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.

濃縮自 Cheng《Field and Wave Electromagnetics》§1-1 至 §1-3。請先讀完本節,再嘗試下方的習題。

Cheng『Field and Wave Electromagnetics』§1-1 〜 §1-3 の要約です。以下の問題に取り組む前に読んでください。


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.

Electromagnetics(電磁學)研究的是靜止與運動中的電荷所產生的效應。電荷有兩種 — 正電荷與負電荷 — 兩者都是 electric field(電場)的來源。運動中的電荷構成電流(current),而電流則產生 magnetic field(磁場)。

電磁気学(Electromagnetics)は、静止および運動している電荷が及ぼす効果を研究する学問です。電荷には正と負の 2 種類があり、どちらも電場(electric field)の源となります。運動する電荷は電流を構成し、電流は磁場(magnetic field)を生じさせます。

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.

Field(場)是某個物理量在空間中的分布,可以隨時間變化,也可以不隨時間變化。隨時間變化的 electric field 必定伴隨 magnetic field,反之亦然 — 兩者彼此耦合(coupled),共同構成 electromagnetic field(電磁場)。在特定條件下,隨時間變化的電磁場會產生從源向外輻射的 waves(波)。

場(field)とは、ある量の空間的な分布のことで、時間の関数である場合もそうでない場合もあります。時間変化する電場には必ず磁場が伴い、その逆も成り立ちます。すなわち両者は結合(coupled)しており、合わせて電磁場(electromagnetic field)を形成します。一定の条件のもとでは、時間に依存する電磁場は波源から放射される波(waves)を生み出します。

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:

為什麼需要場,而不只是電路?Circuit theory(電路理論)只是電磁理論的一個受限特例,僅在網路尺寸遠小於波長時(quasi-static 準靜態範圍)才成立。以下兩種情況是 circuit theory 失效的例子:

なぜ回路だけでなく場が必要なのか。回路理論(circuit theory)は電磁気理論の限定された特殊な場合にすぎず、回路網の寸法が波長よりはるかに小さいとき(準静的領域)にのみ成り立ちます。回路理論が破綻する状況を 2 つ挙げます:

  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. 對講機上的 monopole antenna(單極天線)在 circuit theory 看來是一個開路,但它顯然會輻射 — 只有場論所預測、沿著開路端導體流動的非均勻電流,才能解釋訊號的發射。
  2. 電磁波入射到一面開有小孔的導電牆時,會在牆的另一側產生場,甚至在不正對小孔的位置也有場 — 這是 circuit theory 無法描述的 diffraction(繞射)/shielding(屏蔽)問題。
  1. トランシーバーのモノポールアンテナ(monopole antenna)は、回路理論から見ると開放回路にすぎませんが、明らかに電波を放射します。開放端をもつ導体に沿って流れる不均一な電流(場の理論が予測するもの)だけが、送信を説明できます。
  2. 小さな開口のある導体壁に電磁波が入射すると、壁の反対側、しかも開口の真裏ではない点にも場が生じます。これは回折・遮蔽の問題であり、回路理論では記述できません。

1-2.1   Deductive vs. Inductive Approach

Two approaches exist in developing a scientific subject:

發展一門科學學科有兩種途徑:

科学的な学問分野を構築するには、2 つのアプローチがあります:

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

Cheng 採用 deductive 途徑,因為它更簡潔優雅,而且能有條理地逐步建立整個學科。

Cheng は演繹的アプローチを採用しています。そのほうがより簡潔で、学問を秩序立てて展開できるからです。

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.
  1. 定義(define)與該學科相關的基本量。
  2. 指定(specify)這些量的運算規則(即所需的數學)。
  3. 設定(postulate)基本關係式(公理/定律),其依據為實驗觀察。
  1. その学問に関わる基本量を定義する。
  2. それらの量に対する演算規則(数学)を指定する。
  3. 実験的観察に基づいて、基本的な関係式(公理・法則)を仮定する。

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.

熟悉的例子 — circuit theory:基本量是 \(V, I, R, L, C\);運算規則是代數、常微分方程 (ODE) 與 Laplace transform;基本設定則是 Kirchhoff 的電壓定律與電流定律。

身近な例として回路理論を考えましょう。基本量は \(V, I, R, L, C\)、演算規則は代数・常微分方程式・ラプラス変換であり、基本的な仮定はキルヒホッフの電圧則と電流則です。

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).

對電磁學而言:第 1 章完成步驟 1;第 2 章完成步驟 2(vector algebra & calculus,向量代數與向量微積分);步驟 3 則分三個子步驟引入 — 第 3 章(electrostatics,靜電學)、第 6 章(magnetostatics,靜磁學)與第 7 章(time-varying fields/Maxwell's equations)。

電磁気学では、第 1 章がステップ 1、第 2 章がステップ 2(ベクトル代数 & ベクトル解析)に対応します。ステップ 3 は 3 つの小ステップに分けて導入されます。第 3 章(静電気学)、第 6 章(静磁気学)、第 7 章(時間変化する場・マクスウェルの方程式)です。


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).

電磁模型中的物理量分為兩類:source quantities(源量,即產生場的電荷與電流)與 field quantities(場量,即場本身)。

電磁モデルに現れる量は 2 つに分類されます。源量(source quantities:場を生み出す電荷と電流)と場量(field quantities:場そのもの)です。

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 \(e\):

Electric charge(電荷)是物質的基本性質,其值只能是基本電荷 \(e\) 的正整數倍或負整數倍:

電荷(electric charge)は物質の基本的な性質であり、素電荷 \(e\) の正または負の整数倍としてのみ存在します:

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

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.

電荷守恆原理(principle of conservation of charge)— 如同動量守恆,這是物理學的一條基本設定:電荷既不能被創造,也不能被消滅。孤立系統中正、負電荷的代數和保持不變。此原理在數學上以 equation of continuity(連續方程式,§5-4)表示;Kirchhoff's current law 其實就是這個原理在電路節點上的直接陳述。

電荷保存則(principle of conservation of charge) — 運動量保存則と同様に、これは物理学の基本的な仮定です。電荷は生成も消滅もしません。孤立系の中にある正負の電荷の代数和は不変です。この原理を数学的に表したものが連続の方程式(§5-4)であり、キルヒホッフの電流則は、これを接続点に適用して述べたものにほかなりません。

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:

雖然電荷在微觀上是離散的,但大量電荷集合所產生的電磁效應,可以用平滑化的密度函數來精確描述;這些函數被定義為空間座標的點函數(point functions):

電荷は微視的には離散的ですが、多数の電荷の集まりが及ぼす電磁的効果は、空間座標の点関数(point function)として定義された平滑化された密度関数でよく記述できます:

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

"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.

所謂「夠小」是指能準確反映密度的變化,「夠大」則是指內部仍含有大量離散電荷 — 例如邊長 1 μm 的立方體,體積為 \(10^{-18}\ \text{m}^3\),卻仍含有約 \(10^{11}\) 個原子。

「十分に小さい」とは密度の変化を正確に表せる大きさ、「十分に大きい」とは多数の離散的な電荷を含む大きさという意味です。たとえば一辺 1 μm の立方体の体積は \(10^{-18}\ \text{m}^3\) ですが、それでもおよそ \(10^{11}\) 個の原子を含んでいます。

Current and current density

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

Current(電流)是電荷對時間的變化率:

電流(current)は、電荷の時間に対する変化率です:

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

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

電流流經的是一個有限面積,因此它不是點函數。為了描述 current density(電流密度),定義了兩個向量點函數:

電流は有限の面積を流れるので、点関数ではありません。電流密度(current density)を記述するために、2 つのベクトル点関数が定義されます:


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):

電磁學有 四個 基本的向量場量(vector field quantities),它們都是空間的點函數(在動態情況下也是時間的函數):

電磁気学には 4 つの基本的なベクトル場量があり、いずれも空間の点関数(動的な場合には時間の関数でもあります)です:

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?

何時需要哪一個場?

どの場がいつ必要になるか。

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.

靜態與動態下的耦合:在靜態、穩態或定常(stationary)條件下,\(\{\mathbf{E},\mathbf{D}\}\) 與 \(\{\mathbf{B},\mathbf{H}\}\) 這兩組是彼此獨立的。在隨時間變化的情況下,它們則是耦合的 — 隨時間變化的 \(\mathbf{E}\)/\(\mathbf{D}\) 會產生 \(\mathbf{B}\)/\(\mathbf{H}\),反之亦然。

静的な場合と動的な場合の結合:静的・定常の条件では、2 つの組 \(\{\mathbf{E},\mathbf{D}\}\) と \(\{\mathbf{B},\mathbf{H}\}\) は互いに独立です。時間に依存する場合には両者は結合します。すなわち時間変化する \(\mathbf{E}\)/\(\mathbf{D}\) は \(\mathbf{B}\)/\(\mathbf{H}\) を生み、その逆も成り立ちます。

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):

物質(介質)的性質決定了 \(\mathbf{E}\) 與 \(\mathbf{D}\) 之間、以及 \(\mathbf{B}\) 與 \(\mathbf{H}\) 之間的關係。這些關係稱為該介質的 constitutive relations(組成關係式)。在自由空間中(見 §1-3):

物質(媒質)の性質が、\(\mathbf{E}\) と \(\mathbf{D}\) の関係、および \(\mathbf{B}\) と \(\mathbf{H}\) の関係を決めます。これらの関係を媒質の構成関係式(constitutive relations)と呼びます。自由空間では(§1-3 参照):

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

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:

SI(Système International d'Unités,國際單位制)是一套有理化的 MKSA 系統 — 由四個基本單位出發,電磁學中的其他所有單位都可由此導出:

SI(Système International d'Unités、国際単位系)は有理化された MKSA 系であり、4 つの基本単位から電磁気学のほかのすべての単位が導かれます:

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:

Table 1-1 中的所有單位都是導出單位:

Table 1-1 の単位はすべて組立単位(導出単位)です:

"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).

「Rationalized」(有理化)的意思是:因子 \(4\pi\) 已被吸收到組成常數之中,因此它不會出現在 Maxwell's equations 本身(但仍會出現在許多導出關係式中,例如 Coulomb's law)。

「有理化(rationalized)」とは、係数 \(4\pi\) を構成定数の中に吸収してあるため、マクスウェルの方程式そのものにはこの係数が現れないということです(ただし、クーロンの法則など多くの導出された関係式には現れます)。


1-3.2   The Three Universal Constants

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

自由空間(真空)由 三個 普適常數(universal constants)來描述,而這三者並非彼此獨立:

自由空間(真空)は 3 つの普遍定数(universal constants)で特徴づけられますが、これらは互いに独立ではありません:

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

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

在 SI 中,\(\mu_0\) 被定義為精確的 \(4\pi\times 10^{-7}\) H/m(這是單位制的選擇,而非實驗測得的量)。光速則由公尺的現代定義所固定。於是 permittivity(介電常數)\(\epsilon_0\) 就由下面這條主要恆等式導出:

SI では、\(\mu_0\) は正確に \(4\pi\times 10^{-7}\) H/m と定義されています(これは単位系の取り決めであって、実験で測定された量ではありません)。光速はメートルの現代的な定義によって固定されています。したがって誘電率(permittivity)\(\epsilon_0\) は、次の基本的な恒等式から導かれる量です:

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

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

這條式子將電磁學與光學連結起來 — 光速正是真空之電性與磁性性質的必然結果。

この式は電磁気学と光学を結びつけています。光の速度は、真空の電気的・磁気的性質から導かれる帰結なのです。

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.

摘錄自 Cheng,Field and Wave Electromagnetics,§2-1 至 §2-12。請先閱讀本節,再嘗試下方的習題。

Cheng 著Field and Wave Electromagnetics、§2-1 〜 §2-12 の要約です。下の問題に取り組む前に読んでください。


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.

電磁量不是 scalars(純量:電荷、能量、電位)就是 vectors(向量:\(\mathbf{E}\)、\(\mathbf{B}\)、current density)。在三維空間中,一個向量關係式等於三個純量關係式,因此一套有效率的 vector calculus 是不可或缺的。

電磁気学の諸量はスカラー(scalars)(電荷、エネルギー、電位)かベクトル(vectors)(\(\mathbf{E}\)、\(\mathbf{B}\)、電流密度)のいずれかです。三次元空間では 1 つのベクトル関係式は3 つのスカラー関係式に相当するため、効率的なベクトル解析(vector calculus)が不可欠です。

Three pillars of this chapter:

本章的三大支柱:

本章の 3 つの柱:

  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.
  1. Vector algebra — 向量加法、dot product 與 cross product。
  2. Orthogonal coordinate systems — Cartesian、cylindrical、spherical 三種正交座標系。
  3. Vector calculus — gradient、divergence、curl,以及將它們串連起來的積分定理。
  1. ベクトル代数(Vector algebra) — 加法、内積と外積。
  2. 直交座標系(Orthogonal coordinate systems) — 直角座標、円筒座標、球座標。
  3. ベクトル解析(Vector calculus) — 勾配、発散、回転、およびそれらを結び付ける積分定理。

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

物理定律本身不隨座標系的選擇而改變;選擇座標系只是為了配合特定問題的幾何形狀。

物理法則そのものは座標系の選び方に依存しません。座標系は、与えられた問題の幾何学的形状に合わせるためにのみ選ばれます。


2-2   Vector Addition and Subtraction

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

向量具有大小與方向;unit vector 只指示方向而不帶大小:

ベクトルは大きさと方向を持ちます。単位ベクトル(unit vector)は大きさを持たず方向だけを示します:

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

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

向量加法滿足交換律與結合律;減法定義為 \(\mathbf{A}-\mathbf{B} = \mathbf{A}+(-\mathbf{B})\)。

加法は交換法則と結合法則を満たします。減法は \(\mathbf{A}-\mathbf{B} = \mathbf{A}+(-\mathbf{B})\) と定義されます。

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

2-3   Products of Vectors

Dot (scalar) product — yields a scalar:

Dot (scalar) product — 結果為一純量:

内積(Dot product、スカラー積) — 結果はスカラーになります:

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

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).

兩向量正交時(且僅當此時)內積為零;其值等於一向量在另一向量上的投影乘以後者的大小。滿足交換律與分配律;但不滿足結合律(三重 dot product 沒有意義)。

内積が零になるのは、2 つのベクトルが直交するとき、かつそのときに限ります。その値は、一方のベクトルを他方へ射影した長さに他方の大きさを掛けたものに等しくなります。交換法則と分配法則を満たしますが、結合法則は満たしません(三重の内積には意味がありません)。

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

Cross (vector) product — 結果為一向量,垂直於兩運算元所在的平面,大小等於兩者所張成的平行四邊形面積:

外積(Cross product、ベクトル積) — 結果は 2 つのベクトルが張る平面に垂直なベクトルで、その大きさは両者がつくる平行四辺形の面積に等しくなります:

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

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

方向 \(\mathbf{a}_n\) 由 right-hand rule 決定。Cross product 滿足反交換律與分配律,但不滿足結合律。

方向 \(\mathbf{a}_n\) は右手の法則(right-hand rule)で決まります。外積は反交換的であり、分配法則を満たしますが、結合法則は満たしません。

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

Scalar triple product — 循環置換不改變其值(幾何意義為平行六面體的體積):

スカラー三重積(Scalar triple product) — 巡回置換をしても値は変わりません(幾何学的には平行六面体の体積です):

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

Vector triple product — the BAC-CAB rule:

Vector triple product — 即 BAC-CAB 法則:

ベクトル三重積(Vector triple product) — いわゆる BAC-CAB の規則です:

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

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

括號的位置很重要 — \((\mathbf{A}\times\mathbf{B})\times\mathbf{C}\) 是另一個不同的向量。向量的除法是沒有定義的。

括弧の位置は重要です — \((\mathbf{A}\times\mathbf{B})\times\mathbf{C}\) は別のベクトルになります。ベクトルによる除算は定義されていません。


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:

右手正交座標系的 base vectors \(\mathbf{a}_{u_1},\mathbf{a}_{u_2},\mathbf{a}_{u_3}\) 滿足:

右手系の直交座標系の基底ベクトル(base vectors)\(\mathbf{a}_{u_1},\mathbf{a}_{u_2},\mathbf{a}_{u_3}\) は次を満たします:

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

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

座標 \(u_i\) 本身不一定是長度(例如角度)。Metric coefficient \(h_i\) 將座標的變化量換算為長度:

座標 \(u_i\) 自体は長さとは限りません(例えば角度)。計量係数(metric coefficient) \(h_i\) は座標の変化量を長さに換算します:

\[ d\ell_i = h_i\,du_i \qquad \text{(2-29)} \]
\[ d\mathbf{\ell} = \mathbf{a}_{u_1}(h_1\,du_1) + \mathbf{a}_{u_2}(h_2\,du_2) + \mathbf{a}_{u_3}(h_3\,du_3) \qquad \text{(2-31)} \]
\[ dv = h_1 h_2 h_3\,du_1\,du_2\,du_3 \qquad \text{(2-33)} \]
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):

Cartesian(直角座標,§2-4.1):

直角座標(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. \(\mathbf{a}_r\times\mathbf{a}_\phi=\mathbf{a}_z\).

Cylindrical(圓柱座標,§2-4.2): \((r,\phi,z)\)。適用於 line charges、長導線、coax。\(\mathbf{a}_r\times\mathbf{a}_\phi=\mathbf{a}_z\)。

円筒座標(Cylindrical、§2-4.2): \((r,\phi,z)\)。線電荷、長い導線、同軸線に有用です。\(\mathbf{a}_r\times\mathbf{a}_\phi=\mathbf{a}_z\)。

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

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. \(\mathbf{a}_R\times\mathbf{a}_\theta=\mathbf{a}_\phi\).

Spherical(球座標,§2-4.3): \((R,\theta,\phi)\)。\(R\) 為到原點的距離,\(\theta\) 為自 \(+z\) 軸量起的 polar angle,\(\phi\) 為 azimuth。最適合點源與 far-field 天線問題。\(\mathbf{a}_R\times\mathbf{a}_\theta=\mathbf{a}_\phi\)。

球座標(Spherical、§2-4.3): \((R,\theta,\phi)\)。\(R\) は原点からの距離、\(\theta\) は \(+z\) 軸から測った極角、\(\phi\) は方位角です。点源や遠方界(far-field)のアンテナ問題に最適です。\(\mathbf{a}_R\times\mathbf{a}_\theta=\mathbf{a}_\phi\)。

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

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.

注意:在曲線座標系中,base vectors 會隨位置而變(\(\partial\mathbf{a}_r/\partial\phi = \mathbf{a}_\phi\),\(\partial\mathbf{a}_\phi/\partial\phi = -\mathbf{a}_r\))。在這類座標系中對向量微分時,必須將此納入考量。

注意:曲線座標系では基底ベクトルが位置に依存します(\(\partial\mathbf{a}_r/\partial\phi = \mathbf{a}_\phi\)、\(\partial\mathbf{a}_\phi/\partial\phi = -\mathbf{a}_r\))。このような座標系でベクトルを微分する際には、この点を考慮しなければなりません。


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).

微分面積視為向量 \(d\mathbf{s} = \mathbf{a}_n\,ds\)(封閉曲面取外法線;由輪廓所圍的開放曲面則取 right-hand-rule 法線)。

微小面積はベクトル \(d\mathbf{s} = \mathbf{a}_n\,ds\) として扱います(閉曲面では外向き法線、閉曲線で囲まれた開曲面では右手の法則による法線をとります)。

• 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:

Gradient 是一向量,指向純量 \(V\) 在空間中增加率最大的方向,其大小等於該增加率:

勾配(gradient)は、スカラー \(V\) の空間的な増加率が最大となる方向を向き、その大きさがその増加率に等しいベクトルです:

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

General orthogonal curvilinear form:

一般正交曲線座標形式:

一般の直交曲線座標における形:

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

Cartesian form — also defines the del operator:

Cartesian 形式 — 同時定義了 del operator:

直角座標における形 — ナブラ演算子(del operator)の定義でもあります:

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

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

\(\nabla V\) 永遠垂直於 \(V\) 的等值面。在 electrostatics 中我們使用 \(\mathbf{E}=-\nabla V\) — 負號使 \(V\) 沿著 \(\mathbf{E}\) 的反方向增加。

\(\nabla V\) は常に \(V\) の等値面に垂直です。静電気学では \(\mathbf{E}=-\nabla V\) を用います — 負号により、\(V\) は \(\mathbf{E}\) と逆向きに増加します。


2-7   Divergence of a Vector Field

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

Divergence 度量 \(\mathbf{A}\) 每單位體積的淨向外通量 — 即 flow source density(流源密度):

発散(divergence)は、単位体積あたりの \(\mathbf{A}\) の正味の外向き流束を表します — すなわち湧き出し源の密度(flow source density)です:

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

Coordinate forms:

各座標系形式:

各座標系における形:

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

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 \(\mathbf{B}\)).

處處滿足 \(\nabla\cdot\mathbf{A}=0\) 的場稱為 solenoidal(螺線場) — 通量線自我閉合,沒有源或匯(例如 magnetic flux density \(\mathbf{B}\))。

至るところで \(\nabla\cdot\mathbf{A}=0\) となる場はソレノイダル(solenoidal)です — 流束線は閉じており、湧き出しも吸い込みもありません(例:磁束密度 \(\mathbf{B}\))。


2-8   Divergence Theorem

Also known as Gauss's theorem:

又稱 Gauss's theorem:

ガウスの定理(Gauss's theorem)とも呼ばれます:

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

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.

將 divergence 的體積分轉換為該向量的封閉曲面積分 — 是 Gauss's law 與電荷守恆積分形式背後的主力工具。

発散の体積積分を、そのベクトルの閉曲面積分に変換します — ガウスの法則や電荷保存則の積分形を導く際の中心的な道具です。


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:

Circulation(環流)是 \(\mathbf{A}\) 沿封閉輪廓的線積分。Curl 則是 vortex-source density(渦源密度):

循環(circulation)は閉曲線に沿った \(\mathbf{A}\) の線積分です。回転(curl)は渦源密度(vortex-source density)です:

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

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

\(\mathbf{a}_n\) 的方向是使單位面積環流達到最大的方向(相對於 \(d\mathbf{\ell}\) 的 right-hand rule)。

\(\mathbf{a}_n\) の方向は、単位面積あたりの循環を最大にする方向です(\(d\mathbf{\ell}\) に対する右手の法則)。

Determinant form in Cartesian coordinates:

Cartesian 座標下的行列式形式:

直角座標における行列式の形:

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

General orthogonal curvilinear form:

一般正交曲線座標形式:

一般の直交曲線座標における形:

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

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

處處滿足 \(\nabla\times\mathbf{A}=0\) 的場稱為 irrotational(無旋場,或稱 conservative 保守場)。靜電場永遠是 irrotational。

至るところで \(\nabla\times\mathbf{A}=0\) となる場は非回転(irrotational)(または保存的(conservative))です。静電場は常に非回転です。


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.

將 curl 的曲面積分轉換為沿邊界輪廓的封閉線積分。\(d\mathbf{\ell}\) 與 \(d\mathbf{s}=\mathbf{a}_n\,ds\) 的方向遵循 right-hand rule。

回転の面積分を、境界となる閉曲線に沿った線積分に変換します。\(d\mathbf{\ell}\) と \(d\mathbf{s}=\mathbf{a}_n\,ds\) の向きは右手の法則に従います。

Dictionary. The two integral theorems convert dimensionality:

對照表。這兩個積分定理改變的是積分的維度:

対応表。2 つの積分定理は積分の次元を変換します:

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:

重複施加 del 運算可得兩個對 potential theory 極為重要的恆等式:

ナブラ演算を繰り返し施すと、ポテンシャル論にとってきわめて重要な 2 つの恒等式が得られます:

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

Converse statements (crucial for EM):

逆命題(對 EM 至關重要):

逆の命題(電磁気学にとって決定的に重要):


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.

若一向量場的 divergence 與 curl 在空間中處處給定,且該場在無窮遠處為零,則此向量場即被唯一決定 — 至多相差一個加法常數。

ベクトル場は、その発散と回転が至るところで指定され、かつ無限遠で零になるならば — 加法定数を除いて — 一意に決まります。

Classification of vector fields:

向量場的分類:

ベクトル場の分類:

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:

任何這樣的場都可分解為一個 irrotational 部分與一個 solenoidal 部分:

そのような場はいずれも、非回転部分とソレノイダル部分に分解できます:

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

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.

這正是 Cheng 後續各章的公理基礎 — 只要同時指定 \(\mathbf{E}\)(第 3 章)或 \(\mathbf{B}\)(第 6 章)的 div 與 curl,整個場便隨之確定。這就是 Cheng 選擇以 div 與 curl 作為 postulates,而非從 Coulomb's law 出發的原因。

これが Cheng の以降の全章の公理的な基礎です — \(\mathbf{E}\)(第 3 章)あるいは \(\mathbf{B}\)(第 6 章)の発散と回転の両方を指定すれば、場全体が決まります。Cheng がクーロンの法則から出発するのではなく、発散と回転を公理(postulates)として置くのはこのためです。


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.

濃縮自 Cheng,Field and Wave Electromagnetics,§3-1 至 §3-11。請先讀完本節,再嘗試下方的習題。

Cheng『Field and Wave Electromagnetics』§3-1〜§3-11 の要約です。下の問題に取り組む前にお読みください。


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:

初等教材多由 Coulomb's law 這條實驗定律出發,逐步往上建構。Cheng 則反其道而行:先假設(postulate)自由空間中 \(\mathbf{E}\) 的 divergence 與 curl,再由此推導出 Coulomb's law 與 Gauss's law。理由如下:

初等的な教科書はクーロンの実験法則(Coulomb's law)から出発して理論を積み上げます。これに対して Cheng は、自由空間における \(\mathbf{E}\) の発散(divergence)と回転(curl)を公理として置き、そこからクーロンの法則とガウスの法則を導きます。その根拠は次のとおりです:

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.

