Chapter 1 — The Electromagnetic Model
Key Theory — Chapter 1
Condensed from Cheng, Field and Wave Electromagnetics, §1-1 through §1-3. Read this before attempting the problems below.
1-1 What is Electromagnetics?
Electromagnetics is the study of the effects of electric charges at rest and in motion. There are two kinds of charges — positive and negative — and both are sources of an electric field. Moving charges constitute a current, which gives rise to a magnetic field.
A field is a spatial distribution of a quantity, which may or may not be a function of time. A time-varying electric field is always accompanied by a magnetic field, and vice versa — the two are coupled, forming an electromagnetic field. Under certain conditions, time-dependent electromagnetic fields produce waves that radiate from the source.
Why fields, not just circuits? Circuit theory is a restricted special case of electromagnetic theory, valid only when the dimensions of a network are much smaller than the wavelength (quasi-static regime). Two situations where circuit theory fails:
- 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.
- An electromagnetic wave incident on a conducting wall with a small aperture produces fields on the far side, even at points not directly behind the hole — a diffraction/shielding problem circuit theory cannot describe.
1-2.1 Deductive vs. Inductive Approach
Two approaches exist in developing a scientific subject:
- Inductive: follow the historical path — observe experiments, infer laws and theorems. Reasoning moves from particular phenomena to general principles.
- Deductive (axiomatic): postulate a few fundamental relations as axioms, then derive particular laws as consequences. The model's validity is verified by predictions that agree with experiment.
Cheng uses the deductive approach because it is more elegant and develops the subject in an orderly way.
Three steps to build any theory on an idealized model
- Define the basic quantities germane to the subject.
- Specify the rules of operation (the mathematics) for those quantities.
- Postulate the fundamental relations (axioms/laws), based on experimental observation.
Familiar example — circuit theory: basic quantities are \(V, I, R, L, C\); rules of operation are algebra, ODEs, and Laplace transforms; fundamental postulates are Kirchhoff's voltage and current laws.
For electromagnetics: Chapter 1 covers step 1; Chapter 2 covers step 2 (vector algebra & calculus); step 3 is introduced in three substeps — Chapters 3 (electrostatics), 6 (magnetostatics), and 7 (time-varying fields/Maxwell's equations).
1-2.2 Source Quantities
Quantities in the electromagnetic model fall into two categories: source quantities (the charges and currents that produce fields) and field quantities (the fields themselves).
Electric charge and the charge of an electron
Electric charge is a fundamental property of matter and exists only in positive or negative integer multiples of the elementary charge \(e\):
Principle of conservation of charge — like conservation of momentum, this is a fundamental postulate of physics: charge can neither be created nor destroyed. The algebraic sum of positive and negative charges in an isolated system remains unchanged. This principle is expressed mathematically by the equation of continuity (§5-4), and Kirchhoff's current law is simply an assertion of it applied to a junction.
Volume, surface, and line charge densities
Although charge is discrete microscopically, electromagnetic effects of large aggregates are well described by smoothed-out density functions defined as point functions of space coordinates:
"Small enough" to give accurate variation yet "large enough" to contain many discrete charges — e.g. a cube of side 1 μm has volume \(10^{-18}\ \text{m}^3\) and still contains about \(10^{11}\) atoms.
Current and current density
Current is the rate of change of charge with respect to time:
Current flows through a finite area, so it is not a point function. Two vector point functions are defined to describe current density:
- Volume current density \(\mathbf{J}\) — current per unit area normal to flow direction, units A/m²; direction is the direction of current flow.
- Surface current density \(\mathbf{J}_s\) — used for currents confined to a thin surface layer (e.g. high-frequency currents on good conductors); units A/m.
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):
Which field do you need, and when?
- \(\mathbf{E}\) — the only vector needed for electrostatics in free space; defined as the electric force on a unit test charge.
- \(\mathbf{D}\) — useful when a material medium is present (Ch. 3).
- \(\mathbf{B}\) — the only vector needed for magnetostatics in free space; related to the magnetic force on a moving charge.
- \(\mathbf{H}\) — useful when a magnetic medium is present (Ch. 6).
Static vs. dynamic coupling: under static, steady, or stationary conditions the two pairs \(\{\mathbf{E},\mathbf{D}\}\) and \(\{\mathbf{B},\mathbf{H}\}\) are independent. In time-dependent cases they are coupled — a time-varying \(\mathbf{E}\)/\(\mathbf{D}\) produces \(\mathbf{B}\)/\(\mathbf{H}\), and vice versa.
Constitutive relations
Material (medium) properties determine the relations between \(\mathbf{E}\) and \(\mathbf{D}\), and between \(\mathbf{B}\) and \(\mathbf{H}\). These relations are called the constitutive relations of the medium. In free space (see §1-3):
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:
All units in Table 1-1 are derived:
- Coulomb (C) = A·s
- Volt per meter (V/m) = kg·m/(A·s³)
- Tesla (T) = kg/(A·s²) = V·s/m² = Wb/m²
"Rationalized" means the factor \(4\pi\) has been absorbed into the constitutive constants so that it does not appear in Maxwell's equations themselves (though it does appear in many derived relations such as Coulomb's law).
1-3.2 The Three Universal Constants
Free space (vacuum) is characterized by three universal constants, which are not independent of one another:
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
which links electromagnetism to optics — the velocity of light is a consequence of the electric and magnetic properties of vacuum.
Chapter 1 at a Glance
- Source quantities: \(\rho,\ \rho_s,\ \rho_\ell\) (charge densities), \(I,\ \mathbf{J},\ \mathbf{J}_s\) (current and current densities).
- Field quantities: \(\mathbf{E},\ \mathbf{D}\) (electric) and \(\mathbf{B},\ \mathbf{H}\) (magnetic); 4 vectors total.
- Constitutive relations (free space): \(\mathbf{D}=\epsilon_0\mathbf{E}\), \(\mathbf{B}=\mu_0\mathbf{H}\).
- Universal constants: \(c\), \(\mu_0\), \(\epsilon_0\), with \(c = 1/\sqrt{\epsilon_0\mu_0}\).
- Approach: deductive/axiomatic — define quantities, specify rules, postulate fundamental relations; postulates introduced in Chapters 3, 6, and 7.
- Why not just circuit theory: circuit theory is a quasi-static special case; radiation, diffraction and wave phenomena require the full electromagnetic model.
Problems and Solutions
Review Questions (Tier 1)
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.
(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.
(1) Define basic quantities. (2) Specify rules of operation (the mathematics). (3) Postulate fundamental relations (axioms) verified against experiment.
Meter (m), kilogram (kg), second (s), ampere (A).
\(\vec{E}\) — V/m; \(\vec{D}\) — C/m\(^2\); \(\vec{B}\) — T (V·s/m\(^2\)); \(\vec{H}\) — A/m.
\(c=3{\times}10^8\,\text{m/s}\) (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: \(c=1/\sqrt{\mu_0\epsilon_0}\).
Electric charge \(q\) (or charge density \(\rho_v\)) and current density \(\vec{J}\).
Convert 1 coulomb to statcoulombs (ESU) and abcoulombs (EMU).
→ SolutionExpress c in m/s, cm/s, ft/ns, and km/s.
→ SolutionFind the force between two point charges Q₁ = Q₂ = 1 μC separated by r = 1 m.
→ SolutionDerive ε₀μ₀ = 1/c² and verify numerically.
→ SolutionConvert 1 ampere to statamperes and abamperes.
→ SolutionA plane wave in free space has E₀ = 100 V/m. Find the electric and magnetic energy densities.
→ SolutionFor the wave in Problem 6, find the Poynting vector magnitude.
→ SolutionVerify ∇ × E = −∂B/∂t and ∇ × H = J + ∂D/∂t are dimensionally consistent.
→ SolutionConvert E = 1 kV/m to (a) N/C, (b) V/cm, (c) force on a 1 μC charge.
→ SolutionConvert B = 1 T to Gauss and find H in free space.
→ SolutionDerive η₀ = √(μ₀/ε₀) and verify it equals 120π Ω.
→ SolutionCalculate 1/√(ε₀μ₀) from the SI values of ε₀ and μ₀.
→ SolutionFor E = 1 MV/m in free space, compute u_e in J/m³, erg/cm³, and eV/m³.
→ SolutionExpress k = 1/(4πε₀) in SI units and verify its numerical value.
→ SolutionA plane wave in free space has H = 1 A/m. Find E, power density, and compare E/H to η₀.
→ SolutionChapter 2 — Vector Analysis
Key Theory — Chapter 2
Condensed from Cheng, Field and Wave Electromagnetics, §2-1 through §2-12. Read this before attempting the problems below.
2-1 Introduction — Why Vector Analysis?
Electromagnetic quantities are either scalars (charge, energy, potential) or vectors (\(\mathbf{E}\), \(\mathbf{B}\), current density). In three-dimensional space a vector relation is three scalar relations, so an efficient vector calculus is essential.
Three pillars of this chapter:
- Vector algebra — addition, dot and cross products.
- Orthogonal coordinate systems — Cartesian, cylindrical, spherical.
- Vector calculus — gradient, divergence, curl, and the integral theorems that tie them together.
The physical laws themselves are invariant under choice of coordinates; coordinates are chosen only to match the geometry of a given problem.
2-2 Vector Addition and Subtraction
A vector has magnitude and direction; a unit vector points without magnitude:
Addition is commutative and associative; subtraction is \(\mathbf{A}-\mathbf{B} = \mathbf{A}+(-\mathbf{B})\).
2-3 Products of Vectors
Dot (scalar) product — yields a scalar:
Zero iff vectors are orthogonal; equals the projection of one onto the other times the other's magnitude. Commutative and distributive; not associative (triple dot is meaningless).
Cross (vector) product — yields a vector perpendicular to the plane of the operands, magnitude equals the area of their parallelogram:
Direction \(\mathbf{a}_n\) by right-hand rule. Cross product is anti-commutative, distributive, and not associative.
Scalar triple product — cyclic permutation leaves it unchanged (geometrically, volume of the parallelepiped):
Vector triple product — the BAC-CAB rule:
Parentheses matter — \((\mathbf{A}\times\mathbf{B})\times\mathbf{C}\) is a different vector. Division by a vector is undefined.
2-4 Orthogonal Coordinate Systems
A right-handed orthogonal system has base vectors \(\mathbf{a}_{u_1},\mathbf{a}_{u_2},\mathbf{a}_{u_3}\) satisfying:
Coordinates \(u_i\) may not themselves be lengths (angles, for instance). The metric coefficient \(h_i\) converts a coordinate change to a length:
Cartesian (§2-4.1):
Cylindrical (§2-4.2): \((r,\phi,z)\). Useful for line charges, long wires, coax. \(\mathbf{a}_r\times\mathbf{a}_\phi=\mathbf{a}_z\).
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. \(\mathbf{a}_R\times\mathbf{a}_\theta=\mathbf{a}_\phi\).
Caution: in curvilinear systems the base vectors depend on position (\(\partial\mathbf{a}_r/\partial\phi = \mathbf{a}_\phi\), \(\partial\mathbf{a}_\phi/\partial\phi = -\mathbf{a}_r\)). Differentiation of a vector in such systems must account for this.
2-5 Line, Surface, Volume Integrals
The differential area is treated as a vector \(d\mathbf{s} = \mathbf{a}_n\,ds\) (outward normal for a closed surface, right-hand-rule normal for an open surface bounded by a contour).
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:
General orthogonal curvilinear form:
Cartesian form — also defines the del operator:
\(\nabla V\) is always perpendicular to surfaces of constant \(V\). In electrostatics we use \(\mathbf{E}=-\nabla V\) — the minus sign makes \(V\) increase opposite to \(\mathbf{E}\).
2-7 Divergence of a Vector Field
Divergence measures the net outward flux of \(\mathbf{A}\) per unit volume — the flow source density:
Coordinate forms:
A field with \(\nabla\cdot\mathbf{A}=0\) everywhere is solenoidal — flux lines close on themselves, no sources or sinks (e.g., magnetic flux density \(\mathbf{B}\)).
2-8 Divergence Theorem
Also known as Gauss's theorem:
Converts volume integral of a divergence into a closed surface integral of the vector — the workhorse behind integral forms of Gauss's law and charge conservation.
2-9 Curl of a Vector Field
Circulation is the line integral of \(\mathbf{A}\) around a closed contour. Curl is the vortex-source density:
The direction of \(\mathbf{a}_n\) is that which maximizes the circulation per unit area (right-hand rule relative to \(d\mathbf{\ell}\)).
Determinant form in Cartesian coordinates:
General orthogonal curvilinear form:
A field with \(\nabla\times\mathbf{A}=0\) everywhere is irrotational (or conservative). A static electrostatic field is always irrotational.
2-10 Stokes's Theorem
Converts surface integral of a curl into a closed line integral around the bounding contour. The directions of \(d\mathbf{\ell}\) and \(d\mathbf{s}=\mathbf{a}_n\,ds\) obey the right-hand rule.
Dictionary. The two integral theorems convert dimensionality:
2-11 Two Null Identities
Repeated del operations give two identities of deep importance for potential theory:
Converse statements (crucial for EM):
- If \(\nabla\times\mathbf{E}=0\), then \(\mathbf{E}\) can be written as the gradient of a scalar: \(\mathbf{E} = -\nabla V\) (2-148). This is why electric potential exists in electrostatics.
- If \(\nabla\cdot\mathbf{B}=0\), then \(\mathbf{B}\) can be written as the curl of a vector: \(\mathbf{B} = \nabla\times\mathbf{A}\) (2-152). This is why magnetic vector potential exists.
2-12 Helmholtz's Theorem
A vector field is determined — to within an additive constant — if both its divergence and its curl are specified everywhere, and the field vanishes at infinity.
Classification of vector fields:
Any such field decomposes into an irrotational part and a solenoidal part:
This is the axiomatic foundation for all of Cheng's subsequent chapters — specifying both div and curl of \(\mathbf{E}\) (Chapter 3) or of \(\mathbf{B}\) (Chapter 6) fixes the entire field. That is why Cheng postulates div and curl rather than starting from Coulomb's law.
Chapter 2 at a Glance
- Algebra: \(\mathbf{A}\cdot\mathbf{B}=AB\cos\theta\); \(\mathbf{A}\times\mathbf{B}=\mathbf{a}_n AB\sin\theta\) (right-hand rule). Cyclic scalar triple product; BAC-CAB rule.
- Three coordinate systems: Cartesian \((x,y,z)\), cylindrical \((r,\phi,z)\), spherical \((R,\theta,\phi)\); Table 2-1 has their base vectors, \(h_i\), and \(dv\).
- Del operator \(\nabla\) in Cartesian: \(\mathbf{a}_x\partial_x + \mathbf{a}_y\partial_y + \mathbf{a}_z\partial_z\).
- Gradient \(\nabla V\) — maximum rate of increase; \(dV=\nabla V\cdot d\mathbf{\ell}\); perpendicular to level surfaces.
- Divergence \(\nabla\cdot\mathbf{A}\) — flow-source density; zero \(\Rightarrow\) solenoidal.
- Curl \(\nabla\times\mathbf{A}\) — vortex-source density; zero \(\Rightarrow\) irrotational/conservative.
- Divergence theorem: \(\int_V \nabla\cdot\mathbf{A}\,dv = \oint_S \mathbf{A}\cdot d\mathbf{s}\).
- Stokes's theorem: \(\int_S (\nabla\times\mathbf{A})\cdot d\mathbf{s} = \oint_C \mathbf{A}\cdot d\mathbf{\ell}\).
- Null identities: \(\nabla\times(\nabla V)=0\), \(\nabla\cdot(\nabla\times\mathbf{A})=0\) — allow definition of scalar potential \(V\) and vector potential \(\mathbf{A}\).
- Helmholtz: \(\mathbf{F}=-\nabla V+\nabla\times\mathbf{A}\); divergence + curl uniquely determine a well-behaved field.
Problems and Solutions
Review Questions (Tier 1)
Head-to-tail closure: \(\vec{A}+\vec{B}+\vec{C}=0\). Therefore \(\vec{A}+\vec{B}-\vec{C}=-2\vec{C}\).
When the angle between \(\vec{A}\) and \(\vec{B}\) exceeds \(90^{\circ}\) (\(\cos\theta<0\)).
(a) Parallel: \(\vec{A}\cdot\vec{B}=AB\), \(\vec{A}\times\vec{B}=0\). (b) Perpendicular: \(\vec{A}\cdot\vec{B}=0\), \(|\vec{A}\times\vec{B}|=AB\).
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).
No. The first lies along \(\vec{C}\), the second along \(\vec{A}\); they are equal only in special cases.
No. Only the components of \(\vec{B}\) and \(\vec{C}\) along \(\vec{A}\) must agree; perpendicular components are unconstrained.
No. Only components of \(\vec{B}\) and \(\vec{C}\) perpendicular to \(\vec{A}\) must agree; parallel components are free.
(a) \((\vec{A}\cdot\vec{B})/|\vec{B}|=A\cos\theta\). (b) \((\vec{A}\cdot\vec{B})/|\vec{A}|=B\cos\theta\).
(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.
\(|\vec{F}|=\sqrt{F_1^2+F_2^2+F_3^2}\); unit vector \(\hat{a}_F=\vec{F}/|\vec{F}|\).
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 \((1,r,r\sin\theta)\).
\(\overrightarrow{P_1P_2}=-2\hat{x}-2\hat{y}-\hat{z}\); \(\overrightarrow{P_2P_1}=2\hat{x}+2\hat{y}+\hat{z}\).
\(\vec{A}\cdot\vec{B}=A_xB_x+A_yB_y+A_zB_z\); \(\vec{A}\times\vec{B}=\det\begin{pmatrix}\hat{x}&\hat{y}&\hat{z}\\A_x&A_y&A_z\\B_x&B_y&B_z\end{pmatrix}\).
