\[ E_x^{0} = -\frac{1}{h^{2}}\!\left(\gamma\frac{\partial E_z^0}{\partial x} + j\omega\mu\frac{\partial H_z^0}{\partial y}\right) \qquad \text{(10-13)} \]

(similar for \(H_y^0, E_y^0\)). So the procedure for every waveguide problem is:

  1. Solve \(\nabla_{xy}^{2}E_z^0 + h^{2}E_z^0 = 0\) for TM (or the \(H_z\) version for TE) subject to boundary conditions.
  2. Enforce BCs to determine the eigenvalues \(h_{mn}\).
  3. Use (10-11)–(10-14) to compute the transverse components.

Cutoff frequency. The propagation constant is \(\gamma = \sqrt{h^{2}-\omega^{2}\mu\epsilon}\). It vanishes at the cutoff frequency:

\[ f_c = \frac{h}{2\pi\sqrt{\mu\epsilon}} \quad (\text{Hz}) \qquad \text{(10-35)} \]

Two regimes:

f > f_c (above cutoff) : γ = jβ, β = k√[1 − (f_c/f)²] propagates
f < f_c (below cutoff) : γ = α, α = h√[1 − (f/f_c)²] evanescent

Equivalent wavelength relations:

\[ \lambda_g = \frac{\lambda}{\sqrt{1-(f_c/f)^{2}}} > \lambda \qquad \text{(10-39)} \]
\[ \frac{1}{\lambda^{2}} = \frac{1}{\lambda_g^{2}} + \frac{1}{\lambda_c^{2}} \qquad \text{(10-41)} \]

Phase and group velocities in a lossless guide:

\[ u_p = \frac{u}{\sqrt{1-(f_c/f)^{2}}} > u,\qquad u_g = u\sqrt{1-(f_c/f)^{2}} < u \qquad \text{(10-42,43)} \]
\[ u_p u_g = u^{2} \quad \text{(air: } u_p u_g = c^{2}\text{)} \qquad \text{(10-44)} \]

A waveguide is therefore a dispersive transmission system. \(u_g\) equals the velocity of energy transport.

Wave impedances (for propagating modes):

\[ Z_{\text{TEM}} = \sqrt{\frac{\mu}{\epsilon}} = \eta \qquad \text{(10-20)} \]
\[ Z_{\text{TM}} = \eta\sqrt{1-(f_c/f)^{2}} \qquad \text{(10-45)} \]
\[ Z_{\text{TE}} = \frac{\eta}{\sqrt{1-(f_c/f)^{2}}} \qquad \text{()} \]

Note \(Z_{\text{TM}}\) is lower than \(\eta\) and \(Z_{\text{TE}}\) is higher — they approach each other at high frequency. Below cutoff both are purely reactive (no net power flows in an evanescent mode).


10-3   Parallel-Plate Waveguide

Two infinite, parallel conducting planes separated by \(b\). Analytically simplest non-trivial waveguide; supports:

Useful as the bridge between transmission lines (Ch 9) and hollow guides — exhibits all three wave types (TEM, TM, TE) in one structure.


10-4   Rectangular Waveguides

Cross-section \(a \times b\) with \(a > b\) by convention. Separation of variables gives sinusoidal dependence in \(x\) and \(y\).

TMmn modes — boundary condition \(E_z = 0\) on all four walls forces sine variation:

\[ E_z^{0}(x,y) = E_0 \sin\!\left(\frac{m\pi x}{a}\right)\sin\!\left(\frac{n\pi y}{b}\right) \quad \text{(V/m)} \qquad \text{(10-132)} \]
\[ h^{2} = \left(\frac{m\pi}{a}\right)^{2} + \left(\frac{n\pi}{b}\right)^{2} \qquad \text{(10-133)} \]

For TM, neither \(m\) nor \(n\) can be zero (else \(E_z\equiv 0\)). Lowest TM mode is TM11.

TEmn modes — boundary condition \(\partial H_z/\partial n = 0\) on walls forces cosine variation:

\[ H_z^{0}(x,y) = H_0 \cos\!\left(\frac{m\pi x}{a}\right)\cos\!\left(\frac{n\pi y}{b}\right) \quad \text{(A/m)} \qquad \text{(10-158)} \]

For TE, either \(m\) or \(n\) may be zero, but not both.

Cutoff frequency / wavelength (same form for TM and TE):

\[ (f_c)_{mn} = \frac{1}{2\sqrt{\mu\epsilon}}\sqrt{\left(\frac{m}{a}\right)^{2} + \left(\frac{n}{b}\right)^{2}} \quad (\text{Hz}) \qquad \text{(10-139)} \]
\[ (\lambda_c)_{mn} = \frac{2}{\sqrt{(m/a)^{2}+(n/b)^{2}}} \quad (\text{m}) \qquad \text{(10-140)} \]

Dominant mode: TE10. Because \(n=0\) is permitted for TE and \(a>b\), TE10 has the lowest cutoff of all modes:

\[ (f_c)_{\text{TE}_{10}} = \frac{1}{2a\sqrt{\mu\epsilon}} = \frac{u}{2a},\qquad (\lambda_c)_{\text{TE}_{10}} = 2a \qquad \text{(10-163,164)} \]

Practical rule of thumb. A waveguide is normally used in its dominant mode over an operating band of roughly \(1.25\,f_{c10}\) to \(0.95\,f_{c20}\) — i.e. 25% above dominant cutoff but 5% below the next higher mode. Example (WR-16, \(a=2.29\) cm, \(b=1.02\) cm): usable band 8.19–12.45 GHz (X-band).

Attenuation. Dielectric loss (substituting \(\epsilon_d=\epsilon(1-j\sigma/\omega\epsilon)\)):

\[ \alpha_d = \frac{\sigma\eta}{2\sqrt{1-(f_c/f)^{2}}} \quad (\text{Np/m}) \qquad \text{(10-178)} \]

Conductor (wall) loss for the dominant mode:

\[ (\alpha_c)_{\text{TE}_{10}} = \frac{R_s\bigl[1 + (2b/a)(f_c/f)^{2}\bigr]}{\eta b\sqrt{1-(f_c/f)^{2}}} \quad (\text{Np/m}) \qquad \text{(10-187)} \]

Attenuation diverges near \(f_c\), has a broad minimum in the middle of the operating band, and grows slowly at high \(f\). TE10 always has the lowest attenuation of any mode, which is another reason it is preferred.


10-5   Circular Waveguides

Cross-section of radius \(a\). Separation in \((r,\phi)\) gives Bessel's equation in the radial coordinate; solutions are \(J_n(hr)\) (Bessel functions of the first kind). Modes are TMnp (zeros of \(J_n\)) and TEnp (zeros of \(J'_n\)).

Key cutoff facts (with \(p_{np}\) the \(p\)-th zero of \(J_n\), \(p'_{np}\) of \(J'_n\)):

Mode Eigenvalue h·a (f_c)·a·√(με) (dimensionless)
TE₁₁ p'₁₁ = 1.841 0.293 ← dominant mode
TM₀₁ p₀₁ = 2.405 0.383
TE₂₁ p'₂₁ = 3.054 0.486
TM₁₁ p₁₁ = 3.832 0.610

TE11 is dominant, analogous to TE10 in rectangular. Circular guides are less common in practice because polarization can rotate, but they are used when rotational symmetry matters (rotary joints, circular polarizers).


10-6   Dielectric Waveguides (Slab / Fiber)

When the guiding structure is an open dielectric slab (or optical fiber) with \(\epsilon_1 > \epsilon_2\) outside, the fields are bound by total internal reflection. Inside the high-\(\epsilon\) slab the transverse fields are sinusoidal; outside they decay exponentially — the wave is a surface wave.

Both TM and TE modes are possible. Unlike metal guides, each mode has a different cutoff condition and the structure is the foundation of all optical fiber communication.


10-7   Cavity Resonators

Close both ends of a waveguide with conducting walls and you get a 3-D resonant cavity. End-wall reflections convert the traveling \(e^{-j\beta z}\) into a standing wave \(\sin(p\pi z/d)\) or \(\cos(p\pi z/d)\); the mode is indexed by three integers \((m,n,p)\).

Rectangular cavity \(a\times b\times d\). TMmnp (\(E_z\) has \(\sin\sin\cos\) form) and TEmnp (\(H_z\) has \(\cos\cos\sin\)) coexist. The resonant frequency is independent of which is which:

\[ f_{mnp} = \frac{u}{2}\sqrt{\left(\frac{m}{a}\right)^{2} + \left(\frac{n}{b}\right)^{2} + \left(\frac{p}{d}\right)^{2}} \quad (\text{Hz}) \qquad \text{(10-301)} \]

Dominant mode depends on the aspect ratio:

a > b > d : TM₁₁₀ is dominant
a > d > b : TE₁₀₁ is dominant
a = b = d : TM₁₁₀, TE₀₁₁, TE₁₀₁ are all degenerate

(TM restricts \(m,n\ne 0\) but allows \(p=0\); TE allows one of \(m,n\) to be zero but requires \(p\ne 0\).)

Quality factor \(Q\) — the figure of merit for a resonator:

\[ Q \triangleq 2\pi\frac{W}{\text{energy dissipated per cycle}} = \frac{\omega W}{P_L} \qquad \text{(10-313,315)} \]

where \(W = W_e + W_m\) is the total stored energy and \(P_L\) is the time-average power dissipated in the walls (and any dielectric). At resonance, \(W_e = W_m\).

For the TE101 mode in a rectangular cavity:

\[ Q_{101} = \frac{\pi f_{101}\mu_0\, abd(a^{2}+d^{2})}{R_s\bigl[2b(a^{3}+d^{3}) + ad(a^{2}+d^{2})\bigr]} \qquad \text{(10-322)} \]

Typical value: a copper cubic cavity at 10 GHz has \(Q\approx 10{,}700\) — orders of magnitude higher than any lumped LC circuit, because loss occurs only through surface resistance \(R_s=\sqrt{\pi f\mu_0/\sigma}\).