在 electrostatics 中,電荷靜止不動,場不隨時間變化,也沒有 magnetic field。因此在自由空間裡,四個 field quantities 只需要一個 — \(\mathbf{E}\) — 三個 universal constants 也只需要一個 — \(\epsilon_0\)。

静電気学(electrostatics)では電荷は静止し、場は時間的に変化せず、磁場も存在しません。したがって自由空間では、4 つの場の量のうち \(\mathbf{E}\) だけ、3 つの普遍定数のうち \(\epsilon_0\) だけがあれば十分です。


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:

Electric field intensity(電場強度)定義為作用於靜止 test charge 上、每單位電荷所受的力:

電界強度(electric field intensity)は、静止した試験電荷に働く単位電荷あたりの力として定義されます:

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

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

自由空間中整套 electrostatics 理論建立在兩條 postulates 之上 — 即 \(\mathbf{E}\) 的 divergence 與 curl:

自由空間における静電気学の理論全体は、2 つの公理 — \(\mathbf{E}\) の発散と回転 — の上に成り立っています:

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

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

分別套用 divergence theorem 與 Stokes's theorem,即得積分形式:

それぞれ発散定理(divergence theorem)とストークスの定理(Stokes's theorem)を適用すると、積分形が得られます:

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

Physical reading:

物理意義:

物理的な読み方:

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:

對自由空間中的一個 point charge \(q\),取以其為中心、半徑 \(R\) 的球形 Gaussian surface 並套用 (3-7)。由對稱性可知 \(\mathbf{E}=\mathbf{a}_R E_R\),得 \(E_R(4\pi R^2)=q/\epsilon_0\),因此:

自由空間中の点電荷 \(q\) を中心とする半径 \(R\) の球面をガウス面(Gaussian surface)として (3-7) を適用します。対称性から \(\mathbf{E}=\mathbf{a}_R E_R\) となり、\(E_R(4\pi R^2)=q/\epsilon_0\) が得られるので:

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

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

對於離散電荷系統,由 superposition 得 \(\mathbf{E}=\sum_k \mathbf{E}_k\)。對於 volume charge density 為 \(\rho(\mathbf{R}')\) 的連續分佈:

離散的な電荷の系では、重ね合わせの原理により \(\mathbf{E}=\sum_k \mathbf{E}_k\) となります。体積電荷密度 \(\rho(\mathbf{R}')\) の連続分布に対しては:

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

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:

當能建構一個 Gaussian surface,使其上的 \(\mathbf{E}\cdot d\mathbf{s}\) 為常數時,Gauss's law (3-7) 的威力最大。有三類對稱性允許這麼做:

ガウスの法則 (3-7) が最も威力を発揮するのは、\(\mathbf{E}\cdot d\mathbf{s}\) が一定となるガウス面を構成できるときです。これが可能な対称性は次の 3 種類です:

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

要訣:若對稱性符合,直接用 (3-7)。否則先以積分求出 \(V\),再由 \(\mathbf{E}=-\nabla V\) 求場。

教訓:対称性が合うなら (3-7) を直接使います。そうでなければ、まず積分で \(V\) を求め、次に \(\mathbf{E}=-\nabla V\) とします。


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:

由於 \(\nabla\times\mathbf{E}=0\),依據 null identity \(\nabla\times(\nabla V)\equiv 0\),可將 \(\mathbf{E}\) 寫成某個 scalar 的 gradient:

\(\nabla\times\mathbf{E}=0\) であることと、恒等式 \(\nabla\times(\nabla V)\equiv 0\) から、\(\mathbf{E}\) をスカラーの勾配(gradient)として書くことができます:

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

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:

負號使 \(V\) 沿著與 \(\mathbf{E}\) 相反的方向遞增,與「單位電荷的位能」這個意義一致。兩點之間的 potential difference 與路徑無關:

負号があるために、\(V\) は \(\mathbf{E}\) と逆向きに増加し、単位電荷あたりの位置エネルギーという意味と整合します。2 点間の電位差(potential difference)は経路によりません:

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

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

以無窮遠為參考點,幾種標準電荷分佈的 potential 為:

無限遠を基準にとると、代表的な電荷分布のポテンシャルは次のようになります:

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

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

Field lines 與 equipotentials 互相垂直 — \(\nabla V\) 與 \(V\) 為常數的曲面正交。

電気力線と等電位面は直交します — \(\nabla V\) は \(V\) 一定の面に垂直です。


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:

Conductor 內有束縛鬆散的電子,只要內部存在電場,它們就會自由重新分佈。靜態平衡建立得極快 — 以銅為例,relaxation time 約為 \(\sim 10^{-19}\) s。平衡時:

導体(conductor)には束縛の弱い電子があり、内部に場が存在すれば自由に再分布します。静的な平衡はきわめて速く達成されます — 銅の場合、緩和時間は \(\sim 10^{-19}\) s 程度です。平衡状態では:

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

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

在 conductor / free-space boundary,對狹長矩形迴路套用 (3-8)、對扁平 pillbox 套用 (3-7),得:

導体と自由空間の境界において、細長い長方形の閉路に (3-8) を、薄いピルボックスに (3-7) を適用すると:

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

Consequences:

推論:

その帰結:


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).

在 dielectric 中,電子是受束縛的。外加 \(\mathbf{E}\) 會使它們稍微位移,形成 induced dipoles。巨觀上的量度是 polarization vector(極化向量)\(\mathbf{P}\) — 單位體積的 electric dipole moment (C/m²)。

誘電体(dielectric)では電子は束縛されています。外部から \(\mathbf{E}\) を加えると電子はわずかに変位し、誘起双極子が生じます。これを巨視的に表す量が分極ベクトル(polarization vector)\(\mathbf{P}\) — 単位体積あたりの電気双極子モーメント (C/m²) — です。

Polarized matter behaves as if it carried equivalent bound charges:

極化後的物質,其行為就像帶有等效的 bound charges:

分極した物質は、等価な束縛電荷(bound charges)を帯びているかのように振る舞います:

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

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

任何孤立 dielectric 物體上的總 bound charge 為零。

孤立した誘電体の束縛電荷の総量はゼロです。


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

在物質介質中,divergence postulate (3-4) 必須把 bound charge 納入:\(\nabla\cdot\mathbf{E}=(\rho+\rho_p)/\epsilon_0\)。定義 electric flux density(電通量密度)

物質中では、発散の公理 (3-4) に束縛電荷を含めなければなりません:\(\nabla\cdot\mathbf{E}=(\rho+\rho_p)/\epsilon_0\)。ここで電束密度(electric flux density)を

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

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

便可將 bound charge 吸收進去,右邊只剩下 free charge:

と定義すると、束縛電荷が吸収され、右辺には自由電荷だけが残ります:

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

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

對於線性、等向性(linear, isotropic)介質,\(\mathbf{P}=\epsilon_0\chi_e\mathbf{E}\),故

線形で等方的な媒質では \(\mathbf{P}=\epsilon_0\chi_e\mathbf{E}\) なので、

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

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

其中 \(\epsilon_r\) 是無因次的 dielectric constant(相對介電常數,relative permittivity),\(\chi_e\) 是 electric susceptibility,\(\epsilon\) 是 absolute permittivity (F/m)。

ここで \(\epsilon_r\) は無次元の比誘電率(dielectric constant、relative permittivity)、\(\chi_e\) は電気感受率、\(\epsilon\) は絶対誘電率 (F/m) です。

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

Dielectric strength(介電強度,§3-8.1)是材料在 breakdown 之前所能承受的最大 \(E\)。切勿與 \(\epsilon_r\) 混淆:

絶縁耐力(dielectric strength)(§3-8.1)は、材料が絶縁破壊を起こさずに耐えられる最大の \(E\) です。\(\epsilon_r\) と混同しないでください:

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:

將積分形式 (3-8) 套用於跨越界面的狹長迴路、(3-100) 套用於扁平 pillbox,得:

積分形 (3-8) を境界をまたぐ細長い閉路に、(3-100) を薄いピルボックスに適用すると:

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

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

對於不帶電的 dielectric–dielectric interface(\(\rho_s=0\)):

電荷のない誘電体-誘電体界面(\(\rho_s=0\))では:

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

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 \(\mathbf{D}\).

對於 conductor–dielectric interface(介質 2 為 conductor,\(\mathbf{D}_2=0\)):\(E_{1t}=0\) 且 \(D_{1n}=\rho_s\) — 即特例 (3-71)、(3-72) 推廣到 \(\mathbf{D}\) 的結果。

導体-誘電体界面(媒質 2 が導体、\(\mathbf{D}_2=0\))では \(E_{1t}=0\) かつ \(D_{1n}=\rho_s\) — 特別な場合 (3-71)、(3-72) を \(\mathbf{D}\) に一般化したものです。


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:

在靜態 conductor 系統中,\(\rho_s\)、\(\mathbf{E}\) 與 \(V\) 都與 \(Q\) 成線性比例,因此比值 \(Q/V\) 只取決於幾何形狀與介質:

静的な導体系では \(\rho_s\)、\(\mathbf{E}\)、\(V\) はいずれも \(Q\) に比例するので、比 \(Q/V\) は形状と媒質だけで決まります:

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

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 \(C=Q/V_{12}\).

求雙導體 capacitor 之 \(C\) 的步驟:(1) 選擇座標系;(2) 假設兩導體上分別帶 \(\pm Q\);(3) 由 Gauss's law 或 \(\rho_s\) 求 \(\mathbf{E}\);(4) 積分 \(V_{12}=-\int_{2}^{1}\mathbf{E}\cdot d\mathbf{\ell}\);(5) 取 \(C=Q/V_{12}\)。

2 導体コンデンサの \(C\) を求める手順:(1) 座標系を選ぶ;(2) 導体に \(\pm Q\) を仮定する;(3) ガウスの法則または \(\rho_s\) から \(\mathbf{E}\) を求める;(4) \(V_{12}=-\int_{2}^{1}\mathbf{E}\cdot d\mathbf{\ell}\) を積分する;(5) \(C=Q/V_{12}\) とする。

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

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.

串聯/並聯:並聯的 capacitors 直接相加(\(C_\text{par}=\sum C_k\));串聯的 capacitors 以倒數相加(\(1/C_\text{ser}=\sum 1/C_k\))— 與 resistors 相反,因為串聯時各電容共用同一電荷,並聯時共用同一電壓。

直列/並列:並列のコンデンサは単純に加算され(\(C_\text{par}=\sum C_k\))、直列のコンデンサは逆数の和で合成されます(\(1/C_\text{ser}=\sum 1/C_k\))— 抵抗とは逆です。直列では電荷が共通、並列では電圧が共通だからです。

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.

Electrostatic shielding(靜電屏蔽,§3-10.3):以接地的導體殼包住物體,可使其與外部電荷去耦合 — 耦合電容歸零。這就是敏感電子電路要放在接地金屬機殼內的原因。

静電遮蔽(electrostatic shielding)(§3-10.3):接地した導体殻で物体を囲むと、外部の電荷との結合が断たれ、結合容量はゼロになります。敏感な電子機器が接地された金属筐体に収められるのはこのためです。


3-11   Electrostatic Energy and Forces

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

以電荷與電位表示 — 組裝 \(N\) 個 point charges 所需的功:

電荷と電位による表現 — \(N\) 個の点電荷を集めるのに要する仕事:

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

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

因子 \(\tfrac{1}{2}\) 是為了避免把成對交互作用重複計算。\(V_k\) 是其他所有電荷在 \(Q_k\) 處產生的電位。

係数 \(\tfrac{1}{2}\) は、電荷対の相互作用を二重に数えないためのものです。\(V_k\) は他のすべての電荷が \(Q_k\) の位置につくる電位です。

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

以場量表示 — 代入 \(\rho=\nabla\cdot\mathbf{D}\) 並套用一個 vector identity:

場の量による表現 — \(\rho=\nabla\cdot\mathbf{D}\) を代入し、ベクトル恒等式を適用すると:

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

Energy density:

能量密度 (energy density):

エネルギー密度:

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

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.

注意:(3-165) 只給出交互作用(mutual)能量 — 理想 point charges 的 self-energy 是發散的。(3-170) 與 (3-176) 則包含 self-energy,所以兩者彼此一致,但對同一系統可能大於 (3-165)。

注意:(3-165) が与えるのは相互作用エネルギーだけです — 理想的な点電荷の自己エネルギーは発散します。(3-170) と (3-176) は自己エネルギーを含んでおり、そのため両者は互いに一致しますが、同じ系に対して (3-165) より大きくなることがあります。

Stored energy in a capacitor (equivalent forms):

Capacitor 中儲存的能量(等價形式):

コンデンサに蓄えられるエネルギー(等価な表現):

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

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

由 virtual displacement 求力。對於孤立系統(電荷固定),場所做的功等於儲存能量的減少量,因此:

仮想変位(virtual displacement)による力。孤立した系(電荷一定)では、場のする仕事は蓄積エネルギーの減少量に等しいので:

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

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

符號反轉並不矛盾 — 在 \(V\) 固定的情況下,電池額外供給的能量等於機械功的兩倍。

符号が反転するのは矛盾ではありません — \(V\) 一定の場合、電池が機械的仕事の 2 倍に等しい追加のエネルギーを供給するからです。


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.

節錄自 Cheng, Field and Wave Electromagnetics, §4-1 至 §4-7。請先讀完本節,再嘗試下方的習題。

Cheng, Field and Wave Electromagnetics, §4-1〜§4-7 の要約です。下の問題に取り組む前にお読みください。


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:

第 3 章讓我們能在電荷分佈處處已知的情況下直接計算 \(V\) 與 \(\mathbf{E}\)。實務上這種情況很少見。第 4 章處理的是兩種實際可行的替代做法:

第 3 章では、電荷分布が至るところで既知であるときに \(V\) と \(\mathbf{E}\) を直接計算できました。しかし実際にはそのような状況はまれです。第 4 章では、実用上の 2 つの代替手段を扱います:

  • 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).
  • 簡單幾何形狀的導體附近之點電荷/線電荷。感應的 surface charge 未知 — 使用 method of images(影像法,§4-4)。
  • 保持在指定電位的導體。將 Poisson's/Laplace's equation 當作 boundary-value problem(邊界值問題)求解 — 在合適的座標系中使用 separation of variables(分離變數法,§4-5 至 §4-7)。
  • 単純な形状の導体の近くにある点電荷・線電荷。誘導される表面電荷は未知です — 鏡像法(method of images, §4-4)を用います。
  • 所定の電位に保たれた導体。ポアソン方程式/ラプラス方程式を境界値問題(boundary-value problem)として解きます — 適切な座標系での変数分離法(separation of variables, §4-5〜§4-7)です。

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.

這兩種做法都建立在 uniqueness theorem(唯一性定理,§4-3)之上:任何滿足 Poisson's equation 與 boundary conditions 的 \(V\) 就是唯一的解 — 一個經過驗證的幸運猜測,與嚴格推導同樣嚴謹。

どちらの方法も一意性定理(uniqueness theorem, §4-3)に基づいています:ポアソン方程式と境界条件を満たす \(V\) は、それが唯一の解です — 検証済みの幸運な推測は、導出と同じだけ厳密なのです。


4-2   Poisson's and Laplace's Equations

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

從第 3 章在 linear, isotropic medium(線性、等向性介質)中的基本假設出發(\(\mathbf{D}=\epsilon\mathbf{E}\)、\(\nabla\times\mathbf{E}=0\)、\(\mathbf{E}=-\nabla V\)):

線形・等方性媒質における第 3 章の基本公理(\(\mathbf{D}=\epsilon\mathbf{E}\)、\(\nabla\times\mathbf{E}=0\)、\(\mathbf{E}=-\nabla V\))から出発すると:

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

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

對於 simple medium(簡單介質:均勻、線性、等向性 — \(\epsilon\) 為常數),\(\epsilon\) 可以提到 divergence 運算子之外,得到:

単純媒質(simple medium:均質・線形・等方性 — \(\epsilon\) が定数)では、\(\epsilon\) を発散演算子の外に出すことができ、次式が得られます:

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

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).

Laplace's equation 支配導體之間空間中的電位 — 例如電容器極板之間、同軸電纜導體之間等。一旦求得 \(V\),便有 \(\mathbf{E}=-\nabla V\),而 surface charge 則由 \(\rho_s=\epsilon E_n\) (3-72) 得出。

ラプラス方程式は、導体間の空間における電位を支配します — コンデンサの極板間、同軸線の導体間などです。\(V\) が求まれば \(\mathbf{E}=-\nabla V\) となり、表面電荷は \(\rho_s=\epsilon E_n\) (3-72) から得られます。

Explicit forms of the Laplacian in each coordinate system:

Laplacian 在各座標系中的明確形式:

各座標系におけるラプラシアンの具体的な形は次のとおりです:

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

4-3   Uniqueness Theorem

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

滿足給定 boundary conditions 的 Poisson's equation 之解是唯一的。

与えられた境界条件を満たすポアソン方程式の解は一意です。

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.

為何重要:它為猜測提供了正當性。只要你建構出任何一個 \(V\) — 不論是用影像法、觀察法或有根據的猜測 — 它在區域內滿足 (4-6),且在邊界上符合指定的電位,那它就是唯一可能的答案。

なぜ重要か:これが推測を正当化するからです。鏡像法でも、観察でも、根拠のある推測でも、領域内で (4-6) を満たし、境界上で指定された電位に一致する \(V\) を構成できれば、それが唯一のあり得る答えです。

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:

證明概要:假設 \(V_1\)、\(V_2\) 都是此問題的解;令 \(V_d = V_1 - V_2\)。則區域內 \(\nabla^2 V_d = 0\),且在所有邊界上 \(V_d = 0\)。利用 \(\nabla\cdot(V_d\nabla V_d) = V_d\nabla^2 V_d + |\nabla V_d|^2\) 與 divergence theorem(散度定理):

証明の概略:\(V_1\)、\(V_2\) がともにこの問題の解であるとし、\(V_d = V_1 - V_2\) とおきます。すると領域内で \(\nabla^2 V_d = 0\)、すべての境界上で \(V_d = 0\) です。\(\nabla\cdot(V_d\nabla V_d) = V_d\nabla^2 V_d + |\nabla V_d|^2\) と発散定理(divergence theorem)を用いると:

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

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 \(V_1=V_2\).

非負的被積函數若積分為零,則必須恆為零 — 因此 \(\nabla V_d \equiv 0\),\(V_d\) 為常數;又因邊界上 \(V_d=0\),故處處 \(V_d\equiv 0\)。因此 \(V_1=V_2\)。

非負の被積分関数の積分が零ならば、被積分関数は恒等的に零でなければなりません — したがって \(\nabla V_d \equiv 0\) で \(V_d\) は定数であり、境界上で \(V_d=0\) なので至るところ \(V_d\equiv 0\) です。ゆえに \(V_1=V_2\) となります。


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.

基本想法:用一個等效的 image charge(影像電荷,位於所關心區域之外)取代導體邊界,使邊界上的電位條件自動滿足。根據唯一性,這樣得到的就是原區域中正確的 \(V\) 與 \(\mathbf{E}\)。

考え方:導体境界を、境界上の電位条件が自動的に満たされるような等価な鏡像電荷(image charge、対象領域の外側に置く)で置き換えます。一意性により、これが元の領域における正しい \(V\) と \(\mathbf{E}\) を与えます。

Three standard configurations.

三種標準組態。

3 つの標準的な配置。

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

(a) 接地導體平面上方的點電荷(§4-4.1)。影像:在鏡像位置放置 \(-Q\)。

(a) 接地導体平面の上方にある点電荷(§4-4.1)。鏡像:鏡映位置に \(-Q\) を置きます。

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

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

其中 \(R_\pm\) 為到 \(\pm Q\) 的距離。平面下方的場為零;只有上半空間才具有物理意義。

ここで \(R_\pm\) は \(\pm Q\) までの距離です。平面の下側の場は零であり、物理的に意味を持つのは上半空間だけです。

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

(b) 與導體圓柱平行的線電荷(§4-4.2)。影像:在 inverse point(反演點)處放置一條平行、電性相反的線電荷:

(b) 導体円柱に平行な線電荷(§4-4.2)。鏡像:反転点(inverse point)に、符号が反対の平行な線電荷を置きます:

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

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

應用於 two-wire transmission line(雙線傳輸線,導線半徑 \(a\),中心間距 \(D\)):

二線式伝送線路(two-wire transmission line、導線半径 \(a\)、中心間隔 \(D\))に適用すると:

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

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

(c) 半徑 \(a\) 的導體球附近之點電荷(§4-4.3)。影像電荷位於球內的 inverse point:

(c) 半径 \(a\) の導体球の近くにある点電荷(§4-4.3)。鏡像電荷は球の内部の反転点に置きます:

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

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.

若球體是孤立(未接地)而非接地,則在球心再加入第二個影像電荷 \(Q'=+aQ/d\),使球的總電荷為零。

球が接地ではなく孤立している(接地されていない)場合は、球の総電荷が零になるように、中心に第 2 の鏡像電荷 \(Q'=+aQ/d\) を追加します。

Rules to remember when using images:

使用影像法時要記住的規則:

鏡像法を使うときに覚えておくべき規則:

  • 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).
  • Image charges 必須位於你計算場的區域之外 — 絕不能在其內部。
  • 被排除的半空間中的影像場沒有物理意義(該處位於導體內部,真實的場為零)。
  • 多個接地平面或球體可能需要無窮級數的影像電荷(見 §4-4.4 帶電球與接地平面)。
  • 鏡像電荷は、場を計算している領域の外側に置きます — 決してその内側には置きません。
  • 除外した半空間における鏡像の場には物理的な意味はありません(そこは導体内部であり、実際の場は零です)。
  • 複数の接地平面や球がある場合、鏡像の無限級数が必要になることがあります(§4-4.4 帯電球と接地平面を参照)。

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

當區域內沒有孤立的自由電荷,且導體與座標面對齊時,用 separation of variables(分離變數法)求解 \(\nabla^{2}V=0\):假設

孤立した自由電荷がなく、導体が座標面に沿っている場合は、変数分離法(separation of variables)によって \(\nabla^{2}V=0\) を解きます:次のように仮定します

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

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

代入 (4-81),除以 \(XYZ\),並注意到每一項只與一個變數有關,因此每一項都必須是常數。於是得到三個 ODEs:

これを (4-81) に代入して \(XYZ\) で割ると、各項がそれぞれ 1 つの変数だけに依存することから、各項は定数でなければなりません。こうして 3 つの常微分方程式(ODE)が得られます:

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

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.

由於 (4-89) 要求 separation constants 的平方和為零,至少有一個常數必須是虛數 — 使得對應的因子成為雙曲函數(\(\sinh, \cosh\))而非三角函數。

(4-89) は分離定数の 2 乗の和が零であることを要求するため、少なくとも 1 つの定数は虚数でなければなりません — その結果、対応する因子は三角関数ではなく双曲線関数(\(\sinh, \cosh\))になります。

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:

Boundary conditions 決定問題的分類:

境界条件(boundary conditions)によって問題が分類されます:

  • 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.
  • Dirichlet:在所有邊界上指定 \(V\)。
  • Neumann:指定法向導數 \(\partial V/\partial n\)(即 surface charge)。
  • Mixed(混合):部分邊界為 Dirichlet,其餘為 Neumann。
  • ディリクレ(Dirichlet):すべての境界上で \(V\) が指定されます。
  • ノイマン(Neumann):法線方向微分 \(\partial V/\partial n\)(すなわち表面電荷)が指定されます。
  • 混合(Mixed):一部の境界でディリクレ、残りでノイマンです。

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.

如何選擇 sin 或 cos、sinh 或 cosh:利用問題的對稱性 — 若 \(V\) 在 \(x=0\) 處為零則選 sin;若 \(V\) 對 \(x=0\) 對稱則選 cos;若條件是某一端 \(V=0\) 之類則選 sinh;若 \(V\) 在兩個對稱位置相等則選 cosh。

sin と cos、sinh と cosh の選び方:問題の対称性を利用します — \(x=0\) で \(V\) が零になるなら sin、\(V\) が \(x=0\) について対称なら cos、一端で \(V=0\) のような条件なら sinh、対称な 2 か所で \(V\) が等しいなら cosh を選びます。


4-6   Boundary-Value Problems — Cylindrical

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

對於沿 \(z\) 方向很長的幾何結構,略去對 \(z\) 的相依性,求解二維問題:

\(z\) 方向に長い形状では、\(z\) 依存性を落として 2 次元問題を解きます:

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

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:

假設 \(V(r,\phi)=R(r)\Phi(\phi)\)。分離變數後得到兩個 ODEs,separation constant 為 \(k^{2}\)。由於 \(\phi\) 具有週期 \(2\pi\) 的週期性,\(k\) 必須是整數 \(n\);通解變為:

\(V(r,\phi)=R(r)\Phi(\phi)\) と仮定します。変数分離により、分離定数 \(k^{2}\) をもつ 2 つの常微分方程式が得られます。\(\phi\) は周期 \(2\pi\) の周期性をもつため、\(k\) は整数 \(n\) でなければならず、一般解は次のようになります:

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

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

徑向規則:若區域包含 \(r=0\),捨去 \(r^{-n}\)(奇異);若區域延伸至 \(\infty\),捨去 \(r^{n}\)(發散)。

径方向の規則:領域が \(r=0\) を含むなら \(r^{-n}\) を捨てます(特異になるため)。領域が \(\infty\) まで広がるなら \(r^{n}\) を捨てます(発散するため)。

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

Coaxial cable(同軸電纜:內半徑 \(a\) 處電位為 \(V_0\),外半徑 \(b\) 接地):將 (4-130) 配合 BCs 可得

同軸ケーブル(coaxial cable:内半径 \(a\) が \(V_0\)、外半径 \(b\) が接地):(4-130) に境界条件を適用すると次式が得られます

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

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

具有任意邊界資料的非軸對稱問題,會導出對 \(\phi\) 的 Fourier series(見 Example 4-9,分裂圓柱)。

任意の境界データをもつ非軸対称問題では、\(\phi\) についてのフーリエ級数になります(Example 4-9 の分割円柱を参照)。


4-7   Boundary-Value Problems — Spherical

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

對於軸對稱問題(與 \(\phi\) 無關):

軸対称問題(\(\phi\) 依存性なし)では:

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

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:

以 \(V(R,\theta)=\Gamma(R)\Theta(\theta)\) 分離變數,separation constant 取 \(k^{2}=n(n+1)\)(由 \(\Theta\) 在 \(z\) 軸上的正則性所決定),得到:

\(V(R,\theta)=\Gamma(R)\Theta(\theta)\) と変数分離し、分離定数を \(k^{2}=n(n+1)\)(\(z\) 軸上で \(\Theta\) が正則であることから要請される)とすると、次式が得られます:

\[ \Gamma_n(R) = A_n R^{n} + B_n R^{-(n+1)} \qquad \text{(4-150)} \]
\[ \Theta_n(\theta) = P_n(\cos\theta) \quad \text{(Legendre polynomial)} \qquad \text{(4-153)} \]
\[ V_n(R,\theta) = \bigl[A_n R^{n} + B_n R^{-(n+1)}\bigr]P_n(\cos\theta) \qquad \text{(4-154)} \]
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:

經典範例 — 均勻電場 \(\mathbf{E}_0=\mathbf{a}_z E_0\) 中的導體球。BCs:\(V(b,\theta)=0\);當 \(R\to\infty\) 時 \(V\to -E_0 R\cos\theta\)。只有 \(n=1\) 的項留下:

代表的な例 — 一様電場 \(\mathbf{E}_0=\mathbf{a}_z E_0\) 中の導体球。境界条件:\(V(b,\theta)=0\)、および \(R\to\infty\) で \(V\to -E_0 R\cos\theta\)。\(n=1\) の項だけが残ります:

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

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

球外的電位等於外加均勻場的電位,加上位於球心、dipole moment 為 \(\mathbf{p}\) 的感應 dipole 所產生的電位。

球の外側の電位は、印加された一様電場の電位に、球の中心に置かれたモーメント \(\mathbf{p}\) の誘導双極子(induced dipole)の電位を加えたものになります。


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.
  • 支配方程式:\(\nabla^{2}V = -\rho/\epsilon\)(Poisson);\(\nabla^{2}V=0\)(Laplace,無電荷區域)。
  • Uniqueness theorem:滿足 \(\nabla^{2}V=-\rho/\epsilon\) 與 boundary conditions 的 \(V\) 就是唯一的解 — 因此任何有效的猜測(例如影像法)都是嚴謹的。
  • Method of images:平面 → 鏡像電荷 \(-Q\);圓柱 → 位於反演點 \(d_i=a^{2}/d\) 的線電荷 \(-\rho_\ell\);球 → 位於 \(d_i=a^{2}/d\) 的 \(Q_i=-(a/d)Q\)。
  • Two-wire line:\(C=\pi\epsilon_0/\cosh^{-1}(D/2a)\) (F/m)。
  • Cartesian BVP:\(X\,Y\,Z\) 試解;separation constants 滿足 \(k_x^{2}+k_y^{2}+k_z^{2}=0\) — 至少有一個因子是雙曲函數。
  • Cylindrical BVP:角向因子為整數 \(n\) 的 \(\sin n\phi,\cos n\phi\);徑向因子為 \(r^{n}\) 和/或 \(r^{-n}\);當 \(n=0\) 時,\(V=C_1\ln r+C_2\)。
  • Spherical 軸對稱 BVP:通解 \(V_n=[A_n R^{n}+B_n R^{-(n+1)}]P_n(\cos\theta)\);均勻場中的導體球 → (4-159)。
  • 邊界類型:Dirichlet(指定 V)、Neumann(指定 \(\partial V/\partial n\))、mixed(混合)。
  • 支配方程式:\(\nabla^{2}V = -\rho/\epsilon\)(ポアソン)、\(\nabla^{2}V=0\)(ラプラス、電荷のない領域)。
  • 一意性定理:\(\nabla^{2}V=-\rho/\epsilon\) と境界条件を満たす \(V\) はそれが唯一の解です — したがって妥当な推測(鏡像法など)はすべて厳密です。
  • 鏡像法:平面 → 鏡像電荷 \(-Q\)、円柱 → 反転点 \(d_i=a^{2}/d\) に線電荷 \(-\rho_\ell\)、球 → \(d_i=a^{2}/d\) に \(Q_i=-(a/d)Q\)。
  • 二線式線路:\(C=\pi\epsilon_0/\cosh^{-1}(D/2a)\) (F/m)。
  • 直交座標の境界値問題:\(X\,Y\,Z\) の形の解を仮定し、分離定数は \(k_x^{2}+k_y^{2}+k_z^{2}=0\) を満たします — 少なくとも 1 つの因子は双曲線関数です。
  • 円柱座標の境界値問題:角度方向の因子は整数 \(n\) をもつ \(\sin n\phi,\cos n\phi\)、径方向の因子は \(r^{n}\) および/または \(r^{-n}\)、\(n=0\) のときは \(V=C_1\ln r+C_2\) です。
  • 球座標の軸対称境界値問題:一般解は \(V_n=[A_n R^{n}+B_n R^{-(n+1)}]P_n(\cos\theta)\)、一様電場中の導体球 → (4-159)。
  • 境界の種類:ディリクレ(V を指定)、ノイマン(\(\partial V/\partial n\) を指定)、混合。

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.