A scalar/vector quantity is a single value. A scalar/vector field is a function of position (and possibly time) defined throughout a region.
A vector pointing in the direction of maximum spatial increase of the scalar; magnitude equals that maximum rate of change.
Directional derivative: \(\partial V/\partial \ell=\nabla V\cdot\hat{a}_{\ell}\).
\(\nabla=\hat{x}\dfrac{\partial}{\partial x}+\hat{y}\dfrac{\partial}{\partial y}+\hat{z}\dfrac{\partial}{\partial z}\).
Net outward flux per unit volume at a point: \(\nabla\cdot\vec{A}=\lim_{\Delta v\to 0}\dfrac{1}{\Delta v}\oint\vec{A}\cdot d\vec{S}\).
False. \(\vec{A}=\hat{a}_r/r^2\) is purely radial yet has zero divergence away from the origin (Coulomb-style field).
True. Divergence depends on net flux through a small surface, not on whether flux lines curve.
The volume integral of \(\nabla\cdot\vec{A}\) equals the net outward flux through the enclosing surface: \(\int_V\nabla\cdot\vec{A}\,dV=\oint_S\vec{A}\cdot d\vec{S}\).
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.
Surface integral of curl equals line integral around boundary: \(\int_S(\nabla\times\vec{A})\cdot d\vec{S}=\oint_C\vec{A}\cdot d\vec{\ell}\).
(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.
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 \(\nabla\times\vec{E}\).
A = 2a_x + 3a_y, B = 4a_x − a_y. Find A·B, |A|, |B|, and the angle between them.
→ SolutionA = 2a_ρ + 3a_φ, B = a_ρ + 2a_z. Find A × B.
→ SolutionConvert point P(1, 1, 1) from Cartesian to spherical coordinates.
→ SolutionEvaluate ∬ 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.
→ SolutionFind ∇φ for φ = x²y + yz². Evaluate at P(1, 2, 3).
→ SolutionCalculate ∇·(r² â_r) in spherical coordinates.
→ SolutionVerify the divergence theorem for A = x a_x over the unit cube.
→ SolutionFind ∇ × A for A = ρ² a_φ.
→ SolutionFor A = y a_x, verify Stokes's theorem using the unit square in the xy-plane (counterclockwise).
→ SolutionProve that the divergence of a curl is always zero.
→ SolutionProve the curl of a gradient is always zero.
→ SolutionFind ∇²φ in cylindrical form. Apply it to φ = ρ² cos(2α) (where α is the azimuthal angle).
→ SolutionFind the unit normal to the surface z = x² + y² at point (1, 1, 2).
→ SolutionEvaluate ∮ A·dl for A = −y a_x + x a_y around a circle of radius a in the xy-plane.
→ SolutionDecompose F = x a_x + y a_y + z a_z into irrotational and solenoidal parts.
→ SolutionTextbook Practice Problems
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.
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.
Sample: Convert \(P(2,2,1)\) from Cartesian to cylindrical and spherical.
Sample: Express \(\vec{A}=3\hat{x}+2\hat{y}-\hat{z}\) at \(P(1,\sqrt{3},2)\) in cylindrical components.
Sample: \(V=x^2y+yz^2\). Find \(\nabla V\) at \(P(1,2,-1)\).
Sample: \(V=r^2\sin\theta\cos\phi\) in spherical. Find \(\nabla V\).
Sample: Evaluate \(\nabla\cdot\vec{A}\) for \(\vec{A}=xy\hat{x}+yz\hat{y}+zx\hat{z}\).
Sample: Verify divergence theorem for \(\vec{A}=x\hat{x}+y\hat{y}+z\hat{z}\) over a unit sphere.
Sample: Show that \(\vec{A}\cdot(\vec{B}\times\vec{C})\) equals the volume spanned by the three vectors. Use \(\vec{A}=\hat{x}\), \(\vec{B}=\hat{y}\), \(\vec{C}=\hat{z}\).
Sample: A vector field is given as \(\vec{F}=r\hat{a}_r\) in spherical. Express in Cartesian.
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 \((1,1)\).
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.
Sample: Find a unit normal to the surface \(x^2+y^2-z=0\) at \(P(1,1,2)\).
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, \(0\le z\le 3\).
Sample: Verify Stokes's theorem for \(\vec{A}=y\hat{x}-x\hat{y}\) over a unit disc in the \(xy\)-plane.
Sample: Compute \(\int_S\vec{A}\cdot d\vec{S}\) for \(\vec{A}=\rho\hat{a}_{\rho}\) over the curved surface of a cylinder \(\rho=2\), \(0\le z\le 4\).
Chapter 3 — Static Electric Fields
Key Theory — Chapter 3
Condensed from Cheng, Field and Wave Electromagnetics, §3-1 through §3-11. Read this before attempting the problems below.
3-1 Introduction — Why Postulate, Not Derive?
Elementary texts start from Coulomb's experimental law and build upward. Cheng instead postulates the divergence and curl of \(\mathbf{E}\) in free space, then derives Coulomb's and Gauss's laws from them. Justification:
- Coulomb measured the inverse-square power to limited accuracy; assuming it is exactly 2 is already a postulate.
- By Helmholtz's theorem (§2-12) a vector field in a bounded region is uniquely determined by its divergence and curl. Specifying both of \(\mathbf{E}\) therefore fixes the entire electrostatic theory.
- Coulomb's law, Gauss's law, and scalar potential then follow as consequences, not independent postulates — the axiomatic approach.
In electrostatics, charges are at rest, fields do not change with time, and no magnetic field is present. Only one of the four field quantities — \(\mathbf{E}\) — and one of the three universal constants — \(\epsilon_0\) — are needed in free space.
3-2 Fundamental Postulates of Electrostatics in Free Space
Electric field intensity is defined as the force per unit charge on a stationary test charge:
The entire theory of electrostatics in free space rests on two postulates — the divergence and the curl of \(\mathbf{E}\):
Integral forms follow by the divergence theorem and Stokes's theorem respectively:
Physical reading:
- (3-4) — a static \(\mathbf{E}\) field is not solenoidal unless \(\rho=0\); it has sources (positive charge) and sinks (negative charge).
- (3-5) — a static \(\mathbf{E}\) field is irrotational; its line integral around any closed path is zero. Kirchhoff's voltage law is an assertion of (3-8).
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:
For a system of discrete charges, superposition gives \(\mathbf{E}=\sum_k \mathbf{E}_k\). For a continuous distribution of volume charge density \(\rho(\mathbf{R}')\):
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:
- Spherical: sphere of radius \(R\) for a point charge, charged ball, or shell.
- Cylindrical (line): coaxial cylinder for an infinite line charge or cylindrical charge column.
- Planar: pillbox straddling an infinite sheet of surface charge.
Moral: if symmetry fits, use (3-7) directly. Otherwise find \(V\) first by integration, then \(\mathbf{E}=-\nabla V\).
3-5 Electric Potential
Because \(\nabla\times\mathbf{E}=0\), by the null identity \(\nabla\times(\nabla V)\equiv 0\) we can write \(\mathbf{E}\) as the gradient of a scalar:
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:
Taking the reference at infinity, the potentials of standard charge distributions are:
Field lines and equipotentials are perpendicular — \(\nabla V\) is normal to surfaces of constant \(V\).
3-6 Conductors in Static Electric Field
A conductor has loosely bound electrons that redistribute freely under any interior field. Equilibrium (static) is reached very quickly — for copper the relaxation time is \(\sim 10^{-19}\) s. In equilibrium:
At the conductor / free-space boundary, applying (3-8) to a narrow rectangular contour and (3-7) to a shallow pillbox:
Consequences:
- A conductor's surface is an equipotential.
- All net charge resides on the surface; none in the bulk.
- Field lines meet the surface normally.
- Surface charge concentrates where curvature is highest — basis of the lightning rod.
3-7 Dielectrics — Polarization
In a dielectric, electrons are bound. An applied \(\mathbf{E}\) displaces them slightly, creating induced dipoles. The macroscopic measure is the polarization vector \(\mathbf{P}\) — electric dipole moment per unit volume (C/m²).
Polarized matter behaves as if it carried equivalent bound charges:
The total bound charge on any isolated dielectric body is zero.
3-8 Electric Flux Density \(\mathbf{D}\) and Dielectric Constant
In a material medium the divergence postulate (3-4) must account for bound charge: \(\nabla\cdot\mathbf{E}=(\rho+\rho_p)/\epsilon_0\). Defining the electric flux density
absorbs the bound charge and leaves only free charge on the right:
For a linear, isotropic medium \(\mathbf{P}=\epsilon_0\chi_e\mathbf{E}\), so
where \(\epsilon_r\) is the dimensionless dielectric constant (relative permittivity), \(\chi_e\) the electric susceptibility, and \(\epsilon\) the absolute permittivity (F/m).
Dielectric strength (§3-8.1) is the maximum \(E\) a material withstands before breakdown. Not to be confused with \(\epsilon_r\):
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:
For a charge-free dielectric–dielectric interface (\(\rho_s=0\)):
For a conductor–dielectric interface (medium 2 is a conductor, \(\mathbf{D}_2=0\)): \(E_{1t}=0\) and \(D_{1n}=\rho_s\) — the special case (3-71), (3-72) generalized to \(\mathbf{D}\).
3-10 Capacitance and Capacitors
Because \(\rho_s\), \(\mathbf{E}\), and \(V\) all scale linearly with \(Q\) in a static conductor system, the ratio \(Q/V\) depends only on geometry and the medium:
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 \(C=Q/V_{12}\).
Series / parallel: capacitors in parallel add (\(C_\text{par}=\sum C_k\)); capacitors in series combine as reciprocals (\(1/C_\text{ser}=\sum 1/C_k\)) — the opposite of resistors, because charge is shared in series and voltage in parallel.
Electrostatic shielding (§3-10.3): a grounded conducting shell enclosing a body decouples it from external charges — the coupling capacitance vanishes. This is why sensitive electronics live inside grounded metal enclosures.
3-11 Electrostatic Energy and Forces
In terms of charges and potentials — the work to assemble \(N\) point charges:
The factor \(\tfrac{1}{2}\) avoids double-counting pair interactions. \(V_k\) is the potential at \(Q_k\) due to all other charges.
In terms of field quantities — substituting \(\rho=\nabla\cdot\mathbf{D}\) and applying a vector identity:
Energy density:
Note: (3-165) gives the interaction (mutual) energy only — self-energies of ideal point charges diverge. (3-170) and (3-176) do include self-energy, which is why they agree with each other but can exceed (3-165) for the same system.
Stored energy in a capacitor (equivalent forms):
Force by virtual displacement. For an isolated system (constant charges), work done by the field equals the decrease in stored energy, giving:
The sign flip is not a contradiction — in the constant-\(V\) case the battery supplies additional energy equal to twice the mechanical work.
Chapter 3 at a Glance
- Two postulates fix everything: \(\nabla\cdot\mathbf{E}=\rho/\epsilon_0\), \(\nabla\times\mathbf{E}=0\).
- Integral forms: \(\oint\mathbf{E}\cdot d\mathbf{s}=Q/\epsilon_0\) (Gauss), \(\oint\mathbf{E}\cdot d\mathbf{\ell}=0\) (conservative).
- Scalar potential: \(\mathbf{E}=-\nabla V\); \(V=q/(4\pi\epsilon_0 R)\) for a point charge, referenced at infinity.
- Conductor (static): \(\mathbf{E}=0\) inside, surface equipotential, \(E_t=0\), \(E_n=\rho_s/\epsilon_0\).
- Dielectric: \(\mathbf{D}=\epsilon_0\mathbf{E}+\mathbf{P}=\epsilon\mathbf{E}\); \(\nabla\cdot\mathbf{D}=\rho_\text{free}\).
- BCs: \(E_{1t}=E_{2t}\); \(D_{1n}-D_{2n}=\rho_s\). Dielectric–dielectric: \(\tan\alpha_2/\tan\alpha_1=\epsilon_2/\epsilon_1\).
- Capacitance: \(C=Q/V\); parallel-plate \(\epsilon S/d\); coax \(2\pi\epsilon L/\ln(b/a)\); sphere \(4\pi\epsilon/(1/R_i-1/R_o)\).
- Energy: \(W_e=\tfrac{1}{2}\int\epsilon E^2\,dv=\tfrac{1}{2}CV^2\); force \(\mathbf{F}_Q=-\nabla W_e\) at constant charge.
Problems and Solutions
Review Questions (Tier 1)
\(\nabla\cdot\vec{E}=\rho_v/\epsilon_0\) and \(\nabla\times\vec{E}=0\).
Coulomb's law follows by integrating the postulates for a point charge with spherical symmetry — it is a consequence, not an axiom.
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.
\(\vec{E}=-\nabla V\) so that \(\vec{E}\) points from high to low potential, just as gravity points downhill.
Conductor: \(E_t=0\) and \(D_n=\rho_s\). Dielectric–dielectric (no free \(\rho_s\)): \(E_t\) continuous and \(D_n\) continuous.
\(\vec{P}\) = 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 \(\epsilon_r\).
Parallel-plate \(C=\epsilon A/d\); coaxial \(C=2\pi\epsilon L/\ln(b/a)\); spherical \(C=4\pi\epsilon ab/(b-a)\).
\(W=\tfrac12 CV^2=\tfrac12 QV=Q^2/(2C)\).
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 (\(\vec{F}=+\nabla W\)).
Find E at r = 0.5 m from a point charge Q = 1 μC.
→ Solutionρ_L = 2 nC/m. Find E at ρ = 0.1 m from the line.
→ SolutionSphere of radius a = 0.1 m with total charge Q = 1 μC (uniform ρ_v). Find E inside and outside.
→ SolutionUniform field E = 1000 a_x V/m. Find V_A − V_B where A = (1,0,0) and B = (0,1,0).
→ SolutionPlates: A = 0.01 m², d = 1 mm, εᵣ = 4. Find C, and V for Q = 1 μC.
→ SolutionA conductor has surface charge density ρ_s = 5 μC/m². Find E just outside.
→ SolutionFind P and D for E = 10⁶ V/m in a dielectric with εᵣ = 5.
→ SolutionE₁ = 10³ V/m at θ₁ = 45° to interface. Medium 1: air (ε₁=ε₀), Medium 2: εᵣ=4. Find θ₂.
→ SolutionFor C = 354 pF charged to V = 100 V, find stored energy W.
→ SolutionSame capacitor (C=354pF, V=100V, εᵣ=4, A=0.01m²). Find the attractive force between plates.
→ SolutionThree capacitors: C₁=10pF, C₂=20pF, C₃=30pF. Find C for (a) all series, (b) all parallel, (c) C₁ series with (C₂ ∥ C₃).
→ SolutionA slab (εᵣ=4, thickness 5mm) sits between two others (εᵣ=1, air). Normal D = 10 μC/m². Find E in each region.
→ SolutionIsolated conducting sphere, radius a = 0.1 m, raised to V₀ = 1000 V. Find Q and ρ_s.
→ SolutionA dielectric sphere (εᵣ = 3, radius a) is placed in a uniform field E₀. Find E inside.
→ SolutionExplain why a closed conducting shell provides complete electrostatic shielding. Quantify for copper.
→ SolutionTextbook Practice Problems
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 \(q_2\).
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.
Sample: A spherical volume charge \(\rho_v=\rho_0\) for \(r\le a\). Find \(\vec{E}\) for \(r>a\) and \(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.
Sample: Two parallel infinite plates with surface charges \(+\rho_s\) and \(-\rho_s\). Find \(\vec{E}\) between and outside.
Sample: Find potential at the center of a circular ring of radius \(a\) carrying total charge \(Q\).
Sample: Uniformly charged disc of radius \(a\) with \(\rho_s\). Find \(V\) on axis at distance \(z\).
Sample: Coaxial capacitor: inner \(a=1\,\text{mm}\), outer \(b=5\,\text{mm}\), length \(L=10\,\text{cm}\), \(\epsilon_r=2.25\) (polyethylene). Find \(C\).
Sample: Spherical capacitor with inner radius \(a=2\,\text{cm}\), outer \(b=4\,\text{cm}\). Find \(C\) in vacuum.
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.
Sample: \(V(r)=V_0(a/r)\) for \(r\ge a\). Find \(\vec{E}\) and verify \(\vec{E}=-\nabla V\).
Sample: A grounded conducting sphere of radius \(a\) in a uniform field \(\vec{E}_0=E_0\hat{z}\). Find induced surface charge density.
Sample: Conductor with surface field \(E_n=200\,\text{kV/m}\). Find surface charge density.
Sample: Parallel-plate capacitor with mica (\(\epsilon_r=6\)). Plate area \(100\,\text{cm}^2\), separation \(2\,\text{mm}\). Voltage 100 V. Find \(\vec{D},\vec{E},\vec{P}\).
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 \(\vec{E}_2\).
Sample: Capacitor \(C=100\,\text{pF}\) charged to \(V=500\,\text{V}\). Find stored energy.
Sample: Parallel-plate capacitor, plate area \(A=50\,\text{cm}^2\), separation \(d=1\,\text{mm}\), \(V=200\,\text{V}\). Find force between plates.
Chapter 4 — Solution of Electrostatic Problems
Key Theory — Chapter 4
Condensed from Cheng, Field and Wave Electromagnetics, §4-1 through §4-7. Read this before attempting the problems below.
4-1 Introduction — Two Types of Problem
Chapter 3 let us compute \(V\) and \(\mathbf{E}\) directly when the charge distribution is known everywhere. In practice this rarely happens. Chapter 4 addresses the two practical alternatives:
- Point/line charges near conducting bodies of simple geometry. The induced surface charge is unknown — use the method of images (§4-4).