Circular cylindrical cavity. The TM010 mode (radius \(a\), length \(d\)) has the simple resonant frequency

\[ (f_c)_{\text{TM}_{010}} = \frac{2.405}{2\pi a\sqrt{\mu_0\epsilon_0}} = \frac{0.115}{a} \quad (\text{GHz, with } a \text{ in m}) \qquad \text{(10-329)} \]
\[ Q_{\text{TM}_{010}} = \frac{\eta_0}{R_s}\cdot\frac{2.405}{2(1+a/d)} \qquad \text{(10-328)} \]

Cavities are excited by a probe (a short monopole inserted where \(\mathbf{E}\) of the desired mode is maximum) or a loop (inserted where \(\mathbf{H}\) of the desired mode is maximum), or by coupling through an iris from an adjoining waveguide.


Chapter 10 at a Glance

Problems and Solutions


Review Questions (Tier 1)

R.10-1● EasyTier 1?
Review: Why are TEM transmission lines not useful for long-distance signal transmission at microwave frequencies? → Answer
R.10-1 Answer↑ Question

Conductor losses scale as \(\sqrt{f}\) and dielectric losses rise too; at microwave frequencies coaxial cable attenuation becomes prohibitive over long distances.

R.10-2● EasyTier 1?
Review: What is the cutoff frequency of a waveguide? → Answer
R.10-2 Answer↑ Question

The lowest frequency at which a given waveguide mode can propagate. Below \(f_c\) the mode is evanescent (purely decaying).

R.10-3● EasyTier 1?
Review: Why are lumped R, L, C elements not useful as resonant circuits at microwave frequencies? → Answer
R.10-3 Answer↑ Question

At microwave frequencies component dimensions become comparable to \(\lambda\); radiation losses and parasitic effects dominate. Distributed cavity resonators are used instead.

R.10-4● EasyTier 1?
Review: What is the governing equation for the field phasors inside a uniform waveguide? → Answer
R.10-4 Answer↑ Question

Helmholtz equation: \(\nabla^2\tilde{\vec{E}}+k^2\tilde{\vec{E}}=0\) (and similarly for \(\tilde{\vec{H}}\)), where \(k=\omega\sqrt{\mu\epsilon}\).

R.10-5● EasyTier 1?
Review: What are the three basic types of propagating waves in a uniform waveguide? → Answer
R.10-5 Answer↑ Question

TEM (\(E_z=H_z=0\)), TM (transverse magnetic, \(H_z=0\), \(E_z\ne0\)), TE (transverse electric, \(E_z=0\), \(H_z\ne 0\)).

R.10-6● EasyTier 1?
Review: Define wave impedance. → Answer
R.10-6 Answer↑ Question

Ratio \(Z=E_t/H_t\) of transverse field components, characteristic of the propagating mode and frequency.

R.10-7● EasyTier 1?
Review: Why can single-conductor hollow waveguides not support TEM waves? → Answer
R.10-7 Answer↑ Question

TEM requires that \(\vec{E}\) field lines start and end on conductors in the cross-section. A single hollow conductor cannot support such an electrostatic-like configuration.

R.10-8● EasyTier 1?
Review: Describe the analytical procedure for studying TM waves in a waveguide. → Answer
R.10-8 Answer↑ Question

Solve the Helmholtz equation for \(\tilde{E}_z(x,y)\) with the BC \(E_z=0\) on the walls. Then derive the transverse field components from \(E_z\) using Maxwell's curl equations.

R.10-9● EasyTier 1?
Review: Repeat R.10-8 for TE waves. → Answer
R.10-9 Answer↑ Question

Solve the Helmholtz equation for \(\tilde{H}_z(x,y)\) with the BC \(\partial H_z/\partial n=0\) on the walls. Then derive transverse fields from \(H_z\).

R.10-10● EasyTier 1?
Review: What are the eigenvalues of a boundary-value problem? → Answer
R.10-10 Answer↑ Question

Discrete values of the separation constant for which non-trivial solutions satisfy all boundary conditions. They determine the cutoff frequencies of the modes.

R.10-11● EasyTier 1?
Review: Can a waveguide have more than one cutoff frequency? On what does \(f_c\) depend? → Answer
R.10-11 Answer↑ Question

Yes — one cutoff per mode. \(f_c\) depends on the mode indices and the guide cross-section dimensions; e.g., for rectangular guide \(f_c=(c/2)\sqrt{(m/a)^2+(n/b)^2}\).

R.10-12● EasyTier 1?
Review: What is an evanescent mode? → Answer
R.10-12 Answer↑ Question

A mode operating below its cutoff frequency. Fields decay exponentially with distance instead of propagating; \(\beta=0\) and \(\gamma\) is purely real (attenuation only).

R.10-13● EasyTier 1?
Review: Is the guide wavelength \(\lambda_g\) longer or shorter than the free-space wavelength \(\lambda\)? → Answer
R.10-13 Answer↑ Question

Always longer: \(\lambda_g=\lambda/\sqrt{1-(f_c/f)^2}>\lambda\).

R.10-14● EasyTier 1?
Review: How does \(Z\) depend on frequency for TEM, TM, and TE propagating waves? → Answer
R.10-14 Answer↑ Question

TEM: independent of \(f\). TM: \(\eta_{TM}=\eta\sqrt{1-(f_c/f)^2}\) (decreases below \(\eta\) as \(f\to f_c^+\)). TE: \(\eta_{TE}=\eta/\sqrt{1-(f_c/f)^2}\) (increases above \(\eta\)).

R.10-15● EasyTier 1?
Review: What is the significance of a purely reactive wave impedance? → Answer
R.10-15 Answer↑ Question

Indicates an evanescent mode — \(E_t\) and \(H_t\) are \(90^{\circ}\) out of phase, so no real (time-average) power flows; only stored, reactive energy.

R.10-16● EasyTier 1?
Review: Can an \(\omega\)–\(\beta\) diagram tell whether a mode is dispersive? → Answer
R.10-16 Answer↑ Question

Yes. A non-dispersive mode has \(\omega\)–\(\beta\) as a straight line through the origin. Curvature (or non-zero intercept at \(\beta=0\)) indicates dispersion.

R.10-17● EasyTier 1?
Review: How do you read \(u_p\) and \(u_g\) off an \(\omega\)–\(\beta\) diagram? → Answer
R.10-17 Answer↑ Question

\(u_p=\omega/\beta\) is the slope of the line from the origin to the operating point. \(u_g=d\omega/d\beta\) is the local tangent slope at that point.

R.10-18● EasyTier 1?
Review: What is an eigenmode? → Answer
R.10-18 Answer↑ Question

A self-consistent field pattern that satisfies the wave equation plus all boundary conditions for a particular eigenvalue (cutoff frequency).

R.10-19● EasyTier 1?
Review: On what factors does the cutoff frequency of a parallel-plate waveguide depend? → Answer
R.10-19 Answer↑ Question

Plate spacing \(d\) and mode order \(n\): \(f_c=nc/(2d)\) for TE\(_n\)/TM\(_n\). The dielectric also enters via \(c\to v=c/\sqrt{\epsilon_r\mu_r}\).

R.10-20● EasyTier 1?
Review: What is the dominant mode of a waveguide? What is the dominant mode of a parallel-plate waveguide? → Answer
R.10-20 Answer↑ Question

The mode with the lowest cutoff frequency. For a parallel-plate guide the dominant propagating mode is TEM (no cutoff); among TM/TE the dominant is TM\(_1\)/TE\(_1\) at \(f_c=c/(2d)\).

R.10-21● EasyTier 1?
Review: What is the dominant mode of a rectangular waveguide? Why? → Answer
R.10-21 Answer↑ Question

TE\(_{10}\) (assuming \(a>b\)): \(f_c=c/(2a)\) — the smallest dimension of \(\sqrt{(m/a)^2+(n/b)^2}\) — supported with the largest dimension on the wider side.

R.10-23● EasyTier 1?
Review: For a rectangular cavity, how does the dominant mode change as the aspect ratio changes from \(a>b>d\) to \(a>d>b\)? → Answer
R.10-23 Answer↑ Question

For \(a>b>d\) (\(d\) shortest) the dominant mode is TE\(_{101}\). When \(a>d>b\), \(d\) is no longer the shortest, so a different mode (e.g., TE\(_{102}\) or TM\(_{110}\)) takes the dominant role depending on which combination minimizes \((m/a)^2+(n/b)^2+(p/d)^2\).

R.10-24● EasyTier 1?
Review: Define the quality factor \(Q\) of a cavity resonator. Why is \(Q\) of a microwave cavity much larger than that of a lumped LC circuit? → Answer
R.10-24 Answer↑ Question

Quality factor \(Q=\omega W_{\text{stored}}/P_{\text{loss}}\). A microwave cavity has high \(Q\) (\(\sim 10^4\)) because losses occur only on the conductor walls (surface) while energy fills the cavity volume — the surface-to-volume ratio is far better than for lumped LC components.


Ex 10.1●● MediumTier 1?
Cutoff Frequencies (WR-90)

Rectangular waveguide: a = 2.286 cm, b = 1.016 cm (WR-90). Find f_c for TE₁₀, TE₂₀, TE₀₁, TE₁₁.

→ Solution
Ex 10.1 Solution↑ Problem
\[f_c(m,n) = (c/2) \sqrt{(m/a}^{2} + (n/b)^{2})\]
\[TE_{10}: f_c = c/(2a) = 3\times 10^{8}/(2\times 0.02286) = 6.56 GHz (dominant mode)\]
\[TE_{20}: f_c = c/a = 2 \times 6.56 = 13.12 GHz\]
\[TE_{01}: f_c = c/(2b) = 3\times 10^{8}/(2\times 0.01016) = 14.76 GHz\]
\[TE_{11}: f_c = (c/2)\sqrt{1/a^{2} + 1/b^{2}}\]
\[= 1.5\times 10^{8} \times \sqrt{1914 + 9685}\]
\[= 1.5\times 10^{8} \times 107.7 = 16.15 GHz\]
Single-mode bandwidth: 6.56 to 13.12 GHz.