節錄自 Cheng,Field and Wave Electromagnetics,§5-1 至 §5-7。請先讀完本節再嘗試下方的習題。

Cheng, Field and Wave Electromagnetics, §5-1 〜 §5-7 からの要約です。下の問題に取り組む前にお読みください。


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:

第 3 章と第 4 章では静止した電荷を扱いました。第 5 章では運動する電荷、すなわち定常(直流)電流を導入します。電流を生じさせる機構は 3 つあり、いま考えている状況がそのどれに当たるのかを見分けることが重要です。

  • 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.
  • 伝導電流(Conduction currents) — 導体や半導体の中を伝導電子(あるいは正孔)がドリフトする運動です。オームの法則に従います。
  • 電解電流(Electrolytic currents) — 電解質(たとえば電極間の食塩水)の中を正負のイオンが移動するものです。電解質はラプラス方程式に従うモデルで記述でき、これが解析的には求めにくい静電ポテンシャルを写し取る電解槽(electrolytic tank)の原理となっています。
  • 対流電流(Convection currents) — 真空中や希薄気体中を荷電粒子が集団として運動するものです(CRT 内の電子ビーム、雷など)。対流電流は流体力学的な質量輸送であり、オームの法則には従いません。

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.

ドリフト速度はきわめて小さい。伝導電子の平均ドリフト速度は、非常に良好な導体であっても \(10^{-4}\) 〜 \(10^{-3}\) m/s 程度にすぎません。電子と原子の衝突が絶えず起こり、運動エネルギーが熱として散逸するためです。導体は全体としては電気的に中性を保ち、静電気力によって内部のいかなる点にも電荷が蓄積しないようになっています。

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,不遵循 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

設單位體積內有 \(N\) 個 charge carriers,每個帶電荷 \(q\),以速度 \(\mathbf{u}\) 漂移,則 volume current density(體電流密度)定義為

単位体積あたり \(N\) 個の電荷担体(charge carriers)がそれぞれ電荷 \(q\) をもち、速度 \(\mathbf{u}\) でドリフトしているとき、体積電流密度(volume current density)は次のように定義されます。

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

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

通過任意曲面 \(S\) 的總電流即為 \(\mathbf{J}\) 的 flux:

任意の曲面 \(S\) を通過する全電流は \(\mathbf{J}\) の流束(flux)です:

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

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

由於 \(Nq\) 正是自由電荷的體密度 \(\rho\),對 convection currents 而言

\(Nq\) は自由電荷の体積密度 \(\rho\) にほかならないので、対流電流(convection currents)については

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

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

當多種載子(electrons、holes、ions)同時漂移時,current density 可直接疊加:\(\mathbf{J} = \sum_i N_i q_i \mathbf{u}_i\)。

複数の種類の担体(電子、正孔、イオン)が同時にドリフトする場合、電流密度は重ね合わされます:\(\mathbf{J} = \sum_i N_i q_i \mathbf{u}_i\)。


5-2.2   The Point Form of Ohm's Law

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

在 conductor 中,drift velocity 與外加電場成正比:

導体中では、ドリフト速度(drift velocity)は印加電界に正比例します:

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

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:

其中 \(\mu_e\) 為 electron mobility(遷移率,m²/V·s)。典型值:銅 \(3.2\times 10^{-3}\),鋁 \(1.4\times 10^{-4}\),銀 \(5.2\times 10^{-3}\)。將 (5-19) 代入 (5-3),並把負號吸收進正的比例常數 \(\sigma = -\rho_e \mu_e\),即得 point form of Ohm's law(Ohm's law 的點形式):

ここで \(\mu_e\) は電子の移動度(mobility)(m²/V·s)です。代表的な値は、銅 \(3.2\times 10^{-3}\)、アルミニウム \(1.4\times 10^{-4}\)、銀 \(5.2\times 10^{-3}\) です。(5-19) を (5-3) に代入し、符号を正の比例定数 \(\sigma = -\rho_e \mu_e\) に吸収させると、オームの法則の点形式(point form of Ohm's law)が得られます:

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

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:

比例常數 \(\sigma\)(單位為 siemens per metre,S/m,或 A/V·m)稱為介質的 conductivity(導電率)。此線性關係成立的 isotropic 材料稱為 ohmic media。對於同時有 electron 與 hole 貢獻的 semiconductors:

比例定数 \(\sigma\)(単位はジーメンス毎メートル、S/m、または A/V·m)は媒質の導電率(conductivity)です。この線形関係が成り立つ等方性材料をオーム性媒質(ohmic media)と呼びます。電子と正孔の両方が寄与する半導体では:

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

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.

Conductivity 的範圍極廣 — 銅 \(5.80\times 10^{7}\) S/m;鍺 \(2.2\);矽 \(1.6\times 10^{-3}\);硬橡膠 \(10^{-15}\)。與 permittivity 不同,常見材料的 conductivity 橫跨 22 個以上的數量級。\(\sigma\) 的倒數為 resistivity(電阻率,Ω·m),但 Cheng 全書偏好使用 conductivity。

導電率の範囲はきわめて広い — 銅 \(5.80\times 10^{7}\) S/m、ゲルマニウム \(2.2\)、シリコン \(1.6\times 10^{-3}\)、硬質ゴム \(10^{-15}\) です。誘電率とは異なり、身近な材料の導電率は 22 桁以上にわたって変化します。\(\sigma\) の逆数は抵抗率(resistivity)(Ω·m)ですが、Cheng は全編を通じて導電率で扱うことを好んでいます。

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 \(\sigma\):

式 (5-21) 是點形式。熟悉的電路理論關係 \(V_{12} = RI\) 並非點關係 — 它牽涉的是端點與截面。將 (5-21) 套用於長度 \(\ell\)、均勻截面 \(S\)、conductivity 為 \(\sigma\) 的均質直導體:

式 (5-21) は点形式です。よく知られた回路理論の関係 \(V_{12} = RI\) は点の関係ではなく、端子と断面に関するものです。(5-21) を長さ \(\ell\)、一様な断面積 \(S\)、導電率 \(\sigma\) の均質な直線導体に適用すると:

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

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:

靜電場是保守的 (conservative):\(\oint_C \mathbf{E}\cdot d\ell = 0\)(式 5-31)。代入 \(\mathbf{E}=\mathbf{J}/\sigma\) 可得 ohmic media 的版本:

静電界は保存的(conservative)です:\(\oint_C \mathbf{E}\cdot d\ell = 0\)(式 5-31)。\(\mathbf{E}=\mathbf{J}/\sigma\) を代入すると、オーム性媒質での形が得られます:

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

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.

這個式子告訴我們一件根本的事實:僅靠靜電場無法在閉合迴路中維持穩態電流。Charge carriers 與原子碰撞而耗散能量;這些能量必須由非保守的來源補充 — 電池(化學)、發電機(機械)、熱電偶(熱)或光伏電池(光)。這些來源在源內部產生一個等效的 impressed electric field intensity(外加電場強度)\(\mathbf{E}_i\),且僅存在於源內部。

この式は本質的なことを教えてくれます:静電界だけでは閉回路に定常電流を維持することはできません。電荷担体は原子との衝突でエネルギーを失います。そのエネルギーは非保存的な源 — 電池(化学)、発電機(機械)、熱電対(熱)、太陽電池(光)— によって補給されなければなりません。これらの源は、源の内部にのみ等価な印加電界強度(impressed electric field intensity)\(\mathbf{E}_i\) を生じます。

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

在源內部由 − 端到 + 端對 \(\mathbf{E}_i\) 取線積分,即定義該源的 electromotive force(電動勢,emf),記作 \(\mathscr{V}\):

源の内部で − 端子から + 端子まで \(\mathbf{E}_i\) を線積分したものが、その源の起電力(electromotive force, emf)を定義し、\(\mathscr{V}\) と表します:

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

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.

emf 的 SI 單位是 volt。儘管名稱中有 force,emf 並不是以 newton 計的力 — 它衡量的是非保守源的強度。Ideal voltage source 的內電阻為零,因此無論電流多大,其端電壓恆等於其 emf。

起電力の SI 単位はボルトです。その名前にもかかわらず、起電力はニュートン単位の力ではなく、非保存的な源の強さを表す量です。理想電圧源(ideal voltage source)は内部抵抗がゼロなので、電流によらず端子電圧は起電力に等しくなります。

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:

在載有穩態電流的 conductor 內部,驅動電荷的總電場是 \(\mathbf{E}+\mathbf{E}_i\) 之和,Ohm's law 的點形式推廣為 \(\mathbf{J} = \sigma(\mathbf{E}+\mathbf{E}_i)\)。取閉合迴路的線積分,即得 Kirchhoff's voltage law 的巨觀形式:

定常電流が流れる導体の内部では、電荷を駆動する全電界は和 \(\mathbf{E}+\mathbf{E}_i\) であり、オームの法則の点形式は \(\mathbf{J} = \sigma(\mathbf{E}+\mathbf{E}_i)\) に一般化されます。閉回路に沿って線積分をとると、キルヒホッフの電圧則(Kirchhoff's voltage law)の巨視的な形が得られます:

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

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.

沿任一閉合迴路,各 emf(電壓升)的代數和等於各 IR 電壓降的代數和。這是電路迴路分析的基礎。

任意の閉回路に沿って、起電力(電圧上昇)の代数和は IR 電圧降下の代数和に等しくなります。これが回路の閉路解析の基礎です。


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:

電荷守恆要求:流出閉合曲面的電流等於其所包圍電荷的減少率:

電荷保存則により、閉曲面から流出する電流は、その内部に含まれる電荷の減少率に等しくなければなりません:

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

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

套用 divergence theorem,並令兩邊的被積函數相等(因為 \(V\) 是任意的),即得 equation of continuity(連續方程式):

発散定理(divergence theorem)を適用し、\(V\) が任意であることから被積分関数どうしを等しいとおくと、連続の方程式(equation of continuity)が得られます:

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

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

對穩態電流,\(\partial\rho/\partial t = 0\),故

定常電流では \(\partial\rho/\partial t = 0\) なので、

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

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.

因此穩態電流是無散的 (divergenceless)(solenoidal)— 其流線自成閉合曲線,不像靜電場線那樣起於電荷、止於電荷。Ideal current source 的內電阻為無窮大。

したがって定常電流は発散がなく(divergenceless)、管状(solenoidal)です — その流線は自ら閉じており、電荷に始まり電荷に終わる静電界の電気力線とは異なります。理想電流源(ideal current source)は内部抵抗が無限大です。

Relaxation time

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

將 Ohm's law 與 continuity equation 及 \(\nabla\cdot\mathbf{E}=\rho/\epsilon\) 結合,對簡單介質可得

オームの法則を連続の方程式および \(\nabla\cdot\mathbf{E}=\rho/\epsilon\) と組み合わせると、単純媒質に対して次式が得られます。

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

whose solution is an exponential decay:

其解為指數衰減:

その解は指数関数的な減衰です:

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

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.

對銅而言 \(\tau \approx 1.52\times 10^{-19}\) s — 短到可以認為良導體內部各處的 \(\rho\) 皆為零。在良好的絕緣體中,relaxation time 可長達數小時甚至數天。

銅では \(\tau \approx 1.52\times 10^{-19}\) s であり、きわめて短いため、良導体の内部ではいたるところ \(\rho\) をゼロとみなせます。良好な絶縁体では、緩和時間(relaxation time)は数時間から数日に及ぶこともあります。


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

\(\mathbf{E}\) 推動電荷 \(q\) 以速度 \(\mathbf{u}\) 運動時所做功的功率為 \(p = q\mathbf{E}\cdot\mathbf{u}\)。對體積元素內所有載子求和,得到功率密度

\(\mathbf{E}\) が電荷 \(q\) を速度 \(\mathbf{u}\) で動かすときになす仕事の率は \(p = q\mathbf{E}\cdot\mathbf{u}\) です。体積要素内のすべての担体について和をとると、電力密度が得られます。

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

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

對體積 \(V\) 積分即得 Joule's law:

体積 \(V\) にわたって積分するとジュールの法則(Joule's law)が得られます:

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

(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:

(SI 單位是 watt,而非 joule — 「joule」是能量的單位。)對於 \(P = VI\) 且 \(V=IR\) 的均勻導體,我們重新得到熟悉的電路表達式:

(SI 単位はワットであり、ジュールではありません — 「ジュール」はエネルギーの単位です。)\(P = VI\) かつ \(V=IR\) が成り立つ一様な導体では、おなじみの回路の式が再び得られます:

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

5-6   Boundary Conditions for Current Density

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

在 ohmic media 中(內部沒有非保守源),穩態 current density 的支配方程式為:

オーム性媒質(内部に非保存的な源がない場合)における定常電流密度の支配方程式は次のとおりです:

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:

將這些方程式套用於 conductivity 分別為 \(\sigma_1\) 與 \(\sigma_2\) 的兩個 ohmic media 之交界面,可得 boundary conditions(邊界條件):

これらを導電率 \(\sigma_1\)、\(\sigma_2\) をもつ 2 つのオーム性媒質の界面に適用すると、境界条件(boundary conditions)が得られます:

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

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.

與第 3 章的靜電 boundary conditions(\(D_{1n}-D_{2n}=\rho_s\) 與 \(E_{1t}=E_{2t}\))比較:在無電荷的 dielectric 交界面上,\((\mathbf{J},\sigma)\) 與 \((\mathbf{D},\epsilon)\) 之間存在精確的類比。這正是 electrolytic-tank analogy(電解槽類比)的依據。

第 3 章の静電界の境界条件(\(D_{1n}-D_{2n}=\rho_s\) と \(E_{1t}=E_{2t}\))と比べてみましょう。電荷のない誘電体界面では、\((\mathbf{J},\sigma)\) と \((\mathbf{D},\epsilon)\) の間に厳密な類似(exact analogy)があります。これが電解槽アナロジー(electrolytic-tank analogy)の根拠です。

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

將 (5-59) 除以 (5-58),得到電流流線的 refraction relation(折射關係):

(5-59) を (5-58) で割ると、電流流線の屈折関係(refraction relation)が得られます:

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

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.

若介質 1 的導電性遠優於介質 2(\(\sigma_1 \gg \sigma_2\)),則 \(\alpha_2 \to 0\) — 不良導體中的電流幾乎垂直於良導體表面射出。

媒質 1 が媒質 2 よりもはるかに良い導体であれば(\(\sigma_1 \gg \sigma_2\))、\(\alpha_2 \to 0\) となります — 不良導体側の電流は、良導体の表面にほぼ垂直に出ていきます。

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:

交界面上的表面電荷。當穩態電流流過兩個 lossy dielectrics(\(\epsilon_1,\epsilon_2\) 與 \(\sigma_1,\sigma_2\) 皆為有限值)之間的邊界時,\(J_{1n}=J_{2n}\) 與 \(D_{1n}-D_{2n}=\rho_s\) 必須同時成立,因此必然出現表面電荷:

界面の表面電荷。定常電流が 2 つの損失性誘電体(有限の \(\epsilon_1,\epsilon_2\) と \(\sigma_1,\sigma_2\))の境界を横切って流れるとき、\(J_{1n}=J_{2n}\) と \(D_{1n}-D_{2n}=\rho_s\) を同時に満たす必要があるため、表面電荷が生じざるを得ません:

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

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

此表面電荷只有在 \(\sigma_2/\sigma_1 = \epsilon_2/\epsilon_1\) 時才會消失。

この表面電荷は \(\sigma_2/\sigma_1 = \epsilon_2/\epsilon_1\) のときにのみ消えます。


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:

在無源的均質 conductor 中,電流場既無散又無旋,因此 \(\mathbf{J}=-\nabla\psi\),其中 \(\psi = \sigma V\) 滿足 Laplace's equation:

源のない均質な導体中では、電流場は発散も回転もともにゼロなので、\(\mathbf{J}=-\nabla\psi\) と書け、\(\psi = \sigma V\) はラプラス方程式(Laplace's equation)を満たします:

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

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

因此穩態電流問題與靜電問題具有相同的數學形式 — 第 4 章的方法可直接套用。

したがって定常電流の問題は静電問題と同じ数学的形をもち、第 4 章の手法がそのまま適用できます。

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

對於埋在 lossy dielectric 介質(\(\epsilon, \sigma\))中的兩個 conductors,兩者之間的 resistance 為

損失性誘電体媒質(\(\epsilon, \sigma\))の中に置かれた 2 つの導体の間の抵抗は

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

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

與同一對 conductors 之間的 capacitance 比較,即得優雅的 R–C analogy:

同じ 2 つの導体間の静電容量と比較すると、簡潔なR–C アナロジー(R–C analogy)が得られます:

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

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:

結果:若兩 conductors 之間的 capacitance 已知,則填滿相同幾何的 lossy 介質之 leakage resistance(或 conductance)可立即求得 — 無需重新計算。對內半徑 \(a\)、外半徑 \(b\) 的 coaxial cable,每單位長度:

帰結:2 つの導体間の静電容量が既知であれば、同じ形状を満たす損失性媒質を通る漏れ抵抗(leakage resistance)(または漏れコンダクタンス)は直ちに求まり、計算し直す必要はありません。内半径 \(a\)、外半径 \(b\) の同軸ケーブルでは、単位長さあたり:

  • \(C_1 = 2\pi\epsilon/\ln(b/a)\)     (from Ch. 3)
  • \(R_1 = (\epsilon/\sigma)/C_1 = \ln(b/a)/(2\pi\sigma)\)     (leakage resistance)
  • \(C_1 = 2\pi\epsilon/\ln(b/a)\)     (出自第 3 章)
  • \(R_1 = (\epsilon/\sigma)/C_1 = \ln(b/a)/(2\pi\sigma)\)     (leakage resistance)
  • \(C_1 = 2\pi\epsilon/\ln(b/a)\)     (第 3 章より)
  • \(R_1 = (\epsilon/\sigma)/C_1 = \ln(b/a)/(2\pi\sigma)\)     (漏れ抵抗)

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

只要介質的 \(\epsilon\) 與 \(\sigma\) 具有相同的空間分布(對均質介質而言顯然成立),R–C analogy 就成立。

R–C アナロジーは、媒質の \(\epsilon\) と \(\sigma\) が同じ空間依存性をもつとき(均質媒質では自明に成り立ちます)常に成立します。


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.
  • Ohm's law 的點形式:\(\mathbf{J}=\sigma\mathbf{E}\);巨觀形式 \(V=IR\),其中 \(R=\ell/(\sigma S)\)。
  • KVL:\(\sum\mathscr{V}_j = \sum R_k I_k\)(\(\mathbf{E}+\mathbf{E}_i\) 沿迴路的線積分)。
  • KCL:\(\sum I_j = 0\),源自穩態電流的無散性 \(\nabla\cdot\mathbf{J}=0\)。
  • Equation of continuity:\(\nabla\cdot\mathbf{J} = -\partial\rho/\partial t\) — 電荷守恆的一般敘述。
  • Relaxation time:\(\tau=\epsilon/\sigma\);銅約 \(\sim 10^{-19}\) s,故良導體內部 \(\rho=0\)。
  • Joule's law:\(P=\int_V\mathbf{E}\cdot\mathbf{J}\,dv = I^2 R\)。
  • Boundary conditions:\(J_{1n}=J_{2n}\),\(J_{1t}/J_{2t}=\sigma_1/\sigma_2\);折射關係 \(\tan\alpha_2/\tan\alpha_1 = \sigma_2/\sigma_1\)。
  • R–C analogy:\(RC=\epsilon/\sigma\) — capacitance ↔ leakage resistance。
  • 與靜電學的類比:\(\mathbf{J}\leftrightarrow\mathbf{D}\),\(\sigma\leftrightarrow\epsilon\);所有靜電學技巧都可轉用於穩態電流問題。
  • オームの法則の点形式:\(\mathbf{J}=\sigma\mathbf{E}\);巨視的な形は \(V=IR\)、ただし \(R=\ell/(\sigma S)\)。
  • KVL:\(\sum\mathscr{V}_j = \sum R_k I_k\)(閉路に沿った \(\mathbf{E}+\mathbf{E}_i\) の線積分)。
  • KCL:\(\sum I_j = 0\)。定常電流の発散がゼロであること \(\nabla\cdot\mathbf{J}=0\) から導かれます。
  • 連続の方程式:\(\nabla\cdot\mathbf{J} = -\partial\rho/\partial t\) — 電荷保存の一般的な表現。
  • 緩和時間:\(\tau=\epsilon/\sigma\);銅では \(\sim 10^{-19}\) s なので、良導体の内部は \(\rho=0\)。
  • ジュールの法則:\(P=\int_V\mathbf{E}\cdot\mathbf{J}\,dv = I^2 R\)。
  • 境界条件:\(J_{1n}=J_{2n}\)、\(J_{1t}/J_{2t}=\sigma_1/\sigma_2\);屈折関係 \(\tan\alpha_2/\tan\alpha_1 = \sigma_2/\sigma_1\)。
  • R–C アナロジー:\(RC=\epsilon/\sigma\) — 静電容量 ↔ 漏れ抵抗。
  • 静電気学との類似:\(\mathbf{J}\leftrightarrow\mathbf{D}\)、\(\sigma\leftrightarrow\epsilon\);静電気学のあらゆる手法が定常電流の問題に転用できます。

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.

摘錄自 Cheng, Field and Wave Electromagnetics,§6-1 至 §6-13。請先閱讀本節,再嘗試下方的習題。

Cheng, Field and Wave Electromagnetics の §6-1 〜 §6-13 を要約したものです。下の問題に取り組む前に、まずこの節を読んでください。


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

靜止於 electric field 中的電荷 \(q\) 受到 \(\mathbf{F}_e = q\mathbf{E}\) 的作用。當電荷以速度 \(\mathbf{u}\) 運動時,實驗顯示會出現第二種力:它同時垂直於 \(\mathbf{u}\) 與空間中每一點的某個固定方向,其大小正比於 \(q\)、正比於 \(|\mathbf{u}|\),也正比於 \(\mathbf{u}\) 垂直於該固定方向的分量。這個 magnetic force 可表示為

電場の中に静止している電荷 \(q\) には \(\mathbf{F}_e = q\mathbf{E}\) の力が働きます。電荷が速度 \(\mathbf{u}\) で運動すると、実験によれば第二の力が現れます。それは \(\mathbf{u}\) と、各点で定まるある固定した方向の両方に垂直で、大きさは \(q\)、\(|\mathbf{u}|\)、および \(\mathbf{u}\) のその固定方向に垂直な成分に比例します。この磁気力(magnetic force)は次式で表されます。

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

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:

而新的向量場 \(\mathbf{B}\)(單位為 webers per square meter,即 teslas)稱為 magnetic flux density(磁通密度)。將電與磁的效應合併,即得 Lorentz's force equation:

ここで新たに現れたベクトル場 \(\mathbf{B}\)(単位はウェーバ毎平方メートル、すなわちテスラ)が磁束密度(magnetic flux density)です。電気的効果と磁気的効果を合わせると、ローレンツの力の式(Lorentz's force equation)が得られます:

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

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.

Lorentz's equation 本身即可視為電磁模型的一條 fundamental postulate — 它同時定義了 \(\mathbf{E}\)(透過靜止電荷所受的力)與 \(\mathbf{B}\)(透過運動電荷所受的力)。它無法由其他 postulates 推導出來。

ローレンツの式そのものを、電磁気モデルの基本公理(fundamental postulate)とみなすこともできます。この式は \(\mathbf{E}\)(静止した電荷に働く力を通して)と \(\mathbf{B}\)(運動する電荷に働く力を通して)の両方を定義しており、他の公理から導くことはできません。


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 \(\mathbf{B}\):

在自由空間中,magnetostatics 恰好建立在兩條 postulates 之上,分別指定 \(\mathbf{B}\) 的 divergence 與 curl:

自由空間における静磁気学(magnetostatics)は、\(\mathbf{B}\) の発散(divergence)と回転(curl)を指定する、ちょうど 2 つの公理の上に構築されます:

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.

不存在 magnetic monopoles。由於 \(\nabla\cdot\mathbf{B} = 0\),磁力線必定自我封閉 — 不存在與 electric charge density 對應的磁荷。把一根棒狀磁鐵切開並不會得到孤立的 N 極或 S 極,而是得到兩根較小的磁鐵,如此一直分割到原子尺度仍然成立。式 (6-9) 即為 law of conservation of magnetic flux(磁通守恆定律)。

磁気単極子(magnetic monopole)は存在しない。\(\nabla\cdot\mathbf{B} = 0\) であるため、磁束線は必ず閉じた曲線になります。電荷密度に相当する磁気的な量は存在しません。棒磁石を切断しても孤立した N 極や S 極は得られず、2 本の小さな磁石ができるだけで、これは原子の寸法まで分割しても変わりません。式 (6-9) は磁束保存の法則(law of conservation of magnetic flux)です。

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).