- Conductors held at prescribed potentials. Solve Poisson's/Laplace's equation as a boundary-value problem — separation of variables in the appropriate coordinate system (§4-5 through §4-7).
Both approaches rest on the uniqueness theorem (§4-3): any \(V\) that satisfies Poisson's equation and the boundary conditions is the solution — a lucky guess, verified, is as rigorous as a derivation.
4-2 Poisson's and Laplace's Equations
Starting from the Ch. 3 postulates in a linear, isotropic medium (\(\mathbf{D}=\epsilon\mathbf{E}\), \(\nabla\times\mathbf{E}=0\), \(\mathbf{E}=-\nabla V\)):
For a simple medium (homogeneous, linear, isotropic — \(\epsilon\) constant), \(\epsilon\) comes out of the divergence operator and we obtain:
Laplace's equation governs the potential in the space between conductors — between capacitor plates, between coax conductors, etc. Once \(V\) is found, \(\mathbf{E}=-\nabla V\) and the surface charge follows from \(\rho_s=\epsilon E_n\) (3-72).
Explicit forms of the Laplacian in each coordinate system:
4-3 Uniqueness Theorem
A solution of Poisson's equation satisfying the given boundary conditions is unique.
Why it matters: it justifies guessing. If you construct any \(V\) — by images, by inspection, by educated guess — that satisfies (4-6) inside the region and matches the prescribed potentials on the boundary, you have the only possible answer.
Sketch of proof: suppose \(V_1\), \(V_2\) both solve the problem; let \(V_d = V_1 - V_2\). Then \(\nabla^2 V_d = 0\) inside and \(V_d = 0\) on all boundaries. Using \(\nabla\cdot(V_d\nabla V_d) = V_d\nabla^2 V_d + |\nabla V_d|^2\) and the divergence theorem:
A nonnegative integrand with zero integral must vanish — so \(\nabla V_d \equiv 0\), \(V_d\) is constant, and since \(V_d=0\) on the boundary, \(V_d\equiv 0\) everywhere. Hence \(V_1=V_2\).
4-4 Method of Images
Idea: replace a conducting boundary by an equivalent image charge (located outside the region of interest) such that the boundary potential condition is automatically satisfied. By uniqueness, this gives the correct \(V\) and \(\mathbf{E}\) in the original region.
Three standard configurations.
(a) Point charge above a grounded conducting plane (§4-4.1). Image: \(-Q\) at the mirror position.
where \(R_\pm\) are distances to \(\pm Q\). The field below the plane is zero; only the upper half-space is physical.
(b) Line charge parallel to a conducting cylinder (§4-4.2). Image: a parallel line charge of opposite sign at the inverse point:
Applied to a two-wire transmission line (wires of radius \(a\), center-to-center separation \(D\)):
(c) Point charge near a conducting sphere of radius \(a\) (§4-4.3). Image charge inside the sphere at the inverse point:
If the sphere is isolated (not grounded) rather than grounded, add a second image \(Q'=+aQ/d\) at the center so total sphere charge is zero.
Rules to remember when using images:
- Image charges live outside the region where you are computing the field — never inside it.
- The image field in the excluded half-space has no physical meaning (the real field there is zero, inside the conductor).
- Multiple grounded planes or spheres may require infinite series of images (see §4-4.4 charged sphere and grounded plane).
4-5 Boundary-Value Problems — Cartesian (Separation of Variables)
When no free charge is isolated and conductors align with coordinate surfaces, solve \(\nabla^{2}V=0\) by separation of variables: assume
Substituting into (4-81), dividing by \(XYZ\), and noting each term depends on only one variable, each must be constant. We obtain three ODEs:
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.
Boundary conditions classify the problem:
- Dirichlet: \(V\) specified on all boundaries.
- Neumann: normal derivative \(\partial V/\partial n\) specified (i.e., surface charge).
- Mixed: Dirichlet on some boundaries, Neumann on others.
Choosing sin vs. cos, sinh vs. cosh: use the symmetry of the problem — pick sin if \(V\) vanishes at \(x=0\); cos if \(V\) is symmetric about \(x=0\); sinh for a condition like \(V=0\) at one end; cosh for \(V\) equal at two symmetric locations.
4-6 Boundary-Value Problems — Cylindrical
For geometries long in \(z\), drop the \(z\)-dependence and solve the 2-D problem:
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:
Radial rules: if the region includes \(r=0\), drop \(r^{-n}\) (singular); if the region extends to \(\infty\), drop \(r^{n}\) (blows up).
Coaxial cable (inner radius \(a\) at \(V_0\), outer radius \(b\) grounded): (4-130) with BCs gives
Non-axisymmetric problems with arbitrary boundary data lead to Fourier series in \(\phi\) (see Example 4-9, the split cylinder).
4-7 Boundary-Value Problems — Spherical
For axisymmetric problems (no \(\phi\)-dependence):
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:
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:
The external potential is the applied uniform field plus that of an induced dipole of moment \(\mathbf{p}\) at the sphere's center.
Chapter 4 at a Glance
- Governing equations: \(\nabla^{2}V = -\rho/\epsilon\) (Poisson); \(\nabla^{2}V=0\) (Laplace, charge-free).
- Uniqueness theorem: a \(V\) satisfying \(\nabla^{2}V=-\rho/\epsilon\) and the boundary conditions is the solution — so any valid guess (e.g., images) is rigorous.
- Method of images: plane → mirror charge \(-Q\); cylinder → line charge \(-\rho_\ell\) at inverse point \(d_i=a^{2}/d\); sphere → \(Q_i=-(a/d)Q\) at \(d_i=a^{2}/d\).
- Two-wire line: \(C=\pi\epsilon_0/\cosh^{-1}(D/2a)\) (F/m).
- Cartesian BVP: \(X\,Y\,Z\) ansatz; separation constants satisfy \(k_x^{2}+k_y^{2}+k_z^{2}=0\) — at least one factor is hyperbolic.
- Cylindrical BVP: angular factor \(\sin n\phi,\cos n\phi\) with integer \(n\); radial factor \(r^{n}\) and/or \(r^{-n}\); for \(n=0\), \(V=C_1\ln r+C_2\).
- Spherical axisymmetric BVP: general solution \(V_n=[A_n R^{n}+B_n R^{-(n+1)}]P_n(\cos\theta)\); conducting sphere in uniform field → (4-159).
- Boundary types: Dirichlet (V specified), Neumann (\(\partial V/\partial n\) specified), mixed.
Problems and Solutions
Review Questions (Tier 1)
(a) Simple medium: \(\nabla^2 V=-\rho_v/\epsilon\). (b) Linear isotropic but inhomogeneous: \(\nabla\cdot(\epsilon\nabla V)=-\rho_v\).
(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.
(a) \(\nabla^2 V=0\). (b) \(\partial^2_x V+\partial^2_y V+\partial^2_z V=0\).
Harmonic functions other than zero exist (e.g., \(U=ax+by+c\)). The boundary conditions select among them; only specific BCs force \(U\equiv 0\).
\(E=V/d\) is fixed by the applied voltage (independent of \(\epsilon\)). \(D=\epsilon E\) scales with \(\epsilon\).
\(D=\rho_s=Q/A\) is fixed by the deposited charge. \(E=D/\epsilon\) decreases as \(\epsilon\) increases.
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.
Image is \(-q\) located at the mirror position, distance \(d\) below the plane.
\(q'=-(a/d)q\) located at distance \(a^2/d\) from the sphere's center, on the line from center to original charge.
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.
Parallel plates at x=0 (V=0) and x=d (V=V₀). Solve Laplace's equation for V(x) and find E.
→ SolutionRegion: 0≤x≤a, 0≤y≤b. V=0 on three sides, V = V₀ sin(πx/a) on y=b. Find V(x,y).
→ SolutionQ = 1 μC at height h = 0.5 m above grounded plane (z=0). Find E at z=0 directly below, and ρ_s.
→ SolutionLine 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.
→ SolutionTwo parallel cylindrical conductors: radius a = 1 mm, center-to-center separation D = 10 mm. Find C/L.
→ SolutionBetween plates at x=0 (V=0) and x=d (V=V₀), uniform charge density ρ_v. Solve for V(x).
→ SolutionCoaxial conductors: inner radius a=2mm at V₀=100V, outer radius b=10mm at V=0. Find V(ρ) and E.
→ SolutionSpherical capacitor: inner sphere a=2cm at V₀=100V, outer sphere b=6cm at V=0. Find V(r) and C.
→ SolutionCharge Q at distance d from center of grounded sphere (radius a, a < d). Find image charge.
→ SolutionPoint charge Q at (x₀, y₀) above a grounded plane at y=0 AND grounded plane at x=0 (corner). Find images.
→ SolutionHalf-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).
→ SolutionRectangular 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).
→ SolutionInner conductor radius a=1mm, outer radius b=4mm, εᵣ=2.25. Find E(ρ) and C/L.
→ SolutionCylindrical region 0≤ρ≤a, uniform ρ_v. V(a)=0 (grounded). Find V(ρ).
→ SolutionConducting wedge: V=0 at φ=0 and V=V₀ at φ=α. Find V(φ) and E.
→ SolutionTextbook Practice Problems
Textbook Practice — Approach Hints
Sample: Solve \(\nabla^2 V=0\) between two parallel plates at \(V=0\) (\(z=0\)) and \(V=V_0\) (\(z=d\)).
Sample: Point charge \(+q\) at distance \(d\) above a grounded conducting plane. Find \(\vec{E}\) above the plane.
Sample: Charge \(q\) at \((0,0,h)\) above grounded plane \(z=0\). Find induced surface charge density on plane.
Sample: Concentric cylinders \(a\) and \(b\) with \(V(a)=V_0\), \(V(b)=0\). Solve in cylindrical.
Sample: Solve Laplace in spherical: concentric spheres \(a,b\) with \(V(a)=V_0\), \(V(b)=0\).
Sample: Slab \(0\le x\le d\) with uniform \(\rho_v=\rho_0\), grounded at both faces. Solve Poisson.
Sample: Cylindrical region \(\rho\le a\) has \(\rho_v=\rho_0\), surrounded by vacuum. Find \(V\) inside.
Sample: Charge \(q\) at distance \(d\) from center of grounded sphere of radius \(a\). Find image charge.
Sample: Rectangular trough: \(V=0\) on three sides, \(V=V_0\) on top (\(y=b\)). Width \(a\). Find \(V(x,y)\).
Sample: Square box \(0\le x,y\le a\), \(V=0\) on bottom & sides, \(V=V_0\sin(\pi x/a)\) on top. Find \(V\).
Sample: Cylindrical region with \(V(a,\phi)=V_0\cos\phi\). Find \(V\) inside.
Sample: Coaxial cable, dielectric \(\epsilon_r\), applied voltage \(V_0\) on inner. Solve via cylindrical separation.
Sample: Spherical region with \(V(a,\theta)=V_0\cos\theta\). Find \(V\) inside (\(ra\)).
Sample: Conducting sphere of radius \(a\) in uniform field \(E_0\hat{z}\). Find \(V\) outside.
Sample: \(p\)-\(n\) junction depletion region: charge density \(-eN_A\) for \(-x_p
Sample (comprehensive): Two grounded conducting plates at \(z=\pm d\), point charge \(+q\) between them at \(z=0\).
Chapter 5 — Steady Electric Currents
Key Theory — Chapter 5
Condensed from Cheng, Field and Wave Electromagnetics, §5-1 through §5-7. Read this before attempting the problems below.
5-1 Three Types of Current
Chapters 3 and 4 dealt with charges at rest. Chapter 5 introduces charges in motion — steady (d.c.) currents. Three mechanisms produce electric current, and it is important to recognise which one applies to a given situation:
- Conduction currents — drift motion of conduction electrons (or holes) in conductors and semiconductors. Governed by Ohm's law.
- Electrolytic currents — migration of positive and negative ions in an electrolyte (e.g. a salt solution between electrodes). Electrolytes obey a Laplace-equation model which is the basis of the electrolytic tank used to map hard-to-solve electrostatic potentials.
- Convection currents — bulk motion of charged particles in a vacuum or rarefied gas (electron beams in a CRT, lightning). Convection currents are a hydrodynamic mass transport and are not governed by Ohm's law.
Drift velocity is tiny. The average drift velocity of conduction electrons is only \(10^{-4}\) to \(10^{-3}\) m/s — even in excellent conductors — because of continual electron-atom collisions that dissipate kinetic energy as heat. A conductor remains electrically neutral in bulk; electric forces prevent charge accumulation at any interior point.
第 3、4 章探討的是 at rest 的電荷,第 5 章開始討論 in motion 的電荷 — 即穩態 (d.c.) currents。產生 electric current 的機制有三種,辨識當下情境屬於哪一種非常重要:
- Conduction currents — conductors 與 semiconductors 內 conduction electrons(或 holes)的 drift motion,遵循 Ohm's law。
- Electrolytic currents — electrolyte(例如兩 electrodes 間的鹽溶液)中正、負 ions 的遷移。Electrolytes 遵守 Laplace-equation 模型,這也是 electrolytic tank 的原理基礎 — 用以實驗映射難以解析求得的 electrostatic potentials。
- Convection currents — 真空或稀薄氣體中帶電粒子的整體運動(CRT 中的 electron beams、閃電)。Convection currents 屬於 hydrodynamic mass transport,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
The total current through an arbitrary surface \(S\) is the flux of \(\mathbf{J}\):
Since \(Nq\) is just the free-charge volume density \(\rho\), for convection currents
When several species drift simultaneously (electrons, holes, ions), the current density superposes: \(\mathbf{J} = \sum_i N_i q_i \mathbf{u}_i\).
5-2.2 The Point Form of Ohm's Law
In a conductor the drift velocity is directly proportional to the applied field:
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:
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:
Conductivity spans an enormous range — copper \(5.80\times 10^{7}\) S/m; germanium \(2.2\); silicon \(1.6\times 10^{-3}\); hard rubber \(10^{-15}\). Unlike permittivity, conductivity varies over 22+ orders of magnitude across common materials. The reciprocal of \(\sigma\) is resistivity (Ω·m), but Cheng prefers to work with conductivity throughout.
Equation (5-21) is the point form. The familiar circuit-theory relation \(V_{12} = RI\) is not a point relation — it concerns terminals and a cross section. Applying (5-21) to a straight homogeneous conductor of length \(\ell\), uniform cross section \(S\), conductivity \(\sigma\):
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:
This equation tells us something fundamental: a steady current cannot be maintained in a closed loop by an electrostatic field alone. Charge carriers dissipate energy colliding with atoms; that energy must be resupplied by a non-conservative source — a battery (chemical), generator (mechanical), thermocouple (thermal), or photovoltaic cell (optical). These sources produce an equivalent impressed electric field intensity \(\mathbf{E}_i\) inside the source only.
The line integral of \(\mathbf{E}_i\) from − to + terminal inside the source defines the electromotive force (emf) of the source, denoted \(\mathscr{V}\):
The SI unit for emf is volt. Despite its name, emf is not a force in newtons — it is a measure of the strength of the non-conservative source. An ideal voltage source has zero internal resistance, so its terminal voltage equals its emf regardless of current.
Inside a conductor carrying steady current, the total field that drives the charges is the sum \(\mathbf{E}+\mathbf{E}_i\), and the point form of Ohm's law generalises to \(\mathbf{J} = \sigma(\mathbf{E}+\mathbf{E}_i)\). Taking the closed-loop line integral yields the macroscopic form of Kirchhoff's voltage law:
Around any closed loop, the algebraic sum of the emfs (voltage rises) equals the algebraic sum of the IR voltage drops. This is the basis for loop analysis in circuits.
5-4 Equation of Continuity and Kirchhoff's Current Law
Conservation of charge requires that current leaving a closed surface equal the rate of decrease of the enclosed charge:
Applying the divergence theorem and letting the integrand equal out (since \(V\) is arbitrary) gives the equation of continuity:
For steady currents, \(\partial\rho/\partial t = 0\), so
Steady currents are therefore divergenceless (solenoidal) — their streamlines close on themselves, unlike electrostatic field lines which start and end on charges. An ideal current source has infinite internal resistance.
Relaxation time
Combining Ohm's law with the continuity equation and \(\nabla\cdot\mathbf{E}=\rho/\epsilon\) gives, for a simple medium,
whose solution is an exponential decay:
For copper \(\tau \approx 1.52\times 10^{-19}\) s — so brief that \(\rho\) can be taken as zero everywhere inside a good conductor. In a good insulator the relaxation time can be hours or days.
5-5 Power Dissipation — Joule's Law
The work done by \(\mathbf{E}\) in moving a charge \(q\) at velocity \(\mathbf{u}\) is \(p = q\mathbf{E}\cdot\mathbf{u}\). Summing over all carriers in a volume element gives a power density
Integrating over a volume \(V\) gives Joule's law:
(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:
5-6 Boundary Conditions for Current Density
The governing equations for steady current density in ohmic media (no non-conservative sources inside) are:
Applying these at an interface between two ohmic media with conductivities \(\sigma_1\) and \(\sigma_2\) gives the boundary conditions:
Compare with the electrostatic boundary conditions in Chapter 3 (\(D_{1n}-D_{2n}=\rho_s\) and \(E_{1t}=E_{2t}\)): there is an exact analogy between \((\mathbf{J},\sigma)\) and \((\mathbf{D},\epsilon)\) at charge-free dielectric interfaces. This underlies the electrolytic-tank analogy.
Dividing (5-59) by (5-58) yields the refraction relation for current streamlines:
If medium 1 is a much better conductor than medium 2 (\(\sigma_1 \gg \sigma_2\)), then \(\alpha_2 \to 0\) — current in the poor conductor emerges almost normal to the surface of the good conductor.