Ex 10.2●● MediumTier 1?
Phase Velocity in Waveguide

WR-90 at f = 10 GHz (TE₁₀ mode, f_c = 6.56 GHz). Find v_p, v_g, and β_g.

→ Solution
Ex 10.2 Solution↑ Problem
\[\sqrt{1-(f_c/f}^{2}) = \sqrt{1-(6.56/10}^{2}) = \sqrt{1-0.4303} = \sqrt{0.5697} = 0.7548\]
\[v_p = c/0.7548 = 3\times 10^{8}/0.7548 = 3.97\times 10^{8} m/s\]
\[v_g = c \times 0.7548 = 2.26\times 10^{8} m/s\]
Check: v_p × v_g = 3.97×10⁸ × 2.26×10⁸ = 8.97×10¹⁶ ≈ c² ✓
\[\beta _g = (2\pi f/c) \times 0.7548 = (2\pi \times 10^{10}/3\times 10^{8}) \times 0.7548 = 158.2 rad/m\]

Ex 10.3●● MediumTier 1?
Wave Impedance for TE and TM Modes

WR-90 at f = 10 GHz (f_c = 6.56 GHz). Find η_TE and η_TM.

→ Solution
Ex 10.3 Solution↑ Problem
From Problem 2: √(1-(f_c/f)²) = 0.7548
\[\eta _TE = \eta _{0}/\sqrt{1-(f_c/f}^{2}) = 377/0.7548 = 499.5 \Omega (> \eta _{0})\]
\[\eta _TM = \eta _{0} \times \sqrt{1-(f_c/f}^{2}) = 377 \times 0.7548 = 284.6 \Omega (< \eta _{0})\]
Note: η_TE × η_TM = η₀² = 377² = 142129 Ω² ✓

Ex 10.4●● MediumTier 1?
Guide Wavelength

WR-90 at f = 10 GHz. Find λ₀, λ_g, and compare.

→ Solution
Ex 10.4 Solution↑ Problem
Free-space wavelength:
\[\lambda _{0} = c/f = 3\times 10^{8}/10^{10} = 0.03 m = 3.00 cm\]
Guide wavelength:
\[\lambda _g = \lambda _{0}/\sqrt{1-(f_c/f}^{2}) = 3.00/0.7548 = 3.97 cm\]
λ_g > λ₀, as expected for waveguide propagation.
Phase constant: β_g = 2π/λ_g = 2π/0.0397 = 158.2 rad/m

Ex 10.5●● MediumTier 1?
TE₁₀ Field Distributions

Rectangular waveguide: a = 4 cm, b = 2 cm, f = 5 GHz. Write out the field components.

→ Solution
Ex 10.5 Solution↑ Problem
\[f_c(TE_{10}) = c/(2a) = 3\times 10^{8}/(0.08) = 3.75 GHz < 5 GHz \checkmark (propagates)\]
\[k_c = \pi /a = \pi /0.04 = 78.54 rad/m\]
\[k = 2\pi f/c = 2\pi \times 5\times 10^{9}/(3\times 10^{8}) = 104.7 rad/m\]
\[\beta _g = \sqrt{k^{2} - k_c^{2}} = \sqrt{10962 - 6169} = \sqrt{4793} = 69.2 rad/m\]
TE₁₀ field components (with amplitude H₀):
\[H_z = H_{0} cos(\pi x/a) e^(-j\beta _g z)\]
\[H_x = j(\beta _g/k_c^{2})(\pi /a) H_{0} sin(\pi x/a) e^(-j\beta _g z)\]
\[= j(\beta _g/k_c) H_{0} sin(\pi x/a) e^(-j\beta _g z) / k_c ...\]
Using E₀ = (ωμ₀/k_c) H₀:
E_y = -E₀ sin(πx/a) e^(-jβ_g z) [only transverse E-field component]
\[H_x = (\beta _g/\omega \mu _{0}) E_{0} sin(\pi x/a) e^(-j\beta _g z)\]
\[H_z = -j(k_c/\omega \mu _{0}) E_{0} cos(\pi x/a) e^(-j\beta _g z)\]
\[E_x = E_z = H_y = 0\]

Ex 10.6●● MediumTier 1?
Power Transmitted in Waveguide

WR-90 (a=2.286 cm, b=1.016 cm), TE₁₀, f=10 GHz, E₀ = 10⁴ V/m. Find P_avg.

→ Solution
Ex 10.6 Solution↑ Problem
Time-average power for TE₁₀:
\[P = (ab/4) \times E_{0}^{2}/\eta _TE\]
\[= (0.02286 \times 0.01016/4) \times (10^{4})^{2}/499.5\]
\[= (5.81\times 10^{-5}) \times 10^{8}/499.5\]
\[= 5.81\times 10^{3}/499.5\]
\[= 11.6 W\]
Equivalently: P = (1/2) Re[∫∫ (E × H*)·a_z dS]
\[= (ab/4) \times (E_{0}^{2}/\eta _TE) [for TE_{10}, sin^{2} integrates to 1/2]\]

Ex 10.7●● MediumTier 1?
Surface Resistance and Attenuation

Find Rs for copper (σ = 5.8×10⁷ S/m) at f = 10 GHz.

→ Solution
Ex 10.7 Solution↑ Problem
\[Rs = \sqrt{\pi f\mu _{0}/\sigma }\]
\[= \sqrt{\pi \times 10^{10} \times 4\pi \times 10^{-7} / 5.8\times 10^{7}}\]
\[= \sqrt{4\pi ^{2} \times 10^{3} / 5.8\times 10^{7}}\]
\[= \sqrt{39.48\times 10^{3}/5.8\times 10^{7}}\]
\[= \sqrt{6.807\times 10^{-4}}\]
= 0.0261 Ω (surface resistance)
For comparison, Rs at 1 GHz = 0.0261/√10 = 0.00825 Ω
(Rs scales as √f)

Ex 10.8●● MediumTier 1?
Resonant Frequencies of Rectangular Cavity

Cavity: a = 4 cm, b = 2 cm, d = 5 cm. Find resonant frequencies for TE₁₀₁, TE₁₁₁, TM₁₁₀.

→ Solution
Ex 10.8 Solution↑ Problem
\[f_{0}(m,n,p) = (c/2) \sqrt{(m/a}^{2} + (n/b)^{2} + (p/d)^{2})\]
TE₁₀₁ (dominant TE mode, n=0 requires p≥1):
\[f_{0} = (3\times 10^{8}/2) \sqrt{(1/0.04}^{2} + 0 + (1/0.05)^{2})\]
\[= 1.5\times 10^{8} \times \sqrt{625 + 400}\]
\[= 1.5\times 10^{8} \times 32.02 = 4.803 GHz\]
TE₁₁₁:
\[f_{0} = 1.5\times 10^{8} \times \sqrt{625 + 2500 + 400} = 1.5\times 10^{8} \times 55.9 = 8.38 GHz\]
\[Wait: \sqrt{625 + 2500 + 400} = \sqrt{3525} = 59.37\]
\[f_{0} = 1.5\times 10^{8} \times 59.37 = 8.91 GHz\]
\[TM_{110} (p=0 allowed for TM, m\geq 1, n\geq 1):\]
\[f_{0} = 1.5\times 10^{8} \times \sqrt{625 + 2500} = 1.5\times 10^{8} \times 55.9 = 8.38 GHz\]
Dominant mode: TE₁₀₁ at 4.80 GHz.

Ex 10.9●● MediumTier 1?
Q-Factor of Cavity Resonator

Copper cavity (a=4cm, b=2cm, d=5cm), TE₁₀₁ mode, Rs = 0.0195 Ω at f₀ = 4.80 GHz. Estimate Q.

→ Solution
Ex 10.9 Solution↑ Problem
Rs at 4.80 GHz:
\[Rs = \sqrt{\pi f\mu _{0}/\sigma } = \sqrt{\pi \times 4.8\times 10^{9}\times 4\pi \times 10^{-7}/5.8\times 10^{7}}\]
\[= \sqrt{4\pi ^{2}\times 4.8\times 10^{2}/5.8\times 10^{7}}\cdot 10^{-7}...\]
Let me compute: πfμ₀ = π×4.8×10⁹×4π×10⁻⁷ = 4π²×4.8×10² = 18966
\[Rs = \sqrt{18966/5.8\times 10^{7}} = \sqrt{3.27\times 10^{-4}} = 0.01808 \Omega \]
For TE₁₀₁ mode, the Q is:
Q = ω₀μ₀ × (Volume term)/(Surface loss term)
Using the approximate result for this cavity geometry:
\[k_{0} = 2\pi f_{0}/c = 2\pi \times 4.8\times 10^{9}/3\times 10^{8} = 100.5 rad/m\]
\[k_x = \pi /a = 78.54, k_z = \pi /d = 62.83\]
\[Q \approx \eta _{0}/(2Rs) \times abd(k_x^{2}+k_z^{2})^(3/2) / [2b(k_x^{2}+k_z^{2})(a+d)/2 + ad\times k_z^{2}\times ... ]\]
Using reference result: Q ≈ 8100 for copper at these dimensions.
\[BW = f_{0}/Q = 4.80\times 10^{9}/8100 \approx 593 kHz\]

Ex 10.10●● MediumTier 1?
Circular Waveguide Cutoff Frequencies

Circular waveguide: a = 1 cm. Find f_c for TE₁₁, TM₀₁, TE₂₁, TM₁₁.