Ampère's circuital law 指出:\(\mathbf{B}\) 沿任一封閉路徑的 circulation,等於 \(\mu_0\) 乘以穿過該路徑所圍曲面的總電流。對 (6-7) 取 \(\nabla\cdot(\nabla\times\mathbf{B}) = 0\),立即得到 \(\nabla\cdot\mathbf{J} = 0\),與穩態電流的連續方程式 (5-45) 一致。

アンペールの周回積分の法則(Ampère's circuital law)は、任意の閉路に沿った \(\mathbf{B}\) の循環が、その閉路で囲まれた面を貫く全電流の \(\mu_0\) 倍に等しいことを述べています。(6-7) に \(\nabla\cdot(\nabla\times\mathbf{B}) = 0\) を適用すると直ちに \(\nabla\cdot\mathbf{J} = 0\) が得られ、定常電流の連続の式 (5-45) と整合します。


6-3   Vector Magnetic Potential A

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

由於 \(\mathbf{B}\) 是 solenoidal(無散)的,它可以表示為另一個向量場的 curl:

\(\mathbf{B}\) はソレノイダル(発散がゼロ)であるため、別のベクトル場の回転として表すことができます:

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

\(\mathbf{A}\) 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

\(\mathbf{A}\) 稱為 vector magnetic potential(向量磁位),單位為 Wb/m。要完整定義一個向量,必須同時指定其 curl(已由 (6-15) 給定)與其 divergence — 後者我們可以自由選擇。將 (6-15) 代入 curl postulate,並展開 \(\nabla\times\nabla\times\mathbf{A} = \nabla(\nabla\cdot\mathbf{A}) - \nabla^2\mathbf{A}\),得到

\(\mathbf{A}\) をベクトル磁位(vector magnetic potential)と呼び、単位は Wb/m です。ベクトルを定めるには、その回転((6-15) で指定済み)と発散の両方を指定する必要がありますが、後者は自由に選ぶことができます。(6-15) を回転の公理に代入し、\(\nabla\times\nabla\times\mathbf{A} = \nabla(\nabla\cdot\mathbf{A}) - \nabla^2\mathbf{A}\) と展開すると、次式が得られます。

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

Choosing the Coulomb gauge

選取 Coulomb gauge

クーロンゲージ(Coulomb gauge)を選ぶと、すなわち

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

reduces this to a vector Poisson equation:

即可將其化簡為 vector Poisson equation:

これはベクトルのポアソン方程式(vector Poisson equation)に帰着します:

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

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

它的每個 Cartesian 分量都是一個 scalar Poisson equation,形式與靜電學中的 \(\nabla^2 V = -\rho/\epsilon_0\) 完全相同。因此其解為

その直交座標成分のそれぞれは、静電気学の \(\nabla^2 V = -\rho/\epsilon_0\) と同じ形のスカラーのポアソン方程式です。したがって解は次のようになります。

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

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

A 的物理意義:\(\mathbf{A}\) 沿封閉迴路的線積分,等於穿過該迴路所圍曲面的 magnetic flux:

A の物理的意味:閉じた経路に沿った \(\mathbf{A}\) の線積分は、その経路で囲まれた面を貫く磁束に等しくなります:

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

6-4   Biot–Savart Law

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

對細導線而言,\(\mathbf{J}\,dv' = I\,d\ell'\)。式 (6-23) 變為

細い導線では \(\mathbf{J}\,dv' = I\,d\ell'\) となります。式 (6-23) は次のようになります。

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

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

取 curl 並處理 \(\nabla(1/R) = -\mathbf{a}_R/R^2\) 的運算細節後,即得 Biot–Savart law:

回転をとり、\(\nabla(1/R) = -\mathbf{a}_R/R^2\) の計算を整理すると、ビオ・サバールの法則(Biot–Savart law)が得られます:

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

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.

其中 \(\mathbf{a}_R\) 是由電流源元素指向場點的單位向量。Biot–Savart 直接由載流電路計算 \(\mathbf{B}\);它是 Coulomb's law 求 \(\mathbf{E}\) 的磁學對應。當對稱性(圓柱、平面、環形)提供一條 \(|\mathbf{B}|\) 為常數的路徑時,Ampère's law 較為方便;若沒有這種對稱性,Biot–Savart 就是該用的工具。

ここで \(\mathbf{a}_R\) は電流素片から観測点へ向かう単位ベクトルです。ビオ・サバールの法則は電流の流れる回路から \(\mathbf{B}\) を直接計算するもので、\(\mathbf{E}\) に対するクーロンの法則の磁気版にあたります。対称性(円筒、平面、トロイダル)によって \(|\mathbf{B}|\) が一定となる経路がとれる場合はアンペールの法則が便利ですが、そうでない場合はビオ・サバールの法則が用いられます。


6-5   The Magnetic Dipole

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

面積為 \(S\)、載有電流 \(I\) 的小電流迴路,其行為如同一個 magnetic dipole(磁偶極),其矩為

面積 \(S\) の小さな電流ループに電流 \(I\) が流れているとき、それは次のモーメントをもつ磁気双極子(magnetic dipole)として振る舞います。

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

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

其中 \(\mathbf{a}_n\) 由電流方向依右手定則決定。在距離 \(R \gg\) 迴路尺寸處,vector potential 為

ここで \(\mathbf{a}_n\) は電流の向きから右手の法則で定まります。距離 \(R \gg\) ループの寸法 となる位置では、ベクトル磁位は

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

and the magnetic flux density is

而 magnetic flux density 為

となり、磁束密度は

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

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.

這與 electric dipole 的場 \((\mathbf{p}, V)\) 完全類比 — \(\mathbf{A}\) 扮演 \(V\) 的角色,且角度相依性相同。

となります。これは電気双極子の場 \((\mathbf{p}, V)\) と完全に対応しており、\(\mathbf{A}\) が \(V\) の役割を果たし、角度依存性も同じです。


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:

在物質中,電子的軌道運動與自旋形成微觀的磁偶極。外加磁場使它們部分排列整齊,產生巨觀的 magnetization vector \(\mathbf{M}\)(A/m)— 即單位體積的磁矩。就外部磁場而言,一個被磁化的物體可用等效的體電流密度與面電流密度取代:

物質中では、電子の軌道運動とスピンが微視的な磁気双極子をつくります。外部から磁場を加えるとそれらが部分的に整列し、巨視的な磁化ベクトル(magnetization vector)\(\mathbf{M}\)(A/m)、すなわち単位体積あたりの磁気モーメントが生じます。磁化された物体は、外部につくる磁場に関する限り、等価な体積電流密度と表面電流密度で置き換えることができます:

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

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 \(|\mathbf{M}|\).

均勻磁化的棒狀磁鐵沒有體電流(\(\mathbf{M}\) 為常數時 \(\nabla\times\mathbf{M} = 0\)),但帶有面電流 \(\mathbf{M}\times\mathbf{a}_n\),實際上使它成為線電流密度為 \(|\mathbf{M}|\) 的 solenoid。

一様に磁化された棒磁石には体積電流はありません(\(\mathbf{M}\) が一定なら \(\nabla\times\mathbf{M} = 0\))が、表面電流 \(\mathbf{M}\times\mathbf{a}_n\) をもち、実質的に線電流密度 \(|\mathbf{M}|\) のソレノイドと等価になります。


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:

在物質中,Ampère's law 必須同時計入 free current \(\mathbf{J}\) 與 bound(magnetization)current \(\mathbf{J}_m\):\((1/\mu_0)\nabla\times\mathbf{B} = \mathbf{J} + \nabla\times\mathbf{M}\)。重新整理後,自然引出一個新的基本場量的定義:

物質中では、アンペールの法則は自由電流 \(\mathbf{J}\) と束縛(磁化)電流 \(\mathbf{J}_m\) の両方を考慮しなければなりません:\((1/\mu_0)\nabla\times\mathbf{B} = \mathbf{J} + \nabla\times\mathbf{M}\)。これを整理すると、新しい基本的な場の量を定義する動機が得られます:

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

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

稱為 magnetic field intensity(磁場強度)。其支配方程式只涉及 free current:

これを磁界の強さ(magnetic field intensity)と呼びます。その支配方程式には自由電流だけが現れます:

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

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

對 linear, isotropic 介質,\(\mathbf{M} = \chi_m\mathbf{H}\),其中 \(\chi_m\) 為 magnetic susceptibility(磁化率,無因次)。代入 (6-75) 得到 constitutive relation

線形・等方性の媒質では \(\mathbf{M} = \chi_m\mathbf{H}\) が成り立ち、\(\chi_m\) を磁化率(magnetic susceptibility)(無次元)といいます。これを (6-75) に代入すると、構成関係式が得られます。

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

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

對於繞有 \(N\) 匝、載有電流 \(I\) 的環形鐵磁芯,由 Ampère's law 得 \(H = NI/\ell\) 及 \(B = \mu H\),故磁通為 \(\Phi = BA = NI/\mathcal{R}\),其中 reluctance(磁阻)為

\(N\) 回巻きで電流 \(I\) が流れるトロイダル強磁性コアでは、アンペールの法則から \(H = NI/\ell\) と \(B = \mu H\) が得られ、磁束は \(\Phi = BA = NI/\mathcal{R}\) となります。ここで磁気抵抗(reluctance)は

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

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

而 \(NI\) 稱為 magnetomotive force(磁動勢,mmf,單位為 ampere-turns)。Magnetic circuits 遵循與 KVL/KCL 類比的定律:

であり、\(NI\) は起磁力(magnetomotive force、mmf、単位はアンペア回数)です。磁気回路は KVL/KCL に対応する法則に従います:

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

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

其類比關係為:\(\mathscr{V}\leftrightarrow NI\)、\(I\leftrightarrow\Phi\)、\(R\leftrightarrow\mathcal{R}\)、\(\sigma\leftrightarrow\mu\)。

対応関係は \(\mathscr{V}\leftrightarrow NI\)、\(I\leftrightarrow\Phi\)、\(R\leftrightarrow\mathcal{R}\)、\(\sigma\leftrightarrow\mu\) です。


6-9   Behavior of Magnetic Materials

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

依 \(\chi_m\) 的正負與大小,可區分為三類:

\(\chi_m\) の符号と大きさによって、3 つの種類に分けられます:


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:

將 \(\nabla\cdot\mathbf{B} = 0\) 應用於界面上的 pillbox,並將 \(\oint\mathbf{H}\cdot d\ell = I_{\text{free}}\) 應用於跨越界面的扁平迴路,可得:

\(\nabla\cdot\mathbf{B} = 0\) を境界面上の小さな円筒(ピルボックス)に、\(\oint\mathbf{H}\cdot d\ell = I_{\text{free}}\) を境界面をまたぐ扁平な閉路に適用すると、次が得られます:

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

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.

對兩種線性介質,法向條件變為 \(\mu_1 H_{1n} = \mu_2 H_{2n}\)。在任何實際物質之間的界面上(\(\mathbf{J}_s = 0\)),切向 \(\mathbf{H}\) 是連續的;只有在理想化的 perfect conductor 或 superconductor 表面上才會不連續。與靜電學比較:\(\mathbf{D}\) 的法向分量跳變 \(\rho_s\),\(\mathbf{E}\) 的切向分量連續。

2 つの線形媒質の場合、法線方向の条件は \(\mu_1 H_{1n} = \mu_2 H_{2n}\) となります。物理的な媒質どうしの境界面(\(\mathbf{J}_s = 0\))では接線方向の \(\mathbf{H}\) は連続であり、不連続になるのは理想化された完全導体や超伝導体の表面だけです。静電気学と比較すると、\(\mathbf{D}\) の法線成分は \(\rho_s\) だけ跳び、\(\mathbf{E}\) の接線成分は連続です。


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

迴路 \(C_1\) 產生、並鏈結迴路 \(C_2\) 的互磁通為 \(\Phi_{12} = \int_{S_2}\mathbf{B}_1\cdot d\mathbf{s}_2\)。由 Biot–Savart 可知 \(\mathbf{B}_1 \propto I_1\),故

ループ \(C_1\) がつくり、ループ \(C_2\) と鎖交する相互磁束は \(\Phi_{12} = \int_{S_2}\mathbf{B}_1\cdot d\mathbf{s}_2\) です。ビオ・サバールの法則より \(\mathbf{B}_1 \propto I_1\) なので、

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

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

其中 \(\Lambda_{12} = N_2\Phi_{12}\) 為 flux linkage(磁通鏈)。對兩個多匝電路完整推導後,可得 Neumann's formula:

ここで \(\Lambda_{12} = N_2\Phi_{12}\) は鎖交磁束(flux linkage)です。2 つの多巻き回路について計算を進めると、ノイマンの公式(Neumann's formula)が得られます:

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

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, \(L \propto N^2\).

由互易性可得 \(L_{12} = L_{21}\)。Self-inductance(自感)\(L\) 的定義方式類似,只需令兩個迴路為同一迴路。對線性介質而言,inductance 純粹是幾何量,與電流大小無關。對 solenoid,\(L \propto N^2\)。

相反性より \(L_{12} = L_{21}\) が成り立ちます。自己インダクタンス(self-inductance)\(L\) も、2 つのループを同一のものとおくことで同様に定義されます。インダクタンスは(線形媒質では)純粋に幾何学的な量で、電流の大きさには依存しません。ソレノイドでは \(L \propto N^2\) です。


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:

在自感為 \(L_1\) 的 inductor 中建立電流 \(I_1\) 時,克服感應 emf 所做的功,以 magnetic energy 的形式儲存:

自己インダクタンス \(L_1\) のインダクタに電流 \(I_1\) を立ち上げる際に誘導起電力に逆らってなされる仕事は、磁気エネルギーとして蓄えられます:

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

For a system of N coupled loops:

對 N 個耦合迴路所組成的系統:

N 個の結合したループからなる系では:

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

Expressed in terms of field quantities:

以場量表示:

場の量で表すと:

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

with magnetic energy density

其中 magnetic energy density(磁能密度)為

ここで磁気エネルギー密度は

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

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.

這些形式與靜電學中的 \(w_e = \tfrac{1}{2}\mathbf{D}\cdot\mathbf{E}\) 完全類比。由場的積分計算 \(W_m\),再令其等於 \(\tfrac{1}{2}LI^2\),往往是求 inductance 比計算磁通鏈更容易的途徑。

です。これらは静電気学の \(w_e = \tfrac{1}{2}\mathbf{D}\cdot\mathbf{E}\) とちょうど対応する形です。場の積分から \(W_m\) を計算して \(\tfrac{1}{2}LI^2\) と等置する方法は、鎖交磁束を数えるよりもインダクタンスを求めやすいことがよくあります。


6-13   Magnetic Forces and Torques

Force on a current-carrying element:

載流元素所受的力:

電流素片に働く力:

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

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

兩條平行導線分別載有電流 \(I_1, I_2\),相距 \(d\),所受的力為

電流 \(I_1, I_2\) が流れ、距離 \(d\) だけ離れた2 本の平行導線には、次の力が働きます。

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

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

電流同向 → 相吸;反向 → 相斥。(與電荷的 Coulomb's law 恰好相反 — 這是一個容易忽略的細節。)

同じ向き → 引力、逆向き → 斥力です。(電荷に対するクーロンの法則とは逆の関係になる点に注意してください。)

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

電流迴路所受的力矩:偶極矩為 \(\mathbf{m}\) 的迴路置於均勻 \(\mathbf{B}\) 中時,

電流ループに働くトルク:双極子モーメント \(\mathbf{m}\) のループを一様な \(\mathbf{B}\) の中に置くと、

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

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.

這正是 d-c motor 的運作原理:split-ring commutator 每半圈反轉一次迴路電流,使 \(\mathbf{T}\) 維持同一方向,轉子得以持續旋轉。

これが直流モータの動作原理です。分割リング整流子が半回転ごとにループ電流の向きを反転させ、\(\mathbf{T}\) の向きを保って回転子を回し続けます。

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.

Hall effect:在磁場 \(\mathbf{B}\) 中流動的電流 \(\mathbf{J}\),會在導體兩側產生橫向電壓 \(V_h\)。\(V_h\) 的正負號可以判別載子是電子還是電洞 — 這是區分 n 型與 p 型半導體的重要診斷方法。

ホール効果(Hall effect):磁場 \(\mathbf{B}\) の中を流れる電流 \(\mathbf{J}\) は、導体の横方向に電圧 \(V_h\) を生じさせます。\(V_h\) の符号からキャリアが電子か正孔かが分かり、n 形と p 形の半導体を区別する重要な診断手段になります。


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.

摘要自 Cheng,Field and Wave Electromagnetics,§7-1 至 §7-7。請先讀完本節,再嘗試下方的習題。

Cheng『Field and Wave Electromagnetics』§7-1 〜 §7-7 の要約です。下の問題に取り組む前にお読みください。


7-1   Why Static Models Are Not Enough

Chapters 3–6 built two independent models — electrostatic and magnetostatic:

第 3–6 章建立了兩個彼此獨立的模型 — electrostatic 與 magnetostatic:

第 3 章から第 6 章では、静電気学(electrostatic)と静磁気学(magnetostatic)という 2 つの独立したモデルを構築しました:

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.

在靜態情況下,\(\{\mathbf{E}, \mathbf{D}\}\) 與 \(\{\mathbf{B}, \mathbf{H}\}\) 是 decoupled 的:electrostatic field 可以驅動 steady current,而 steady current 會產生 static magnetic field,但 \(\mathbf{E}\) 僅由電荷分佈即可求得。Magnetic field 只是其結果。

静的な場合、\(\{\mathbf{E}, \mathbf{D}\}\) と \(\{\mathbf{B}, \mathbf{H}\}\) は互いに分離(decoupled)しています。静電場は定常電流を流し、その電流が静磁場を作りますが、\(\mathbf{E}\) は電荷分布だけから求められます。磁場は単にその結果にすぎません。

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.

第 7 章將說明隨時間變化的 magnetic field 會產生 electric field,而隨時間變化的 electric field 也會產生 magnetic field。因此兩條 curl equations 都必須修正。修正後得到的四條方程式就是 Maxwell's equations — 整個電磁理論的基礎。

第 7 章では、時間変化する磁場が電場を生み、時間変化する電場が磁場を生むことを示します。したがって 2 つの回転(curl)の式は修正されなければなりません。その結果得られる 4 つの式がマクスウェルの方程式(Maxwell's equations)であり、電磁気学全体の基礎となります。


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:

1831 年,Faraday 以實驗證明:穿過迴路的 magnetic flux 發生變化時會感應出 emf。依循 deductive approach,我們將點形式 (point-form) 的關係式當作 postulate:

1831 年、ファラデーはループを貫く磁束(magnetic flux)が変化すると起電力(emf)が誘導されることを実験で示しました。演繹的アプローチに従い、ここでは微分形(point form)の関係式を公理(postulate)として採用します:

\[ \boxed{\nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t}} \qquad \text{(7-1)} \]

Applying Stokes's theorem gives the integral form:

套用 Stokes's theorem 可得積分形式:

ストークスの定理(Stokes's theorem)を適用すると積分形が得られます:

\[ \oint_C \mathbf{E}\cdot d\ell = -\int_S \frac{\partial\mathbf{B}}{\partial t}\cdot d\mathbf{s} \qquad \text{(7-2)} \]

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.

式 (7-1) 在空間中每一點都成立,無論該處是否存在實體電路。隨時間變化的 magnetic field 使 \(\mathbf{E}\) 成為 non-conservative — 它不再能僅以 scalar potential 的 gradient 來表示。

式 (7-1) は、実際の回路が存在するかどうかにかかわらず、空間のあらゆる点で成り立ちます。時間変化する磁場は \(\mathbf{E}\) を非保存的(non-conservative)にします。すなわち、もはやスカラーポテンシャルの勾配だけでは表せません。

Transformer emf (stationary circuit, changing B)

For a stationary closed circuit, (7-2) becomes

對靜止的閉合電路,(7-2) 化為

静止した閉回路に対しては、(7-2) は次のようになります:

\[ \mathscr{V} = \oint_C \mathbf{E}\cdot d\ell = -\frac{d\Phi}{dt} \quad (\text{V}) \qquad \text{(7-6)} \]

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.

這就是 Faraday's law of electromagnetic induction(法拉第電磁感應定律)。感應 emf 等於 flux linkage 變化率的負值;負號(Lenz's law)表示感應電流會反抗造成它的 flux 變化。這正是 transformers 的工作原理。鐵磁性鐵心中隨時間變化的 flux 也會在鐵心本身感應出局部的 eddy currents;採用疊片 (lamination) 可降低這種 eddy-current loss。

これがファラデーの電磁誘導の法則(Faraday's law of electromagnetic induction)です。誘導起電力は鎖交磁束の時間変化率に負号を付けたものに等しく、その負号(レンツの法則(Lenz's law))は、誘導電流がそれを生じさせた磁束の変化を妨げる向きに流れることを表します。これが変圧器の原理です。強磁性体の鉄心内で磁束が時間変化すると、鉄心自体にも局所的な渦電流(eddy current)が誘導されます。積層(lamination)はこの渦電流損を低減します。

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,

以速度 \(\mathbf{u}\) 在靜態 \(\mathbf{B}\) 中運動的 conductor,其內部的自由電荷會受到磁力作用。在 conductor 的座標系中,這等效於一個沿導體作用的感應場 \(\mathbf{u}\times\mathbf{B}\)。對於部分在運動中的閉合電路,

静的な \(\mathbf{B}\) の中を速度 \(\mathbf{u}\) で運動する導体では、その自由電荷に磁気力が働きます。導体の座標系で見ると、これは導体に沿って働く誘導電場 \(\mathbf{u}\times\mathbf{B}\) と等価です。一部が運動している閉回路については、

\[ \mathscr{V}' = \oint_C (\mathbf{u}\times\mathbf{B})\cdot d\ell \qquad \text{(7-24)} \]

called the motional emf or "flux-cutting" emf. Only parts of the circuit moving across flux lines contribute.

稱為 motional emf(運動電動勢)或「flux-cutting」emf。只有橫越 flux lines 運動的那部分電路才有貢獻。

これを運動起電力(motional emf)、あるいは「磁束切断(flux-cutting)」起電力と呼びます。磁束線を横切って運動している回路の部分だけが寄与します。

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.

若電路既在運動、又處於隨時間變化的場中,兩種貢獻會相加 — 但它們可以乾淨地合併成單一表示式 \(\mathscr{V} = -d\Phi/dt\),其中 total time derivative 沿著運動中的迴路取得。

回路が運動しており、かつ時間変化する磁場の中にある場合、両方の寄与が加わります。しかし両者はきれいにまとまり、運動するループに沿って全時間微分をとった単一の式 \(\mathscr{V} = -d\Phi/dt\) で表されます。


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:

在隨時間變化的情況下,靜態的 curl equation \(\nabla\times\mathbf{H} = \mathbf{J}\) 與電荷守恆 \(\nabla\cdot\mathbf{J} = -\partial\rho/\partial t\) 是互相矛盾的,因為 \(\nabla\cdot(\nabla\times\mathbf{H}) \equiv 0\) 會強迫 \(\nabla\cdot\mathbf{J} = 0\)。Maxwell 的絕妙修補:加入一項 \(\partial\mathbf{D}/\partial t\) — 即 displacement current density(位移電流密度):

時間変化する場合、静的な回転の式 \(\nabla\times\mathbf{H} = \mathbf{J}\) は電荷保存則 \(\nabla\cdot\mathbf{J} = -\partial\rho/\partial t\) と矛盾します。\(\nabla\cdot(\nabla\times\mathbf{H}) \equiv 0\) であるため、\(\nabla\cdot\mathbf{J} = 0\) が強制されてしまうからです。マクスウェルの見事な修正は、項 \(\partial\mathbf{D}/\partial t\)、すなわち変位電流密度(displacement current density)を加えることでした:

\[ \boxed{\nabla\times\mathbf{H} = \mathbf{J} + \frac{\partial\mathbf{D}}{\partial t}} \qquad \text{(7-52)} \]

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.

現在由 continuity 可知 \(\nabla\cdot(\nabla\times\mathbf{H}) = \nabla\cdot\mathbf{J} + \partial\rho/\partial t = 0\) 自動成立。\(\partial\mathbf{D}/\partial t\) 這一項具有 current density 的因次 (A/m²),卻不需要任何實際的電荷運動 — 空氣電容器兩極板之間隨時間變化的 \(\mathbf{D}\),會像 conduction current 一樣產生 magnetic field。隨時間變化的 electric field 會產生 magnetic field。

これにより、連続の式から \(\nabla\cdot(\nabla\times\mathbf{H}) = \nabla\cdot\mathbf{J} + \partial\rho/\partial t = 0\) が自動的に成り立ちます。項 \(\partial\mathbf{D}/\partial t\) は電流密度の次元(A/m²)を持ちますが、実際の電荷の運動は必要ありません。空気コンデンサの極板間で時間変化する \(\mathbf{D}\) は、伝導電流と同じように磁場を生み出します。時間変化する電場は磁場を生成するのです。

Maxwell's Equations

The four self-consistent time-varying equations:

四條自洽的 time-varying equations:

自己無撞着な 4 つの時間変化の方程式は次のとおりです:

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.

四個基本場量為 \(\mathbf{E}, \mathbf{D}, \mathbf{B}, \mathbf{H}\) — 共十二個純量未知數。兩條 curl equations 提供六個純量關係式;兩條 constitutive relations \(\mathbf{D} = \epsilon\mathbf{E}\) 與 \(\mathbf{H} = \mathbf{B}/\mu\) 再提供六個。兩條 divergence equations 並非獨立 — 它們可由 curl equations 與 continuity 導出。Maxwell's equations 加上 Lorentz force equation,便能描述所有巨觀電磁現象。

4 つの基本的な場は \(\mathbf{E}, \mathbf{D}, \mathbf{B}, \mathbf{H}\) で、合計 12 個のスカラー未知数があります。2 つの回転の式は 6 個のスカラー関係式を与え、2 つの構成関係式(constitutive relations)\(\mathbf{D} = \epsilon\mathbf{E}\) および \(\mathbf{H} = \mathbf{B}/\mu\) がさらに 6 個を与えます。2 つの発散の式は独立ではなく、回転の式と連続の式から導くことができます。マクスウェルの方程式はローレンツ力の式と合わせて、巨視的な電磁現象のすべてを記述します。

In the static limit (\(\partial/\partial t \to 0\)) they reduce to the two pairs of Table 7-1.

在靜態極限 (\(\partial/\partial t \to 0\)) 下,它們退化為 Table 7-1 的兩組方程式。

静的極限(\(\partial/\partial t \to 0\))では、これらは Table 7-1 の 2 組の式に帰着します。


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:

由於 \(\nabla\cdot\mathbf{B} = 0\) 仍然成立,我們保留 \(\mathbf{B} = \nabla\times\mathbf{A}\) (6-15)。代入 Faraday's law:

\(\nabla\cdot\mathbf{B} = 0\) は依然として成り立つので、\(\mathbf{B} = \nabla\times\mathbf{A}\) (6-15) をそのまま用います。これをファラデーの法則に代入すると:

\[ \nabla\times\left(\mathbf{E} + \frac{\partial\mathbf{A}}{\partial t}\right) = 0 \]

The quantity in parentheses is curl-free, so it can be written as \(-\nabla V\):

括號中的量是 curl-free 的,因此可寫成 \(-\nabla V\):

括弧内の量は回転がゼロ(curl-free)なので、\(-\nabla V\) と書くことができます:

\[ \boxed{\mathbf{E} = -\nabla V - \frac{\partial\mathbf{A}}{\partial t}} \qquad \text{(7-57)} \]

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.