Surface charge at the interface. When steady current flows across a boundary between two lossy dielectrics (finite \(\epsilon_1,\epsilon_2\) and \(\sigma_1,\sigma_2\)), the simultaneous requirements \(J_{1n}=J_{2n}\) and \(D_{1n}-D_{2n}=\rho_s\) force a surface charge:
which vanishes only when \(\sigma_2/\sigma_1 = \epsilon_2/\epsilon_1\).
5-7 Resistance Calculations and the R–C Analogy
In a source-free homogeneous conductor the current field is both divergenceless and curl-free, so \(\mathbf{J}=-\nabla\psi\) with \(\psi = \sigma V\) satisfying Laplace's equation:
A steady-current problem therefore has the same mathematical form as an electrostatic problem — methods of Chapter 4 apply directly.
For two conductors embedded in a lossy dielectric medium (\(\epsilon, \sigma\)), the resistance between them is
Comparing with the capacitance between the same two conductors yields the elegant R–C analogy:
Consequence: if the capacitance between two conductors is already known, the leakage resistance (or conductance) through a lossy medium filling the same geometry follows immediately — no recomputation needed. For a coaxial cable of inner radius \(a\), outer radius \(b\), per unit length:
- \(C_1 = 2\pi\epsilon/\ln(b/a)\) (from Ch. 3)
- \(R_1 = (\epsilon/\sigma)/C_1 = \ln(b/a)/(2\pi\sigma)\) (leakage resistance)
The R–C analogy holds whenever \(\epsilon\) and \(\sigma\) of the medium have the same spatial dependence (trivially true for a homogeneous medium).
Chapter 5 at a Glance
- Point form of Ohm's law: \(\mathbf{J}=\sigma\mathbf{E}\); macroscopic form \(V=IR\) with \(R=\ell/(\sigma S)\).
- KVL: \(\sum\mathscr{V}_j = \sum R_k I_k\) (line integral of \(\mathbf{E}+\mathbf{E}_i\) around a loop).
- KCL: \(\sum I_j = 0\), from the divergenceless steady current \(\nabla\cdot\mathbf{J}=0\).
- Equation of continuity: \(\nabla\cdot\mathbf{J} = -\partial\rho/\partial t\) — general statement of charge conservation.
- Relaxation time: \(\tau=\epsilon/\sigma\); in copper \(\sim 10^{-19}\) s, so good conductors have \(\rho=0\) inside.
- Joule's law: \(P=\int_V\mathbf{E}\cdot\mathbf{J}\,dv = I^2 R\).
- Boundary conditions: \(J_{1n}=J_{2n}\), \(J_{1t}/J_{2t}=\sigma_1/\sigma_2\); refraction \(\tan\alpha_2/\tan\alpha_1 = \sigma_2/\sigma_1\).
- R–C analogy: \(RC=\epsilon/\sigma\) — capacitance ↔ leakage resistance.
- Analogy with electrostatics: \(\mathbf{J}\leftrightarrow\mathbf{D}\), \(\sigma\leftrightarrow\epsilon\); every electrostatic technique transfers to steady-current problems.
Problems and Solutions
Review Questions (Tier 1)
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).
\(\vec{J}=\sigma\vec{E}\). Holds in linear, isotropic conductors in steady state when \(\sigma\) is constant and the medium is non-dispersive.
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.
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).
\(\nabla\cdot\vec{J}=-\partial\rho_v/\partial t\). Integrating over a node volume in steady state gives \(\sum I_{\text{in}}=\sum I_{\text{out}}\), which is KCL.
\(\tau=\epsilon/\sigma\) — the time constant for free charge to redistribute. A material behaves as a good conductor on time scales \(t\gg\tau\) (or frequencies \(\omega\ll 1/\tau\)).
Point form: \(p=\vec{J}\cdot\vec{E}=\sigma E^2\,\text{[W/m}^3\text{]}\). Macroscopic: \(P=I^2R=V^2/R\).
\(J_{1n}=J_{2n}\) and \(E_{1t}=E_{2t}\), giving \(\tan\theta_1/\tan\theta_2=\sigma_1/\sigma_2\).
For two electrodes in a lossy dielectric with the same geometry: \(RC=\epsilon/\sigma\). Holds when conductor field shape and current field shape coincide.
A copper wire (diameter 2 mm) carries I = 10 A. Find the current density J.
→ SolutionCopper (σ = 5.8×10⁷ S/m) wire: length L = 1 m, diameter d = 2 mm. Find R.
→ SolutionTwo batteries in series: V₁=12V (internal r₁=0.5Ω) and V₂=9V (internal r₂=1Ω), external R=10Ω.
Find I and terminal voltages.
→ SolutionCircuit: V=24V, R₁=4Ω, R₂=8Ω in series. Verify KVL.
→ SolutionNode with currents: I₁=5A (in), I₂=3A (in), I₃=? (out), I₄=2A (out). Find I₃.
→ SolutionResistor: R = 100 Ω, V = 50 V. Find P, I, and energy in 1 hour.
→ SolutionTruncated cone: conductivity σ, small radius a, large radius b, length L, axis along z. Find R.
→ SolutionMedium 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.
→ SolutionCylindrical shell: inner radius a=1cm, outer radius b=3cm, length L=10cm, σ = 10³ S/m, current flows radially. Find R.
→ SolutionUniform conductivity sphere, radius a, uniform E₀ applied externally. Find J inside.
→ SolutionCharge density ρ_v = ρ₀ e^(-t/τ) uniformly distributed in a medium (σ, ε). Find J and verify continuity.
→ SolutionTwo spherical electrodes (radius a = 5 mm) buried in earth (σ = 10⁻² S/m), separation D = 1 m (D >> a). Find R.
→ SolutionFind equivalent resistance: R₁=6Ω, R₂=3Ω in parallel, that combination in series with R₃=4Ω.
→ SolutionConductor tapers linearly from radius a to 2a over length L. Uniform current I flows axially. Find J(z) and E(z).
→ SolutionTwo-wire line: conductor radius a=1mm, separation D=10mm, σ_c=5.8×10⁷ S/m, length L=100m, I=10A. Find P_loss.
→ SolutionTextbook Practice Problems
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 \(\vec{E}\).
Sample: Aluminum wire (\(\sigma=3.5\!\cdot\!10^7\) S/m), \(L=100\,\text{m}\), radius 2 mm. Find resistance.
Sample: Time-varying point charge \(\rho_v(\vec{r},t)\). Show continuity equation in steady state.
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.
Sample: At interface \(\sigma_1\ne\sigma_2\): angle of \(\vec{J}\) on each side.
Sample: Truncated cone of length \(L\), radii \(a\) and \(b\), conductivity \(\sigma\). Find resistance end-to-end.
Sample: Two cylindrical resistors in parallel: \(R_1=10\,\Omega\), \(R_2=20\,\Omega\). Find equivalent.
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.
Sample: Battery EMF \(\mathcal{E}=12\,\text{V}\), internal resistance \(0.5\,\Omega\), external \(5.5\,\Omega\). Find current and terminal voltage.
Sample: Loop with two batteries (12 V, 6 V opposing) and resistors \(2\,\Omega\), \(4\,\Omega\) in series. KVL gives current.
Sample: Conductor with \(\sigma=10^7\), time-constant for charge decay?
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.
Sample: Resistor \(R=10\,\Omega\), current \(I=2\,\text{A}\). Find dissipated power.
Sample: Coaxial geometry filled with imperfect dielectric \(\sigma\). Inner \(a\), outer \(b\), length \(L\). Find leakage resistance.
Sample: Spherical resistor: inner radius \(a\), outer \(b\), conductivity \(\sigma\). Find \(R\) between concentric surfaces.
Sample: Two parallel hemispherical electrodes buried in earth (\(\sigma\)). Find ground resistance.
Sample (synthesis): Resistance between two small spheres of radius \(a\) separated by \(d\gg a\) in conducting medium \(\sigma\).
Chapter 6 — Static Magnetic Fields
Key Theory — Chapter 6
Condensed from Cheng, Field and Wave Electromagnetics, §6-1 through §6-13. Read this before attempting the problems below.
6-1 The Magnetic Force and the Definition of B
A charge \(q\) at rest in an electric field experiences \(\mathbf{F}_e = q\mathbf{E}\). When the charge moves with velocity \(\mathbf{u}\), experiment shows a second force appears: one perpendicular to both \(\mathbf{u}\) and a fixed direction at each point, with magnitude proportional to \(q\), to \(|\mathbf{u}|\), and to the component of \(\mathbf{u}\) perpendicular to that fixed direction. This magnetic force is expressed as
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:
Lorentz's equation can itself be taken as a fundamental postulate of the electromagnetic model — it defines both \(\mathbf{E}\) (via the force on a stationary charge) and \(\mathbf{B}\) (via the force on a moving one). It cannot be derived from other postulates.
6-2 Fundamental Postulates of Magnetostatics in Free Space
In free space, magnetostatics is built on exactly two postulates specifying the divergence and curl of \(\mathbf{B}\):
No magnetic monopoles. Because \(\nabla\cdot\mathbf{B} = 0\), magnetic flux lines always close on themselves — there is no magnetic analog of electric charge density. Cutting a bar magnet doesn't produce an isolated N or S pole; it produces two smaller magnets, and this continues down to atomic dimensions. Equation (6-9) is the law of conservation of magnetic flux.
Ampère's circuital law says the circulation of \(\mathbf{B}\) around any closed path equals \(\mu_0\) times the total current through the enclosed surface. Taking \(\nabla\cdot(\nabla\times\mathbf{B}) = 0\) in (6-7) immediately gives \(\nabla\cdot\mathbf{J} = 0\), consistent with steady-current continuity (5-45).
6-3 Vector Magnetic Potential A
Because \(\mathbf{B}\) is solenoidal, it can be expressed as the curl of another vector field:
\(\mathbf{A}\) 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
Choosing the Coulomb gauge
reduces this to a vector Poisson equation:
Each Cartesian component is a scalar Poisson equation identical in form to \(\nabla^2 V = -\rho/\epsilon_0\) from electrostatics. The solution is therefore
Physical meaning of A: the line integral of \(\mathbf{A}\) around a closed contour equals the magnetic flux through the enclosed surface:
6-4 Biot–Savart Law
For a thin wire, \(\mathbf{J}\,dv' = I\,d\ell'\). Equation (6-23) becomes
Taking the curl and doing the \(\nabla(1/R) = -\mathbf{a}_R/R^2\) bookkeeping yields Biot–Savart law:
where \(\mathbf{a}_R\) is the unit vector from source element to field point. Biot–Savart computes \(\mathbf{B}\) directly from a current-carrying circuit; it is the magnetic analogue of Coulomb's law for \(\mathbf{E}\). Ampère's law is more convenient when symmetry (cylindrical, planar, toroidal) provides a path over which \(|\mathbf{B}|\) is constant; Biot–Savart is the tool when it doesn't.
6-5 The Magnetic Dipole
A small current loop of area \(S\) carrying current \(I\) behaves as a magnetic dipole with moment
where \(\mathbf{a}_n\) follows the right-hand rule from the current direction. At distances \(R \gg\) loop size, the vector potential is
and the magnetic flux density is
which is exactly analogous to the field of an electric dipole \((\mathbf{p}, V)\) — with \(\mathbf{A}\) playing the role of \(V\), and the same angular dependence.
6-6 Magnetization and Equivalent Current Densities
In a material, orbital and spin motions of electrons create microscopic magnetic dipoles. An applied field partially aligns them, yielding a macroscopic magnetization vector \(\mathbf{M}\) (A/m) — magnetic moment per unit volume. A magnetized body can be replaced, as far as its external field is concerned, by an equivalent volume current density and surface current density:
Uniformly magnetized bar magnets have no volume current (\(\nabla\times\mathbf{M} = 0\) for constant \(\mathbf{M}\)) but carry a surface current \(\mathbf{M}\times\mathbf{a}_n\) that effectively turns them into solenoids of lineal current density \(|\mathbf{M}|\).
6-7 Magnetic Field Intensity H and the Constitutive Relation
In matter, Ampère's law must account for both free current \(\mathbf{J}\) and bound (magnetization) current \(\mathbf{J}_m\): \((1/\mu_0)\nabla\times\mathbf{B} = \mathbf{J} + \nabla\times\mathbf{M}\). Rearranging motivates the definition of a new fundamental field:
called the magnetic field intensity. Its governing equation involves only free current:
For linear, isotropic media, \(\mathbf{M} = \chi_m\mathbf{H}\) with magnetic susceptibility \(\chi_m\) (dimensionless). Substituting into (6-75) yields the constitutive relation
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
and \(NI\) is the magnetomotive force (mmf, in ampere-turns). Magnetic circuits obey KVL/KCL analogues:
The analogy: \(\mathscr{V}\leftrightarrow NI\), \(I\leftrightarrow\Phi\), \(R\leftrightarrow\mathcal{R}\), \(\sigma\leftrightarrow\mu\).
6-9 Behavior of Magnetic Materials
Three classes are distinguished by the sign and magnitude of \(\chi_m\):
- Diamagnetic (\(\chi_m \sim -10^{-5}\), \(\mu_r \lesssim 1\)) — induced moments oppose the applied field (Lenz's law at the atomic scale). Bismuth, copper, silver, gold, water. No permanent magnetism; effect vanishes when field is removed.
- Paramagnetic (\(\chi_m \sim +10^{-5}\), \(\mu_r \gtrsim 1\)) — existing molecular dipoles partially align with the field but are disrupted by thermal agitation. Aluminium, magnesium, titanium, tungsten. Temperature-dependent.
- Ferromagnetic (\(\mu_r\) from 50 to over \(10^6\)) — strong coupling between atoms forms aligned domains. Iron, nickel, cobalt, and alloys. Nonlinear B–H relation and hysteresis: as H cycles, B traces a loop. Important parameters: remanence \(B_r\) (B at H=0, enables permanent magnets), coercivity \(H_c\) (H needed to drive B back to zero), and the Curie temperature above which coupling collapses and the material becomes paramagnetic. "Soft" magnetic materials have narrow loops (low loss — transformers); "hard" materials have wide loops (large \(H_c\) — permanent magnets).
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:
For two linear media, the normal condition becomes \(\mu_1 H_{1n} = \mu_2 H_{2n}\). The tangential \(\mathbf{H}\) is continuous at any interface between physical media (\(\mathbf{J}_s = 0\)); it is discontinuous only at the surface of an idealised perfect conductor or superconductor. Compare with electrostatics: \(\mathbf{D}\) normal jumps by \(\rho_s\), \(\mathbf{E}\) tangential is continuous.
6-11 Inductance
Mutual flux from loop \(C_1\) linking \(C_2\) is \(\Phi_{12} = \int_{S_2}\mathbf{B}_1\cdot d\mathbf{s}_2\). By Biot–Savart, \(\mathbf{B}_1 \propto I_1\), so
where \(\Lambda_{12} = N_2\Phi_{12}\) is the flux linkage. Carrying the algebra through for two multi-turn circuits yields Neumann's formula:
Reciprocity gives \(L_{12} = L_{21}\). Self-inductance \(L\) is defined analogously by setting both loops equal. Inductance is a purely geometrical quantity (for a linear medium), independent of the current magnitude. For a solenoid, \(L \propto N^2\).
6-12 Magnetic Energy
The work done against induced emf to build up current \(I_1\) in an inductor of self-inductance \(L_1\) is stored as magnetic energy:
For a system of N coupled loops:
Expressed in terms of field quantities:
with magnetic energy density
These forms are exact analogues of \(w_e = \tfrac{1}{2}\mathbf{D}\cdot\mathbf{E}\) in electrostatics. Computing \(W_m\) from the field integral and equating to \(\tfrac{1}{2}LI^2\) is often an easier route to inductance than counting flux linkages.
6-13 Magnetic Forces and Torques
Force on a current-carrying element:
Two parallel wires carrying currents \(I_1, I_2\) separated by distance \(d\) experience
Same direction → attraction; opposite directions → repulsion. (Opposite sense to Coulomb's law for charges — a subtle point.)
Torque on a current loop of dipole moment \(\mathbf{m}\) in a uniform \(\mathbf{B}\):
This is the operating principle of the d-c motor: a split-ring commutator reverses the loop current every half turn to keep \(\mathbf{T}\) aligned and the rotor spinning.
Hall effect: a current \(\mathbf{J}\) flowing in a magnetic field \(\mathbf{B}\) develops a transverse voltage \(V_h\) across the conductor. The sign of \(V_h\) reveals whether carriers are electrons or holes — an essential diagnostic for distinguishing n- and p-type semiconductors.
Chapter 6 at a Glance
- Defining equation: Lorentz force \(\mathbf{F} = q(\mathbf{E} + \mathbf{u}\times\mathbf{B})\).
- Postulates (free space): \(\nabla\cdot\mathbf{B} = 0\), \(\nabla\times\mathbf{B} = \mu_0\mathbf{J}\); integral form gives flux conservation and Ampère's circuital law.
- In matter: \(\mathbf{H} = \mathbf{B}/\mu_0 - \mathbf{M}\); Ampère's law becomes \(\nabla\times\mathbf{H} = \mathbf{J}\); constitutive relation \(\mathbf{B} = \mu\mathbf{H}\).
- Vector potential: \(\mathbf{B}=\nabla\times\mathbf{A}\), Coulomb gauge \(\nabla\cdot\mathbf{A}=0\), \(\nabla^2\mathbf{A}=-\mu_0\mathbf{J}\); \(\Phi = \oint\mathbf{A}\cdot d\ell\).
- Biot–Savart: \(\mathbf{B} = (\mu_0 I/4\pi)\oint d\ell'\times\mathbf{a}_R/R^2\).