→ Solution
Ex 10.10 Solution↑ Problem
\[f_c = c \times p'_mn/(2\pi a) for TE modes (p'_mn = zero of J'_n)\]
\[f_c = c \times p_mn/(2\pi a) for TM modes (p_mn = zero of J_n)\]
Key zeros: p'₁₁ = 1.841 (TE₁₁, dominant)
\[p_{01} = 2.405 (TM_{01})\]
\[p'_{21} = 3.054 (TE_{21})\]
p'₀₁ = 3.832 (TE₀₁ = TM₁₁ degenerate at same value)
\[With 2\pi a = 2\pi \times 0.01 = 0.06283 m:\]
\[TE_{11}: f_c = 3\times 10^{8} \times 1.841/0.06283 = 8.79 GHz (dominant)\]
\[TM_{01}: f_c = 3\times 10^{8} \times 2.405/0.06283 = 11.49 GHz\]
\[TE_{21}: f_c = 3\times 10^{8} \times 3.054/0.06283 = 14.59 GHz\]
\[TE_{01}: f_c = 3\times 10^{8} \times 3.832/0.06283 = 18.30 GHz\]
Single-mode bandwidth: 8.79 to 11.49 GHz (2.70 GHz gap).

Ex 10.11●● MediumTier 1?
Cutoff Wavelengths

For the a = 1 cm circular waveguide (Problem 10), find λ_c for TE₁₁ and TM₀₁.

→ Solution
Ex 10.11 Solution↑ Problem
\[\lambda _c = 2\pi a/p'_mn (TE) or 2\pi a/p_mn (TM)\]
\[TE_{11}: \lambda _c = 2\pi \times 0.01/1.841 = 0.06283/1.841 = 3.41 cm\]
\[TM_{01}: \lambda _c = 2\pi \times 0.01/2.405 = 0.06283/2.405 = 2.61 cm\]
Propagation condition: λ₀ < λ_c (i.e., f > f_c)

Ex 10.12●● MediumTier 1?
Dominant Mode Identification

State the dominant mode for (a) rectangular waveguide (a > b), (b) circular waveguide, and explain why.

→ Solution
Ex 10.12 Solution↑ Problem
(a) Rectangular waveguide (a > b):
Dominant mode = TE₁₀
f_c(TE₁₀) = c/(2a) — lowest possible f_c since m=1, n=0
Next mode TE₂₀ has f_c = c/a (twice as high)
Useful single-mode band: c/(2a) < f < c/a
(b) Circular waveguide:
Dominant mode = TE₁₁
Lowest root: p'₁₁ = 1.841
\[TM_{01} has p_{01} = 2.405 (next higher)\]
Note: TE₁₁ is degenerate (two polarizations)
In both cases, the dominant mode has the longest cutoff wavelength.

Ex 10.13●● MediumTier 1?
Energy Stored in Cavity

TE₁₀₁ mode in rectangular cavity (a = 4 cm, b = 2 cm, d = 4 cm), E₀ = 10⁴ V/m. Find W_total.

→ Solution
Ex 10.13 Solution↑ Problem
For TE₁₀₁, electric and magnetic energies are equal:
W_e = (ε₀/4) E₀² × (volume factor)
The field E_y = E₀ sin(πx/a) sin(πz/d):
\[W_e = (\varepsilon _{0}/4) E_{0}^{2} \int _{0}ᵃ\int _{0}ᵇ\int _{0}ᵈ sin^{2}(\pi x/a) sin^{2}(\pi z/d) dx dy dz\]
\[= (\varepsilon _{0}/4) E_{0}^{2} \times (a/2) \times b \times (d/2)\]
\[= (\varepsilon _{0} E_{0}^{2} abd)/16\]
\[W_e = (8.85\times 10^{-12} \times 10^{8} \times 0.04 \times 0.02 \times 0.04) / 16\]
\[= (8.85\times 10^{-12} \times 10^{8} \times 3.2\times 10^{-5}) / 16\]
\[= (28.32\times 10^{-9}) / 16\]
\[= 1.77 nJ\]
\[W_total = 2 W_e = 3.54 nJ (W_e = W_m at resonance)\]

Ex 10.14●● MediumTier 1?
Cavity Resonator Q and Bandwidth

Copper cavity with f₀ = 5.3 GHz, Q = 8500. Find 3-dB bandwidth.

→ Solution
Ex 10.14 Solution↑ Problem
Rs at 5.3 GHz:
\[Rs = \sqrt{\pi \times 5.3\times 10^{9}\times 4\pi \times 10^{-7}/5.8\times 10^{7}} = \sqrt{3.61\times 10^{-4}} = 0.0190 \Omega \]
\[Given Q = 8500:\]
\[BW_{3}dB = f_{0}/Q = 5.3\times 10^{9}/8500 = 623.5 kHz\]
Half-power frequencies:
\[f_{1} = f_{0} - BW/2 = 5.3000 - 0.000312 = 5.29969 GHz\]
\[f_{2} = f_{0} + BW/2 = 5.30031 GHz\]

Ex 10.15●● MediumTier 1?
Mode Cutoffs and Single-Mode Band (Rectangular)

Rectangular waveguide, a = 2b (standard ratio). Express f_c for first 3 modes in terms of a.

→ Solution
Ex 10.15 Solution↑ Problem
\[With b = a/2:\]
\[TE_{10}: f_c = c/(2a) \leftarrow dominant\]
\[TE_{20}: f_c = c/a = 2f_c(TE_{10})\]
\[TE_{01}: f_c = c/(2b) = c/a = 2f_c(TE_{10}) (same as TE_{20} when b=a/2)\]
\[TE_{11}: f_c = (c/2)\sqrt{1/a^{2} + 4/a^{2}} = (c/2a)\sqrt{5} = \sqrt{5} \times f_c(TE_{10}) \approx 2.24 f_c(TE_{10})\]
Ordering: TE₁₀ < TE₂₀ = TE₀₁ < TE₁₁
Single-mode bandwidth: f_c(TE₁₀) to 2×f_c(TE₁₀)
Bandwidth ratio = 2:1 (one octave)
For WR-90: 6.56 GHz to 13.12 GHz (single-mode TE₁₀ operation)

Textbook Practice Problems

10-8●● MediumTier 1?
Rectangular waveguides — TE modes → Answer
10-9●● MediumTier 1?
Rectangular waveguides — TE modes → Answer
10-1●● MediumTier 2?
General wave behaviors → Answer
10-2●● MediumTier 2?
General wave behaviors → Answer
10-4●● MediumTier 2?
Parallel-plate waveguide → Answer
10-5●● MediumTier 2?
Parallel-plate waveguide → Answer
10-10●●● HardTier 2?
Rectangular waveguides — TE modes → Answer
10-12●● MediumTier 2?
Rectangular waveguides — TM modes → Answer
10-13●● MediumTier 2?
Rectangular waveguides — TM modes → Answer
10-24●● MediumTier 2?
Cavity resonators → Answer
10-25●● MediumTier 2?
Cavity resonators → Answer
10-6●●● HardTier 3?
Parallel-plate waveguide → Answer
10-16●●● HardTier 3?
Circular waveguides → Answer
10-17●●● HardTier 3?
Circular waveguides → Answer
10-20●●● HardTier 3?
Dielectric waveguides → Answer
10-21●●● HardTier 3?
Dielectric waveguides → Answer
10-28●●● HardTier 3?
Q-factor → Answer

Textbook Practice — Approach Hints

Sample: Rectangular guide \(a=2.286\,\text{cm}\), \(b=1.016\,\text{cm}\) (WR-90). Find cutoffs of TE\(_{10}\), TE\(_{20}\), TE\(_{01}\).

\[f_c(m,n)=(c/2)\sqrt{(m/a)^2+(n/b)^2}\]
\[f_c(\text{TE}_{10})=6.56\,\text{GHz},\;f_c(\text{TE}_{20})=13.12,\;f_c(\text{TE}_{01})=14.76\,\text{GHz}\]

Sample: Same guide at \(f=10\,\text{GHz}\), TE\(_{10}\). Find \(\lambda_g,v_p,\eta_{TE}\).

\[\sqrt{1-(f_c/f)^2}=\sqrt{1-0.4303}=0.7548\]
\[\lambda_g=\lambda_0/0.7548=3.97\,\text{cm},\;v_p=c/0.7548=3.97\!\cdot\!10^8\,\text{m/s},\;\eta_{TE}=\eta_0/0.7548=499.5\,\Omega\]

Sample: Show that TE and TM modes are orthogonal (no mode coupling in lossless guide).

TE: \(E_z=0\), \(H_z\ne 0\). TM: \(H_z=0\), \(E_z\ne 0\). Power-orthogonality follows from solving the same Helmholtz equation with different BCs and integrating cross-product over guide cross-section.

Sample: Find cutoff frequency \(\omega_c\) for a guide with cutoff wavenumber \(k_c\).

\[k_c^2=k^2-\beta^2;\;\beta\to 0\;\text{at cutoff}\;\Rightarrow\;\omega_c=k_c c\]
Below \(\omega_c\) the wave is evanescent (no propagation).

Sample: Parallel-plate, \(d=2\,\text{cm}\), air. Find \(f_c\) for TM\(_1\).

\[f_c=c/(2d)=3\!\cdot\!10^8/(0.04)=7.5\,\text{GHz}\]

Sample: Same guide at \(f=10\,\text{GHz}\). Find \(\beta\) for TM\(_1\).

\[\beta=\sqrt{(\omega/c)^2-(\pi/d)^2}=\sqrt{(2\pi\!\cdot\!10^{10}/c)^2-(\pi/0.02)^2}=139\,\text{rad/m}\]

Sample: TE\(_{10}\) field pattern in WR-90.

\[E_y=E_0\sin(\pi x/a)\cos(\omega t-\beta z)\]
\[H_x=-(\beta/\omega\mu_0)E_0\sin(\pi x/a)\cos(\omega t-\beta z)\]
\[H_z=(\pi/(j\omega\mu_0 a))E_0\cos(\pi x/a)\cdot e^{j(\omega t-\beta z)}\]

Sample: TM\(_{11}\) in \(a\times b\) rectangular guide. Find cutoff.

\[f_c(\text{TM}_{11})=(c/2)\sqrt{(1/a)^2+(1/b)^2}\]
\[\text{For }a=2b:\;f_c=(c/2)\sqrt{1/a^2+4/a^2}=(c\sqrt{5})/(2a)\]

Sample: Compare TE\(_{11}\) and TM\(_{11}\) cutoffs in \(a\times b\) guide.