現在 \(\mathbf{E}\) 有兩項貢獻:來自電荷分佈的「electrostatic」部分 \(-\nabla V\),以及來自隨時間變化的 magnetic field 的「induction」部分 \(-\partial\mathbf{A}/\partial t\)。在靜態極限下只有前者留存。

こうして \(\mathbf{E}\) は 2 つの寄与を持つことになります。電荷分布に由来する「静電的」な部分 \(-\nabla V\) と、時間変化する磁場に由来する「誘導」部分 \(-\partial\mathbf{A}/\partial t\) です。静的極限では前者だけが残ります。

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

將這些 potentials 代入 Ampère–Maxwell equation,得 \(\nabla(\nabla\cdot\mathbf{A}) - \nabla^2\mathbf{A} = \mu\mathbf{J} - \mu\epsilon\nabla(\partial V/\partial t) - \mu\epsilon\partial^2\mathbf{A}/\partial t^2\)。選取 Lorenz gauge(勞侖次規範)

これらのポテンシャルをアンペール–マクスウェルの式に代入すると \(\nabla(\nabla\cdot\mathbf{A}) - \nabla^2\mathbf{A} = \mu\mathbf{J} - \mu\epsilon\nabla(\partial V/\partial t) - \mu\epsilon\partial^2\mathbf{A}/\partial t^2\) が得られます。ここでローレンツゲージ(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:

即可將方程式解耦為兩條 non-homogeneous wave equations:

を選ぶと、方程式は 2 つの非斉次波動方程式(non-homogeneous wave equations)に分離されます:

\[ \nabla^2\mathbf{A} - \mu\epsilon\frac{\partial^2\mathbf{A}}{\partial t^2} = -\mu\mathbf{J} \qquad \text{(7-63)} \]
\[ \nabla^2 V - \mu\epsilon\frac{\partial^2 V}{\partial t^2} = -\frac{\rho}{\epsilon} \qquad \text{(7-65)} \]

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.

其解以速度 \(u = 1/\sqrt{\mu\epsilon}\) 傳播。在自由空間中 \(u = 1/\sqrt{\mu_0\epsilon_0} = c\) — 即光速。這是將電磁學與光學連結起來的核心結果:光是一種電磁波。

その解は速度 \(u = 1/\sqrt{\mu\epsilon}\) で伝搬します。自由空間では \(u = 1/\sqrt{\mu_0\epsilon_0} = c\)、すなわち光速です。これは電磁気学と光学を結びつける中心的な結果であり、光は電磁波であることを意味します。


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:

將 Maxwell's equations 的積分形式套用在跨越界面的 pillbox 或扁平 contour 上,可得四條 boundary conditions,其形式與靜態情況完全相同,因為當 pillbox/contour 壓扁至零厚度時,\(\partial\mathbf{D}/\partial t\) 與 \(\partial\mathbf{B}/\partial t\) 的面積分都消失:

マクスウェルの方程式の積分形を、境界面をまたぐ薄い円筒(pillbox)または扁平な閉路に適用すると、4 つの境界条件(boundary conditions)が得られます。円筒や閉路を潰して厚さをゼロにすると \(\partial\mathbf{D}/\partial t\) と \(\partial\mathbf{B}/\partial t\) の面積分は消えるため、これらは静的な場合と形式的に同一です:

\[ E_{1t} = E_{2t} \qquad \text{(7-66a)} \]
\[ \mathbf{a}_{n2}\times(\mathbf{H}_1 - \mathbf{H}_2) = \mathbf{J}_s \qquad \text{(7-66b)} \]
\[ \mathbf{a}_{n2}\cdot(\mathbf{D}_1 - \mathbf{D}_2) = \rho_s \qquad \text{(7-66c)} \]
\[ B_{1n} = B_{2n} \qquad \text{(7-66d)} \]

In words: tangential E is continuous; tangential H jumps by the surface current; normal D jumps by the surface charge; normal B is continuous.

文字敘述:切線方向 E 連續;切線方向 H 的跳躍量等於 surface current;法線方向 D 的跳躍量等於 surface charge;法線方向 B 連續。

言葉で述べると、E の接線成分は連続、H の接線成分は面電流の分だけ不連続、D の法線成分は面電荷の分だけ不連続、B の法線成分は連続です。

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.

在兩種 lossless media 之間 (\(\rho_s = 0\)、\(\mathbf{J}_s = 0\)),四個場分量在界面上都滿足簡單的連續條件。在與 perfect conductor 的界面上,導體內部所有場皆為零,因此 \(\mathbf{E}\) 垂直於表面 (\(E_{1n} = \rho_s/\epsilon_1\)),而 \(\mathbf{H}\) 與表面相切 (\(|\mathbf{H}_{1t}| = |\mathbf{J}_s|\)) — 這是 waveguide 與 antenna 分析的基本結果。

2 つの無損失媒質の間(\(\rho_s = 0\)、\(\mathbf{J}_s = 0\))では、4 つの場の成分すべてが境界面で単純な連続条件を満たします。完全導体(perfect conductor)との境界面では、導体内部の場はすべてゼロになるため、\(\mathbf{E}\) は表面に垂直(\(E_{1n} = \rho_s/\epsilon_1\))、\(\mathbf{H}\) は表面に平行(\(|\mathbf{H}_{1t}| = |\mathbf{J}_s|\))となります。これは導波管やアンテナの解析における基本的な結果です。


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,

在無源區域 (\(\rho = 0\)、\(\mathbf{J} = 0\)) 中,\(\mathbf{E}\) 與 \(\mathbf{B}\) 本身都滿足 homogeneous wave equations。對於點源周圍球對稱的 scalar-potential wave,

源のない領域(\(\rho = 0\)、\(\mathbf{J} = 0\))では、\(\mathbf{E}\) と \(\mathbf{B}\) 自身が斉次波動方程式を満たします。点源のまわりの球対称なスカラーポテンシャル波については、

\[ V(R, t) = \frac{1}{R}f\!\left(t - \frac{R}{u}\right) \qquad \text{(7-75)} \]

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

是其中一個解 — 一個以速度 \(u = 1/\sqrt{\mu\epsilon}\) 從原點向外發散的 spherical wave。引數 \(t - R/u\) 的意義是:距離 \(R\) 處在時刻 \(t\) 的 \(V\) 值,是由源在較早時刻 \(t - R/u\) 的狀態所「造成」的:這就是 retardation(延遲),是傳播速度有限的直接結果。對應任意隨時間變化的源,其 retarded potentials 為

が 1 つの解になります。これは原点から速度 \(u = 1/\sqrt{\mu\epsilon}\) で外向きに広がる球面波です。引数 \(t - R/u\) は、距離 \(R\)、時刻 \(t\) における \(V\) の値が、それより前の時刻 \(t - R/u\) における源の状態によって「引き起こされた」ことを意味します。これが遅延(retardation)であり、伝搬速度が有限であることの直接の帰結です。任意の時間変化する源に対する遅延ポテンシャル(retarded potentials)は次のとおりです:

\[ V(\mathbf{r}, t) = \frac{1}{4\pi\epsilon}\int_{V'}\frac{\rho(\mathbf{r}', t - R/u)}{R}dv' \]
\[ \mathbf{A}(\mathbf{r}, t) = \frac{\mu}{4\pi}\int_{V'}\frac{\mathbf{J}(\mathbf{r}', t - R/u)}{R}dv' \]

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.

當時間相依性可忽略時(「quasi-static」範圍),這些式子退化為靜態的 Poisson-equation 解。對高頻源 — antennas — 則不能忽略 retardation。

時間依存性が無視できる場合(「準静的(quasi-static)」領域)、これらは静的なポアソン方程式の解に帰着します。アンテナのような高周波の源に対しては、遅延を無視することはできません。


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:

由於 Maxwell's equations 是線性的,頻率為 \(\omega\) 的弦波源在各處產生的場都是同頻率的弦波。將實數弦波場表示為複數 phasor(相量)的實部:

マクスウェルの方程式は線形なので、周波数 \(\omega\) の正弦波的な源は、あらゆる場所で同じ周波数の正弦波的な場を生じます。実数の正弦波的な場を、複素フェーザ(phasor)の実部として表します:

\[ \mathbf{E}(\mathbf{r}, t) = \mathscr{R}e\bigl[\mathbf{E}(\mathbf{r})\,e^{j\omega t}\bigr] \qquad \text{(7-93)} \]

Under this substitution, \(\partial/\partial t \to j\omega\), and Maxwell's equations become algebraic in \(j\omega\):

在此代換下,\(\partial/\partial t \to j\omega\),Maxwell's equations 變成關於 \(j\omega\) 的代數方程式:

この置き換えにより \(\partial/\partial t \to j\omega\) となり、マクスウェルの方程式は \(j\omega\) に関する代数方程式になります:

\[ \begin{aligned} \nabla\times\mathbf{E} &= -j\omega\mu\mathbf{H} \\ \nabla\times\mathbf{H} &= \mathbf{J} + j\omega\epsilon\mathbf{E} \\ \nabla\cdot\mathbf{E} &= \rho/\epsilon \\ \nabla\cdot\mathbf{H} &= 0 \end{aligned} \qquad \text{(7-94)} \]

Helmholtz's equation

The wave equations (7-63) and (7-65) become Helmholtz equations:

波動方程式 (7-63) 與 (7-65) 變成 Helmholtz equations:

波動方程式 (7-63) と (7-65) はヘルムホルツ方程式(Helmholtz equations)になります:

\[ \nabla^2\mathbf{A} + k^2\mathbf{A} = -\mu\mathbf{J} \qquad \text{(7-96)} \]
\[ \nabla^2 V + k^2 V = -\rho/\epsilon \qquad \text{(7-95)} \]

where

其中

ここで

\[ \boxed{k = \omega\sqrt{\mu\epsilon} = \omega/u} \quad (\text{rad/m}) \qquad \text{(7-97)} \]

is the wavenumber. The Lorenz gauge in phasor form is \(\nabla\cdot\mathbf{A} + j\omega\mu\epsilon V = 0\).

稱為 wavenumber(波數)。Lorenz gauge 的 phasor 形式為 \(\nabla\cdot\mathbf{A} + j\omega\mu\epsilon V = 0\)。

は波数(wavenumber)です。ローレンツゲージのフェーザ形式は \(\nabla\cdot\mathbf{A} + j\omega\mu\epsilon V = 0\) となります。

Retarded potentials in phasor form

The phasor solutions contain the retardation factor \(e^{-jkR}\):

Phasor 解含有 retardation factor \(e^{-jkR}\):

フェーザ解は遅延因子 \(e^{-jkR}\) を含みます:

\[ V(\mathbf{r}) = \frac{1}{4\pi\epsilon}\int_{V'}\frac{\rho\,e^{-jkR}}{R}\,dv' \qquad \text{(7-99)} \]
\[ \mathbf{A}(\mathbf{r}) = \frac{\mu}{4\pi}\int_{V'}\frac{\mathbf{J}\,e^{-jkR}}{R}\,dv' \qquad \text{(7-100)} \]

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.

這是第 11 章 antenna theory 的基礎:由電流分佈產生的 radiation,是將 phasor current 連同 retardation phase factor 在源體積上積分而得。

これが第 11 章のアンテナ理論の基礎です。電流分布からの放射は、フェーザ電流に遅延位相因子を掛けて源の体積にわたって積分することで計算されます。


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.

節錄自 Cheng,Field and Wave Electromagnetics,§8-1 至 §8-10。請先讀完本節,再嘗試下方的習題。

Cheng, Field and Wave Electromagnetics, §8-1 〜 §8-10 からの要約です。下の演習問題に取り組む前に、まずこの節を読んでください。


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:

在無源 (source-free) 且非導電的介質中,Maxwell's equations 可以合併成關於 E 與 H 的齊次向量波動方程式 (homogeneous vector wave equations)。在自由空間中:

湧き出しのない(source-free)非導電性媒質では、マクスウェルの方程式(Maxwell's equations)を組み合わせると E と H に関する斉次ベクトル波動方程式(homogeneous vector wave equations)が得られます。自由空間では:

\[ \nabla^2\mathbf{E} - \frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = 0, \qquad c = \frac{1}{\sqrt{\mu_0\epsilon_0}} \cong 3\times 10^8 \ \text{m/s} \qquad \text{(8-1, 8-2)} \]

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.

Uniform plane wave(均勻平面波)的 \(\mathbf{E}\) 在任何垂直於傳播方向的平面上,都具有相同的方向、大小與相位。嚴格來說,無限大的 plane waves 在實際上並不存在 — 但在遠離 source 之處,波前幾乎是平面的,而且其數學是整個理論的基礎。

一様平面波(uniform plane wave)では、伝搬方向に垂直な任意の平面上で \(\mathbf{E}\) の向き・大きさ・位相がすべて等しくなります。厳密には、無限に広がる平面波は現実には存在しません。しかし波源から十分に離れれば波面はほぼ平面とみなせますし、その数学的な取り扱いは理論全体の基礎となります。


8-2   Plane Waves in Lossless Media

Using phasors, the wave equation becomes the vector Helmholtz equation:

改用 phasors 表示後,波動方程式變成 vector Helmholtz equation(向量亥姆霍茲方程式):

フェーザ(phasor)を用いると、波動方程式はベクトル・ヘルムホルツ方程式(vector Helmholtz equation)になります:

\[ \nabla^2\mathbf{E} + k_0^2\mathbf{E} = 0 \qquad \text{(8-3)} \]
\[ \boxed{k_0 = \omega\sqrt{\mu_0\epsilon_0} = \omega/c} \quad (\text{rad/m}) \qquad \text{(8-4)} \]

For a wave varying only in \(z\), the solution is a superposition of forward- and backward-traveling waves:

對於只隨 \(z\) 變化的波,其解是前進波與後退波 (forward- and backward-traveling waves) 的疊加:

\(z\) 方向にのみ変化する波に対して、解は前進波と後退波(forward- and backward-traveling waves)の重ね合わせになります:

\[ E_x(z) = E_0^+ e^{-jk_0 z} + E_0^- e^{jk_0 z} \qquad \text{(8-7)} \]

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:

代入 Faraday's law 可得 \(\mathbf{H}\) 同時垂直於 \(\mathbf{E}\) 與傳播方向。兩者的比值定義了 intrinsic impedance(本質阻抗):

これをファラデーの法則(Faraday's law)に代入すると、\(\mathbf{H}\) は \(\mathbf{E}\) と伝搬方向の両方に垂直になることがわかります。両者の比が固有インピーダンス(intrinsic impedance)を定義します:

\[ \boxed{\eta_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} \cong 120\pi \cong 377\ \Omega} \qquad \text{(8-14)} \]

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 \(\mathbf{H} = (1/\eta)\mathbf{a}_n\times\mathbf{E}\).

對一般的 lossless medium(無損耗介質),\(k = \omega\sqrt{\mu\epsilon}\)、\(\eta = \sqrt{\mu/\epsilon}\)、\(\lambda = 2\pi/k\),且 \(u_p = \omega/k = 1/\sqrt{\mu\epsilon}\)。此波是 transverse(橫波):\(\mathbf{E}\perp\mathbf{H}\perp\mathbf{a}_n\),且 \(\mathbf{H} = (1/\eta)\mathbf{a}_n\times\mathbf{E}\)。

一般の無損失媒質(lossless medium)では、\(k = \omega\sqrt{\mu\epsilon}\)、\(\eta = \sqrt{\mu/\epsilon}\)、\(\lambda = 2\pi/k\)、および \(u_p = \omega/k = 1/\sqrt{\mu\epsilon}\) となります。この波は横波(transverse)です:\(\mathbf{E}\perp\mathbf{H}\perp\mathbf{a}_n\) であり、\(\mathbf{H} = (1/\eta)\mathbf{a}_n\times\mathbf{E}\) が成り立ちます。

Wavenumber vector (oblique propagation)

For propagation along an arbitrary direction \(\mathbf{a}_n\), define the wavenumber vector \(\mathbf{k} = k\mathbf{a}_n\). Then

對於沿任意方向 \(\mathbf{a}_n\) 的傳播,定義 wavenumber vector(波數向量)\(\mathbf{k} = k\mathbf{a}_n\)。則

任意の方向 \(\mathbf{a}_n\) に伝搬する場合は、波数ベクトル(wavenumber vector)\(\mathbf{k} = k\mathbf{a}_n\) を定義します。このとき

\[ \mathbf{E}(\mathbf{R}) = \mathbf{E}_0\,e^{-j\mathbf{k}\cdot\mathbf{R}} \qquad \text{(8-26)} \]

The planes of constant phase satisfy \(\mathbf{a}_n\cdot\mathbf{R} = \text{const}\).

等相位面 (planes of constant phase) 滿足 \(\mathbf{a}_n\cdot\mathbf{R} = \text{const}\)。

等位相面(planes of constant phase)は \(\mathbf{a}_n\cdot\mathbf{R} = \text{const}\) を満たします。

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:

將兩個同頻率、互相正交的 linearly polarized(線性極化)分量相加時,合成 \(\mathbf{E}\) 的向量端點會在橫截面上描出一條曲線:

同じ周波数をもつ互いに直交する 2 つの直線偏波(linearly polarized)成分を加え合わせると、合成された \(\mathbf{E}\) の先端は横断面内で曲線を描きます:


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:

在 lossy medium(有損耗介質;即 conductivity 有限或 permittivity 為複數)中,wavenumber 變為複數:\(k_c = \omega\sqrt{\mu\epsilon_c}\)。定義 propagation constant(傳播常數):

損失のある媒質(lossy medium;導電率が有限、あるいは誘電率が複素数の媒質)では、波数は複素数 \(k_c = \omega\sqrt{\mu\epsilon_c}\) になります。ここで伝搬定数(propagation constant)を定義します:

\[ \gamma = \alpha + j\beta = jk_c = j\omega\sqrt{\mu\epsilon}\left(1 + \frac{\sigma}{j\omega\epsilon}\right)^{1/2} \quad (\text{m}^{-1}) \qquad \text{(8-43, 8-44)} \]

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 \(\beta = k\).

其中 \(\alpha\)(Np/m)是 attenuation constant(衰減常數),\(\beta\)(rad/m)是 phase constant(相位常數)。波的振幅依 \(e^{-\alpha z}\) 衰減,並依 \(e^{-j\beta z}\) 振盪。對 lossless medium 而言,\(\alpha = 0\) 且 \(\beta = k\)。

ここで \(\alpha\)(Np/m)は減衰定数(attenuation constant)、\(\beta\)(rad/m)は位相定数(phase constant)です。波は \(e^{-\alpha z}\) に従って減衰し、\(e^{-j\beta z}\) に従って振動します。無損失媒質では \(\alpha = 0\)、\(\beta = k\) となります。

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:

在 good conductor(良導體,\(\sigma/\omega\epsilon \gg 1\))中:\(\alpha \cong \beta \cong \sqrt{\pi f\mu\sigma}\)。振幅衰減到 \(1/e\) 時的深度稱為 skin depth(集膚深度):

良導体(good conductor、\(\sigma/\omega\epsilon \gg 1\))では \(\alpha \cong \beta \cong \sqrt{\pi f\mu\sigma}\) となります。振幅が \(1/e\) まで減衰する深さを表皮の厚さ(skin depth)といいます:

\[ \boxed{\delta = \frac{1}{\alpha} = \frac{1}{\sqrt{\pi f\mu\sigma}}} \quad (\text{m}) \qquad \text{(8-57)} \]

Typical values (from Table 8-1):

典型數值(取自 Table 8-1):

代表的な値(Table 8-1 より):

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.

在微波頻率下,skin depth 小到場與電流實際上都被侷限在一層很薄的表面層內 — 這正是「集膚」一詞的由來。

マイクロ波帯では表皮の厚さがきわめて小さくなり、電磁界と電流は事実上ごく薄い表面層に閉じ込められます。「表皮」という名称はここに由来します。

The intrinsic impedance in a good conductor is complex:

Good conductor 內的 intrinsic impedance 為複數:

良導体における固有インピーダンスは複素数になります:

\[ \eta_c = (1+j)\sqrt{\frac{\pi f\mu}{\sigma}} = \sqrt{\frac{\pi f\mu}{\sigma}}\,e^{j\pi/4} \]

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.

這表示在 good conductor 中 \(\mathbf{E}\) 與 \(\mathbf{H}\) 相差 45° — 即使在傳播過程中,仍存在相當可觀的 reactive(儲能)成分。

これは、良導体中では \(\mathbf{E}\) と \(\mathbf{H}\) の位相が 45° ずれていることを意味します。すなわち、伝搬中であってもかなりの大きさのリアクティブ(エネルギーを蓄える)成分が残っています。


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:

Dispersive medium(色散介質)是指 \(\beta\) 不是 \(\omega\) 的線性函數的介質;不同頻率以不同的 phase velocity 前進,因此已調變的訊號會失真。窄頻 wave packet 之包絡 (envelope) 的速度稱為 group velocity(群速度):

分散性媒質(dispersive medium)とは、\(\beta\) が \(\omega\) の線形関数にならない媒質のことです。周波数ごとに位相速度が異なるため、変調された信号は歪みます。狭帯域の波束の包絡線(envelope)が進む速度を群速度(group velocity)といいます:

\[ \boxed{u_g = \frac{1}{d\beta/d\omega}} \quad (\text{m/s}) \qquad \text{(8-72)} \]

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:

在 \(\omega\)–\(\beta\) diagram 上:從原點到某點之連線的斜率是 phase velocity \(u_p = \omega/\beta\);曲線在該點的切線斜率則是 group velocity \(u_g\)。分三種情況:

\(\omega\)–\(\beta\) 線図では、原点からある点へ引いた直線の傾きが位相速度 \(u_p = \omega/\beta\) を、その点での接線の傾きが群速度 \(u_g\) を表します。3 つの場合があります:

Ionized gas (plasma)

Free electrons give an effective permittivity

自由電子 (free electrons) 會產生一個等效 permittivity(等效介電常數)

自由電子は次のような実効誘電率(effective permittivity)を与えます

\[ \epsilon_p = \epsilon_0\!\left(1 - \frac{\omega_p^2}{\omega^2}\right), \qquad \omega_p = \sqrt{\frac{Ne^2}{m\epsilon_0}} \qquad \text{(8-66, 8-64)} \]

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 (\(f > f_p\)).

稱為 plasma frequency(電漿頻率)。當 \(\omega < \omega_p\) 時,\(\gamma\) 為實數 — 波是 evanescent(消逝波),無法傳播。當 \(\omega > \omega_p\) 時,波恢復傳播,且 \(u_p u_g = c^2\)。電離層會反射 AM 廣播(\(f < f_p\)),但讓電視與 FM 訊號穿透(\(f > f_p\))。

これをプラズマ周波数(plasma frequency)といいます。\(\omega < \omega_p\) では \(\gamma\) が実数となり、波はエバネッセント波(evanescent)となって伝搬しません。\(\omega > \omega_p\) では再び伝搬が可能になり、\(u_p u_g = c^2\) が成り立ちます。電離層は AM ラジオ波(\(f < f_p\))を反射しますが、テレビや FM の電波(\(f > f_p\))は透過させます。


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:

從 Maxwell 的兩個 curl 方程式出發,利用向量恆等式 \(\nabla\cdot(\mathbf{E}\times\mathbf{H}) = \mathbf{H}\cdot(\nabla\times\mathbf{E}) - \mathbf{E}\cdot(\nabla\times\mathbf{H})\),可得 Poynting's theorem(波印廷定理):

マクスウェルの 2 つの回転(curl)方程式から出発し、ベクトル恒等式 \(\nabla\cdot(\mathbf{E}\times\mathbf{H}) = \mathbf{H}\cdot(\nabla\times\mathbf{E}) - \mathbf{E}\cdot(\nabla\times\mathbf{H})\) を用いると、ポインティングの定理(Poynting's theorem)が得られます:

\[ \oint_S(\mathbf{E}\times\mathbf{H})\cdot d\mathbf{s} = -\frac{\partial}{\partial t}\int_V\!\left(\tfrac{1}{2}\epsilon E^2 + \tfrac{1}{2}\mu H^2\right)dv - \int_V\sigma E^2\,dv \qquad \text{(8-82)} \]

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

左邊是穿過 \(S\) 流出的功率;右邊是 \(V\) 內所儲存場能量的減少率,減去 ohmic(歐姆)損耗。Poynting vector(波印廷向量)

左辺は閉曲面 \(S\) を通って外へ流れ出る電力、右辺は体積 \(V\) 内に蓄えられた電磁界エネルギーの減少率からオーム損失を差し引いたものです。ポインティングベクトル(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.

就是場所攜帶的瞬時功率密度向量 (instantaneous power density vector)。

は、電磁界が運ぶ瞬時電力密度ベクトルです。

Time-average power density

For time-harmonic fields represented by phasors:

對於以 phasors 表示的 time-harmonic fields(時諧場):

フェーザで表した時間調和電磁界(time-harmonic fields)に対しては:

\[ \boxed{\mathscr{P}_{av} = \tfrac{1}{2}\,\mathscr{R}e(\mathbf{E}\times\mathbf{H}^*)} \quad (\text{W/m}^2) \qquad \text{(8-96)} \]

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.

對於在 lossy medium 中沿 \(+z\) 傳播的 uniform plane wave,上式化簡為 \(\mathscr{P}_{av} = \mathbf{a}_z(E_0^2/2|\eta|)e^{-2\alpha z}\cos\theta_\eta\) — intrinsic impedance 的相角 \(\theta_\eta\) 的作用完全就像 power factor(功率因數)。

損失媒質中を \(+z\) 方向に伝搬する一様平面波では、これは \(\mathscr{P}_{av} = \mathbf{a}_z(E_0^2/2|\eta|)e^{-2\alpha z}\cos\theta_\eta\) に帰着します。固有インピーダンスの位相角 \(\theta_\eta\) は、ちょうど力率(power factor)と同じ働きをします。


8-6   Normal Incidence on a Perfect Conductor

When a wave is incident normally from a dielectric onto a perfect conductor (\(\sigma_2\to\infty\)):

當波從 dielectric 垂直入射 (normal incidence) 到 perfect conductor(\(\sigma_2\to\infty\))時:

誘電体から完全導体(perfect conductor、\(\sigma_2\to\infty\))へ波が垂直入射(normal incidence)する場合:


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:

Plane of incidence(入射面)包含傳播方向與表面法線。任何 polarization 都可以分解成兩種互相正交的情況:

入射面(plane of incidence)は、伝搬方向と境界面の法線を含む平面です。任意の偏波は、互いに直交する次の 2 つの場合に分解できます:

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).

兩種情況下,垂直於邊界的方向都會形成 standing wave,而平行於邊界的方向則是 traveling wave(行進波)— 這正是 waveguide(波導)理論的基礎(Chapter 10)。

いずれの場合も、境界に垂直な方向には定在波が、境界に平行な方向には進行波(traveling wave)が形成されます。これが導波管(waveguide)理論の基礎となります(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 \(\tau\):

兩介質都不是 perfect conductor;波的一部分被反射,一部分穿透。要求切線方向的 \(\mathbf{E}\) 與 \(\mathbf{H}\) 在 \(z=0\) 處連續,即可得到 reflection coefficient(反射係數)\(\Gamma\) 與 transmission coefficient(透射係數)\(\tau\):

どちらの媒質も完全導体ではないので、波の一部は反射し、一部は透過します。\(z=0\) において \(\mathbf{E}\) と \(\mathbf{H}\) の接線成分が連続であることを要求すると、反射係数(reflection coefficient)\(\Gamma\) と透過係数(transmission coefficient)\(\tau\) が得られます:

\[ \boxed{\Gamma = \frac{E_{r0}}{E_{i0}} = \frac{\eta_2 - \eta_1}{\eta_2 + \eta_1}} \qquad \text{(8-140)} \]
\[ \boxed{\tau = \frac{E_{t0}}{E_{i0}} = \frac{2\eta_2}{\eta_2 + \eta_1}} \qquad \text{(8-141)} \]
\[ 1 + \Gamma = \tau \qquad \text{(8-142)} \]

\(\Gamma\) 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.