- Magnetic dipole: \(\mathbf{m}=IS\mathbf{a}_n\); \(\mathbf{B}\) falls off as \(1/R^3\) with the same angular pattern as an electric dipole.
- Boundary conditions: \(B_n\) continuous; \(H_t\) jumps by \(J_s\) (zero for ordinary media).
- Inductance: \(L = \Lambda/I\) — geometry only; Neumann formula for mutual inductance.
- Energy: \(W_m = \tfrac{1}{2}LI^2 = \tfrac{1}{2}\int\mathbf{H}\cdot\mathbf{B}\,dv\); density \(w_m = B^2/(2\mu)\).
- Forces: \(\mathbf{F} = I\int d\ell\times\mathbf{B}\), torque \(\mathbf{T}=\mathbf{m}\times\mathbf{B}\); parallel wires attract if currents are codirectional.
- Materials: diamagnetic, paramagnetic, ferromagnetic; hysteresis with remanence \(B_r\) and coercivity \(H_c\); Curie temperature; Hall effect.
Problems and Solutions
Part A — Biot-Savart, Ampère's Law, Magnetostatics (Sections 6-1 to 6-7)
Infinite wire carrying I = 10 A along the z-axis. Find B at ρ = 0.05 m.
→ SolutionCircular loop of radius a = 0.1 m, current I = 5 A. Find B at center and along axis.
→ SolutionSolenoid: N=500 turns, L=0.25m, I=2A, air core. Find B inside.
→ SolutionFind A for an infinite straight wire (I along z). Verify B = ∇×A.
→ SolutionTwo coaxial loops (Helmholtz coil), each radius a=0.1m, separated by d=0.1m=a, each carrying I=5A. Find B at midpoint.
→ SolutionTwo parallel wires 0.1 m apart, each carrying I = 100 A in the same direction. Find force per unit length.
→ SolutionIron with μᵣ = 5000, H = 500 A/m. Find B, M, and magnetization current density.
→ SolutionUniformly magnetized sphere: M = M₀ â_z. Find equivalent bound current densities.
→ SolutionToroid: N = 200 turns, mean radius R = 0.1 m, current I = 3 A, μᵣ = 500 (iron core). Find B and H.
→ SolutionIdeal solenoid: n = 1000 turns/m, I = 2 A. Find B inside and outside.
→ SolutionSurface current K = K₀ â_x (A/m) on z=0 plane. Find H above and below.
→ SolutionStraight wire of length 2L carrying current I along z-axis, centered at origin. Find B at point (ρ, 0, 0).
→ SolutionCircular current loop: I = 5 A, radius a = 0.02 m. Find magnetic dipole moment m.
→ SolutionFor a solenoid (B = B₀ â_z inside, 0 outside), find A using ∇×A = B.
→ SolutionRectangular 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.
→ SolutionPart B — Magnetic Circuits, Inductance, Energy, Forces (Sections 6-8 to 6-13)
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.
→ SolutionSame core as Problem 16. Find reluctance ratio R_gap/R_iron.
→ SolutionFor 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.
→ SolutionB₁ = 0.5 T at θ₁ = 60° to normal, in iron (μᵣ = 1000). Find B₂ in air.
→ SolutionSolenoid: N=500 turns, L=0.25m, radius r=1cm, air core. Find L.
→ SolutionTwo coaxial solenoids: both length l=0.25m, inner (N₁=500, r₁=1cm), outer (N₂=200, r₂=2cm). Find M.
→ SolutionSolenoid (L=0.395 mH) carrying I=5A. Find stored energy W.
→ SolutionRectangular loop (a=0.1m, b=0.05m), I=2A, in field B = 0.3 â_z T. Loop in xy-plane. Find torque.
→ SolutionMagnetic dipole m = 0.05 â_x A·m² in field B = 2 â_z T. Find torque and stable equilibrium position.
→ SolutionTwo bar magnets: pole strength p = 0.01 Wb, separation r = 0.1 m (between same poles). Find force.
→ SolutionToroid: N=500 turns, mean radius R=5cm, cross-sectional radius r=1cm, μᵣ=1. Find L.
→ SolutionToroid gap with B = 1 T in gap. Find the magnetic pressure (force per unit area) trying to close the gap.
→ SolutionFor the toroid in Problem 26 (L=0.314 mH), carrying I=2A. Find W and verify with field integral.
→ SolutionCopper 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.
→ SolutionStraight wire, length L = 0.5 m, I = 20 A, in field B = 0.4 â_z T. Wire is along â_x. Find force.
→ SolutionTextbook Practice Problems
Textbook Practice — Approach Hints
Sample: Show \(\nabla\cdot\vec{B}=0\) for \(\vec{B}=B_0(\hat{x}\sin y-\hat{y}\cos x)\).
Sample: Finite straight wire on \(z\)-axis from \(-L\) to \(+L\), current \(I\). Find \(\vec{B}\) at perpendicular distance \(\rho\).
Sample: Circular loop radius \(a=5\,\text{cm}\), current \(I=10\,\text{A}\). Find \(\vec{B}\) at center and on axis at \(z=10\,\text{cm}\).
Sample: Use Ampere's law for an infinite straight wire \(I=10\,\text{A}\) at \(\rho=5\,\text{cm}\).
Sample: Toroid with \(N=200\) turns, mean radius \(r=5\,\text{cm}\), \(I=2\,\text{A}\). Find \(H\) inside.
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 \(\vec{H}_2,\vec{B}_2\).
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 \(\vec{H}_2\).
Sample: Solenoid: \(N=400\), length \(20\,\text{cm}\), area \(5\,\text{cm}^2\), air core. Find \(L\).
Sample: Toroid: \(N=500\), mean radius \(5\,\text{cm}\), area \(2\,\text{cm}^2\), \(\mu_r=200\). Find \(L\).
Sample: Verify \(\nabla\times\vec{H}=\vec{J}\) for an infinite straight wire on \(z\)-axis carrying \(I\).
Sample: Long solenoid, \(n=1000\) turns/m, \(I=2\,\text{A}\). Find \(\vec{B}\) inside.
Sample: Vector potential of finite straight wire: \(\vec{A}=A_z(\rho)\hat{z}\). Set up integral.
Sample: Magnetic dipole at origin, \(\vec{m}=m\hat{z}\). Find \(\vec{A}\) in far field.
Sample: Current loop area \(A=10\,\text{cm}^2\), \(I=1\,\text{A}\). Find \(\vec{m}\) and \(\vec{B}\) on axis at \(z=20\,\text{cm}\).
Sample: Compare electric and magnetic dipoles in far field.
Sample: Bar magnet uniformly magnetized \(\vec{M}=M_0\hat{z}\), length \(L\), area \(A\). Find equivalent surface current.
Sample: Permanent magnet: \(M=8\!\cdot\!10^5\,\text{A/m}\), neglecting external field. Find \(\vec{B}\) inside.
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.
Sample: Square loop side \(L=10\,\text{cm}\), current \(I=2\,\text{A}\). Find \(\vec{B}\) at center.
Sample: Toroidal core \(N=500\), \(I=1\,\text{A}\), \(\mu_r=1000\), mean length \(30\,\text{cm}\), area \(4\,\text{cm}^2\). Find flux.
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.
Sample: Iron at operating point \(H=200\,\text{A/m}\), \(B=0.8\,\text{T}\). Find relative permeability.
Sample: Soft-iron core at \(H=100\), \(B=1.2\). Find \(\mu_r\) and total energy density.
Sample: Two coaxial solenoids: \(N_1=200\), \(N_2=300\), common length \(L=10\,\text{cm}\), area \(4\,\text{cm}^2\). Find \(M\).
Sample: Inductor \(L=10\,\text{mH}\), current \(I=2\,\text{A}\). Find stored energy.
Sample: Solenoid with \(B=1\,\text{T}\), volume \(V=10^{-3}\,\text{m}^3\). Find energy.
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.
Sample: Two parallel wires \(1\,\text{cm}\) apart, currents \(5\,\text{A}\) and \(3\,\text{A}\) same direction. Find force per meter.
Sample: Helmholtz coils: two coaxial loops radius \(a\), separation \(a\), current \(I\) each. Find \(B_z\) at midpoint.
Sample: Solenoid with iron plunger: virtual work for force.
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.
Sample: Magnetic dipole \(\vec{m}=10^{-2}\hat{z}\,\text{A·m}^2\) in \(\vec{B}=0.1\hat{x}\,\text{T}\). Find energy.
Chapter 7 — Time-Varying Fields and Maxwell's Equations
Key Theory — Chapter 7
Condensed from Cheng, Field and Wave Electromagnetics, §7-1 through §7-7. Read this before attempting the problems below.
7-1 Why Static Models Are Not Enough
Chapters 3–6 built two independent models — electrostatic and magnetostatic:
In the static case \(\{\mathbf{E}, \mathbf{D}\}\) and \(\{\mathbf{B}, \mathbf{H}\}\) are decoupled: an electrostatic field can drive a steady current, which produces a static magnetic field, but \(\mathbf{E}\) can be found from the charge distribution alone. The magnetic field is only a consequence.
Chapter 7 shows that a time-varying magnetic field produces an electric field, and a time-varying electric field produces a magnetic field. The two curl equations must therefore be modified. The resulting four equations are Maxwell's equations — the foundation of all electromagnetic theory.
7-2 Faraday's Law — Fundamental Postulate
In 1831 Faraday showed experimentally that a changing magnetic flux through a loop induces an emf. Following the deductive approach, we take the point-form relation as postulate:
Applying Stokes's theorem gives the integral form:
Equation (7-1) applies at every point of space whether or not a physical circuit exists. A time-varying magnetic field makes \(\mathbf{E}\) non-conservative — it can no longer be expressed as the gradient of a scalar potential alone.
Transformer emf (stationary circuit, changing B)
For a stationary closed circuit, (7-2) becomes
This is Faraday's law of electromagnetic induction. The induced emf equals the negative rate of change of flux linkage; the minus sign (Lenz's law) asserts that the induced current opposes the flux change that created it. This is the principle of transformers. Time-varying flux in a ferromagnetic core also induces local eddy currents in the core itself; lamination reduces this eddy-current loss.
Motional emf (moving conductor in static B)
A conductor moving with velocity \(\mathbf{u}\) through a static \(\mathbf{B}\) experiences magnetic forces on its free charges. In the conductor's frame, these are equivalent to an induced field \(\mathbf{u}\times\mathbf{B}\) acting along it. For a closed circuit partially in motion,
called the motional emf or "flux-cutting" emf. Only parts of the circuit moving across flux lines contribute.
For a circuit that both moves and lies in a time-varying field, both contributions add — but they combine cleanly into the single expression \(\mathscr{V} = -d\Phi/dt\) with the total time derivative taken along the moving loop.
7-3 Displacement Current and the Modification of Ampère's Law
The static curl equation \(\nabla\times\mathbf{H} = \mathbf{J}\) is inconsistent with charge conservation \(\nabla\cdot\mathbf{J} = -\partial\rho/\partial t\) in the time-varying case, because \(\nabla\cdot(\nabla\times\mathbf{H}) \equiv 0\) would force \(\nabla\cdot\mathbf{J} = 0\). Maxwell's brilliant repair: add a term \(\partial\mathbf{D}/\partial t\) — the displacement current density:
Now \(\nabla\cdot(\nabla\times\mathbf{H}) = \nabla\cdot\mathbf{J} + \partial\rho/\partial t = 0\) automatically, by continuity. The term \(\partial\mathbf{D}/\partial t\) has the dimension of current density (A/m²) but requires no actual charge motion — a time-varying \(\mathbf{D}\) between the plates of an air capacitor produces a magnetic field just as a conduction current would. A time-varying electric field generates a magnetic field.
Maxwell's Equations
The four self-consistent time-varying equations:
The four fundamental fields are \(\mathbf{E}, \mathbf{D}, \mathbf{B}, \mathbf{H}\) — twelve scalar unknowns total. The two curl equations supply six scalar relations; the two constitutive relations \(\mathbf{D} = \epsilon\mathbf{E}\) and \(\mathbf{H} = \mathbf{B}/\mu\) supply six more. The two divergence equations are not independent — they can be derived from the curl equations and continuity. Together with Lorentz's force equation, Maxwell's equations describe all macroscopic electromagnetic phenomena.
In the static limit (\(\partial/\partial t \to 0\)) they reduce to the two pairs of Table 7-1.
7-4 Potential Functions for Time-Varying Fields
Since \(\nabla\cdot\mathbf{B} = 0\) still holds, we retain \(\mathbf{B} = \nabla\times\mathbf{A}\) (6-15). Substituting into Faraday's law:
The quantity in parentheses is curl-free, so it can be written as \(-\nabla V\):
Now \(\mathbf{E}\) has two contributions: an "electrostatic" part \(-\nabla V\) from charge distribution, and an "induction" part \(-\partial\mathbf{A}/\partial t\) from time-varying magnetic field. In the static limit only the first survives.
Lorenz gauge and wave equations
Substituting the potentials into the Ampère–Maxwell equation gives \(\nabla(\nabla\cdot\mathbf{A}) - \nabla^2\mathbf{A} = \mu\mathbf{J} - \mu\epsilon\nabla(\partial V/\partial t) - \mu\epsilon\partial^2\mathbf{A}/\partial t^2\). Choosing the Lorenz gauge
decouples the equations into two non-homogeneous wave equations:
Their solutions travel with velocity \(u = 1/\sqrt{\mu\epsilon}\). In free space \(u = 1/\sqrt{\mu_0\epsilon_0} = c\) — the speed of light. This is the central result linking electromagnetism to optics: light is an electromagnetic wave.
7-5 Electromagnetic Boundary Conditions
The integral forms of Maxwell's equations applied to a pillbox or flat contour straddling an interface give four boundary conditions that are identical in form to those of the static case, because \(\partial\mathbf{D}/\partial t\) and \(\partial\mathbf{B}/\partial t\) have no surface integrals when the pillbox/contour is collapsed:
In words: tangential E is continuous; tangential H jumps by the surface current; normal D jumps by the surface charge; normal B is continuous.
Between two lossless media (\(\rho_s = 0\), \(\mathbf{J}_s = 0\)) all four field components have simple continuity conditions across the interface. At an interface with a perfect conductor, all fields inside vanish, so \(\mathbf{E}\) is normal to the surface (\(E_{1n} = \rho_s/\epsilon_1\)) and \(\mathbf{H}\) is tangential (\(|\mathbf{H}_{1t}| = |\mathbf{J}_s|\)) — fundamental results for waveguide and antenna analysis.
7-6 Wave Equations and Their Solutions
In a source-free region (\(\rho = 0\), \(\mathbf{J} = 0\)), \(\mathbf{E}\) and \(\mathbf{B}\) themselves satisfy homogeneous wave equations. For a spherically symmetric scalar-potential wave about a point source,
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
These reduce to the static Poisson-equation solutions when time dependence is negligible ("quasi-static" regime). For high-frequency sources — antennas — retardation cannot be ignored.
7-7 Time-Harmonic (Phasor) Electromagnetics
Because Maxwell's equations are linear, sinusoidal sources of frequency \(\omega\) produce sinusoidal fields of the same frequency everywhere. Represent a real sinusoidal field as the real part of a complex phasor:
Under this substitution, \(\partial/\partial t \to j\omega\), and Maxwell's equations become algebraic in \(j\omega\):
Helmholtz's equation
The wave equations (7-63) and (7-65) become Helmholtz equations:
where
is the wavenumber. The Lorenz gauge in phasor form is \(\nabla\cdot\mathbf{A} + j\omega\mu\epsilon V = 0\).
Retarded potentials in phasor form
The phasor solutions contain the retardation factor \(e^{-jkR}\):
These are the foundation for antenna theory in Chapter 11: the radiation from a current distribution is computed by integrating the phasor current over the source volume with the retardation phase factor.
Chapter 7 at a Glance
- New postulate: \(\nabla\times\mathbf{E} = -\partial\mathbf{B}/\partial t\) — a changing B makes E non-conservative.
- Faraday's law: \(\mathscr{V} = -d\Phi/dt\); transformer emf (stationary, changing B) + motional emf (moving, static B) both contained in this single total derivative.
- Displacement current: \(\partial\mathbf{D}/\partial t\) added to Ampère's law to save charge conservation; a changing E makes H, even with no free current.
- Maxwell's equations (Table 7-2): four coupled equations in \(\mathbf{E}, \mathbf{D}, \mathbf{B}, \mathbf{H}\); static model is a special case.
- Potentials: \(\mathbf{B} = \nabla\times\mathbf{A}\), \(\mathbf{E} = -\nabla V - \partial\mathbf{A}/\partial t\). In the Lorenz gauge, both \(V\) and \(\mathbf{A}\) obey non-homogeneous wave equations with propagation speed \(u = 1/\sqrt{\mu\epsilon}\).
- Free-space speed: \(1/\sqrt{\mu_0\epsilon_0} = c\) — the speed of light. Light is an electromagnetic wave.
- Boundary conditions: same as static cases — tangential E continuous, tangential H jumps by \(\mathbf{J}_s\), normal D jumps by \(\rho_s\), normal B continuous.
- Retardation: \(V(\mathbf{r},t) = (1/4\pi\epsilon)\int\rho(\mathbf{r}', t-R/u)/R\,dv'\) — the source's state at the earlier time reaches the field point now.
- Phasor form: \(\partial/\partial t \to j\omega\), wave equations become Helmholtz \(\nabla^2 + k^2\), wavenumber \(k = \omega\sqrt{\mu\epsilon}\), retarded potentials acquire the factor \(e^{-jkR}/R\).
Problems and Solutions
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.
→ SolutionA 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.
→ SolutionParallel-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.
→ SolutionGiven 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.