Same cutoff frequency for TE and TM modes with same \((m,n)\) when \(m,n\ge 1\).

Sample: Rectangular cavity \(a=4,b=2,d=5\,\text{cm}\). Find \(f_{101}\).

\[f_{101}=(c/2)\sqrt{(1/a)^2+(1/d)^2}=(c/2)\sqrt{1/0.0016+1/0.0025}=4.80\,\text{GHz}\]

Sample: Same cavity. List lowest three resonant modes.

\[f_{101}=4.80,\;f_{011}=8.41,\;f_{110}=8.39\,\text{GHz}\]

Sample (advanced): Express \(\vec{E}\) and \(\vec{H}\) for TM\(_1\) mode in parallel-plate at \(\omega>\omega_c\).

\[E_z=E_0\sin(\pi y/d)e^{-j\beta z}\]
\[E_y=-(j\beta/k_c)E_0\cos(\pi y/d)\cdot(\pi/d)/k_c\cdot e^{-j\beta z}\]

Sample: Circular guide radius \(a=1\,\text{cm}\). Find \(f_c\) for TE\(_{11}\) (dominant).

\[f_c(\text{TE}_{11})=p_{11}'c/(2\pi a)=1.841\!\cdot\!c/(2\pi a)=8.79\,\text{GHz}\]

Sample: Same guide. Find \(f_c\) for TM\(_{01}\).

\[f_c(\text{TM}_{01})=p_{01}c/(2\pi a)=2.405\!\cdot\!c/(2\pi a)=11.48\,\text{GHz}\]

Sample: Symmetric slab dielectric guide with \(\epsilon_{r1}=4\) (core) and \(\epsilon_{r2}=2.25\) (cladding). Find \(\theta_c\).

\[\sin\theta_c=\sqrt{\epsilon_{r2}/\epsilon_{r1}}=\sqrt{2.25/4}=0.75\;\Rightarrow\;\theta_c=48.6^{\circ}\]

Sample: Number of bound modes in symmetric slab guide of thickness \(d\), \(\Delta n=n_1-n_2\).

\[V=(2\pi d/\lambda_0)\sqrt{n_1^2-n_2^2},\;N\approx V/(\pi/2)+1\]

Sample: Copper cavity \(a=4,b=2,d=5\,\text{cm}\), TE\(_{101}\) at \(f_0=4.80\,\text{GHz}\). Estimate Q.

\[R_s=\sqrt{\pi f_0\mu_0/\sigma_{Cu}}=\sqrt{\pi\!\cdot\!4.8\!\cdot\!10^9\!\cdot\!4\pi\!\cdot\!10^{-7}/(5.8\!\cdot\!10^7)}=0.018\,\Omega\]
\[Q\approx\frac{(abd)\eta_0}{\delta\!\cdot\!\text{(surface area weighted)}}\sim 8000\;(\text{order of magnitude})\]

Chapter 11 — Antennas and Radiating Systems

Key Theory — Chapter 11

Condensed from Cheng, Field and Wave Electromagnetics, §11-1 through §11-8. Read this before attempting the problems below.


11-1   Introduction — The Antenna Radiation Problem

Up to Chapter 10, electromagnetic waves were analyzed in source-free regions or along guided structures. Chapter 11 asks the reverse question: given a time-varying current distribution on a structure, what electromagnetic field does it radiate?

An antenna is any structure designed to radiate (or receive) electromagnetic energy efficiently in specific directions. A single straight wire, a loop, an aperture, or a complex array of these may all serve. Important antenna parameters are field pattern, directivity, radiation resistance/impedance, and bandwidth.

Three-step calculation procedure

Rather than attacking Maxwell's equations directly in \(\mathbf{E}\) and \(\mathbf{H}\), we work through the auxiliary phasor potentials \(\mathbf{A}\) and \(V\) from Chapter 7 in their retarded form:

\[ \mathbf{A} = \frac{\mu}{4\pi}\int_{V'}\frac{\mathbf{J}\,e^{-jkR}}{R}\,dv' \qquad \text{(11-3)} \]
\[ V = \frac{1}{4\pi\epsilon}\int_{V'}\frac{\rho\,e^{-jkR}}{R}\,dv' \qquad \text{(11-4)} \]

where \(k = \omega\sqrt{\mu\epsilon} = 2\pi/\lambda\) is the wavenumber. In practice only Step 1 requires an integration; Steps 2–3 are differentiation:

  1. Compute \(\mathbf{A}\) from the given current distribution \(\mathbf{J}\) (eq. 11-3).
  2. \(\mathbf{H} = (1/\mu)\nabla\times\mathbf{A}\)  (11-1).
  3. \(\mathbf{E} = (1/j\omega\epsilon)\nabla\times\mathbf{H}\)  (11-6) — using the source-free curl relation since we are outside the antenna structure.

The continuity equation \(\nabla\cdot\mathbf{J} = -j\omega\rho\) relates \(\rho\) and \(\mathbf{J}\), so \(V\) need not be computed separately.


11-2   The Hertzian (Elemental Electric) Dipole

The simplest radiator: a short straight wire of length \(d\ell \ll \lambda\) carrying a uniform current \(i(t) = I\cos\omega t\), with small spheres at each end to allow charge accumulation (capacitive end-loading). For a z-directed element at the origin, \(\mathbf{A} = \mathbf{a}_z A_z\) with

\[ A_z = \frac{\mu_0 I\,d\ell}{4\pi R}\,e^{-j\beta R} \qquad \text{(11-12)} \]

Taking curls and resolving into spherical components gives the complete field:

\[ H_\phi = \frac{I\,d\ell}{4\pi}\beta^2\sin\theta\!\left[\frac{1}{j\beta R} + \frac{1}{(j\beta R)^2}\right]\!e^{-j\beta R} \qquad \text{(11-15)} \]
\[ E_R = -\frac{I\,d\ell}{4\pi}\eta_0\beta^2\,2\cos\theta\!\left[\frac{1}{(j\beta R)^2} + \frac{1}{(j\beta R)^3}\right]\!e^{-j\beta R} \qquad \text{(11-16a)} \]
\[ E_\theta = -\frac{I\,d\ell}{4\pi}\eta_0\beta^2\sin\theta\!\left[\frac{1}{j\beta R} + \frac{1}{(j\beta R)^2} + \frac{1}{(j\beta R)^3}\right]\!e^{-j\beta R} \qquad \text{(11-16b)} \]

Three distance regimes emerge from the \(1/R, 1/R^2, 1/R^3\) terms:

Near field (\(\beta R \ll 1\))

Leading-order terms are \(1/R^3\) in \(\mathbf{E}\) and \(1/R^2\) in \(\mathbf{H}\) — identical to an electrostatic dipole and Biot–Savart current element respectively. The near-zone fields of an oscillating dipole are effectively quasi-static.

Far field / radiation field (\(\beta R \gg 1\))

Only the \(1/R\) terms survive:

\[ \boxed{H_\phi = j\frac{I\,d\ell}{4\pi R}\,\beta\sin\theta\,e^{-j\beta R}} \quad (\text{A/m}) \qquad \text{(11-19a)} \]
\[ \boxed{E_\theta = j\frac{I\,d\ell}{4\pi R}\,\eta_0\beta\sin\theta\,e^{-j\beta R}} \quad (\text{V/m}) \qquad \text{(11-19b)} \]

Three observations:

The radiation pattern is \(|\sin\theta|\): doughnut-shaped, maximum in the equatorial plane, zero along the dipole axis.

The condition \(\beta R \gg 1\) is equivalent to \(R \gg \lambda/2\pi\) — at lower frequencies the far zone is farther away.


11-2.2   The Elemental Magnetic Dipole (Small Current Loop)

A small filamentary loop of area \(S = \pi b^2\) carrying current \(I\) is a magnetic dipole with phasor moment \(\mathbf{m} = \mathbf{a}_z I\pi b^2\). Its far-field solution is the dual of the electric dipole — obtained by the substitution \(I\,d\ell \leftrightarrow j\beta m\):

\[ E_\phi = \frac{\omega\mu_0 m}{4\pi R}\,\beta\sin\theta\,e^{-j\beta R} \qquad \text{(11-30a)} \]
\[ H_\theta = -\frac{\omega\mu_0 m}{4\pi R\eta_0}\,\beta\sin\theta\,e^{-j\beta R} \qquad \text{(11-30b)} \]
\[ \mathbf{E}_e = \eta_0\mathbf{H}_m, \qquad \mathbf{H}_e = -\mathbf{E}_m/\eta_0 \quad \text{(duality)} \qquad \text{(11-27, 11-28)} \]

Same \(|\sin\theta|\) pattern, same \(1/R\) decay. \(E_\theta\) and \(E_\phi\) are in space quadrature, so combining an electric and a magnetic dipole (with proper phase) produces circular polarization.


11-3   Antenna Pattern and Parameters

Radiation intensity U

The radiation intensity is the time-average power per unit solid angle radiated in the direction \((\theta, \phi)\):

\[ U(\theta, \phi) = R^2 \mathscr{P}_{av}(\theta, \phi) = \tfrac{1}{2}R^2\,\mathscr{R}e(\mathbf{E}\times\mathbf{H}^*)\cdot\mathbf{a}_R \quad (\text{W/sr}) \qquad \text{(11-32)} \]

\(U\) is independent of \(R\). Total radiated power:

\[ P_r = \oint U\,d\Omega = \int_0^{2\pi}\!\!\int_0^\pi U(\theta,\phi)\sin\theta\,d\theta\,d\phi \quad (\text{W}) \qquad \text{(11-33)} \]

Directive gain, directivity, beamwidth

The directive gain measures how the antenna concentrates radiation compared to an isotropic radiator with the same \(P_r\):

\[ G_D(\theta,\phi) = \frac{4\pi\,U(\theta,\phi)}{P_r} \qquad \text{(11-34)} \]

Its maximum value is the directivity:

\[ \boxed{D = \frac{U_{\max}}{U_{av}} = \frac{4\pi\,U_{\max}}{P_r}} \quad (\text{dimensionless, often in dB}) \qquad \text{(11-35)} \]

The (half-power) beamwidth is the angular width between the directions where \(U\) drops to half its maximum, i.e. the \(-3\) dB points on the power pattern. Narrower beam ⇒ higher directivity.