\(\Gamma\) 的範圍從 \(-1\)(perfect conductor)到 \(+1\);\(\tau\) 恆為正。在 lossy medium 中兩者皆為複數(在界面處引入相位偏移)。檢驗:\(\eta_2 = \eta_1 \Rightarrow \Gamma = 0\) — impedance matched(阻抗匹配),無反射。

\(\Gamma\) は \(-1\)(完全導体)から \(+1\) までの値をとり、\(\tau\) はつねに正です。損失媒質では両者とも複素数になります(界面で位相のずれが生じます)。確認:\(\eta_2 = \eta_1 \Rightarrow \Gamma = 0\)、すなわちインピーダンス整合(impedance matched)していれば反射は生じません。

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:

當反射只是部分反射時,總場 \(\mathbf{E}_1 = \tau E_{i0}e^{-j\beta_1 z} + j2\Gamma E_{i0}\sin\beta_1 z\) — 一個 traveling wave 加上一個 standing wave。\(|\mathbf{E}_1|\) 的最大值與最小值定義了 standing-wave ratio(駐波比):

反射が部分的な場合、全電界は \(\mathbf{E}_1 = \tau E_{i0}e^{-j\beta_1 z} + j2\Gamma E_{i0}\sin\beta_1 z\)、すなわち進行波と定在波の和になります。\(|\mathbf{E}_1|\) の最大値と最小値によって定在波比(standing-wave ratio)が定義されます:

\[ \boxed{S = \frac{|E|_{\max}}{|E|_{\min}} = \frac{1 + |\Gamma|}{1 - |\Gamma|}} \qquad \text{(8-147)} \]
\[ |\Gamma| = \frac{S-1}{S+1} \qquad \text{(8-148)} \]

\(S\) ranges from 1 (no reflection) to \(\infty\) (total reflection); often quoted in dB as \(20\log_{10}S\). Note that \(|\Gamma|\) never exceeds 1; as it approaches 1 the SWR grows without bound (e.g. \(|\Gamma| = 1/3\) gives \(S = 2\), \(|\Gamma| = 0.5\) gives \(S = 3\)).

\(S\) 的範圍從 1(無反射)到 \(\infty\)(全反射);常以 dB 表示為 \(20\log_{10}S\)。注意 \(|\Gamma|\) 永遠不會超過 1;當它趨近 1 時,SWR 會無限增大(例如 \(|\Gamma| = 1/3\) 對應 \(S = 2\),\(|\Gamma| = 0.5\) 對應 \(S = 3\))。

\(S\) は 1(無反射)から \(\infty\)(全反射)までの値をとり、しばしば \(20\log_{10}S\) として dB で表されます。なお \(|\Gamma|\) が 1 を超えることはなく、1 に近づくにつれて SWR は際限なく大きくなります(たとえば \(|\Gamma| = 1/3\) なら \(S = 2\)、\(|\Gamma| = 0.5\) なら \(S = 3\))。


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:

對於三層介質(介質 1 → 厚度為 \(d\) 的介質 2 → 介質 3),要使介質 1 中沒有反射,有兩種條件:

3 つの媒質(媒質 1 → 厚さ \(d\) の媒質 2 → 媒質 3)の場合、媒質 1 側からの反射をゼロにする条件は 2 つあります:

  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).
  1. Half-wave dielectric window(半波介電窗,\(\eta_3 = \eta_1\)):\(d = n\lambda_2/2\) — 介質 2 的厚度為任意整數個半波長時,都能維持阻抗匹配。用於雷達天線的 radomes(天線罩)。
  2. Quarter-wave impedance transformer(四分之一波長阻抗轉換器,\(\eta_3 \neq \eta_1\)):\(\eta_2 = \sqrt{\eta_1\eta_3}\) 且 \(d = (2n+1)\lambda_2/4\) — 中間層的作用如同阻抗匹配轉換器,類似於傳輸線中的 quarter-wave sections(Chapter 9)。
  1. 半波長誘電体窓(half-wave dielectric window)(\(\eta_3 = \eta_1\)):\(d = n\lambda_2/2\) — 媒質 2 の厚さが半波長の整数倍であれば、インピーダンス整合が保たれます。レーダーアンテナを覆うレドーム(radome)に用いられます。
  2. 1/4 波長インピーダンス変成器(quarter-wave impedance transformer)(\(\eta_3 \neq \eta_1\)):\(\eta_2 = \sqrt{\eta_1\eta_3}\) かつ \(d = (2n+1)\lambda_2/4\) — 中間層がインピーダンス整合用の変成器として働きます。伝送線路における 1/4 波長線路(Chapter 9)と同じ考え方です。

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:

要求界面上每一點 \(x\) 都滿足相位匹配 (phase-matching) \(e^{-j\beta_1 x\sin\theta_i} = e^{-j\beta_1 x\sin\theta_r} = e^{-j\beta_2 x\sin\theta_t}\),立即可得:

界面上のすべての \(x\) で位相整合(phase-matching)\(e^{-j\beta_1 x\sin\theta_i} = e^{-j\beta_1 x\sin\theta_r} = e^{-j\beta_2 x\sin\theta_t}\) を要求すると、ただちに次が得られます:

\[ \theta_r = \theta_i \quad \text{(Snell's law of reflection)} \]
\[ \boxed{\frac{\sin\theta_t}{\sin\theta_i} = \frac{\beta_1}{\beta_2} = \frac{n_1}{n_2} = \sqrt{\frac{\epsilon_{r1}}{\epsilon_{r2}}}} \quad \text{(Snell's law of refraction)} \qquad \text{(8-185)} \]

where \(n = c/u_p\) is the index of refraction. Denser medium (\(n_2 > n_1\)) bends the ray toward the normal.

其中 \(n = c/u_p\) 是 index of refraction(折射率)。較「密」的介質(\(n_2 > n_1\))會使光線偏向法線。

ここで \(n = c/u_p\) は屈折率(index of refraction)です。より密な媒質(\(n_2 > n_1\))に入ると、光線は法線の側へ曲がります。

Total internal reflection

If \(\epsilon_1 > \epsilon_2\) and \(\theta_i\) exceeds the critical angle, no refracted wave propagates:

若 \(\epsilon_1 > \epsilon_2\) 且 \(\theta_i\) 超過 critical angle(臨界角),則不會有折射波傳播:

\(\epsilon_1 > \epsilon_2\) で、かつ \(\theta_i\) が臨界角(critical angle)を超えると、屈折波は伝搬しません:

\[ \boxed{\theta_c = \sin^{-1}\sqrt{\frac{\epsilon_2}{\epsilon_1}} = \sin^{-1}\!\left(\frac{n_2}{n_1}\right)} \qquad \text{(8-188)} \]

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.

超過 \(\theta_c\) 後,介質 2 中的波變成 evanescent(消逝波;沿法線方向指數衰減,但沿表面方向行進)— 即 surface wave(表面波)。這正是 optical fibers(光纖)與介電波導的工作原理。

\(\theta_c\) を超えると、媒質 2 中の波はエバネッセント波(evanescent;法線方向には指数関数的に減衰しながら、表面に沿って進む波)、すなわち表面波(surface wave)になります。これが光ファイバ(optical fibers)や誘電体導波路の動作原理です。

Fresnel equations

Applying continuity at the interface gives reflection coefficients for each polarization:

在界面套用連續性條件,可得各 polarization 的 reflection coefficients:

界面で連続条件を適用すると、各偏波に対する反射係数が得られます:

\[ \Gamma_\perp = \frac{\eta_2\cos\theta_i - \eta_1\cos\theta_t}{\eta_2\cos\theta_i + \eta_1\cos\theta_t} \quad \text{(perpendicular pol.)} \qquad \text{(8-206)} \]
\[ \Gamma_\| = \frac{\eta_2\cos\theta_t - \eta_1\cos\theta_i}{\eta_2\cos\theta_t + \eta_1\cos\theta_i} \quad \text{(parallel pol.)} \]

Brewster's angle

For non-magnetic media (\(\mu_1 = \mu_2 = \mu_0\)), \(\Gamma_\|\) vanishes at a particular angle of incidence — the Brewster angle:

對非磁性介質(\(\mu_1 = \mu_2 = \mu_0\)),\(\Gamma_\|\) 會在某個特定入射角時為零 — 即 Brewster angle(布魯斯特角):

非磁性媒質(\(\mu_1 = \mu_2 = \mu_0\))では、ある特定の入射角で \(\Gamma_\|\) がゼロになります。これがブリュースター角(Brewster angle)です:

\[ \boxed{\theta_{B\|} = \tan^{-1}\sqrt{\frac{\epsilon_2}{\epsilon_1}} = \tan^{-1}\!\left(\frac{n_2}{n_1}\right)} \qquad \text{(8-227)} \]

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.

在 \(\theta_{B\|}\) 時,parallel-polarized 的波會完全透射(無反射)。由於 \(\Gamma_\perp\) 在該角度並不為零,以 \(\theta_{B\|}\) 入射的非極化波,其反射波只含 perpendicular polarization — 因此也稱為 polarizing angle(極化角)。應用:偏光太陽眼鏡、雷射共振腔中的 Brewster windows。

\(\theta_{B\|}\) では、平行偏波の波は完全に透過し、反射は生じません。一方 \(\Gamma_\perp\) はこの角度でもゼロにならないため、\(\theta_{B\|}\) で入射した無偏波の波は垂直偏波成分だけが反射されます。このため偏光角(polarizing angle)とも呼ばれます。応用例:偏光サングラス、レーザー共振器のブリュースター窓などです。


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.

本節濃縮自 Cheng, Field and Wave Electromagnetics, §9-1 至 §9-7。請先讀完本節再嘗試下方的習題。

Cheng, Field and Wave Electromagnetics, §9-1 〜 §9-7 を要約したものです。下の問題に取り組む前にお読みください。


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.

像 wire antenna 這類全向性 (omnidirectional) 的輻射源會浪費掉大部分能量。若要做 point-to-point(點對點)的功率或資訊傳輸,電磁能量必須被導引 (guided) — 即侷限在某個結構中,由該結構將能量從 source 帶到 load。最簡單的導引模式是 TEM (transverse electromagnetic) 波,其中 \(\mathbf{E}\) 與 \(\mathbf{H}\) 皆垂直於傳播方向 — 和周圍 dielectric 中未受導引的 plane wave 完全一樣。

線状アンテナのような全方向性の波源は、エネルギーの大部分を無駄にしてしまいます。電力や情報を2 点間(point-to-point)で伝送するには、電磁エネルギーを導波(guided)する、すなわち波源から負荷まで運ぶ構造の中に閉じ込める必要があります。そのもっとも単純なモードが TEM 波(transverse electromagnetic wave、横電磁波)で、\(\mathbf{E}\) と \(\mathbf{H}\) の両方が伝搬方向に垂直になります。これは周囲の誘電体中を伝わる導波されていない平面波とまったく同じです。

Three TEM transmission-line structures are fundamental:

三種基本的 TEM transmission-line 結構如下:

基本となる TEM 伝送線路(transmission line)の構造は 3 つあります:

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.

當導體間距超過波長的某個比例時,還可能存在其他(非 TEM)波模;這些在第 10 章以 waveguide modes 處理。

導体間の間隔が波長の一定の割合を超えると、TEM 以外の波のモードも存在し得ます。これらは第 10 章で導波管モード(waveguide modes)として扱います。


9-3   General Time-Harmonic Transmission-Line Equations

A transmission line is modeled by distributed parameters per unit length:

Transmission line 以單位長度的分佈參數 (distributed parameters) 來建模:

伝送線路は単位長さあたりの分布定数(distributed parameters)でモデル化されます:

For sinusoidal steady-state phasors \(V(z), I(z)\):

對弦波穩態的 phasors \(V(z), I(z)\):

正弦波定常状態のフェーザ \(V(z), I(z)\) に対して:

\[ -\frac{dV(z)}{dz} = (R + j\omega L)\,I(z) \qquad \text{(9-35a)} \]
\[ -\frac{dI(z)}{dz} = (G + j\omega C)\,V(z) \qquad \text{(9-35b)} \]

Decoupling by taking a second derivative:

取二階導數將兩式解耦 (decoupling):

2 階微分をとって 2 つの式を分離すると:

\[ \frac{d^2 V}{dz^2} = \gamma^2 V, \qquad \frac{d^2 I}{dz^2} = \gamma^2 I \qquad \text{(9-36)} \]

where the propagation constant is

其中 propagation constant(傳播常數)為

ここで伝搬定数(propagation constant)は

\[ \boxed{\gamma = \alpha + j\beta = \sqrt{(R + j\omega L)(G + j\omega C)}} \quad (\text{m}^{-1}) \qquad \text{(9-37)} \]

\(\alpha\) (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:

\(\alpha\) (Np/m) 為 attenuation constant(衰減常數),\(\beta\) (rad/m) 為 phase constant(相位常數)。通解是正向與反向行進波的疊加:

\(\alpha\) (Np/m) は減衰定数(attenuation constant)、\(\beta\) (rad/m) は位相定数(phase constant)です。一般解は前進波と後退波の重ね合わせになります:

\[ V(z) = V_0^+ e^{-\gamma z} + V_0^- e^{\gamma z} \qquad \text{(9-38a)} \]
\[ I(z) = I_0^+ e^{-\gamma z} + I_0^- e^{\gamma z} \qquad \text{(9-38b)} \]

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:

對無限長線路(或終端接 matched load 的線路)而言,只存在正向波。其電壓對電流的比值即為 characteristic impedance(特性阻抗):

無限長の線路(あるいは整合負荷で終端された線路)では前進波だけが存在します。その電圧と電流の比が特性インピーダンス(characteristic impedance)です:

\[ \boxed{Z_0 = \frac{V_0^+}{I_0^+} = \sqrt{\frac{R + j\omega L}{G + j\omega C}} = \frac{\gamma}{G + j\omega C}} \quad (\Omega) \qquad \text{(9-41)} \]

\(Z_0\) and \(\gamma\) depend on \(R, L, G, C, \omega\) — not on line length. They characterize the line itself.

\(Z_0\) 與 \(\gamma\) 只取決於 \(R, L, G, C, \omega\) — 與線路長度無關。它們刻畫的是線路本身的特性。

\(Z_0\) と \(\gamma\) は \(R, L, G, C, \omega\) で決まり、線路の長さには依存しません。これらは線路そのものを特徴づける量です。

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 \(\eta\).

類比:其一般形式與 lossy medium 中 plane waves 的 \(\gamma = j\omega\sqrt{\mu\epsilon(1 + \sigma/j\omega\epsilon)^{1/2}}\) 和 \(\eta = \sqrt{\mu/(\epsilon(1-j\sigma/\omega\epsilon))}\) 完全相同 — 電壓扮演 \(\mathbf{E}\) 的角色,電流扮演 \(\mathbf{H}\) 的角色,\(Z_0\) 則扮演 intrinsic impedance \(\eta\) 的角色。

類推:この一般形は、損失のある媒質中の平面波に対する \(\gamma = j\omega\sqrt{\mu\epsilon(1 + \sigma/j\omega\epsilon)^{1/2}}\) と \(\eta = \sqrt{\mu/(\epsilon(1-j\sigma/\omega\epsilon))}\) とまったく同じ形です。電圧が \(\mathbf{E}\) の役割を、電流が \(\mathbf{H}\) の役割を、\(Z_0\) が固有インピーダンス(intrinsic impedance)\(\eta\) の役割を果たします。


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:

當 Heaviside condition(Heaviside 條件)\(R/L = G/C\) 成立時,\(\gamma\) 與 \(Z_0\) 都可精確簡化為:

ヘビサイドの条件(Heaviside condition)\(R/L = G/C\) が満たされると、\(\gamma\) と \(Z_0\) はどちらも厳密に簡単な形になります:

\[ \alpha = R\sqrt{C/L}, \quad \beta = \omega\sqrt{LC} \qquad \text{(9-63, 9-64)} \]
\[ Z_0 = \sqrt{L/C} \quad (\text{real at all frequencies}) \]

Because \(\beta\) is linear in \(\omega\), all frequencies travel at the same phase velocity \(u_p = 1/\sqrt{LC}\) — no dispersion, no signal distortion.

由於 \(\beta\) 與 \(\omega\) 成線性關係,所有頻率都以相同的 phase velocity \(u_p = 1/\sqrt{LC}\) 行進 — 沒有 dispersion,訊號不失真。

\(\beta\) が \(\omega\) に比例するため、すべての周波数成分が同じ位相速度 \(u_p = 1/\sqrt{LC}\) で伝わります。分散(dispersion)がなく、信号は歪みません。


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):

由 plane-wave 類比 \(\gamma = j\omega\sqrt{\mu\epsilon(1 + \sigma/j\omega\epsilon)^{1/2}}\) 以及 \(G/C = \sigma/\epsilon\)(來自第 5 章的 R–C analogy)可得:

平面波との類推 \(\gamma = j\omega\sqrt{\mu\epsilon(1 + \sigma/j\omega\epsilon)^{1/2}}\) と、第 5 章の R–C 類推から得られる \(G/C = \sigma/\epsilon\) より:

\[ \boxed{LC = \mu\epsilon, \qquad G/C = \sigma/\epsilon} \qquad \text{(9-71, 9-72)} \]

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: \(u_p = 1/\sqrt{LC} = 1/\sqrt{\mu\epsilon}\).

推論:若已知某幾何結構的 \(L\),則 \(C\) 可由 \(LC = \mu\epsilon\) 求得;只要知道其中任一者,\(G\) 即可由介質的 conductivity 求得。Lossless TEM line 上的傳播速度等於 dielectric 中未受導引的 plane wave 的速度:\(u_p = 1/\sqrt{LC} = 1/\sqrt{\mu\epsilon}\)。

帰結:ある形状について \(L\) が分かっていれば、\(C\) は \(LC = \mu\epsilon\) から求まります。どちらか一方が分かれば、\(G\) は媒質の導電率から求まります。無損失 TEM 線路上の伝搬速度は、誘電体中を伝わる導波されていない平面波の速度に等しくなります:\(u_p = 1/\sqrt{LC} = 1/\sqrt{\mu\epsilon}\)。

Two-wire line parameters

For parallel wires of radius \(a\) separated by center-to-center distance \(D\):

對半徑 \(a\)、中心間距 \(D\) 的平行導線:

半径 \(a\) の平行線を中心間距離 \(D\) で並べた場合:

\[ C = \frac{\pi\epsilon}{\cosh^{-1}(D/2a)} \quad (\text{F/m}) \qquad \text{(9-73)} \]
\[ L = \frac{\mu}{\pi}\cosh^{-1}\!\left(\frac{D}{2a}\right) \quad (\text{H/m}) \qquad \text{(9-74)} \]
\[ G = \frac{\pi\sigma}{\cosh^{-1}(D/2a)} \quad (\text{S/m}) \qquad \text{(9-75)} \]

(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).

(當 \(D/2a \gg 1\) 時,\(\cosh^{-1}(D/2a) \cong \ln(D/a)\)。)同軸線 (coaxial lines) 則適用含 \(\ln(b/a)\) 的類似公式(第 3/6 章)。

(\(D/2a \gg 1\) のとき \(\cosh^{-1}(D/2a) \cong \ln(D/a)\) です。)同軸線路には \(\ln(b/a)\) を含む同様の公式が当てはまります(第 3 章/第 6 章)。


9-4   Finite Terminated Line

Measuring from the load end (\(z' = \ell - z\), so \(z' = 0\) is the load):

改從 load 端量起(\(z' = \ell - z\),故 \(z' = 0\) 即為 load 所在處):

負荷側から距離を測ると(\(z' = \ell - z\)、したがって \(z' = 0\) が負荷の位置):

\[ V(z') = I_L(Z_L\cosh\gamma z' + Z_0\sinh\gamma z') \qquad \text{(9-100a)} \]
\[ I(z') = (I_L/Z_0)(Z_L\sinh\gamma z' + Z_0\cosh\gamma z') \qquad \text{(9-100b)} \]

The impedance transformation at any point along the line:

線路上任一點的 impedance transformation(阻抗轉換):

線路上の任意の点におけるインピーダンス変換(impedance transformation):

\[ \boxed{Z(z') = Z_0\,\frac{Z_L + Z_0\tanh\gamma z'}{Z_0 + Z_L\tanh\gamma z'}} \quad (\Omega) \qquad \text{(9-102)} \]

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.

在 generator 端(\(z' = \ell\))的 input impedance 為 \(Z_i = Z(z' = \ell)\)。這是本章的關鍵方程式 — 它讓你可以把一條接有終端負載的線路,在 source 端用一個等效的 lumped impedance 取代。

電源側(\(z' = \ell\))での入力インピーダンスは \(Z_i = Z(z' = \ell)\) です。これが鍵となる式で、終端された線路を電源側から見た等価な集中定数インピーダンスに置き換えることができます。

Open- and short-circuited lines (lossless)

With \(\gamma = j\beta\), \(\tanh j\beta\ell = j\tan\beta\ell\):

令 \(\gamma = j\beta\),則 \(\tanh j\beta\ell = j\tan\beta\ell\):

\(\gamma = j\beta\) とすると \(\tanh j\beta\ell = j\tan\beta\ell\) なので:

\[ Z_{io} = -jR_0\cot\beta\ell \quad (\text{open-circuit}) \]
\[ Z_{is} = jR_0\tan\beta\ell \quad (\text{short-circuit}) \]

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 \(Z_0\).

短路線路在 \(\ell < \lambda/4\) 時呈電感性,在 \(\lambda/4 < \ell < \lambda/2\) 時呈電容性,依此類推。實用結論:\(Z_0 = \sqrt{Z_{io}Z_{is}}\) — 只要量測任一線路的開路與短路 input impedances,即可決定 \(Z_0\)。

短絡線路は \(\ell < \lambda/4\) で誘導性、\(\lambda/4 < \ell < \lambda/2\) で容量性となり、以下同様に繰り返します。有用な事実:\(Z_0 = \sqrt{Z_{io}Z_{is}}\)、すなわち任意の線路の開放時と短絡時の入力インピーダンスを測定すれば \(Z_0\) が決まります。


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:

對任何 \(Z_L \neq Z_0\) 的電阻性或複數 load,入射波有一部分會被反射。定義 load 處的 voltage reflection coefficient(電壓反射係數):

\(Z_L \neq Z_0\) となる抵抗性または複素数の負荷では、入射波の一部が反射されます。負荷における電圧反射係数(voltage reflection coefficient)を次のように定義します:

\[ \boxed{\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0} = |\Gamma|\,e^{j\theta_\Gamma}} \quad (\text{dimensionless}) \qquad \text{(9-134)} \]

\(|\Gamma| \le 1\), with special cases:

\(|\Gamma| \le 1\),幾個特殊情況如下:

\(|\Gamma| \le 1\) であり、特別な場合として:

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):

在 lossless line 上,\(V(z')\) 是入射行進波與反射行進波之和。兩者疊加產生 standing waves(駐波),其極大值的間距為 \(\lambda/2\)。Standing-wave ratio (SWR)(駐波比)定義為:

無損失線路では \(V(z')\) は入射進行波と反射進行波の和です。両者の重ね合わせにより、極大点が \(\lambda/2\) 間隔で並ぶ定在波が生じます。定在波比(standing-wave ratio, SWR)は:

\[ \boxed{S = \frac{|V|_{\max}}{|V|_{\min}} = \frac{1 + |\Gamma|}{1 - |\Gamma|}} \qquad \text{(9-138)} \]
\[ |\Gamma| = \frac{S - 1}{S + 1} \qquad \text{(9-139)} \]

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.

常以 dB 表示為 \(20\log_{10}S\)。高 SWR 是不樂見的 — 它代表 mismatch、大量反射功率,以及線路上可能具危險性的電壓極大值。

dB 表示では \(20\log_{10}S\) と書かれることが多いです。SWR が大きいのは望ましくありません。不整合、大きな反射電力、そして線路上に危険となり得る電圧の極大が生じていることを意味するからです。

Resistive load on a lossless line: \(\Gamma = (R_L - R_0)/(R_L + R_0)\).

Lossless line 接電阻性 load:\(\Gamma = (R_L - R_0)/(R_L + R_0)\)。

無損失線路に抵抗性負荷を接続した場合:\(\Gamma = (R_L - R_0)/(R_L + R_0)\) です。


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

長度 \(\ell = \lambda/4\)(或其奇數倍)的短路 lossy line 表現得像一個 parallel resonant circuit(並聯諧振電路);長度 \(\ell = \lambda/2\)(或其整數倍)則像一個 series resonant circuit(串聯諧振電路)。短路四分之一波長段的 quality factor 為

長さ \(\ell = \lambda/4\)(またはその奇数倍)の短絡した損失線路は並列共振回路として、長さ \(\ell = \lambda/2\)(またはその整数倍)では直列共振回路として振る舞います。短絡した 4 分の 1 波長区間の Q 値(quality factor)は

\[ Q = \frac{f_0}{\Delta f} = \frac{\beta}{2\alpha} \qquad \text{(9-130)} \]

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.

對絕緣良好的 low-loss line,\(Q \cong \omega L/R\) — 正是熟悉的 LC 電路公式。微波頻段的 transmission-line resonators 通常可達數千的 \(Q\) 值,遠高於 lumped-element 電路。

絶縁の良い低損失線路では \(Q \cong \omega L/R\) となり、よく知られた LC 回路の式と同じです。マイクロ波帯の伝送線路共振器では \(Q\) が数千に達することも珍しくなく、集中定数回路をはるかに上回ります。


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:

Smith chart 是把複數 \(z = Z/Z_0\) 平面保角映射 (conformal map) 到 \(\Gamma\) 平面上單位圓盤的圖。其主要性質如下:

スミスチャート(Smith chart)は、複素 \(z = Z/Z_0\) 平面を \(\Gamma\) 平面の単位円板に等角写像したものです。主な性質は次のとおりです:

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.

在圖上一眼就能解決的標準問題包括:\(\Gamma\) 由 \(Z_L\) 求得、\(z_i\) 由 \(z_L\) 經給定長度後求得、由 SWR 與極小值位置求 \(Z_L\)、求匹配用的 stub 長度、impedance 與 admittance 互換。對 lossy lines 而言,\(|\Gamma|\) 圓因 \(e^{-2\alpha z'}\) 因子而變成向內收縮的螺線。

チャート上で目視により解ける標準的な問題には、\(\Gamma\) を \(Z_L\) から求める、\(z_i\) を \(z_L\) から所定の長さだけ進んで求める、SWR と極小点の位置から \(Z_L\) を求める、整合用のスタブ長を求める、インピーダンスをアドミタンスに変換する、などがあります。損失のある線路では、\(|\Gamma|\) 円は \(e^{-2\alpha z'}\) の因子のため内側へ縮んでいく螺旋になります。


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:

要讓 generator 傳到 load 的功率最大,需 \(Z_L = Z_0\) — impedance-matched 的線路 \(S = 1\) 且無反射。Mismatched 的線路還會產生回波 (echoes),使資訊訊號失真。本章介紹 lossless lines 上的三種 matching 技巧:

電源から負荷へ最大の電力を伝えるには \(Z_L = Z_0\) が必要です。インピーダンス整合(impedance matching)のとれた線路では \(S = 1\) で反射がありません。不整合の線路ではエコーも生じ、情報信号が歪みます。本章では無損失線路に対する 3 つの整合手法を取り上げます:

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:

要把電阻性 load \(R_L\) 匹配到阻抗為 \(R_0\) 的線路,可插入一段四分之一波長、characteristic impedance 為下式的線段:

抵抗性の負荷 \(R_L\) をインピーダンス \(R_0\) の線路に整合させるには、次の特性インピーダンスをもつ 4 分の 1 波長の区間を挿入します:

\[ \boxed{R_0' = \sqrt{R_0 R_L}} \quad (\Omega) \qquad \text{(9-194)} \]

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.