→ SolutionGiven A = A₀ sin(kz − ωt) a_x, V = 0. Find E and H.
→ SolutionE_i = 10 cos(ωt − kz) a_x V/m incident on perfect conductor at z = 0. Find reflected field and surface current.
→ SolutionFind E(z,t) for a +z propagating wave at f = 1 GHz, amplitude 50 V/m in free space.
→ SolutionConvert E(x,t) = 10 cos(ωt − 2x) a_x − 5 sin(ωt − 2x) a_y to phasor form.
→ SolutionCopper: σ = 5.8×10⁷ S/m, μᵣ = 1. Find δ at 60 Hz, 1 MHz, 1 GHz.
→ SolutionNon-magnetic dielectric with εᵣ = 4. Find phase velocity, wavelength at 500 MHz, and wave impedance.
→ SolutionE = 100 cos(ωt − kz) a_x V/m in free space. Find H, instantaneous S, and time-average ⟨S⟩.
→ SolutionRectangular loop (0.2 m × 0.1 m) in B = 0.5 cos(1000t) a_z T. Find induced EMF.
→ SolutionBar 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.
→ SolutionB = B₀ sin(kx) cos(ωt) a_z. Find E using Faraday's law.
→ SolutionE = 50 cos(2π×10⁸t − πz/1.5) a_x V/m. Find f, ω, k, λ, v_p, H.
→ SolutionTextbook Practice Problems
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.
Sample: Single-turn loop in \(B(t)=B_0 e^{-t/\tau}\), area \(A\), \(\tau=1\,\text{ms}\). Find EMF.
Sample: Transformer: \(N_1=200\) primary, \(N_2=1000\) secondary, primary \(V_1=120\,\text{V}\) (60 Hz). Find \(V_2\).
Sample: Coil in time-varying field \(\vec{B}=B_0\cos(\omega t)\hat{z}\), \(N\) turns, area \(A\). Find phasor EMF.
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.
Sample: Loop moving with \(\vec{v}\) in non-uniform \(\vec{B}(\vec{r},t)\). Total EMF.
Sample: Write Maxwell's equations in differential form (SI units).
Sample: Parallel-plate capacitor: \(V(t)=V_0\sin\omega t\), \(C=10\,\text{nF}\). Find conduction and displacement currents.
Sample: Boundary between two dielectrics (\(\epsilon_1,\epsilon_2\)) with no surface current/charge. State all four boundary conditions.
Sample: Surface current \(\vec{K}=K_0\hat{x}\,\text{A/m}\) on plane \(z=0\). Apply BC.
Sample: Source-free region. Derive vector wave equation for \(\vec{E}\).
Sample: Convert \(\vec{E}(z,t)=10\hat{x}\cos(\omega t-\beta z+\pi/4)\,\text{V/m}\) to phasor.
Sample: Phasor Maxwell's curl equation in lossy medium.
Sample: Coil in increasing flux. Apply Lenz's law to determine current direction.
Sample: Express \(\vec{E},\vec{B}\) from \(V(\vec{r},t)\) and \(\vec{A}(\vec{r},t)\).
Sample: Apply Lorenz gauge to derive wave equations for \(V\) and \(\vec{A}\).
Sample: Plane-wave solution \(\vec{E}=E_0\hat{x}\cos(\omega t-kz)\). Find \(\vec{H}\) and check Maxwell.
Chapter 8 — Plane Electromagnetic Waves
Key Theory — Chapter 8
Condensed from Cheng, Field and Wave Electromagnetics, §8-1 through §8-10. Read this before attempting the problems below.
8-1 Introduction — The Free-Space Wave Equation
In a source-free non-conducting medium, Maxwell's equations combine to give homogeneous vector wave equations in E and H. In free space:
A uniform plane wave has \(\mathbf{E}\) with the same direction, magnitude, and phase across any plane perpendicular to the direction of propagation. Strictly, infinite plane waves don't exist in practice — but far from a source the wavefront is nearly planar, and their mathematics is fundamental.
8-2 Plane Waves in Lossless Media
Using phasors, the wave equation becomes the vector Helmholtz equation:
For a wave varying only in \(z\), the solution is a superposition of forward- and backward-traveling waves:
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:
For a general lossless medium, \(k = \omega\sqrt{\mu\epsilon}\), \(\eta = \sqrt{\mu/\epsilon}\), \(\lambda = 2\pi/k\), and \(u_p = \omega/k = 1/\sqrt{\mu\epsilon}\). The wave is transverse: \(\mathbf{E}\perp\mathbf{H}\perp\mathbf{a}_n\), with \(\mathbf{H} = (1/\eta)\mathbf{a}_n\times\mathbf{E}\).
Wavenumber vector (oblique propagation)
For propagation along an arbitrary direction \(\mathbf{a}_n\), define the wavenumber vector \(\mathbf{k} = k\mathbf{a}_n\). Then
The planes of constant phase satisfy \(\mathbf{a}_n\cdot\mathbf{R} = \text{const}\).
Polarization
When two orthogonal linearly polarized components of the same frequency are added, the tip of the resultant \(\mathbf{E}\) traces a curve in the transverse plane:
- Linear — components are in phase or 180° out of phase: \(\mathbf{E}(0,t) = (\mathbf{a}_x E_{10} + \mathbf{a}_y E_{20})\cos\omega t\).
- Elliptical — 90° out of phase, unequal amplitudes: \([E_2(0,t)/E_{20}]^2 + [E_1(0,t)/E_{10}]^2 = 1\).
- Circular — 90° out of phase, equal amplitudes (\(E_{10} = E_{20}\)). Right- or left-hand depending on rotation sense.
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:
where \(\alpha\) (Np/m) is the attenuation constant and \(\beta\) (rad/m) is the phase constant. The wave decays as \(e^{-\alpha z}\) and oscillates as \(e^{-j\beta z}\). For a lossless medium \(\alpha = 0\) and \(\beta = k\).
Good conductors and skin depth
In a good conductor (\(\sigma/\omega\epsilon \gg 1\)): \(\alpha \cong \beta \cong \sqrt{\pi f\mu\sigma}\). The depth at which the amplitude falls to \(1/e\) is the skin depth:
Typical values (from Table 8-1):
At microwave frequencies the skin depth is so small that fields and currents are effectively confined to a thin surface layer — hence the name.
The intrinsic impedance in a good conductor is complex:
meaning \(\mathbf{E}\) and \(\mathbf{H}\) are 45° out of phase in a good conductor — a substantial reactive (energy-storing) component remains even during propagation.
8-4 Group Velocity and Dispersion
A dispersive medium is one in which \(\beta\) is not a linear function of \(\omega\); different frequencies travel at different phase velocities, so a modulated signal distorts. The velocity of the envelope of a narrow-band wave packet is the group velocity:
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:
- No dispersion (\(du_p/d\omega = 0\)): \(u_g = u_p\).
- Normal dispersion (\(du_p/d\omega < 0\)): \(u_g < u_p\).
- Anomalous dispersion (\(du_p/d\omega > 0\)): \(u_g > u_p\).
Ionized gas (plasma)
Free electrons give an effective permittivity
called the plasma frequency. For \(\omega < \omega_p\), \(\gamma\) is real — the wave is evanescent, no propagation. For \(\omega > \omega_p\), propagation resumes and \(u_p u_g = c^2\). The ionosphere reflects AM radio (\(f < f_p\)) but transmits TV and FM (\(f > f_p\)).
8-5 Flow of Power — The Poynting Vector
Starting from Maxwell's curl equations and using the vector identity \(\nabla\cdot(\mathbf{E}\times\mathbf{H}) = \mathbf{H}\cdot(\nabla\times\mathbf{E}) - \mathbf{E}\cdot(\nabla\times\mathbf{H})\), one obtains Poynting's theorem:
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
is the instantaneous power density vector carried by the field.
Time-average power density
For time-harmonic fields represented by phasors:
For a uniform plane wave in a lossy medium propagating in \(+z\), this reduces to \(\mathscr{P}_{av} = \mathbf{a}_z(E_0^2/2|\eta|)e^{-2\alpha z}\cos\theta_\eta\) — the phase angle \(\theta_\eta\) of the intrinsic impedance acts exactly like a power factor.
8-6 Normal Incidence on a Perfect Conductor
When a wave is incident normally from a dielectric onto a perfect conductor (\(\sigma_2\to\infty\)):
- The tangential \(\mathbf{E}\) must vanish at the surface, forcing \(E_{r0} = -E_{i0}\).
- Total \(\mathbf{E}\) in medium 1 is a pure standing wave: \(\mathbf{E}_1 = -\mathbf{a}_x\,j2E_{i0}\sin\beta_1 z\).
- \(\mathbf{H}\) has a maximum at the conductor, \(\mathbf{E}\) a zero; they are in space and time quadrature.
- \(\mathscr{P}_{av} = 0\) — no average power crosses the boundary; all energy is reflected.
- Nulls of \(\mathbf{E}\) occur at \(z = -n\lambda/2\) (multiples of half-wavelength from the wall).
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:
- Perpendicular (horizontal, E-) polarization: \(\mathbf{E}\) perpendicular to plane of incidence. Boundary condition forces \(\theta_r = \theta_i\) (Snell's law of reflection) and \(E_{r0} = -E_{i0}\).
- Parallel (vertical, H-) polarization: \(\mathbf{E}\) lies in the plane of incidence; \(\mathbf{H}\) is perpendicular.
In both cases a standing wave forms perpendicular to the boundary and a traveling wave parallel to it — this is the foundation of waveguide theory (Chapter 10).
8-8 Normal Incidence on a Dielectric Boundary
Neither medium is a perfect conductor; part of the wave reflects, part transmits. Requiring continuity of tangential \(\mathbf{E}\) and \(\mathbf{H}\) at \(z=0\) yields the reflection coefficient \(\Gamma\) and transmission coefficient \(\tau\):
\(\Gamma\) ranges from \(-1\) (perfect conductor) to \(+1\); \(\tau\) is always positive. In a lossy medium both are complex (introducing a phase shift at the interface). Check: \(\eta_2 = \eta_1 \Rightarrow \Gamma = 0\) — impedance matched, no reflection.
Standing-wave ratio (SWR)
When reflection is partial, the total \(\mathbf{E}_1 = \tau E_{i0}e^{-j\beta_1 z} + j2\Gamma E_{i0}\sin\beta_1 z\) — a traveling wave plus a standing wave. The maxima and minima of \(|\mathbf{E}_1|\) define the standing-wave ratio:
\(S\) ranges from 1 (no reflection) to \(\infty\) (total reflection); often quoted in dB as \(20\log_{10}S\). Each 1-unit increase in \(\Gamma\) corresponds to a roughly doubling of SWR.
8-9 Multiple Dielectric Interfaces — Quarter-Wave Transformer
For three media (medium 1 → medium 2 of thickness \(d\) → medium 3), there are two conditions for zero reflection from medium 1:
- 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.
- Quarter-wave impedance transformer (\(\eta_3 \neq \eta_1\)): \(\eta_2 = \sqrt{\eta_1\eta_3}\) and \(d = (2n+1)\lambda_2/4\) — the middle layer acts as an impedance-matching transformer, analogous to quarter-wave sections in transmission lines (Chapter 9).
8-10 Oblique Incidence on a Dielectric Boundary
Snell's laws
Phase-matching \(e^{-j\beta_1 x\sin\theta_i} = e^{-j\beta_1 x\sin\theta_r} = e^{-j\beta_2 x\sin\theta_t}\) at every \(x\) on the interface immediately gives:
where \(n = c/u_p\) is the index of refraction. Denser medium (\(n_2 > n_1\)) bends the ray toward the normal.
Total internal reflection
If \(\epsilon_1 > \epsilon_2\) and \(\theta_i\) exceeds the critical angle, no refracted wave propagates:
Beyond \(\theta_c\) the wave in medium 2 becomes evanescent (exponentially attenuated along the normal while traveling along the surface) — a surface wave. This is the operating principle of optical fibers and dielectric waveguides.
Fresnel equations
Applying continuity at the interface gives reflection coefficients for each polarization:
Brewster's angle
For non-magnetic media (\(\mu_1 = \mu_2 = \mu_0\)), \(\Gamma_\|\) vanishes at a particular angle of incidence — the Brewster angle:
At \(\theta_{B\|}\), a parallel-polarized wave is totally transmitted (no reflection). Because \(\Gamma_\perp\) does not vanish there, an unpolarized wave incident at \(\theta_{B\|}\) reflects with perpendicular polarization only — the polarizing angle. Applications: polarizing sunglasses, Brewster windows in laser cavities.
Chapter 8 at a Glance
- Wave equation (free space): \(\nabla^2\mathbf{E} - (1/c^2)\partial^2\mathbf{E}/\partial t^2 = 0\); phasor form \(\nabla^2\mathbf{E} + k^2\mathbf{E} = 0\) with \(k = \omega\sqrt{\mu\epsilon}\).
- Intrinsic impedance: \(\eta = \sqrt{\mu/\epsilon}\); free space \(\eta_0 = 120\pi \cong 377\ \Omega\).
- Transverse wave: \(\mathbf{H} = (1/\eta)\mathbf{a}_n\times\mathbf{E}\); \(\mathbf{E}\perp\mathbf{H}\perp\mathbf{a}_n\).
- Polarization: linear, elliptical, circular — depends on relative amplitude and phase of two orthogonal components.
- Lossy media: \(\gamma = \alpha + j\beta\); skin depth \(\delta = 1/\sqrt{\pi f\mu\sigma}\) in a good conductor; \(\eta_c\) has 45° phase angle.
- Group vs. phase velocity: \(u_g = 1/(d\beta/d\omega)\); in lossless non-dispersive media \(u_g = u_p\).
- Plasma cutoff at \(\omega_p\): evanescent below, propagating above; \(u_p u_g = c^2\).
- Poynting vector: \(\mathscr{P} = \mathbf{E}\times\mathbf{H}\); time-avg \(\mathscr{P}_{av} = \tfrac{1}{2}\mathscr{R}e(\mathbf{E}\times\mathbf{H}^*)\).
- Normal incidence, perfect conductor: total reflection, standing wave, nulls at \(\lambda/2\) spacings.
- Normal incidence, dielectric: \(\Gamma = (\eta_2-\eta_1)/(\eta_2+\eta_1)\), \(\tau = 2\eta_2/(\eta_2+\eta_1)\), \(1+\Gamma=\tau\); SWR \(S = (1+|\Gamma|)/(1-|\Gamma|)\).
- Quarter-wave transformer: \(\eta_2 = \sqrt{\eta_1\eta_3}\), \(d = \lambda_2/4\).
- Snell's laws: \(\theta_r = \theta_i\); \(\sin\theta_t/\sin\theta_i = n_1/n_2\).
- Critical angle \(\theta_c = \sin^{-1}(n_2/n_1)\) — total internal reflection for \(\theta_i > \theta_c\); basis of optical fiber guidance.
- Brewster angle \(\theta_B = \tan^{-1}(n_2/n_1)\) — no reflection for parallel polarization; functions as a polarizer.
Problems and Solutions
Find the intrinsic impedance in (a) free space, (b) a non-magnetic dielectric with εᵣ = 9.
→ SolutionMedium: εᵣ = 2.5, σ = 0.01 S/m, μᵣ = 1, f = 1 GHz. Find α, β, and skin depth δ.
→ SolutionDetermine the polarization of E = 3 cos(ωt − kz) a_x + 4 sin(ωt − kz) a_y.
→ SolutionPlasma with ωₚ = 2π×10⁹ rad/s (fₚ = 1 GHz). Find phase and group velocities at f = 3 GHz.
→ SolutionE = 50 cos(ωt − kz) a_x V/m propagates in a lossless medium with εᵣ = 4, μᵣ = 1.
Find ⟨S⟩.
→ SolutionPlane wave from air (η₁ = 377 Ω) normally incident on glass (εᵣ = 4, μᵣ = 1, η₂ = 188.5 Ω).
Find Γ, τ, R, T.
→ SolutionFrom Problem 6 (Γ = −0.333), find the SWR in medium 1.
→ SolutionFind Brewster's angle for TM (parallel) polarization at an air–glass interface (n₁ = 1, n₂ = 2).
State whether it exists for TE polarization.
→ SolutionGlass (n₁ = 1.5) to air (n₂ = 1) interface. Find the critical angle for total internal reflection.
→ SolutionNormal incidence from air into medium with εᵣ = 9 (η₂ = 125.7 Ω). Find Γ, τ, R, T.
→ SolutionTE wave from air (n₁ = 1) into glass (n₂ = 2) at θᵢ = 30°. Find θₜ, Γ_TE, τ_TE.
→ SolutionRectangular waveguide with cutoff frequency f_c = 5 GHz, operating at f = 10 GHz. Find v_p, v_g, λ_g.
→ SolutionFrom Problem 11 (Γ_TE = −0.382, θᵢ = 30°, θₜ = 14.5°). Find R and T.
→ SolutionTM (parallel) wave, same geometry as Problem 11 (θᵢ = 30°, air to glass n₂ = 2). Find Γ_TM.
→ SolutionFind α, β, and skin depth in copper (σ = 5.8×10⁷ S/m, μᵣ = 1) at f = 100 MHz.
→ SolutionTextbook Practice Problems
Textbook Practice — Approach Hints
Sample: Plane wave at 10 GHz in free space. Find \(\beta,\lambda,\eta\).
Sample: \(\vec{E}=100\hat{x}\cos(\omega t-\beta z)\,\text{V/m}\) in air, \(f=300\,\text{MHz}\). Find \(\vec{H}\).
Sample: Wet earth: \(\sigma=10^{-2}\,\text{S/m}\), \(\epsilon_r=15\), \(f=1\,\text{MHz}\). Find \(\alpha,\beta\).