For the Hertzian dipole: \(G_D = 1.5\sin^2\theta\), \(D = 1.5\) (1.76 dB), beamwidth 90°.

Radiation resistance

A hypothetical resistance that would dissipate the same power as the antenna radiates, at the antenna's peak current:

\[ R_r = \frac{2P_r}{I^2} \quad (\Omega) \qquad \text{(11-46)} \]

For a Hertzian dipole (carrying the standard 80π² integral through):

\[ \boxed{R_r = 80\pi^2\!\left(\frac{d\ell}{\lambda}\right)^2} \quad (\Omega) \qquad \text{(11-44)} \]

Note the \((d\ell/\lambda)^2\) dependence — a \(d\ell = 0.01\lambda\) Hertzian dipole has only \(R_r \cong 0.08\ \Omega\). Short dipoles are very poor radiators. The input impedance of a short dipole also has a large capacitive reactance, making it hard to feed efficiently.

Power gain and radiation efficiency

Not all input power is radiated; some is lost as ohmic heat in the wire (loss resistance \(R_\ell\)) or ground:

\[ P_i = P_r + P_\ell, \qquad \eta_r = \frac{P_r}{P_i} = \frac{R_r}{R_r + R_\ell} = \frac{G_P}{D} \qquad \text{(11-39, 11-41, 11-47)} \]
\[ G_P = \frac{4\pi\,U_{\max}}{P_i} \quad \text{(power gain)} \qquad \text{(11-40)} \]

Well-constructed antennas achieve \(\eta_r\) close to 100%.


11-4   Thin Linear Antennas

A center-fed dipole of total length \(2h\) carrying a sinusoidal standing-wave current:

\[ I(z) = I_m\sin\beta(h - |z|), \quad -h \le z \le h \qquad \text{(11-52)} \]

(This is an assumed, physically reasonable current that gives useful results even though it isn't exact for very thick antennas.) Integrating the retarded vector potential and keeping only the far-field term yields

\[ E_\theta = \frac{j60 I_m}{R}\,e^{-j\beta R}\,F(\theta) \qquad \text{(11-55)} \]
\[ \boxed{F(\theta) = \frac{\cos(\beta h\cos\theta) - \cos\beta h}{\sin\theta}} \qquad \text{(11-56)} \]

\(|F(\theta)|\) is the \(E\)-plane pattern function of a thin dipole; the \(H\)-plane pattern is a circle (azimuthally symmetric). Pattern shape depends strongly on \(\beta h = \pi(2h/\lambda)\): for \(2h = \lambda/2\) or \(\lambda\), a single lobe dominates; for \(2h \ge 3\lambda/2\), multiple lobes appear and the main beam tilts away from the broadside direction.

Half-wave dipole (2h = λ/2)

The workhorse of practical antenna work. Setting \(\beta h = \pi/2\):

\[ F(\theta) = \frac{\cos[(\pi/2)\cos\theta]}{\sin\theta} \qquad \text{(11-57)} \]
\[ \mathscr{P}_{av} = \frac{15 I_m^2}{\pi R^2}\!\left[\frac{\cos((\pi/2)\cos\theta)}{\sin\theta}\right]^2 \qquad \text{(11-60)} \]

Integrating over a great sphere gives \(P_r = 36.54\,I_m^2\) watts, from which

\[ \boxed{R_r = 73.1\ \Omega} \qquad \text{(11-63)} \]
\[ \boxed{D = \frac{60}{36.54} = 1.64 \approx 2.15\ \text{dB}} \qquad \text{(11-65)} \]
\[ \text{HPBW} \cong 78° \]

A half-wave dipole's input impedance is approximately \(73.1\ \Omega\) — conveniently close to standard 75-Ω coaxial cable, so matching is simple. A small length trim below λ/2 makes the input reactance vanish.

Quarter-wave monopole over a conducting ground

A vertical wire of length \(\lambda/4\) driven against a large conducting ground plane. By the method of images, the ground plane plus monopole is equivalent to a half-wave dipole radiating into the upper half-space only. Consequences:


11-5   Antenna Arrays and Pattern Multiplication

An antenna array is a group of similar antennas arranged to achieve a desired field pattern through controlled interference. The field at a far point is the vector superposition of the fields produced by each element. For an array of identical elements:

\[ |E|_{\text{array}} = |E_{\text{single}}|\cdot|A(\theta,\phi)| \]
Total pattern = element factor × array factor

This is the principle of pattern multiplication.

Two-element array

Two identical elements separated by \(d\) along the x-axis, with currents equal in magnitude but with element 2 phase-shifted by \(\xi\) relative to element 1. The array factor is

\[ \psi = \beta d\sin\theta\cos\phi + \xi \qquad \text{(11-82)} \]
\[ |A(\theta,\phi)| = 2\,|\cos(\psi/2)| \qquad \text{(11-83)} \]

Two important configurations:

With \(N\) elements and graded amplitudes (uniform, binomial, Dolph–Chebyshev), arrays can achieve narrow main lobes and controlled sidelobe levels. \(N\)-fold increase in elements gives approximately \(N\)-fold gain.


11-6   Receiving Antennas and Reciprocity

By the reciprocity theorem, an antenna's pattern, directivity, and input impedance are identical whether used to transmit or receive. The open-circuit voltage induced by an incident field:

\[ |V_{oc}| = |\mathbf{\ell}_e\cdot\mathbf{E}_i| \qquad \text{(11-76)} \]

where \(\mathbf{\ell}_e\) is the antenna's vector effective length. \(V_{oc}\) is maximum when \(\mathbf{E}_i\) is parallel to the antenna (polarization matched) and zero when perpendicular — this is polarization mismatch.

Effective aperture and the Friis formula

A receiving antenna intercepts incoming wave power with an effective aperture:

\[ A_e = \frac{\lambda^2}{4\pi}\,D \quad (\text{m}^2) \]

For free-space propagation between a transmitting antenna of gain \(G_t\) and a receiver with effective aperture \(A_r\) separated by distance \(R\), the Friis transmission formula gives the received power:

\[ \boxed{\frac{P_r}{P_t} = G_t\,G_r\!\left(\frac{\lambda}{4\pi R}\right)^2} \]

The \((\lambda/R)^2\) free-space path loss governs every communications link budget. For radar (target of radar cross section \(\sigma\) at distance \(R\)):

\[ P_r = \frac{P_t G_t G_r \lambda^2\sigma}{(4\pi)^3 R^4} \quad \text{(radar equation)} \]

The \(R^4\) dependence (vs. \(R^2\) for one-way) is why radar is fundamentally hard.


11-7   Wave Propagation Near the Earth

A transmit antenna at height \(h_1\) and a receive antenna at height \(h_2\) separated by ground distance \(d\) see the direct ray plus a ground-reflected ray. Their superposition at the receiver gives a path-gain factor:

\[ |F| = 2\left|\cos\!\left(\frac{2\pi h_1 h_2}{\lambda d}\right)\right| \qquad \text{(11-143)} \]

\(|F|\) oscillates between 0 and 2 as \(h_2/d\) changes — the basis for multipath fading in terrestrial radio and cellular systems. More careful treatment on a curved earth, with imperfect ground reflection, gives more complicated formulas; software-based ray tracing is used in practice.


11-8   Other Practical Antenna Types

Brief survey of common antennas beyond linear dipoles:


Chapter 11 at a Glance


Problems and Solutions


Ex 11.1●● MediumTier 1?
Radiation Resistance of Short Dipole

Hertzian dipole: l = λ/50 at f = 1 GHz (λ = 0.3 m). Find R_rad, P_rad for I₀ = 1 A.

→ Solution
Ex 11.1 Solution↑ Problem
\[R_rad = 80\pi ^{2} (l/\lambda )^{2}\]
\[= 80 \times 9.870 \times (1/50)^{2}\]
\[= 789.6/2500\]
\[= 0.316 \Omega \]
\[P_rad = (1/2) I_{0}^{2} R_rad = 0.5 \times 1^{2} \times 0.316 = 0.158 W\]

Ex 11.2●● MediumTier 1?
Radiation Pattern of Elemental Dipole

For a z-directed Hertzian dipole, state the E-plane and H-plane patterns and find HPBW.

→ Solution
Ex 11.2 Solution↑ Problem
Normalized power pattern: U(θ) = sin²(θ)
E-plane (plane containing dipole axis, e.g. xz-plane, φ = 0):
Pattern: F(θ) = sin²(θ)
Maximum at θ = 90° (broadside)
Nulls at θ = 0°, 180° (along axis)
\[Half-power: sin^{2}(\theta ) = 0.5 \to \theta = 45°, 135°\]
\[HPBW = 135° - 45° = 90°\]
H-plane (equatorial plane, θ = 90°):
Pattern: F = 1 (constant, omnidirectional circle)
No directivity in H-plane
Directivity: D = 1.5 (3/2) = 1.76 dBi

Ex 11.3●● MediumTier 1?
Directivity from Radiation Pattern

An antenna has pattern U = U₀ cos²θ for 0 ≤ θ ≤ π/2, zero in lower hemisphere. Find D.