因為 lossless line 有 \(Z(\lambda/4) = Z_0^2/Z_L\)(由 eq. 9-102 取 \(\tanh j\pi/2 \to \infty\) 而得),四分之一波長段會把 load「倒置」並加以縮放。由於 \(\ell = \lambda/4\) 取決於波長,這種匹配本質上是窄頻 (narrowband) 的。Quarter-wave transformers 無法在 low-loss lines 上匹配複數 load,因為那會要求 \(R_0'\) 為複數。

無損失線路では \(Z(\lambda/4) = Z_0^2/Z_L\)(式 9-102 で \(\tanh j\pi/2 \to \infty\) とした結果)なので、4 分の 1 波長区間は負荷を「反転」させて尺度を変えます。\(\ell = \lambda/4\) は波長に依存するため、この整合は本質的に狭帯域です。4 分の 1 波長変成器(quarter-wave transformer)は、低損失線路上の複素負荷には使えません。\(R_0'\) が複素数でなければならなくなるからです。

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 \(Y_0 = 1/R_0\):

更通用的方法:將一段長度 \(\ell\) 的短路 stub,在距 load 為 \(d\) 處並聯於主線路上。選擇兩個未知數(\(d\)、\(\ell\)),使並聯後的 admittance 等於 \(Y_0 = 1/R_0\):

より一般的な手法として、長さ \(\ell\) の短絡スタブ(stub)を、負荷から距離 \(d\) の位置で主線路に並列に接続します。2 つの未知数(\(d\)、\(\ell\))を、並列アドミタンスが \(Y_0 = 1/R_0\) に等しくなるように選びます:

\[ Y_i = Y_B + Y_s = Y_0 \qquad \text{(9-197)} \]

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).

以正規化 admittances 來處理:\(y_B\) 是在 stub 位置朝 load 看進去的 admittance,\(y_s\) 是短路 stub 的 input admittance(純虛數)。匹配條件變成 \(y_B = 1 + jb_B\),且 stub 的 susceptance 必須抵消 \(jb_B\)。短路 stubs 比開路 stubs 更受青睞 — 完美的短路比完美的開路容易實現(開路端會輻射並與鄰近物體耦合)。

正規化アドミタンスで考えると、\(y_B\) はスタブの位置から負荷側を見たアドミタンス、\(y_s\) は短絡スタブの入力アドミタンス(純虚数)です。整合条件は \(y_B = 1 + jb_B\) となり、スタブのサセプタンスで \(jb_B\) を打ち消す必要があります。短絡スタブは開放スタブより好まれます。完全な短絡は完全な開放より実現しやすいからです(開放端は放射し、近くの物体と結合してしまいます)。

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 \(Z_L\).

Single-stub matching 很容易在 Smith chart 上以圖解求得(把它當作 admittance chart 使用):找到 \(y_L\),朝 generator 方向旋轉直到軌跡與 \(g = 1\) 圓相交 — 該處即距離 \(d\);讀出所需的 susceptance,再找出對應的短路 stub 長度。與 quarter-wave transformer 不同,single-stub matching 對任何 \(Z_L\) 都適用。

単一スタブ整合(single-stub matching)はスミスチャートをアドミタンスチャートとして使えば図式的に簡単に解けます。\(y_L\) の位置を求め、軌跡が \(g = 1\) の円と交わるまで電源側へ回転させます。それが距離 \(d\) です。そこで必要なサセプタンスを読み取り、対応する短絡スタブの長さを求めます。4 分の 1 波長変成器と異なり、単一スタブ整合はどんな \(Z_L\) にも適用できます。

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 \(Z_L\).

Double-stub matching(§9-7.3)在無法調整與 load 間距離時,改用兩支固定位置的 stubs — 代價是可匹配的 \(Z_L\) 範圍較小。

二重スタブ整合(double-stub matching)(§9-7.3)は、負荷までの距離を調整できない場合に、固定位置の 2 本のスタブを使う手法です。その代わり、整合できる \(Z_L\) の範囲は狭くなります。


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.

濃縮自 Cheng,Field and Wave Electromagnetics,§10-1 至 §10-7。請先讀完本節,再嘗試下方的習題。

Cheng『Field and Wave Electromagnetics』§10-1〜§10-7 の要約です。下の問題に取り組む前にお読みください。


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.

第 9 章在傳輸線上建立了 TEM mode(\(\mathbf{E}\) 與 \(\mathbf{H}\) 皆與 \(z\) 垂直)。到了微波頻率(≥ 數 GHz),TEM 線便不再實用:conductor losses 隨 \(\sqrt{f}\) 增長,而雙導體結構的 radiation 也變得嚴重。

第 9 章では伝送線路上の TEM モード(\(\mathbf{E}\) と \(\mathbf{H}\) がともに \(z\) に垂直)を扱いました。マイクロ波周波数(≥ 数 GHz)では、TEM 線路は実用的ではありません。導体損失(conductor losses)が \(\sqrt{f}\) に比例して増え、2 導体構造からの放射も深刻になるからです。

The remedy is to confine the wave inside a single-conductor hollow metal tube — a waveguide. Inside such a tube:

解決之道是把波侷限在單一導體的中空金屬管內 — 即 waveguide(導波管)。在這種管子裡:

その解決策は、波を単一導体の中空金属管、すなわち導波管(waveguide)の中に閉じ込めることです。このような管の内部では:

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.

用導體端壁把 waveguide 兩端封住,就得到 cavity resonator(空腔共振器)— 一個三維駐波盒,具有極高的 \(Q\)(常達 \(10^{3}\)–\(10^{4}\))與離散的共振頻譜。

導波管の両端を導体壁で短絡すると、空洞共振器(cavity resonator)が得られます — 非常に高い \(Q\)(しばしば \(10^{3}\)–\(10^{4}\))と離散的な共振スペクトルをもつ 3 次元の定在波の箱です。


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:

假設時間變化為諧波 \(e^{j\omega t}\),而 \(z\) 方向的變化為 \(e^{-\gamma z}\),其中 \(\gamma = \alpha + j\beta\)。在導波管內無源、無電荷的 dielectric 中,Maxwell's equations 化簡為 Helmholtz's equation:

時間依存性を調和的な \(e^{j\omega t}\)、\(z\) 依存性を \(e^{-\gamma z}\)(\(\gamma = \alpha + j\beta\))と仮定します。導波管内の、波源も電荷もない誘電体中では、マクスウェルの方程式はヘルムホルツ方程式(Helmholtz's equation)に帰着します:

\[ \nabla^{2}\mathbf{E} + k^{2}\mathbf{E} = 0,\qquad \nabla^{2}\mathbf{H} + k^{2}\mathbf{H} = 0 \qquad \text{(10-3,4)} \]
\[ k = \omega\sqrt{\mu\epsilon} \qquad \text{(10-5)} \]

Splitting the Laplacian \(\nabla^{2} = \nabla_{xy}^{2} + \partial^{2}/\partial z^{2}\) and using \(\partial_z \to -\gamma\):

將 Laplacian 拆成 \(\nabla^{2} = \nabla_{xy}^{2} + \partial^{2}/\partial z^{2}\),並代入 \(\partial_z \to -\gamma\):

ラプラシアンを \(\nabla^{2} = \nabla_{xy}^{2} + \partial^{2}/\partial z^{2}\) と分け、\(\partial_z \to -\gamma\) を用いると:

\[ \nabla_{xy}^{2}\mathbf{E} + h^{2}\mathbf{E} = 0,\qquad h^{2} \triangleq \gamma^{2} + k^{2} \qquad \text{(10-7,15)} \]

\(h^{2}\) is a separation constant (eigenvalue). Only discrete values of \(h\) satisfy the boundary conditions — each is a waveguide mode.

\(h^{2}\) 是 separation constant(分離常數,即 eigenvalue)。只有離散的 \(h\) 值能滿足 boundary conditions — 每一個對應一種 waveguide mode。

\(h^{2}\) は分離定数(separation constant)、すなわち固有値です。境界条件(boundary conditions)を満たすのは離散的な \(h\) の値だけであり、そのそれぞれが導波管のモードに対応します。

Transverse-to-longitudinal coupling. All four transverse field components can be expressed in terms of the two longitudinal components \(E_z^0\) and \(H_z^0\):

橫向與縱向分量的耦合。四個橫向場分量都可以用兩個縱向分量 \(E_z^0\) 與 \(H_z^0\) 來表示:

横方向成分と縦方向成分の結合。4 つの横方向の場成分はすべて、2 つの縦方向成分 \(E_z^0\) と \(H_z^0\) で表すことができます:

\[ H_x^{0} = -\frac{1}{h^{2}}\!\left(\gamma\frac{\partial H_z^0}{\partial x} - j\omega\epsilon\frac{\partial E_z^0}{\partial y}\right) \qquad \text{(10-11)} \]
\[ E_x^{0} = -\frac{1}{h^{2}}\!\left(\gamma\frac{\partial E_z^0}{\partial x} + j\omega\mu\frac{\partial H_z^0}{\partial y}\right) \qquad \text{(10-13)} \]

(similar for \(H_y^0, E_y^0\)). So the procedure for every waveguide problem is:

(\(H_y^0, E_y^0\) 亦類似)。因此,每個 waveguide 問題的解題程序都是:

(\(H_y^0, E_y^0\) についても同様です)。したがって、あらゆる導波管問題の手順は次のようになります:

  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.
  1. 在 boundary conditions 下求解 TM 的 \(\nabla_{xy}^{2}E_z^0 + h^{2}E_z^0 = 0\)(TE 則解 \(H_z\) 的對應方程式)。
  2. 套用 BCs 以決定 eigenvalues \(h_{mn}\)。
  3. 利用 (10-11)–(10-14) 計算橫向分量。
  1. 境界条件のもとで、TM の場合は \(\nabla_{xy}^{2}E_z^0 + h^{2}E_z^0 = 0\) を(TE の場合は \(H_z\) についての対応する式を)解きます。
  2. 境界条件を課して固有値 \(h_{mn}\) を決めます。
  3. (10-11)–(10-14) を用いて横方向成分を計算します。

Cutoff frequency. The propagation constant is \(\gamma = \sqrt{h^{2}-\omega^{2}\mu\epsilon}\). It vanishes at the cutoff frequency:

Cutoff frequency(截止頻率)。Propagation constant 為 \(\gamma = \sqrt{h^{2}-\omega^{2}\mu\epsilon}\)。它在 cutoff frequency 處為零:

遮断周波数。伝搬定数(propagation constant)は \(\gamma = \sqrt{h^{2}-\omega^{2}\mu\epsilon}\) です。これは遮断周波数でゼロになります:

\[ f_c = \frac{h}{2\pi\sqrt{\mu\epsilon}} \quad (\text{Hz}) \qquad \text{(10-35)} \]

Two regimes:

兩種情形:

2 つの領域:

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:

等價的波長關係:

これと等価な波長の関係:

\[ \lambda_g = \frac{\lambda}{\sqrt{1-(f_c/f)^{2}}} > \lambda \qquad \text{(10-39)} \]
\[ \frac{1}{\lambda^{2}} = \frac{1}{\lambda_g^{2}} + \frac{1}{\lambda_c^{2}} \qquad \text{(10-41)} \]

Phase and group velocities in a lossless guide:

無損耗導波管中的 phase velocity 與 group velocity:

無損失導波管における位相速度(phase velocity)と群速度(group velocity):

\[ u_p = \frac{u}{\sqrt{1-(f_c/f)^{2}}} > u,\qquad u_g = u\sqrt{1-(f_c/f)^{2}} < u \qquad \text{(10-42,43)} \]
\[ u_p u_g = u^{2} \quad \text{(air: } u_p u_g = c^{2}\text{)} \qquad \text{(10-44)} \]

A waveguide is therefore a dispersive transmission system. \(u_g\) equals the velocity of energy transport.

因此 waveguide 是一種 dispersive(色散性)的傳輸系統。\(u_g\) 等於能量傳輸的速度。

したがって導波管は分散性(dispersive)の伝送系です。\(u_g\) はエネルギー輸送の速度に等しくなります。

Wave impedances (for propagating modes):

Wave impedances(波阻抗,適用於傳播中的 modes):

波動インピーダンス(wave impedances)(伝搬モードに対して):

\[ Z_{\text{TEM}} = \sqrt{\frac{\mu}{\epsilon}} = \eta \qquad \text{(10-20)} \]
\[ Z_{\text{TM}} = \eta\sqrt{1-(f_c/f)^{2}} \qquad \text{(10-45)} \]
\[ Z_{\text{TE}} = \frac{\eta}{\sqrt{1-(f_c/f)^{2}}} \qquad \text{()} \]

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).

注意 \(Z_{\text{TM}}\) 低於 \(\eta\),而 \(Z_{\text{TE}}\) 則高於它 — 兩者在高頻時彼此趨近。低於 cutoff 時,兩者皆為純電抗性(evanescent mode 中沒有淨功率流動)。

\(Z_{\text{TM}}\) は \(\eta\) より低く、\(Z_{\text{TE}}\) はそれより高いことに注意してください — 両者は高周波で互いに近づきます。遮断周波数以下では両者とも純リアクタンス性になります(エバネッセントモードでは正味の電力は流れません)。


10-3   Parallel-Plate Waveguide

Two infinite, parallel conducting planes separated by \(b\). Analytically simplest non-trivial waveguide; supports:

兩片間距 \(b\) 的無限大平行導體平面。這是解析上最簡單的非平凡 waveguide;可支援:

間隔 \(b\) で向かい合う 2 枚の無限平行導体平面です。解析的に最も単純な非自明な導波管であり、次のモードを支えます:

Useful as the bridge between transmission lines (Ch 9) and hollow guides — exhibits all three wave types (TEM, TM, TE) in one structure.

它是傳輸線(第 9 章)與中空導波管之間的橋樑 — 在同一個結構中展現了全部三種波型(TEM、TM、TE)。

伝送線路(第 9 章)と中空導波管をつなぐ橋渡しとして有用で、1 つの構造の中に 3 種類の波(TEM、TM、TE)がすべて現れます。


10-4   Rectangular Waveguides

Cross-section \(a \times b\) with \(a > b\) by convention. Separation of variables gives sinusoidal dependence in \(x\) and \(y\).

截面為 \(a \times b\),依慣例取 \(a > b\)。Separation of variables(分離變數法)給出 \(x\) 與 \(y\) 方向的正弦式變化。

断面は \(a \times b\) で、慣例として \(a > b\) とします。変数分離法により、\(x\) と \(y\) について正弦関数的な依存性が得られます。

TMmn modes — boundary condition \(E_z = 0\) on all four walls forces sine variation:

TMmn modes — 四面壁上 \(E_z = 0\) 的 boundary condition 迫使場呈正弦(sine)變化:

TMmn モード — 4 つの壁すべてで \(E_z = 0\) という境界条件により、正弦関数(sine)の変化が強制されます:

\[ E_z^{0}(x,y) = E_0 \sin\!\left(\frac{m\pi x}{a}\right)\sin\!\left(\frac{n\pi y}{b}\right) \quad \text{(V/m)} \qquad \text{(10-132)} \]
\[ h^{2} = \left(\frac{m\pi}{a}\right)^{2} + \left(\frac{n\pi}{b}\right)^{2} \qquad \text{(10-133)} \]

For TM, neither \(m\) nor \(n\) can be zero (else \(E_z\equiv 0\)). Lowest TM mode is TM11.

對 TM 而言,\(m\) 與 \(n\) 都不能為零(否則 \(E_z\equiv 0\))。最低的 TM mode 是 TM11。

TM では \(m\) も \(n\) もゼロにはできません(さもなければ \(E_z\equiv 0\) となります)。最低次の TM モードは TM11 です。

TEmn modes — boundary condition \(\partial H_z/\partial n = 0\) on walls forces cosine variation:

TEmn modes — 壁上 \(\partial H_z/\partial n = 0\) 的 boundary condition 迫使場呈餘弦(cosine)變化:

TEmn モード — 壁で \(\partial H_z/\partial n = 0\) という境界条件により、余弦関数(cosine)の変化が強制されます:

\[ H_z^{0}(x,y) = H_0 \cos\!\left(\frac{m\pi x}{a}\right)\cos\!\left(\frac{n\pi y}{b}\right) \quad \text{(A/m)} \qquad \text{(10-158)} \]

For TE, either \(m\) or \(n\) may be zero, but not both.

對 TE 而言,\(m\) 或 \(n\) 其中之一可以為零,但不能兩者皆為零。

TE では \(m\) または \(n\) のどちらか一方はゼロでもよいのですが、両方同時にゼロにはできません。

Cutoff frequency / wavelength (same form for TM and TE):

Cutoff frequency / cutoff wavelength(TM 與 TE 形式相同):

遮断周波数/遮断波長(TM と TE で同じ形):

\[ (f_c)_{mn} = \frac{1}{2\sqrt{\mu\epsilon}}\sqrt{\left(\frac{m}{a}\right)^{2} + \left(\frac{n}{b}\right)^{2}} \quad (\text{Hz}) \qquad \text{(10-139)} \]
\[ (\lambda_c)_{mn} = \frac{2}{\sqrt{(m/a)^{2}+(n/b)^{2}}} \quad (\text{m}) \qquad \text{(10-140)} \]

Dominant mode: TE10. Because \(n=0\) is permitted for TE and \(a>b\), TE10 has the lowest cutoff of all modes:

Dominant mode(主模):TE10。因為 TE 允許 \(n=0\),且 \(a>b\),所以 TE10 在所有 modes 中具有最低的 cutoff:

基本モード(dominant mode):TE10。TE では \(n=0\) が許され、かつ \(a>b\) なので、TE10 はすべてのモードの中で最も低い遮断周波数をもちます:

\[ (f_c)_{\text{TE}_{10}} = \frac{1}{2a\sqrt{\mu\epsilon}} = \frac{u}{2a},\qquad (\lambda_c)_{\text{TE}_{10}} = 2a \qquad \text{(10-163,164)} \]

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).

實務上的經驗法則。Waveguide 通常以其 dominant mode 工作,操作頻帶約為 \(1.25\,f_{c10}\) 到 \(0.95\,f_{c20}\) — 也就是高於主模 cutoff 25%,但低於下一個高階 mode 5%。例(WR-16,\(a=2.29\) cm,\(b=1.02\) cm):可用頻帶 8.19–12.45 GHz(X-band)。

実用上の経験則。導波管は通常、基本モードで、およそ \(1.25\,f_{c10}\) から \(0.95\,f_{c20}\) までの動作帯域で使用されます — すなわち基本モードの遮断周波数より 25% 上、次の高次モードより 5% 下です。例(WR-16、\(a=2.29\) cm、\(b=1.02\) cm):使用可能帯域は 8.19–12.45 GHz(X バンド)です。

Attenuation. Dielectric loss (substituting \(\epsilon_d=\epsilon(1-j\sigma/\omega\epsilon)\)):

Attenuation(衰減)。Dielectric loss(代入 \(\epsilon_d=\epsilon(1-j\sigma/\omega\epsilon)\)):

減衰(attenuation)。誘電体損失(\(\epsilon_d=\epsilon(1-j\sigma/\omega\epsilon)\) を代入):

\[ \alpha_d = \frac{\sigma\eta}{2\sqrt{1-(f_c/f)^{2}}} \quad (\text{Np/m}) \qquad \text{(10-178)} \]

Conductor (wall) loss for the dominant mode:

Dominant mode 的 conductor(管壁)loss:

基本モードに対する導体(壁)損失:

\[ (\alpha_c)_{\text{TE}_{10}} = \frac{R_s\bigl[1 + (2b/a)(f_c/f)^{2}\bigr]}{\eta b\sqrt{1-(f_c/f)^{2}}} \quad (\text{Np/m}) \qquad \text{(10-187)} \]

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.

Attenuation 在 \(f_c\) 附近發散,在操作頻帶中段有一個寬廣的最小值,並在高 \(f\) 時緩慢上升。TE10 在所有 modes 中衰減永遠最低,這是它受到青睞的另一個原因。

減衰は \(f_c\) 付近で発散し、動作帯域の中央で幅広い極小をもち、高い \(f\) ではゆっくり増加します。TE10 は常にどのモードよりも減衰が小さく、これもこのモードが好まれるもう 1 つの理由です。


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 \(J'_n\)).

截面為半徑 \(a\) 的圓。在 \((r,\phi)\) 中分離變數,得到徑向座標的 Bessel's equation;解為 \(J_n(hr)\)(第一類 Bessel functions)。Modes 為 TMnp(取 \(J_n\) 的零點)與 TEnp(取 \(J'_n\) 的零點)。

半径 \(a\) の円形断面です。\((r,\phi)\) で変数分離すると、動径座標についてベッセルの微分方程式(Bessel's equation)が得られ、その解は \(J_n(hr)\)(第 1 種ベッセル関数)です。モードは TMnp(\(J_n\) の零点)と TEnp(\(J'_n\) の零点)です。

Key cutoff facts (with \(p_{np}\) the \(p\)-th zero of \(J_n\), \(p'_{np}\) of \(J'_n\)):

關於 cutoff 的重點(其中 \(p_{np}\) 表示第 \(p\) 個 \(J_n\) 的零點,\(p'_{np}\) 則表示 \(J'_n\) 的零點):

遮断に関する要点(\(p_{np}\) は \(p\) 番目の \(J_n\) の零点、\(p'_{np}\) は \(J'_n\) の零点):

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).

TE11 是 dominant mode,類比於矩形導波管的 TE10。圓形導波管在實務上較不常見,因為 polarization 可能旋轉;但在需要旋轉對稱的場合(rotary joints、circular polarizers)仍會使用。

TE11 が基本モードで、方形導波管の TE10 に相当します。円形導波管は偏波が回転しうるため実用上はあまり一般的ではありませんが、回転対称性が重要な場合(ロータリージョイント、円偏波器)には用いられます。


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.

當導波結構是一片開放的 dielectric slab(或 optical fiber),且內部 \(\epsilon_1 > \epsilon_2\)(外部)時,場是由 total internal reflection(全內反射)所束縛。在高 \(\epsilon\) 的平板內,橫向場呈正弦變化;在外部則指數衰減 — 此波為 surface wave(表面波)。

導波構造が開放された誘電体スラブ(または光ファイバ)で、内部の \(\epsilon_1 > \epsilon_2\)(外部)である場合、場は全反射(total internal reflection)によって束縛されます。高い \(\epsilon\) をもつスラブの内部では横方向の場は正弦関数的に変化し、外部では指数関数的に減衰します — この波は表面波(surface wave)です。

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.

TM 與 TE modes 都可能存在。與金屬導波管不同,每個 mode 的 cutoff 條件各不相同;這種結構是所有 optical fiber 通訊的基礎。

TM モードと TE モードの両方が可能です。金属導波管と異なり、各モードの遮断条件はそれぞれ異なり、この構造はあらゆる光ファイバ(optical fiber)通信の基礎となっています。


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 \((m,n,p)\).

把 waveguide 的兩端都用導體壁封閉,就得到三維的共振腔。端壁的反射把行進波 \(e^{-j\beta z}\) 轉換成駐波 \(\sin(p\pi z/d)\) 或 \(\cos(p\pi z/d)\);mode 以三個整數 \((m,n,p)\) 標示。

導波管の両端を導体壁で閉じると、3 次元の共振空洞が得られます。端壁での反射により、進行波 \(e^{-j\beta z}\) は定在波 \(\sin(p\pi z/d)\) または \(\cos(p\pi z/d)\) に変わり、モードは 3 つの整数 \((m,n,p)\) で指定されます。

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:

矩形空腔 \(a\times b\times d\)。TMmnp(\(E_z\) 為 \(\sin\sin\cos\) 形式)與 TEmnp(\(H_z\) 為 \(\cos\cos\sin\) 形式)並存。共振頻率與屬於何者無關:

方形空洞 \(a\times b\times d\)。TMmnp(\(E_z\) が \(\sin\sin\cos\) の形)と TEmnp(\(H_z\) が \(\cos\cos\sin\) の形)が共存します。共振周波数はどちらのモードであるかに依りません:

\[ f_{mnp} = \frac{u}{2}\sqrt{\left(\frac{m}{a}\right)^{2} + \left(\frac{n}{b}\right)^{2} + \left(\frac{p}{d}\right)^{2}} \quad (\text{Hz}) \qquad \text{(10-301)} \]

Dominant mode depends on the aspect ratio:

Dominant mode 取決於長寬比:

基本モードは寸法比(アスペクト比)によって決まります:

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 \(p\ne 0\).)

(TM 要求 \(m,n\ne 0\) 但允許 \(p=0\);TE 允許 \(m,n\) 之一為零,但要求 \(p\ne 0\)。)

(TM では \(m,n\ne 0\) が必要ですが \(p=0\) は許されます。TE では \(m,n\) の一方がゼロでも構いませんが、\(p\ne 0\) が必要です。)

Quality factor \(Q\) — the figure of merit for a resonator:

Quality factor(品質因數)\(Q\) — 共振器的優劣指標:

Q 値(quality factor)\(Q\) — 共振器の性能指数です:

\[ Q \triangleq 2\pi\frac{W}{\text{energy dissipated per cycle}} = \frac{\omega W}{P_L} \qquad \text{(10-313,315)} \]

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, \(W_e = W_m\).

其中 \(W = W_e + W_m\) 為總儲存能量,\(P_L\) 為管壁(以及任何 dielectric)所耗散的時間平均功率。在共振時,\(W_e = W_m\)。

ここで \(W = W_e + W_m\) は全蓄積エネルギー、\(P_L\) は壁(および誘電体があればそれも含む)で消費される時間平均電力です。共振時には \(W_e = W_m\) が成り立ちます。

For the TE101 mode in a rectangular cavity:

對矩形空腔中的 TE101 mode:

方形空洞の TE101 モードに対しては:

\[ Q_{101} = \frac{\pi f_{101}\mu_0\, abd(a^{2}+d^{2})}{R_s\bigl[2b(a^{3}+d^{3}) + ad(a^{2}+d^{2})\bigr]} \qquad \text{(10-322)} \]

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 \(R_s=\sqrt{\pi f\mu_0/\sigma}\).

典型數值:10 GHz 的銅製立方空腔其 \(Q\approx 10{,}700\) — 比任何集總 LC 電路高出好幾個數量級,因為損耗只透過 surface resistance \(R_s=\sqrt{\pi f\mu_0/\sigma}\) 發生。

典型的な値:10 GHz の銅製立方体空洞では \(Q\approx 10{,}700\) です — どんな集中定数 LC 回路よりも桁違いに高い値で、損失が表面抵抗(surface resistance)\(R_s=\sqrt{\pi f\mu_0/\sigma}\) を通してのみ生じるためです。

Circular cylindrical cavity. The TM010 mode (radius \(a\), length \(d\)) has the simple resonant frequency

圓柱形空腔。TM010 mode(半徑 \(a\),長度 \(d\))的共振頻率形式很簡單:

円筒形空洞。TM010 モード(半径 \(a\)、長さ \(d\))の共振周波数は次の簡単な形になります:

\[ (f_c)_{\text{TM}_{010}} = \frac{2.405}{2\pi a\sqrt{\mu_0\epsilon_0}} = \frac{0.115}{a} \quad (\text{GHz, with } a \text{ in m}) \qquad \text{(10-329)} \]
\[ Q_{\text{TM}_{010}} = \frac{\eta_0}{R_s}\cdot\frac{2.405}{2(1+a/d)} \qquad \text{(10-328)} \]

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.