Sample: Copper at 1 GHz: \(\sigma=5.8\!\cdot\!10^7\). Find skin depth.
Sample: Plane wave \(E_0=10\,\text{V/m}\) in free space. Find time-average Poynting.
Sample: Circularly polarized wave \(E_0=5\,\text{V/m}\) in air. Find \(\langle\vec{S}\rangle\).
Sample: Plane wave normal incidence on perfect conductor. Show \(\Gamma=-1\).
Sample: Standing wave from above. Find positions of \(E\) nulls and maxima.
Sample: Air to glass (\(n=1.5\)), normal incidence. Find \(\Gamma,\tau\) and powers.
Sample: Verify power conservation for above.
Sample: Polyethylene (\(\epsilon_r=2.25\)) at \(f=1\,\text{GHz}\). Find \(v_p,\lambda,\eta\).
Sample: \(\vec{E}=E_0(\hat{x}+j\hat{y})e^{-j\beta z}\). Identify polarization.
Sample: \(E_x=3\cos\omega t,\;E_y=4\cos(\omega t-\pi/2)\) at \(z=0\). Identify polarization.
Sample: Dispersive medium: \(\beta(\omega)=\omega\sqrt{\mu\epsilon(\omega)}\) with \(\epsilon=\epsilon_0(2+\omega/\omega_0)\). Find \(v_p\) and \(v_g\).
Sample: Plasma with \(\omega_p=2\pi\!\cdot\!10^9\,\text{rad/s}\). At \(\omega=2\omega_p\), find \(v_p,v_g\).
Sample: Plane wave hits perfect conductor at \(\theta_i=30^{\circ}\), perpendicular polarization. Find reflection.
Sample: Same wave, parallel polarization.
Sample: Air-to-water interface (\(n_2=1.33\)), \(\theta_i=45^{\circ}\). Find \(\theta_t\) via Snell.
Sample: Find Brewster angle for air-glass (\(n=1.5\)).
Chapter 9 — Theory and Applications of Transmission Lines
Key Theory — Chapter 9
Condensed from Cheng, Field and Wave Electromagnetics, §9-1 through §9-7. Read this before attempting the problems below.
9-1 Guided TEM Waves on Transmission Lines
An omnidirectional source like a wire antenna wastes most of its energy. For point-to-point transmission of power or information, electromagnetic energy must be guided — confined to a structure that carries it from source to load. The simplest such mode is the TEM (transverse electromagnetic) wave, in which both \(\mathbf{E}\) and \(\mathbf{H}\) are perpendicular to the direction of propagation — exactly like an unguided plane wave in the surrounding dielectric.
Three TEM transmission-line structures are fundamental:
- Parallel-plate line — two parallel conductors with a dielectric slab between. At microwave frequencies these are fabricated as striplines/microstrips on printed-circuit boards.
- Two-wire line — a pair of parallel wires of radius \(a\) separated by \(D\). Overhead power and telephone lines, TV feeder lines.
- Coaxial line — inner conductor and outer sheath separated by a dielectric. Fields are confined entirely within the dielectric; no stray radiation, minimal interference. Telephone/TV cables, RF instrument cables.
Other (non-TEM) wave modes can exist when the conductor separation exceeds a certain fraction of a wavelength; these are treated as waveguide modes in Chapter 10.
9-3 General Time-Harmonic Transmission-Line Equations
A transmission line is modeled by distributed parameters per unit length:
- \(R\) — series resistance (Ω/m)
- \(L\) — series inductance (H/m)
- \(G\) — shunt conductance (S/m)
- \(C\) — shunt capacitance (F/m)
For sinusoidal steady-state phasors \(V(z), I(z)\):
Decoupling by taking a second derivative:
where the propagation constant is
\(\alpha\) (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:
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:
\(Z_0\) and \(\gamma\) depend on \(R, L, G, C, \omega\) — not on line length. They characterize the line itself.
Analogy: the general form is identical to \(\gamma = j\omega\sqrt{\mu\epsilon(1 + \sigma/j\omega\epsilon)^{1/2}}\) and \(\eta = \sqrt{\mu/(\epsilon(1-j\sigma/\omega\epsilon))}\) for plane waves in a lossy medium — voltage plays the role of \(\mathbf{E}\), current the role of \(\mathbf{H}\), \(Z_0\) the role of intrinsic impedance \(\eta\).
9-3.2 Three Practical Cases
Lossless line (R = G = 0)
Low-loss line (R ≪ ωL, G ≪ ωC)
Distortionless line (R/L = G/C)
When the Heaviside condition \(R/L = G/C\) is satisfied, both \(\gamma\) and \(Z_0\) simplify exactly:
Because \(\beta\) is linear in \(\omega\), all frequencies travel at the same phase velocity \(u_p = 1/\sqrt{LC}\) — no dispersion, no signal distortion.
Universal Relation: LC = με
From the plane-wave analogy \(\gamma = j\omega\sqrt{\mu\epsilon(1 + \sigma/j\omega\epsilon)^{1/2}}\) and \(G/C = \sigma/\epsilon\) (from the R–C analogy of Chapter 5):
Consequence: if \(L\) is known for a given geometry, \(C\) follows from \(LC = \mu\epsilon\); if either is known, \(G\) follows from the medium conductivity. The velocity of propagation on a lossless TEM line equals the velocity of an unguided plane wave in the dielectric: \(u_p = 1/\sqrt{LC} = 1/\sqrt{\mu\epsilon}\).
Two-wire line parameters
For parallel wires of radius \(a\) separated by center-to-center distance \(D\):
(When \(D/2a \gg 1\), \(\cosh^{-1}(D/2a) \cong \ln(D/a)\).) Analogous formulas with \(\ln(b/a)\) apply to coaxial lines (Chapter 3/6).
9-4 Finite Terminated Line
Measuring from the load end (\(z' = \ell - z\), so \(z' = 0\) is the load):
The impedance transformation at any point along the line:
The input impedance at the generator end (\(z' = \ell\)) is \(Z_i = Z(z' = \ell)\). This is the key equation — it lets you replace a terminated line by an equivalent lumped impedance at the source.
Open- and short-circuited lines (lossless)
With \(\gamma = j\beta\), \(\tanh j\beta\ell = j\tan\beta\ell\):
Short-circuited lines appear inductive for \(\ell < \lambda/4\), capacitive for \(\lambda/4 < \ell < \lambda/2\), and so on. Useful fact: \(Z_0 = \sqrt{Z_{io}Z_{is}}\) — measuring open- and short-circuit input impedances of any line determines \(Z_0\).
9-4.2 Reflection Coefficient and Standing Wave Ratio
For any resistive or complex load \(Z_L \neq Z_0\), part of the incident wave reflects. Define the voltage reflection coefficient at the load:
\(|\Gamma| \le 1\), with special cases:
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):
Often quoted in dB as \(20\log_{10}S\). A high SWR is undesirable — it indicates mismatch, large reflected power, and potentially dangerous voltage maxima on the line.
Resistive load on a lossless line: \(\Gamma = (R_L - R_0)/(R_L + R_0)\).
- \(R_L > R_0\) → \(\Gamma > 0\), voltage maximum at load, minima at \(z' = n\lambda/2\).
- \(R_L < R_0\) → \(\Gamma < 0\), voltage minimum at load, maxima at \(z' = n\lambda/2\).
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
For a well-insulated low-loss line, \(Q \cong \omega L/R\) — the familiar LC-circuit expression. Transmission-line resonators at microwave frequencies routinely achieve \(Q\)'s in the thousands, far beyond lumped-element circuits.
9-6 The Smith Chart
The Smith chart is a conformal map of the complex \(z = Z/Z_0\) plane onto the unit disk in the \(\Gamma\) plane. Its key properties:
- Every impedance maps to a unique point in the \(|\Gamma| \le 1\) disk.
- Constant-\(r\) circles (\(\mathscr{R}e\{z\} = r\)) and constant-\(x\) circles (\(\mathscr{I}m\{z\} = x\)) form the chart's two orthogonal families.
- Moving along a lossless line by a distance \(\Delta z'\) corresponds to rotating on a constant-\(|\Gamma|\) circle centered at the origin, by an angle \(2\beta\Delta z' = 4\pi\Delta z'/\lambda\). Clockwise rotation corresponds to movement "toward the generator."
- A half-wavelength (\(\Delta z' = \lambda/2\)) is one full revolution; a quarter-wavelength (\(\Delta z' = \lambda/4\)) is half a revolution — which places \(z\) and \(y = 1/z\) diametrically opposite on the \(|\Gamma|\)-circle. This is how the Smith chart doubles as an admittance chart.
- The short-circuit point is at the extreme left (\(z = 0\)), the open-circuit point at the extreme right (\(z \to \infty\)), and matched point at the center (\(z = 1\)).
Standard problems solved by inspection on the chart: find \(\Gamma\) from \(Z_L\), find \(z_i\) from \(z_L\) after a given length, find \(Z_L\) from SWR and minimum location, find stub lengths for matching, convert impedance to admittance. For lossy lines, the \(|\Gamma|\)-circle becomes a spiral contracting inward due to the \(e^{-2\alpha z'}\) factor.
9-7 Transmission-Line Impedance Matching
Maximum power transfer from generator to load requires \(Z_L = Z_0\) — an impedance-matched line has \(S = 1\) and no reflections. Mismatched lines also cause echoes that distort information signals. The chapter presents three matching techniques on lossless lines:
Quarter-wave transformer
To match a resistive load \(R_L\) to a line of impedance \(R_0\), insert a quarter-wavelength section of characteristic impedance:
Because \(Z(\lambda/4) = Z_0^2/Z_L\) for a lossless line (from eq. 9-102 with \(\tanh j\pi/2 \to \infty\)), the quarter-wave section "inverts" the load and scales it. Since \(\ell = \lambda/4\) depends on wavelength, this match is inherently narrowband. Quarter-wave transformers don't work for complex loads on low-loss lines because \(R_0'\) would need to be complex.
Single-stub matching
A more general technique: attach a short-circuited line stub of length \(\ell\) in parallel with the main line at a distance \(d\) from the load. Two unknowns (\(d\), \(\ell\)) are chosen so the parallel admittance equals \(Y_0 = 1/R_0\):
Working in normalized admittances, \(y_B\) is the admittance at the stub location looking toward the load, and \(y_s\) is the input admittance of the short-circuited stub (purely imaginary). The match condition becomes \(y_B = 1 + jb_B\), and the stub susceptance must cancel \(jb_B\). Short-circuited stubs are preferred over open-circuited ones — a perfect short is easier to realize than a perfect open (open ends radiate and couple to neighboring objects).
Single-stub matching is easily solved graphically on the Smith chart (using it as an admittance chart): locate \(y_L\), rotate toward generator until the trajectory crosses the \(g = 1\) circle — that's distance \(d\); read off the needed susceptance and find the corresponding short-circuit stub length. Unlike the quarter-wave transformer, single-stub matching works for any \(Z_L\).
Double-stub matching (§9-7.3) uses two stubs at fixed distances when the distance to the load cannot be adjusted — at the cost of a smaller range of matchable \(Z_L\).
Chapter 9 at a Glance
- Transmission-line equations: \(-dV/dz = (R+j\omega L)I\), \(-dI/dz = (G+j\omega C)V\).
- Propagation constant: \(\gamma = \alpha + j\beta = \sqrt{(R+j\omega L)(G+j\omega C)}\).
- Characteristic impedance: \(Z_0 = \sqrt{(R+j\omega L)/(G+j\omega C)}\).
- Lossless line: \(Z_0 = R_0 = \sqrt{L/C}\), \(\beta = \omega\sqrt{LC}\), \(u_p = 1/\sqrt{LC}\).
- Low-loss: \(\alpha \cong \tfrac{1}{2}(R/R_0 + GR_0)\), \(\beta \cong \omega\sqrt{LC}\).
- Distortionless condition: \(R/L = G/C\) — no dispersion.
- Universal relation: \(LC = \mu\epsilon\), \(G/C = \sigma/\epsilon\); TEM velocity equals plane-wave velocity in the dielectric.
- Input impedance: \(Z(z') = Z_0(Z_L + Z_0\tanh\gamma z')/(Z_0 + Z_L\tanh\gamma z')\).
- Open/short reactances (lossless): \(X_{io} = -R_0\cot\beta\ell\), \(X_{is} = R_0\tan\beta\ell\); \(Z_0 = \sqrt{Z_{io}Z_{is}}\).
- Reflection coefficient: \(\Gamma = (Z_L - Z_0)/(Z_L + Z_0)\); \(|\Gamma| \le 1\).
- SWR: \(S = (1 + |\Gamma|)/(1 - |\Gamma|)\); \(|\Gamma| = (S-1)/(S+1)\).
- Q of TL resonator: \(Q = \beta/(2\alpha)\); quarter-wave shorted = parallel resonance.
- Smith chart: constant-\(|\Gamma|\) circles are trajectories along a lossless line; \(\lambda/4\) is 180° rotation; \(z\) and \(y\) are diametrically opposite.
- Quarter-wave transformer match: \(R_0' = \sqrt{R_0 R_L}\) (resistive loads).
- Single-stub matching: short-circuited parallel stub of length \(\ell\) at distance \(d\) chosen to cancel the susceptance; works for any load.
Problems and Solutions
Find Z₀ for a coaxial line with inner radius a = 1 mm, outer radius b = 5 mm, εᵣ = 2.25.
→ SolutionLine 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.
→ SolutionZ₀ = 50 Ω, Z_L = 100 + j50 Ω. Find Γ and SWR.
→ SolutionZ₀ = 75 Ω, Z_L = 150 Ω (real), l = λ/8. Find Z_in.
→ SolutionZ₀ = 50 Ω, Z_L = 25 − j50 Ω. Find Γ and SWR.
→ SolutionMatch Z_L = 200 Ω to Z₀ = 50 Ω using a λ/4 transformer section. Find Z₀' and verify Z_in.
→ SolutionZ₀ = 50 Ω, Z_L = 75 + j25 Ω. Find |Γ| and ∠Γ.
→ SolutionTransmission line: α = 0.01 Np/m, length l = 100 m. Find power loss in dB.
→ SolutionZ₀ = 50 Ω, Z_L = 100 Ω (real). Find short-circuit stub position d and length l for matching.
→ SolutionZ₀ = 50 Ω, Z_g = 50 Ω, Z_L = ∞ (open), V_g = 100 V step. Find V at load for 0 < t < 3T.
→ SolutionZ₀ = 50 Ω, Z_g = 200 Ω, Z_L = ∞ (open), V_g = 100 V step, T = 0.5 μs. Trace voltage at load.
→ SolutionZ₀ = 50 Ω, Z_L = 25 − j50 Ω, stub separation = λ/4. Find stub lengths (SC stubs).
→ SolutionZ₀ = 100 Ω, Z_L = 50 Ω, l = 3λ/8. Find Z_in.
→ SolutionZ₀ = 50 Ω, Z_s = 50 Ω, Z_L = 75 Ω, V_s = 100∠0° V (peak), lossless λ/4 line.
→ SolutionFrom Problem 7 (|Γ| = 0.277). Find return loss and mismatch loss.
→ SolutionTextbook Practice Problems
Textbook Practice — Approach Hints
Sample: Coax with \(L'=0.4\,\mu\text{H/m}\), \(C'=100\,\text{pF/m}\). Find \(Z_0\) and \(v\).
Sample: Air-filled coax with \(a=1\,\text{mm}\), \(b=4\,\text{mm}\). Find \(Z_0\).
Sample: Lossless line with \(L'=0.5\,\mu\text{H/m}\), \(C'=200\,\text{pF/m}\) at \(f=100\,\text{MHz}\). Find \(\beta,\lambda\).
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 \(\Gamma\).
Sample: Lossy line: \(R'=2\,\Omega/\text{m}\), \(G'=10^{-3}\,\text{S/m}\), \(L'=0.5\,\mu\text{H/m}\), \(C'=80\,\text{pF/m}\), \(f=100\,\text{MHz}\).
Sample: \(Z_0=50\,\Omega\), \(Z_L=100+j50\,\Omega\). Find \(\Gamma\) and VSWR.
Sample: Measured VSWR=3 on \(Z_0=50\,\Omega\) line; voltage min at \(z=-0.1\lambda\). Find \(Z_L\).
Sample: \(Z_0=50\), \(Z_L=100\), line length \(\ell=\lambda/4\). Find \(Z_{in}\).
Sample: Same line but \(\ell=\lambda/8\), \(Z_L=100\). Find \(Z_{in}\).
Sample: \(Z_L=25+j50\), \(Z_0=50\). Find \(\Gamma\) on Smith chart and rotate \(0.2\lambda\) toward generator.
Sample: Move \(0.1\lambda\) along Smith chart (TWG): how much rotation?
Sample: Single short-circuited stub matching \(Z_L=100\) to \(Z_0=50\). Find stub length and position.
Sample: Quarter-wave transformer matches \(Z_L=200\) to \(Z_0=50\).
Sample: Two-wire line, conductor radius \(a\), separation \(D\gg a\). Find \(L',C'\).
Sample: Low-loss line with \(R'=1\,\Omega/\text{m}\), \(Z_0=50\,\Omega\), \(G'\approx 0\). Find \(\alpha\).
Sample: Step source \(V_0=10\,\text{V}\), source impedance \(Z_s=Z_0=50\), load \(Z_L=100\). Find voltages on line.
Sample: Lossless line, length \(L\), propagation delay \(\tau\). Step input arrives at \(t=\tau\) at load.
Sample (comprehensive): Match \(Z_L=75-j25\) to \(50\,\Omega\) at \(1\,\text{GHz}\) using a single short-circuited stub.