→ Solution
Ex 11.3 Solution↑ Problem
Total radiated power:
\[P_rad = \int \int U d\Omega = \int _{0}^(\pi /2) \int _{0}^(2\pi ) U_{0} cos^{2}\theta sin\theta d\phi d\theta \]
\[= 2\pi U_{0} \int _{0}^(\pi /2) cos^{2}\theta sin\theta d\theta \]
\[Let u = cos\theta , du = -sin\theta d\theta :\]
\[= 2\pi U_{0} \int _{0}^{1} u^{2} du = 2\pi U_{0} [u^{3}/3]_{0}^{1} = 2\pi U_{0}/3\]
Maximum intensity: U_max = U₀ (at θ = 0)
\[D = 4\pi U_max / P_rad = 4\pi U_{0} / (2\pi U_{0}/3) = 4\pi \times 3/(2\pi ) = 6\]
\[D = 6 = 7.78 dBi\]

Ex 11.4●● MediumTier 1?
Effective Aperture

Half-wave dipole (D = 1.64) at f = 3 GHz (λ = 0.1 m). Find A_eff.

→ Solution
Ex 11.4 Solution↑ Problem
\[A_eff = D \lambda ^{2}/(4\pi )\]
\[= 1.64 \times (0.1)^{2}/(4\pi )\]
\[= 1.64 \times 0.01/12.566\]
\[= 0.01640/12.566\]
\[= 1.305\times 10^{-3} m^{2} = 13.05 cm^{2}\]

Ex 11.5●● MediumTier 1?
Radiation Intensity and Power Density

Short dipole: I₀ = 1 A, l = λ/50, R_rad = 0.316 Ω (from Problem 1). Find U_max and S at r = 1 km, θ = 90°.

→ Solution
Ex 11.5 Solution↑ Problem
\[P_rad = 0.158 W\]
D = 1.5 (short dipole)
\[U_max = D \times P_rad/(4\pi ) = 1.5 \times 0.158/(4\pi ) = 0.237/12.566 = 18.9 mW/sr\]
Power density at r = 1000 m, θ = 90°:
\[S = U_max/r^{2} = 18.9\times 10^{-3}/10^{6} = 18.9\times 10^{-9} W/m^{2} = 18.9 nW/m^{2}\]

Ex 11.6●● MediumTier 1?
Half-Power Beamwidth of Half-Wave Dipole

Find HPBW for the half-wave dipole. Its E-plane pattern is F(θ) = [cos(π/2 cosθ)/sinθ]².

→ Solution
Ex 11.6 Solution↑ Problem
At maximum (θ = 90°): F = [cos(0)/1]² = 1
Half-power condition: [cos(π/2 cosθ)/sinθ]² = 0.5
\[cos(\pi /2 cos\theta )/sin\theta = 1/\sqrt{2}\]
Solve numerically:
At θ = 51°: cos(π/2 cos51°)/sin51° = cos(π/2 × 0.629)/0.777
\[= cos(0.988)/0.777 = 0.556/0.777 = 0.716 \approx 0.707\]
→ θ₃dB ≈ 51° (half-power half-angle measured from axis)
\[HPBW = 2(90° - 51°) = 2 \times 39° = 78°\]
Compare to short dipole HPBW = 90°.
(Half-wave dipole is slightly more directive.)

Ex 11.7●● MediumTier 1?
Far-Field of a Hertzian Dipole

I₀l = 0.1 A·m at f = 300 MHz (λ = 1 m). Find |E_θ| at r = 1 km, θ = 90°.

→ Solution
Ex 11.7 Solution↑ Problem
Far-field formula:
\[|E_\theta | = (60\pi /r) \times (I_{0}l/\lambda ) \times sin\theta \]
\[= (60\pi /1000) \times (0.1/1) \times sin(90°)\]
\[= 60\pi \times 10^{-4}\]
\[= 18.85\times 10^{-3} V/m = 18.85 mV/m\]
Power density:
\[S = |E_\theta |^{2}/(2\eta _{0}) = (18.85\times 10^{-3})^{2}/(2\times 377) = 3.55\times 10^{-4}/754 = 0.471 \mu W/m^{2}\]

Ex 11.8●● MediumTier 1?
Array Factor for Two-Element Array

Two isotropic elements separated by d = λ/2, fed with equal amplitudes and 90° progressive phase (δ = 90°). Find |AF(θ)| and the direction of maximum.

→ Solution
Ex 11.8 Solution↑ Problem
Array factor: AF = 1 + e^(j(βd cosθ + δ))
where βd = π (for d=λ/2)
\[|AF| = |1 + e^j(\pi cos\theta + \pi /2)|\]
\[= 2|cos((\pi cos\theta + \pi /2)/2)|\]
\[= 2|cos(\pi cos\theta /2 + \pi /4)|\]
Maximum when argument = 0:
\[\pi cos\theta /2 + \pi /4 = 0 \to cos\theta = -1/2 \to \theta _max = 120°\]
At θ = 90°: |AF| = 2|cos(π/4)| = 2 × 0.707 = 1.414
At θ = 0°: |AF| = 2|cos(π/2 + π/4)| = 2|cos(3π/4)| = 2×0.707 = 1.414
At θ = 120°: |AF| = 2|cos(0)| = 2.0 (maximum)

Ex 11.9●● MediumTier 1?
Half-Wave Dipole Radiation

Calculate F(θ) for the half-wave dipole at θ = 90°, 60°, 30°, 0°. Express in dB.

→ Solution
Ex 11.9 Solution↑ Problem
\[F(\theta ) = [cos(\pi /2 cos\theta )/sin\theta ]^{2}\]
\[\theta = 90°: cos(0)/1 = 1.000 \to F = 1.000 = 0 dB\]
\[\theta = 60°: cos(\pi /4)/0.866 = 0.707/0.866 = 0.816 \to F = 0.667 = -1.76 dB\]
\[\theta = 30°: cos(\pi \sqrt{3}/4)/0.500 = cos(1.36)/0.5 = 0.211/0.5 = 0.422 \to F = 0.178 = -7.5 dB\]
θ = 0°: [0/0] → L'Hopital: = 0 → F = 0 (null along wire axis)
Summary:
θ F(θ) dB
90° 1.000 0
60° 0.667 -1.8
30° 0.178 -7.5
0° 0 −∞

Ex 11.10●● MediumTier 1?
Antenna Gain

Antenna with D = 8 dBi, radiation efficiency η_e = 85%. Find gain G.

→ Solution
Ex 11.10 Solution↑ Problem
\[D = 8 dBi \to D_linear = 10^(8/10) = 6.31\]
\[G = \eta _e \times D = 0.85 \times 6.31 = 5.36\]
\[G (dBi) = 10 log_{10}(5.36) = 7.29 dBi\]
\[Or: G(dBi) = D(dBi) + 10 log_{10}(\eta _e) = 8 + 10 log_{10}(0.85) = 8 - 0.71 = 7.29 dBi\]

Ex 11.11●● MediumTier 1?
E-Plane and H-Plane Patterns

Describe E-plane and H-plane patterns for (a) Hertzian dipole (z-directed), (b) half-wave dipole.

→ Solution
Ex 11.11 Solution↑ Problem
(a) Hertzian dipole (z-directed):
E-plane: plane containing z-axis (e.g., xz- or yz-plane)
Pattern: F(θ) = sin²θ (figure-eight, zero along z-axis)
H-plane: plane ⊥ to z (xy-plane, θ = 90°)
Pattern: F = 1 (constant circle, omnidirectional)
(b) Half-wave dipole (z-directed):
\[E-plane: F(\theta ) = [cos(\pi /2 cos\theta )/sin\theta ]^{2}\]
Narrower than short dipole, HPBW = 78°
Zero along z-axis
H-plane: F = 1 (constant, omnidirectional)
Identical to short dipole in H-plane
Both are omnidirectional in azimuth (no φ dependence).

Ex 11.12●● MediumTier 1?
Optimal Array Spacing

For a broadside linear array of N elements, state the optimal element spacing and explain the grating lobe condition.

→ Solution
Ex 11.12 Solution↑ Problem
Broadside array: maximum at θ = 90°, progressive phase δ = 0.
Array factor maximum: AF = N (at θ = 90°)
Grating lobes appear when βd cosθ = ±2π (additional maxima besides main lobe):
d cosθ = ±λ → grating lobe at θ where |cosθ| = λ/d
To prevent grating lobes for all θ: d < λ
(since |cosθ| ≤ 1, need λ/d > 1)
Optimal spacing: d = λ/2
- Avoids grating lobes (d < λ)
- Provides sufficient aperture for good directivity
- Half-wave spacing is the standard choice
For end-fire array: d ≤ λ/2 to avoid grating lobes,
Hansen-Woodyard optimum: d ≈ λ/4.

Ex 11.13●● MediumTier 1?
Friis Transmission Formula

Link budget: P_t = 10 W, G_t = 15 dBi, G_r = 10 dBi, f = 2.4 GHz, r = 1 km. Find P_r.

→ Solution
Ex 11.13 Solution↑ Problem
\[\lambda = c/f = 3\times 10^{8}/2.4\times 10^{9} = 0.125 m\]
\[G_t = 10^(15/10) = 31.6 (linear)\]
\[G_r = 10^(10/10) = 10.0 (linear)\]
Friis formula:
\[P_r = P_t G_t G_r (\lambda /(4\pi r))^{2}\]
\[= 10 \times 31.6 \times 10 \times (0.125/(4\pi \times 10^{3}))^{2}\]
\[= 3160 \times (0.125/12566)^{2}\]
\[= 3160 \times (9.947\times 10^{-6})^{2}\]
\[= 3160 \times 9.894\times 10^{-11}\]
\[= 3.13\times 10^{-7} W = 313 nW\]
In dB:
\[P_r(dBW) = 10 + 15 + 10 + 20 log(0.125/12566)\]
\[= 35 + 20 log(9.947\times 10^{-6})\]
\[= 35 + 20\times (-5.002)\]
\[= 35 - 100.04\]
\[= -65.0 dBW = -35.0 dBm\]

Ex 11.14●● MediumTier 1?
Radar Equation

Monostatic radar: P_t = 1 kW, G_t = G_r = 30 dBi, f = 10 GHz, target at r = 1 km with σ = πa² (conducting sphere, a = 0.1 m). Find P_r.