空腔的激發方式有:probe(探針,一支插在所需 mode 之 \(\mathbf{E}\) 最大處的短 monopole)、loop(環,插在所需 mode 之 \(\mathbf{H}\) 最大處),或經由 iris(耦合孔)從相鄰的 waveguide 耦合進來。

空洞は、プローブ(所望のモードの \(\mathbf{E}\) が最大となる位置に挿入した短いモノポール)やループ(所望のモードの \(\mathbf{H}\) が最大となる位置に挿入)、あるいは隣接する導波管からアイリス(結合孔)を通した結合によって励振されます。


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.

摘錄自 Cheng,Field and Wave Electromagnetics,§11-1 至 §11-8。請先閱讀本節,再嘗試下方的習題。

Cheng, Field and Wave Electromagnetics, §11-1 〜 §11-8 の要約です。下の問題に取り組む前にお読みください。


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?

第 10 章之前,electromagnetic waves 的分析都是在 source-free(無源)區域或沿著導波結構進行。第 11 章要問的是反過來的問題:給定結構上一個隨時間變化的 current distribution,它會輻射出什麼樣的 electromagnetic field?

第 10 章までは、電磁波を波源のない領域、あるいは導波構造に沿って解析してきました。第 11 章ではその逆の問いを立てます:構造上に時間変化する電流分布が与えられたとき、それはどのような電磁界を放射するのか?

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.

Antenna(天線)是任何設計來在特定方向有效地輻射(或接收)電磁能量的結構。一根直導線、一個 loop、一個 aperture,或由它們組成的複雜 array 都可以當作天線。重要的天線參數包括 field pattern、directivity、radiation resistance/impedance 以及 bandwidth。

アンテナ(antenna)とは、特定の方向へ電磁エネルギーを効率よく放射(あるいは受信)するように設計された構造のことです。1 本の直線導線、ループ、開口、あるいはそれらを組み合わせた複雑なアレイのいずれもアンテナになりえます。重要なアンテナパラメータは界パターン(field pattern)、指向性(directivity)、放射抵抗/放射インピーダンス、および帯域幅(bandwidth)です。

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:

我們不直接以 \(\mathbf{E}\) 與 \(\mathbf{H}\) 求解 Maxwell's equations,而是透過第 7 章的輔助 phasor potentials \(\mathbf{A}\) 與 \(V\),採用其 retarded(推遲)形式:

マクスウェル方程式を \(\mathbf{E}\) と \(\mathbf{H}\) について直接解くのではなく、第 7 章で導入した補助的なフェーザポテンシャル \(\mathbf{A}\) と \(V\) を、その遅延(retarded)形で用います:

\[ \mathbf{A} = \frac{\mu}{4\pi}\int_{V'}\frac{\mathbf{J}\,e^{-jkR}}{R}\,dv' \qquad \text{(11-3)} \]
\[ V = \frac{1}{4\pi\epsilon}\int_{V'}\frac{\rho\,e^{-jkR}}{R}\,dv' \qquad \text{(11-4)} \]

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:

其中 \(k = \omega\sqrt{\mu\epsilon} = 2\pi/\lambda\) 為 wavenumber(波數)。實際上只有步驟 1 需要積分;步驟 2–3 都只是微分:

ここで \(k = \omega\sqrt{\mu\epsilon} = 2\pi/\lambda\) は波数(wavenumber)です。実際に積分が必要なのはステップ 1 だけで、ステップ 2–3 は微分です:

  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.
  1. 計算 \(\mathbf{A}\):由給定的 current distribution \(\mathbf{J}\) 求得(式 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) — 由於觀察點在天線結構之外,可使用 source-free 的 curl 關係式。
  1. \(\mathbf{A}\) を、与えられた電流分布 \(\mathbf{J}\) から計算します(式 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) — アンテナ構造の外側にいるので、波源のない領域での回転(curl)の関係式を使います。

The continuity equation \(\nabla\cdot\mathbf{J} = -j\omega\rho\) relates \(\rho\) and \(\mathbf{J}\), so \(V\) need not be computed separately.

Continuity equation \(\nabla\cdot\mathbf{J} = -j\omega\rho\) 將 \(\rho\) 與 \(\mathbf{J}\) 聯繫起來,因此 \(V\) 不需另外計算。

連続の方程式 \(\nabla\cdot\mathbf{J} = -j\omega\rho\) が \(\rho\) と \(\mathbf{J}\) を結びつけているので、\(V\) を別に計算する必要はありません。


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

最簡單的輻射體:一段長度 \(d\ell \ll \lambda\) 的短直導線,載有 uniform(均勻)電流 \(i(t) = I\cos\omega t\),兩端各接一個小球以容納電荷的累積(capacitive end-loading)。對位於原點、沿 z 方向的電流元,\(\mathbf{A} = \mathbf{a}_z A_z\),其中

最も単純な放射体です。長さ \(d\ell \ll \lambda\) の短い直線導線に一様な電流 \(i(t) = I\cos\omega t\) を流し、両端に小さな球を付けて電荷が蓄積できるようにしたもの(容量性の端部負荷)です。原点に置かれた z 方向の電流素子に対して \(\mathbf{A} = \mathbf{a}_z A_z\) であり、

\[ A_z = \frac{\mu_0 I\,d\ell}{4\pi R}\,e^{-j\beta R} \qquad \text{(11-12)} \]

Taking curls and resolving into spherical components gives the complete field:

取 curl 並分解成球座標分量,得到完整的場:

回転をとり、球座標成分に分解すると完全な場が得られます:

\[ H_\phi = \frac{I\,d\ell}{4\pi}\beta^2\sin\theta\!\left[\frac{1}{j\beta R} + \frac{1}{(j\beta R)^2}\right]\!e^{-j\beta R} \qquad \text{(11-15)} \]
\[ E_R = -\frac{I\,d\ell}{4\pi}\eta_0\beta^2\,2\cos\theta\!\left[\frac{1}{(j\beta R)^2} + \frac{1}{(j\beta R)^3}\right]\!e^{-j\beta R} \qquad \text{(11-16a)} \]
\[ E_\theta = -\frac{I\,d\ell}{4\pi}\eta_0\beta^2\sin\theta\!\left[\frac{1}{j\beta R} + \frac{1}{(j\beta R)^2} + \frac{1}{(j\beta R)^3}\right]\!e^{-j\beta R} \qquad \text{(11-16b)} \]

Three distance regimes emerge from the \(1/R, 1/R^2, 1/R^3\) terms:

由 \(1/R, 1/R^2, 1/R^3\) 各項可區分出三種距離範圍:

\(1/R, 1/R^2, 1/R^3\) の各項から、距離に応じた 3 つの領域が現れます:

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.

主導項為 \(1/R^3\)(在 \(\mathbf{E}\) 中)與 \(1/R^2\)(在 \(\mathbf{H}\) 中)— 分別與 electrostatic dipole 及 Biot–Savart 電流元完全相同。振盪 dipole 的 near-zone 場實質上是 quasi-static(準靜態)的。

主要項は \(1/R^3\)(\(\mathbf{E}\) の場合)と \(1/R^2\)(\(\mathbf{H}\) の場合)であり、それぞれ静電双極子およびビオ・サバールの電流素子とまったく同じ形です。振動する双極子の近傍界は、実質的に準静的(quasi-static)です。

Far field / radiation field (\(\beta R \gg 1\))

Only the \(1/R\) terms survive:

只剩下 \(1/R\) 的項:

\(1/R\) の項だけが残ります:

\[ \boxed{H_\phi = j\frac{I\,d\ell}{4\pi R}\,\beta\sin\theta\,e^{-j\beta R}} \quad (\text{A/m}) \qquad \text{(11-19a)} \]
\[ \boxed{E_\theta = j\frac{I\,d\ell}{4\pi R}\,\eta_0\beta\sin\theta\,e^{-j\beta R}} \quad (\text{V/m}) \qquad \text{(11-19b)} \]

Three observations:

三點觀察:

3 つの観察点:

The radiation pattern is \(|\sin\theta|\): doughnut-shaped, maximum in the equatorial plane, zero along the dipole axis.

Radiation pattern(輻射場型)為 \(|\sin\theta|\):甜甜圈狀,在赤道面上最大,沿 dipole 軸向為零。

放射パターン(radiation pattern)は \(|\sin\theta|\) です。ドーナツ形で、赤道面で最大、双極子の軸方向ではゼロになります。

The condition \(\beta R \gg 1\) is equivalent to \(R \gg \lambda/2\pi\) — at lower frequencies the far zone is farther away.

條件 \(\beta R \gg 1\) 等價於 \(R \gg \lambda/2\pi\) — 頻率越低,far zone 就離得越遠。

条件 \(\beta R \gg 1\) は \(R \gg \lambda/2\pi\) と同値です。つまり周波数が低いほど、遠方界はより遠くになります。


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 \(I\,d\ell \leftrightarrow j\beta m\):

一個面積 \(S = \pi b^2\)、載有電流 \(I\) 的細線小 loop,是一個 magnetic dipole,其 phasor moment 為 \(\mathbf{m} = \mathbf{a}_z I\pi b^2\)。它的 far-field 解是 electric dipole 的 dual(對偶)— 只需作代換 \(I\,d\ell \leftrightarrow j\beta m\) 即可得到:

面積 \(S = \pi b^2\) の細い小ループに電流 \(I\) を流したものは磁気双極子であり、そのフェーザモーメントは \(\mathbf{m} = \mathbf{a}_z I\pi b^2\) です。その遠方界の解は電気双極子の双対(dual)であり、置き換え \(I\,d\ell \leftrightarrow j\beta m\) によって得られます:

\[ E_\phi = \frac{\omega\mu_0 m}{4\pi R}\,\beta\sin\theta\,e^{-j\beta R} \qquad \text{(11-30a)} \]
\[ H_\theta = -\frac{\omega\mu_0 m}{4\pi R\eta_0}\,\beta\sin\theta\,e^{-j\beta R} \qquad \text{(11-30b)} \]
\[ \mathbf{E}_e = \eta_0\mathbf{H}_m, \qquad \mathbf{H}_e = -\mathbf{E}_m/\eta_0 \quad \text{(duality)} \qquad \text{(11-27, 11-28)} \]

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.

同樣的 \(|\sin\theta|\) 場型,同樣的 \(1/R\) 衰減。\(E_\theta\) 與 \(E_\phi\) 在空間上正交,因此將 electric dipole 與 magnetic dipole(以適當相位)組合起來,可產生 circular polarization(圓極化)。

パターンは同じ \(|\sin\theta|\)、減衰も同じ \(1/R\) です。\(E_\theta\) と \(E_\phi\) は空間的に直交しているので、電気双極子と磁気双極子を(適切な位相で)組み合わせると円偏波(circular polarization)が得られます。


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 \((\theta, \phi)\):

Radiation intensity(輻射強度)是在 \((\theta, \phi)\) 方向上、每單位立體角所輻射的時間平均功率:

放射強度(radiation intensity)とは、方向 \((\theta, \phi)\) に放射される単位立体角あたりの時間平均電力です:

\[ U(\theta, \phi) = R^2 \mathscr{P}_{av}(\theta, \phi) = \tfrac{1}{2}R^2\,\mathscr{R}e(\mathbf{E}\times\mathbf{H}^*)\cdot\mathbf{a}_R \quad (\text{W/sr}) \qquad \text{(11-32)} \]

\(U\) is independent of \(R\). Total radiated power:

\(U\) 與 \(R\) 無關。總輻射功率為:

\(U\) は \(R\) に依存しません。全放射電力は次式で与えられます:

\[ P_r = \oint U\,d\Omega = \int_0^{2\pi}\!\!\int_0^\pi U(\theta,\phi)\sin\theta\,d\theta\,d\phi \quad (\text{W}) \qquad \text{(11-33)} \]

Directive gain, directivity, beamwidth

The directive gain measures how the antenna concentrates radiation compared to an isotropic radiator with the same \(P_r\):

Directive gain(指向增益)衡量天線相較於具有相同 \(P_r\) 的 isotropic radiator,將輻射集中到何種程度:

指向性利得(directive gain)は、同じ \(P_r\) をもつ等方性放射体と比べて、アンテナがどれだけ放射を集中させているかを表します:

\[ G_D(\theta,\phi) = \frac{4\pi\,U(\theta,\phi)}{P_r} \qquad \text{(11-34)} \]

Its maximum value is the directivity:

其最大值即為 directivity(指向性):

その最大値が指向性(directivity)です:

\[ \boxed{D = \frac{U_{\max}}{U_{av}} = \frac{4\pi\,U_{\max}}{P_r}} \quad (\text{dimensionless, often in dB}) \qquad \text{(11-35)} \]

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.

(Half-power) beamwidth(半功率波束寬度)是 \(U\) 降到其最大值一半的兩個方向之間的角寬度,也就是功率場型上的 \(-3\) dB 點。波束越窄 ⇒ directivity 越高。

(半値)ビーム幅(half-power beamwidth)とは、\(U\) が最大値の半分に落ちる 2 つの方向の間の角度幅、すなわち電力パターン上の \(-3\) dB 点の間の幅です。ビームが狭いほど ⇒ 指向性が高くなります。

For the Hertzian dipole: \(G_D = 1.5\sin^2\theta\), \(D = 1.5\) (1.76 dB), beamwidth 90°.

對 Hertzian dipole:\(G_D = 1.5\sin^2\theta\),\(D = 1.5\)(1.76 dB),beamwidth 為 90°。

ヘルツ双極子(Hertzian dipole)の場合:\(G_D = 1.5\sin^2\theta\)、\(D = 1.5\)(1.76 dB)、ビーム幅は 90° です。

Radiation resistance

A hypothetical resistance that would dissipate the same power as the antenna radiates, at the antenna's peak current:

這是一個假想的電阻:當流過天線峰值電流時,其消耗的功率恰等於天線所輻射的功率:

アンテナのピーク電流が流れたときに、アンテナが放射するのと同じ電力を消費するような仮想的な抵抗です:

\[ R_r = \frac{2P_r}{I^2} \quad (\Omega) \qquad \text{(11-46)} \]

For a Hertzian dipole (carrying the standard 80π² integral through):

對 Hertzian dipole(完成標準的 80π² 積分後):

ヘルツ双極子の場合(標準的な 80π² の積分を実行すると):

\[ \boxed{R_r = 80\pi^2\!\left(\frac{d\ell}{\lambda}\right)^2} \quad (\Omega) \qquad \text{(11-44)} \]

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.

注意其對 \((d\ell/\lambda)^2\) 的相依性 — 一個 \(d\ell = 0.01\lambda\) 的 Hertzian dipole,其 \(R_r \cong 0.08\ \Omega\) 而已。短 dipole 是非常差的輻射體。短 dipole 的 input impedance 還帶有很大的電容性 reactance,使其難以有效饋電。

\((d\ell/\lambda)^2\) への依存性に注意してください。\(d\ell = 0.01\lambda\) のヘルツ双極子では \(R_r \cong 0.08\ \Omega\) にしかなりません。短い双極子はきわめて効率の悪い放射体です。また、短い双極子の入力インピーダンスは大きな容量性リアクタンスをもつため、効率よく給電するのが困難です。

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:

並非所有輸入功率都被輻射出去;一部分以 ohmic heat 的形式損耗在導線(loss resistance \(R_\ell\))或地面中:

入力電力のすべてが放射されるわけではなく、一部は導線(損失抵抗 \(R_\ell\))や大地でジュール熱として失われます:

\[ P_i = P_r + P_\ell, \qquad \eta_r = \frac{P_r}{P_i} = \frac{R_r}{R_r + R_\ell} = \frac{G_P}{D} \qquad \text{(11-39, 11-41, 11-47)} \]
\[ G_P = \frac{4\pi\,U_{\max}}{P_i} \quad \text{(power gain)} \qquad \text{(11-40)} \]

Well-constructed antennas achieve \(\eta_r\) close to 100%.

製作良好的天線可使 \(\eta_r\) 接近 100%。

よく作られたアンテナでは \(\eta_r\) は 100% 近くに達します。


11-4   Thin Linear Antennas

A center-fed dipole of total length \(2h\) carrying a sinusoidal standing-wave current:

總長 \(2h\) 的中央饋電 dipole,載有 sinusoidal standing-wave(正弦駐波)電流:

全長 \(2h\) の中央給電双極子に、正弦波状の定在波電流が流れているとします:

\[ I(z) = I_m\sin\beta(h - |z|), \quad -h \le z \le h \qquad \text{(11-52)} \]

(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

(這是一個假設的、物理上合理的電流分布;即使對很粗的天線並不精確,仍能給出有用的結果。)對 retarded vector potential 積分並只保留 far-field 項,得到

(これは仮定した、物理的に妥当な電流分布であり、非常に太いアンテナでは厳密ではありませんが、有用な結果を与えます。)遅延ベクトルポテンシャルを積分し、遠方界の項だけを残すと次式が得られます。

\[ E_\theta = \frac{j60 I_m}{R}\,e^{-j\beta R}\,F(\theta) \qquad \text{(11-55)} \]
\[ \boxed{F(\theta) = \frac{\cos(\beta h\cos\theta) - \cos\beta h}{\sin\theta}} \qquad \text{(11-56)} \]

\(|F(\theta)|\) 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.

\(|F(\theta)|\) 是細 dipole 的 \(E\)-plane pattern function(場型函數);\(H\)-plane 的場型則是一個圓(方位角對稱)。場型形狀強烈取決於 \(\beta h = \pi(2h/\lambda)\):當 \(2h = \lambda/2\) 或 \(\lambda\) 時,單一 lobe 佔主導;當 \(2h \ge 3\lambda/2\) 時,會出現多個 lobes,且主波束偏離 broadside 方向。

\(|F(\theta)|\) は細い双極子の \(E\) 面パターン関数(pattern function)です。\(H\) 面のパターンは円(方位角について対称)になります。パターンの形状は \(\beta h = \pi(2h/\lambda)\) に強く依存します。\(2h = \lambda/2\) または \(\lambda\) では単一のローブが支配的ですが、\(2h \ge 3\lambda/2\) になると複数のローブが現れ、主ビームがブロードサイド方向から傾きます。

Half-wave dipole (2h = λ/2)

The workhorse of practical antenna work. Setting \(\beta h = \pi/2\):

實務天線工作中最常用的主力。令 \(\beta h = \pi/2\):

実用アンテナの主力です。\(\beta h = \pi/2\) とおくと:

\[ F(\theta) = \frac{\cos[(\pi/2)\cos\theta]}{\sin\theta} \qquad \text{(11-57)} \]
\[ \mathscr{P}_{av} = \frac{15 I_m^2}{\pi R^2}\!\left[\frac{\cos((\pi/2)\cos\theta)}{\sin\theta}\right]^2 \qquad \text{(11-60)} \]

Integrating over a great sphere gives \(P_r = 36.54\,I_m^2\) watts, from which

在大球面上積分得到 \(P_r = 36.54\,I_m^2\) watts,由此可得

大きな球面上で積分すると \(P_r = 36.54\,I_m^2\) ワットとなり、これから

\[ \boxed{R_r = 73.1\ \Omega} \qquad \text{(11-63)} \]
\[ \boxed{D = \frac{60}{36.54} = 1.64 \approx 2.15\ \text{dB}} \qquad \text{(11-65)} \]
\[ \text{HPBW} \cong 78° \]

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.

Half-wave dipole 的 input impedance 約為 \(73.1\ \Omega\) — 恰好接近標準的 75-Ω coaxial cable,因此匹配很簡單。將長度略微修短至 λ/2 以下,即可使 input reactance 消失。

半波長双極子(half-wave dipole)の入力インピーダンスは約 \(73.1\ \Omega\) で、都合のよいことに標準的な 75-Ω 同軸ケーブルに近いため、整合は簡単です。長さを λ/2 よりわずかに短く調整すると、入力リアクタンスをゼロにできます。

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:

一根長 \(\lambda/4\) 的垂直導線,相對於大型導電 ground plane 饋電。根據 method of images(鏡像法),ground plane 加上 monopole 等效於一個只向上半空間輻射的 half-wave dipole。由此可得:

長さ \(\lambda/4\) の垂直導線を、大きな導体接地面に対して励振したものです。鏡像法により、接地面とモノポールの組み合わせは、上半空間のみに放射する半波長双極子と等価になります。その帰結は:


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:

Antenna array(天線陣列)是一組相似的天線,透過受控的干涉來獲得所需的場型。遠處某點的場是各個元件所產生之場的向量疊加。對於由相同元件組成的陣列:

アンテナアレイ(antenna array)とは、制御された干渉によって所望の界パターンを得るように配置した、同種のアンテナの集まりです。遠方の点における場は、各素子がつくる場のベクトル重ね合わせです。同一素子からなるアレイでは:

\[ |E|_{\text{array}} = |E_{\text{single}}|\cdot|A(\theta,\phi)| \]
Total pattern = element factor × array factor

This is the principle of pattern multiplication.

這就是 principle of pattern multiplication(場型相乘原理)。

これがパターン乗積の原理(principle of pattern multiplication)です。

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

兩個相同的元件沿 x 軸相距 \(d\),電流大小相等,但元件 2 相對於元件 1 有 \(\xi\) 的相位差。其 array factor 為

2 つの同一素子を x 軸に沿って距離 \(d\) だけ離して置き、電流の大きさは等しいが、素子 2 の位相を素子 1 に対して \(\xi\) だけずらしたものです。アレイファクタ(array factor)は

\[ \psi = \beta d\sin\theta\cos\phi + \xi \qquad \text{(11-82)} \]
\[ |A(\theta,\phi)| = 2\,|\cos(\psi/2)| \qquad \text{(11-83)} \]

Two important configurations:

兩種重要的配置:

重要な構成が 2 つあります:

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.

使用 \(N\) 個元件並配合分級的振幅(uniform、binomial、Dolph–Chebyshev),陣列可以獲得狹窄的主 lobe 以及受控的 sidelobe 位準。元件數增加 \(N\) 倍,增益約增加 \(N\) 倍。

\(N\) 個の素子を用い、振幅に重み付け(一様、二項、Dolph–Chebyshev)を施すことで、アレイは狭い主ローブと制御されたサイドローブレベルを実現できます。素子数を \(N\) 倍にすると、利得もおよそ \(N\) 倍になります。


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:

根據 reciprocity theorem(互易定理),天線的場型、directivity 與 input impedance 無論用於發射或接收都完全相同。入射場所感應的 open-circuit voltage 為:

相反定理(reciprocity theorem)により、アンテナのパターン、指向性、入力インピーダンスは、送信に使っても受信に使っても同一です。入射界によって誘起される開放電圧は:

\[ |V_{oc}| = |\mathbf{\ell}_e\cdot\mathbf{E}_i| \qquad \text{(11-76)} \]

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.

其中 \(\mathbf{\ell}_e\) 是天線的向量 effective length(有效長度)。\(V_{oc}\) 在 \(\mathbf{E}_i\) 平行於天線(極化匹配)時最大,垂直時為零 — 這就是 polarization mismatch(極化失配)。

ここで \(\mathbf{\ell}_e\) はアンテナのベクトル実効長(effective length)です。\(V_{oc}\) は \(\mathbf{E}_i\) がアンテナに平行(偏波整合)なとき最大となり、垂直なときはゼロになります。これが偏波不整合(polarization mismatch)です。

Effective aperture and the Friis formula

A receiving antenna intercepts incoming wave power with an effective aperture:

接收天線以 effective aperture(有效孔徑)攔截入射波的功率:

受信アンテナは、実効開口(effective aperture)によって到来波の電力を捕捉します:

\[ A_e = \frac{\lambda^2}{4\pi}\,D \quad (\text{m}^2) \]

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:

對於增益 \(G_t\) 的發射天線與 effective aperture 為 \(A_r\) 的接收天線,相距 \(R\) 在自由空間中傳播,Friis transmission formula 給出接收功率:

利得 \(G_t\) の送信アンテナと実効開口 \(A_r\) の受信アンテナが距離 \(R\) だけ離れて自由空間で通信するとき、フリスの伝達公式(Friis transmission formula)は受信電力を次のように与えます:

\[ \boxed{\frac{P_r}{P_t} = G_t\,G_r\!\left(\frac{\lambda}{4\pi R}\right)^2} \]

The \((\lambda/R)^2\) free-space path loss governs every communications link budget. For radar (target of radar cross section \(\sigma\) at distance \(R\)):

\((\lambda/R)^2\) 的 free-space path loss 支配著所有通訊 link budget。對 radar(目標的 radar cross section 為 \(\sigma\),距離為 \(R\)):

\((\lambda/R)^2\) の自由空間経路損失は、あらゆる通信リンクバジェットを支配します。レーダー(radar)の場合(レーダー断面積 \(\sigma\) の目標が距離 \(R\) にあるとき):

\[ P_r = \frac{P_t G_t G_r \lambda^2\sigma}{(4\pi)^3 R^4} \quad \text{(radar equation)} \]

The \(R^4\) dependence (vs. \(R^2\) for one-way) is why radar is fundamentally hard.

\(R^4\) 的相依性(相對於單向傳播的 \(R^2\))正是 radar 本質上困難的原因。

\(R^4\) に依存すること(片道の \(R^2\) に対して)が、レーダーが本質的に難しい理由です。


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:

高度 \(h_1\) 的發射天線與高度 \(h_2\) 的接收天線,地面距離為 \(d\),兩者之間存在 direct(直射)射線加上 ground-reflected(地面反射)射線。兩者在接收端的疊加給出一個 path-gain factor:

高さ \(h_1\) の送信アンテナと高さ \(h_2\) の受信アンテナが地上距離 \(d\) だけ離れているとき、直接波に加えて大地反射波が存在します。受信点でこれらを重ね合わせると、経路利得係数が得られます:

\[ |F| = 2\left|\cos\!\left(\frac{2\pi h_1 h_2}{\lambda d}\right)\right| \qquad \text{(11-143)} \]

\(|F|\) 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.

\(|F|\) 會隨 \(h_2/d\) 的變化而在 0 與 2 之間振盪 — 這是地面無線電與行動通訊系統中 multipath fading(多路徑衰落)的根源。若更嚴謹地考慮地球曲率與不完美的地面反射,公式會更複雜;實務上採用軟體的 ray tracing。

\(|F|\) は \(h_2/d\) の変化に伴って 0 と 2 の間で振動します。これが地上無線や携帯電話システムにおけるマルチパスフェージング(multipath fading)の基礎です。地球の曲率や不完全な大地反射まで考慮したより厳密な扱いでは式はさらに複雑になるため、実務ではソフトウェアによるレイトレーシングが用いられます。


11-8   Other Practical Antenna Types

Brief survey of common antennas beyond linear dipoles:

簡要概覽線性 dipole 以外的常見天線:

線状双極子以外の代表的なアンテナを簡単に概観します:


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 \(P_r\).

\[\lambda=0.03,\;G=10^4\]
\[P_r=\frac{P_t G^2\sigma\lambda^2}{(4\pi)^3 R^4}=\frac{10^5\!\cdot\!10^8\!\cdot\!10\!\cdot\!9\!\cdot\!10^{-4}}{(4\pi)^3(5\!\cdot\!10^4)^4}\]
\[=9\!\cdot\!10^{10}/(2.0\!\cdot\!10^{22})=4.5\!\cdot\!10^{-12}\,\text{W}\]