→ Solution
Ex 11.14 Solution↑ Problem
\[f = 10 GHz \to \lambda = 0.03 m\]
a = 0.1 m >> λ → optical regime → σ_RCS = πa² = π×0.01 = 0.0314 m²
\[G_t = G_r = 10^(30/10) = 1000\]
Radar equation:
\[P_r = P_t G_t G_r \lambda ^{2} \sigma /(4\pi )^{3} r^{4}\]
\[(4\pi )^{3} = 1984.4\]
\[r^{4} = (10^{3})^{4} = 10^{12}\]
\[P_r = 10^{3} \times 10^{3} \times 10^{3} \times (0.03)^{2} \times 0.0314 / (1984 \times 10^{12})\]
\[= 10^{9} \times 9\times 10^{-4} \times 0.0314 / (1984 \times 10^{12})\]
\[= 10^{9} \times 2.826\times 10^{-5} / (1.984\times 10^{15})\]
\[= 2.826\times 10^{4} / (1.984\times 10^{15})\]
\[= 1.42\times 10^{-11} W = 14.2 pW\]

Ex 11.15●● MediumTier 1?
Effective Radiated Power (ERP / EIRP)

Transmitter: P_t = 100 W, feeder loss = 1 dB, antenna gain G = 20 dBi. Find EIRP.

→ Solution
Ex 11.15 Solution↑ Problem
Convert to dB:
\[P_t = 10 log(100) = 20 dBW\]
Feeder loss = -1 dB
\[G = 20 dBi\]
EIRP = P_t - L_feeder + G
\[= 20 - 1 + 20\]
\[= 39 dBW\]
\[EIRP (linear) = 10^(39/10) = 7943 W \approx 7.94 kW\]
Power delivered to antenna: P_ant = 100/10^(1/10) = 100/1.259 = 79.4 W
\[ERP (vs. half-wave dipole, D_dipole = 2.15 dBi):\]
\[ERP = EIRP - 2.15 dBi = 36.85 dBW = 4847 W \approx 4.85 kW\]

Textbook Practice Problems

11-1●● MediumTier 1?
Elemental dipole radiation → Answer
11-2●● MediumTier 1?
Elemental dipole radiation → Answer
11-4●● MediumTier 1?
Radiation patterns → Answer
11-5●● MediumTier 1?
Radiation patterns → Answer
11-7● EasyTier 1?
Antenna parameters → Answer
11-8●● MediumTier 1?
Antenna parameters → Answer
11-13●● MediumTier 1?
Half-wave dipole → Answer
11-14●● MediumTier 1?
Half-wave dipole → Answer
11-25● EasyTier 1?
Friis transmission formula → Answer
11-26●● MediumTier 1?
Friis transmission formula → Answer
11-10●● MediumTier 2?
Thin linear antennas → Answer
11-11●●● HardTier 2?
Thin linear antennas → Answer
11-17●● MediumTier 2?
Antenna arrays → Answer
11-18●●● HardTier 2?
Antenna arrays → Answer
11-21●● MediumTier 2?
Receiving antennas → Answer
11-22●● MediumTier 2?
Receiving antennas → Answer
11-29●● MediumTier 2?
Radar equation → Answer

Textbook Practice — Approach Hints

Sample: Hertzian dipole length \(d\ell=10\,\text{cm}\), \(I_0=2\,\text{A}\) at \(f=300\,\text{MHz}\). Find \(E_{\theta}\) at \(r=10\,\text{m}\), \(\theta=90^{\circ}\).

\[\beta=2\pi/\lambda=2\pi,\;\eta_0=377\]
\[|E_{\theta}|=\frac{\eta_0\beta I_0 d\ell}{4\pi r}\sin\theta=\frac{377\!\cdot\!2\pi\!\cdot\!2\!\cdot\!0.1}{4\pi\!\cdot\!10}=3.77\,\text{V/m}\]

Sample: Same dipole. Find radiated power and \(R_r\).

\[P_{rad}=\tfrac12\!\cdot\!80\pi^2\!\cdot\!(d\ell/\lambda)^2 I_0^2=80\pi^2(0.1)^2(2^2)/2=158\,\text{W}\]
\[R_r=80\pi^2(d\ell/\lambda)^2=7.9\,\Omega\]

Sample: Hertzian dipole pattern.

\[F(\theta)=\sin\theta,\;F^2(\theta)=\sin^2\theta\]
Maximum at \(\theta=90^{\circ}\) (broadside); nulls at \(\theta=0,180^{\circ}\) (along axis).

Sample: Half-wave dipole pattern.

\[F(\theta)=\frac{\cos((\pi/2)\cos\theta)}{\sin\theta}\]
Slightly narrower than Hertzian; peak at \(\theta=90^{\circ}\); nulls along axis.

Sample: Hertzian dipole. Find directivity.

\[D=4\pi U_{max}/P_{rad}=4\pi\cdot(\eta_0\beta^2 I_0^2 d\ell^2/(32\pi^2))/P_{rad}=1.5\]
\[D_{dB}=10\log 1.5=1.76\,\text{dBi}\]

Sample: Half-wave dipole at \(f=1\,\text{GHz}\). Find effective aperture.

\[\lambda=0.3\,\text{m},\;G=1.64\]
\[A_e=G\lambda^2/(4\pi)=1.64\!\cdot\!0.09/(4\pi)=0.0117\,\text{m}^2\]

Sample: Half-wave dipole at \(f=300\,\text{MHz}\), \(I_0=1\,\text{A}\). Find \(R_r,P_{rad}\).

\[R_r=73\,\Omega\]
\[P_{rad}=\tfrac12 I_0^2 R_r=36.5\,\text{W}\]

Sample: Same antenna. Find \(|E_{\theta}|\) at \(r=100\,\text{m}\), \(\theta=90^{\circ}\).

\[|E_{\theta}|=\eta_0|I_0|/(2\pi r)\!\cdot\!F_{max}=377\!\cdot\!1/(2\pi\!\cdot\!100)\!\cdot\!1=0.6\,\text{V/m}\]

Sample: Friis: \(P_t=10\,\text{W}\), \(G_t=20\,\text{dBi}=100\), \(G_r=10\,\text{dBi}=10\), \(f=10\,\text{GHz}\), \(R=10\,\text{km}\). Find \(P_r\).

\[\lambda=0.03\,\text{m}\]
\[P_r=P_t G_t G_r(\lambda/(4\pi R))^2=10\!\cdot\!100\!\cdot\!10\!\cdot\!(0.03/(4\pi\!\cdot\!10^4))^2=5.7\!\cdot\!10^{-9}\,\text{W}\]

Sample: Required \(P_t\) for \(P_r=10^{-12}\,\text{W}\) in above link?

\[P_t=P_r/[G_t G_r(\lambda/(4\pi R))^2]=10^{-12}/(5.7\!\cdot\!10^{-10})=1.75\!\cdot\!10^{-3}\,\text{W}\!\approx\!1.75\,\text{mW}\]

Sample: Sinusoidal current \(I(z')=I_0\sin(\beta(\ell/2-|z'|))\) on dipole length \(\ell\). Far field?

\[E_{\theta}=\frac{j\eta_0 I_0 e^{-j\beta r}}{2\pi r}F(\theta)\]
\[F(\theta)=[\cos(\beta\ell\cos\theta/2)-\cos(\beta\ell/2)]/\sin\theta\]

Sample: Full-wave dipole (\(\ell=\lambda\)). Pattern.

\[F(\theta)=[\cos(\pi\cos\theta)+1]/\sin\theta\]
Has 4 lobes; narrower than half-wave; useful \(D\approx 2.4\).

Sample: Two-element broadside array, spacing \(d=\lambda/2\), equal currents in phase. Find array factor.

\[\text{AF}=2\cos(\pi\cos\theta/2)\]
\[\text{Maximum at }\theta=90^{\circ}\;(\text{broadside});\text{ nulls at }\theta=0,180^{\circ}\]

Sample: \(N=4\) uniform array, spacing \(d=\lambda/4\), progressive phase \(-\pi/2\) (end-fire).

\[\psi=\beta d\cos\theta+\alpha=(\pi/2)(\cos\theta-1)\]
\[\text{AF}=\sin(N\psi/2)/\sin(\psi/2),\;\text{peak at }\theta=0^{\circ}\]

Sample: Receiving antenna with \(G_r=10\,\text{dBi}\) at \(f=2\,\text{GHz}\). Find \(A_e\).

\[\lambda=0.15\,\text{m},\;G=10\;(=10\,\text{dB})\]
\[A_e=G\lambda^2/(4\pi)=10(0.0225)/(4\pi)=0.0179\,\text{m}^2\]

Sample: Antenna with \(R_r=73\,\Omega\) matched to \(Z_L=R_r\). Incident plane wave \(|E|=1\,\text{mV/m}\). Find received power.

\[P_r=A_e|E|^2/(2\eta_0)\;\Rightarrow\;\text{depends on }A_e\text{ from gain}\]
For half-wave dipole: \(A_e=0.013\,\text{m}^2\) at \(f=300\,\text{MHz}\), giving \(P_r\approx 1.7\!\cdot\!10^{-11}\,\text{W}\).

Sample: Radar: \(P_t=100\,\text{kW}\), \(G=40\,\text{dBi}\), \(\sigma=10\,\text{m}^2\), \(f=10\,\text{GHz}\), target at \(R=50\,\text{km}\). Find \(P_r\).

\[\lambda=0.03,\;G=10^4\]
\[P_r=\frac{P_t G^2\sigma\lambda^2}{(4\pi)^3 R^4}=\frac{10^5\!\cdot\!10^8\!\cdot\!10\!\cdot\!9\!\cdot\!10^{-4}}{(4\pi)^3(5\!\cdot\!10^4)^4}\]
\[=9\!\cdot\!10^{10}/(2.0\!\cdot\!10^{22})=4.5\!\cdot\!10^{-12}\,\text{W}\]