Chapter 10 — Waveguides and Cavity Resonators
Key Theory — Chapter 10
Condensed from Cheng, Field and Wave Electromagnetics, §10-1 through §10-7. Read this before attempting the problems below.
10-1 Introduction — Why Waveguides?
Chapter 9 developed the TEM mode (both \(\mathbf{E}\) and \(\mathbf{H}\) transverse to \(z\)) on transmission lines. At microwave frequencies (≥ a few GHz), TEM lines are impractical: conductor losses scale like \(\sqrt{f}\) and radiation off two-conductor structures becomes severe.
The remedy is to confine the wave inside a single-conductor hollow metal tube — a waveguide. Inside such a tube:
- TEM cannot exist (proved from Ampère's law — no inner conductor means no longitudinal conduction current to support transverse \(\mathbf{H}\) loops).
- Only TM waves (\(H_z=0\), \(E_z\ne 0\)) and TE waves (\(E_z=0\), \(H_z\ne 0\)) propagate.
- Each mode has a cutoff frequency \(f_c\) — below \(f_c\), the mode is evanescent; above \(f_c\), it propagates. Waveguides are high-pass filters.
Short a waveguide off with conducting end walls and you get a cavity resonator — a 3-D standing-wave box with very high \(Q\) (often \(10^{3}\)–\(10^{4}\)) and a discrete resonance spectrum.
10-2 General Wave Behavior Along a Uniform Guide
Assume harmonic time dependence \(e^{j\omega t}\) and \(z\)-dependence \(e^{-\gamma z}\) with \(\gamma = \alpha + j\beta\). In a source-free, charge-free dielectric inside the guide, Maxwell's equations reduce to Helmholtz's equation:
Splitting the Laplacian \(\nabla^{2} = \nabla_{xy}^{2} + \partial^{2}/\partial z^{2}\) and using \(\partial_z \to -\gamma\):
\(h^{2}\) is a separation constant (eigenvalue). Only discrete values of \(h\) satisfy the boundary conditions — each is a waveguide mode.
Transverse-to-longitudinal coupling. All four transverse field components can be expressed in terms of the two longitudinal components \(E_z^0\) and \(H_z^0\):
(similar for \(H_y^0, E_y^0\)). So the procedure for every waveguide problem is:
- 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.
- Enforce BCs to determine the eigenvalues \(h_{mn}\).
- Use (10-11)–(10-14) to compute the transverse components.
Cutoff frequency. The propagation constant is \(\gamma = \sqrt{h^{2}-\omega^{2}\mu\epsilon}\). It vanishes at the cutoff frequency:
Two regimes:
Equivalent wavelength relations:
Phase and group velocities in a lossless guide:
A waveguide is therefore a dispersive transmission system. \(u_g\) equals the velocity of energy transport.
Wave impedances (for propagating modes):
Note \(Z_{\text{TM}}\) is lower than \(\eta\) and \(Z_{\text{TE}}\) is higher — they approach each other at high frequency. Below cutoff both are purely reactive (no net power flows in an evanescent mode).
10-3 Parallel-Plate Waveguide
Two infinite, parallel conducting planes separated by \(b\). Analytically simplest non-trivial waveguide; supports:
- TEM mode (since two conductors are present), same as parallel-plate transmission line. No cutoff.
- TMm and TEm modes with \(h = m\pi/b\), giving \(f_{c,m}=m/(2b\sqrt{\mu\epsilon})\).
Useful as the bridge between transmission lines (Ch 9) and hollow guides — exhibits all three wave types (TEM, TM, TE) in one structure.
10-4 Rectangular Waveguides
Cross-section \(a \times b\) with \(a > b\) by convention. Separation of variables gives sinusoidal dependence in \(x\) and \(y\).
TMmn modes — boundary condition \(E_z = 0\) on all four walls forces sine variation:
For TM, neither \(m\) nor \(n\) can be zero (else \(E_z\equiv 0\)). Lowest TM mode is TM11.
TEmn modes — boundary condition \(\partial H_z/\partial n = 0\) on walls forces cosine variation:
For TE, either \(m\) or \(n\) may be zero, but not both.
Cutoff frequency / wavelength (same form for TM and TE):
Dominant mode: TE10. Because \(n=0\) is permitted for TE and \(a>b\), TE10 has the lowest cutoff of all modes:
Practical rule of thumb. A waveguide is normally used in its dominant mode over an operating band of roughly \(1.25\,f_{c10}\) to \(0.95\,f_{c20}\) — i.e. 25% above dominant cutoff but 5% below the next higher mode. Example (WR-16, \(a=2.29\) cm, \(b=1.02\) cm): usable band 8.19–12.45 GHz (X-band).
Attenuation. Dielectric loss (substituting \(\epsilon_d=\epsilon(1-j\sigma/\omega\epsilon)\)):
Conductor (wall) loss for the dominant mode:
Attenuation diverges near \(f_c\), has a broad minimum in the middle of the operating band, and grows slowly at high \(f\). TE10 always has the lowest attenuation of any mode, which is another reason it is preferred.
10-5 Circular Waveguides
Cross-section of radius \(a\). Separation in \((r,\phi)\) gives Bessel's equation in the radial coordinate; solutions are \(J_n(hr)\) (Bessel functions of the first kind). Modes are TMnp (zeros of \(J_n\)) and TEnp (zeros of \(J'_n\)).
Key cutoff facts (with \(p_{np}\) the \(p\)-th zero of \(J_n\), \(p'_{np}\) of \(J'_n\)):
TE11 is dominant, analogous to TE10 in rectangular. Circular guides are less common in practice because polarization can rotate, but they are used when rotational symmetry matters (rotary joints, circular polarizers).
10-6 Dielectric Waveguides (Slab / Fiber)
When the guiding structure is an open dielectric slab (or optical fiber) with \(\epsilon_1 > \epsilon_2\) outside, the fields are bound by total internal reflection. Inside the high-\(\epsilon\) slab the transverse fields are sinusoidal; outside they decay exponentially — the wave is a surface wave.
Both TM and TE modes are possible. Unlike metal guides, each mode has a different cutoff condition and the structure is the foundation of all optical fiber communication.
10-7 Cavity Resonators
Close both ends of a waveguide with conducting walls and you get a 3-D resonant cavity. End-wall reflections convert the traveling \(e^{-j\beta z}\) into a standing wave \(\sin(p\pi z/d)\) or \(\cos(p\pi z/d)\); the mode is indexed by three integers \((m,n,p)\).
Rectangular cavity \(a\times b\times d\). TMmnp (\(E_z\) has \(\sin\sin\cos\) form) and TEmnp (\(H_z\) has \(\cos\cos\sin\)) coexist. The resonant frequency is independent of which is which:
Dominant mode depends on the aspect ratio:
(TM restricts \(m,n\ne 0\) but allows \(p=0\); TE allows one of \(m,n\) to be zero but requires \(p\ne 0\).)
Quality factor \(Q\) — the figure of merit for a resonator:
where \(W = W_e + W_m\) is the total stored energy and \(P_L\) is the time-average power dissipated in the walls (and any dielectric). At resonance, \(W_e = W_m\).
For the TE101 mode in a rectangular cavity:
Typical value: a copper cubic cavity at 10 GHz has \(Q\approx 10{,}700\) — orders of magnitude higher than any lumped LC circuit, because loss occurs only through surface resistance \(R_s=\sqrt{\pi f\mu_0/\sigma}\).
Circular cylindrical cavity. The TM010 mode (radius \(a\), length \(d\)) has the simple resonant frequency
Cavities are excited by a probe (a short monopole inserted where \(\mathbf{E}\) of the desired mode is maximum) or a loop (inserted where \(\mathbf{H}\) of the desired mode is maximum), or by coupling through an iris from an adjoining waveguide.
Chapter 10 at a Glance
- Three wave types: TEM (both \(E_z, H_z = 0\)) — needs two conductors; TM (\(H_z=0\)); TE (\(E_z=0\)).
- Hollow waveguides (single conductor) support only TM and TE — never TEM.
- Cutoff: \(f_c = h/(2\pi\sqrt{\mu\epsilon})\). Above \(f_c\) propagates, below \(f_c\) evanescent. Waveguide = high-pass filter.
- Velocities: \(u_p > u > u_g\); \(u_p u_g = u^{2}\). Dispersive.
- Rectangular guide TEmn / TMmn: \(f_c = (u/2)\sqrt{(m/a)^{2}+(n/b)^{2}}\). Dominant mode TE10: \(f_{c10}=u/(2a)\), \(\lambda_c=2a\).
- Circular guide: Bessel-function modes; dominant is TE11 (eigenvalue \(p'_{11}=1.841\)).
- Dielectric guide / fiber: total internal reflection; surface wave decays exponentially outside.
- Cavity resonator: \(f_{mnp}=(u/2)\sqrt{(m/a)^{2}+(n/b)^{2}+(p/d)^{2}}\); indexed by three integers. \(Q=\omega W/P_L\); typical microwave metal cavity \(Q \sim 10^{4}\).
Problems and Solutions
Review Questions (Tier 1)
Conductor losses scale as \(\sqrt{f}\) and dielectric losses rise too; at microwave frequencies coaxial cable attenuation becomes prohibitive over long distances.
The lowest frequency at which a given waveguide mode can propagate. Below \(f_c\) the mode is evanescent (purely decaying).
At microwave frequencies component dimensions become comparable to \(\lambda\); radiation losses and parasitic effects dominate. Distributed cavity resonators are used instead.
Helmholtz equation: \(\nabla^2\tilde{\vec{E}}+k^2\tilde{\vec{E}}=0\) (and similarly for \(\tilde{\vec{H}}\)), where \(k=\omega\sqrt{\mu\epsilon}\).
TEM (\(E_z=H_z=0\)), TM (transverse magnetic, \(H_z=0\), \(E_z\ne0\)), TE (transverse electric, \(E_z=0\), \(H_z\ne 0\)).
Ratio \(Z=E_t/H_t\) of transverse field components, characteristic of the propagating mode and frequency.
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.
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.
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 \(H_z\).
Discrete values of the separation constant for which non-trivial solutions satisfy all boundary conditions. They determine the cutoff frequencies of the modes.
Yes — one cutoff per mode. \(f_c\) depends on the mode indices and the guide cross-section dimensions; e.g., for rectangular guide \(f_c=(c/2)\sqrt{(m/a)^2+(n/b)^2}\).
A mode operating below its cutoff frequency. Fields decay exponentially with distance instead of propagating; \(\beta=0\) and \(\gamma\) is purely real (attenuation only).
Always longer: \(\lambda_g=\lambda/\sqrt{1-(f_c/f)^2}>\lambda\).
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 \(\eta\)).
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.
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.
\(u_p=\omega/\beta\) 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.
A self-consistent field pattern that satisfies the wave equation plus all boundary conditions for a particular eigenvalue (cutoff frequency).
Plate spacing \(d\) and mode order \(n\): \(f_c=nc/(2d)\) for TE\(_n\)/TM\(_n\). The dielectric also enters via \(c\to v=c/\sqrt{\epsilon_r\mu_r}\).
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 \(f_c=c/(2d)\).
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.
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 \((m/a)^2+(n/b)^2+(p/d)^2\).
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.
Rectangular waveguide: a = 2.286 cm, b = 1.016 cm (WR-90). Find f_c for TE₁₀, TE₂₀, TE₀₁, TE₁₁.
→ SolutionWR-90 at f = 10 GHz (TE₁₀ mode, f_c = 6.56 GHz). Find v_p, v_g, and β_g.
→ SolutionWR-90 at f = 10 GHz (f_c = 6.56 GHz). Find η_TE and η_TM.
→ SolutionWR-90 at f = 10 GHz. Find λ₀, λ_g, and compare.
→ SolutionRectangular waveguide: a = 4 cm, b = 2 cm, f = 5 GHz. Write out the field components.
→ SolutionWR-90 (a=2.286 cm, b=1.016 cm), TE₁₀, f=10 GHz, E₀ = 10⁴ V/m. Find P_avg.
→ SolutionFind Rs for copper (σ = 5.8×10⁷ S/m) at f = 10 GHz.
→ SolutionCavity: a = 4 cm, b = 2 cm, d = 5 cm. Find resonant frequencies for TE₁₀₁, TE₁₁₁, TM₁₁₀.
→ SolutionCopper cavity (a=4cm, b=2cm, d=5cm), TE₁₀₁ mode, Rs = 0.0195 Ω at f₀ = 4.80 GHz. Estimate Q.
→ SolutionCircular waveguide: a = 1 cm. Find f_c for TE₁₁, TM₀₁, TE₂₁, TM₁₁.
→ SolutionFor the a = 1 cm circular waveguide (Problem 10), find λ_c for TE₁₁ and TM₀₁.
→ SolutionState the dominant mode for (a) rectangular waveguide (a > b), (b) circular waveguide, and explain why.
→ SolutionTE₁₀₁ mode in rectangular cavity (a = 4 cm, b = 2 cm, d = 4 cm), E₀ = 10⁴ V/m. Find W_total.
→ SolutionCopper cavity with f₀ = 5.3 GHz, Q = 8500. Find 3-dB bandwidth.
→ SolutionRectangular waveguide, a = 2b (standard ratio). Express f_c for first 3 modes in terms of a.
→ SolutionTextbook Practice Problems
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\(_{01}\).
Sample: Same guide at \(f=10\,\text{GHz}\), TE\(_{10}\). Find \(\lambda_g,v_p,\eta_{TE}\).
Sample: Show that TE and TM modes are orthogonal (no mode coupling in lossless guide).
Sample: Find cutoff frequency \(\omega_c\) for a guide with cutoff wavenumber \(k_c\).
Sample: Parallel-plate, \(d=2\,\text{cm}\), air. Find \(f_c\) for TM\(_1\).
Sample: Same guide at \(f=10\,\text{GHz}\). Find \(\beta\) for TM\(_1\).
Sample: TE\(_{10}\) field pattern in WR-90.
Sample: TM\(_{11}\) in \(a\times b\) rectangular guide. Find cutoff.
Sample: Compare TE\(_{11}\) and TM\(_{11}\) cutoffs in \(a\times b\) guide.
Sample: Rectangular cavity \(a=4,b=2,d=5\,\text{cm}\). Find \(f_{101}\).
Sample: Same cavity. List lowest three resonant modes.
Sample (advanced): Express \(\vec{E}\) and \(\vec{H}\) for TM\(_1\) mode in parallel-plate at \(\omega>\omega_c\).
Sample: Circular guide radius \(a=1\,\text{cm}\). Find \(f_c\) for TE\(_{11}\) (dominant).
Sample: Same guide. Find \(f_c\) for TM\(_{01}\).
Sample: Symmetric slab dielectric guide with \(\epsilon_{r1}=4\) (core) and \(\epsilon_{r2}=2.25\) (cladding). Find \(\theta_c\).
Sample: Number of bound modes in symmetric slab guide of thickness \(d\), \(\Delta n=n_1-n_2\).
Sample: Copper cavity \(a=4,b=2,d=5\,\text{cm}\), TE\(_{101}\) at \(f_0=4.80\,\text{GHz}\). Estimate Q.
Chapter 11 — Antennas and Radiating Systems
Key Theory — Chapter 11
Condensed from Cheng, Field and Wave Electromagnetics, §11-1 through §11-8. Read this before attempting the problems below.
11-1 Introduction — The Antenna Radiation Problem
Up to Chapter 10, electromagnetic waves were analyzed in source-free regions or along guided structures. Chapter 11 asks the reverse question: given a time-varying current distribution on a structure, what electromagnetic field does it radiate?
An antenna is any structure designed to radiate (or receive) electromagnetic energy efficiently in specific directions. A single straight wire, a loop, an aperture, or a complex array of these may all serve. Important antenna parameters are field pattern, directivity, radiation resistance/impedance, and bandwidth.
Three-step calculation procedure
Rather than attacking Maxwell's equations directly in \(\mathbf{E}\) and \(\mathbf{H}\), we work through the auxiliary phasor potentials \(\mathbf{A}\) and \(V\) from Chapter 7 in their retarded form:
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:
- Compute \(\mathbf{A}\) from the given current distribution \(\mathbf{J}\) (eq. 11-3).
- \(\mathbf{H} = (1/\mu)\nabla\times\mathbf{A}\) (11-1).
- \(\mathbf{E} = (1/j\omega\epsilon)\nabla\times\mathbf{H}\) (11-6) — using the source-free curl relation since we are outside the antenna structure.
The continuity equation \(\nabla\cdot\mathbf{J} = -j\omega\rho\) relates \(\rho\) and \(\mathbf{J}\), so \(V\) need not be computed separately.
11-2 The Hertzian (Elemental Electric) Dipole
The simplest radiator: a short straight wire of length \(d\ell \ll \lambda\) carrying a uniform current \(i(t) = I\cos\omega t\), with small spheres at each end to allow charge accumulation (capacitive end-loading). For a z-directed element at the origin, \(\mathbf{A} = \mathbf{a}_z A_z\) with
Taking curls and resolving into spherical components gives the complete field:
Three distance regimes emerge from the \(1/R, 1/R^2, 1/R^3\) terms:
Near field (\(\beta R \ll 1\))
Leading-order terms are \(1/R^3\) in \(\mathbf{E}\) and \(1/R^2\) in \(\mathbf{H}\) — identical to an electrostatic dipole and Biot–Savart current element respectively. The near-zone fields of an oscillating dipole are effectively quasi-static.
Far field / radiation field (\(\beta R \gg 1\))
Only the \(1/R\) terms survive:
Three observations:
- \(E_\theta\) and \(H_\phi\) are in time phase and space quadrature — exactly like a plane wave.
- Their ratio is \(E_\theta/H_\phi = \eta_0 = 120\pi\ \Omega\) — intrinsic impedance of free space.
- Amplitudes fall as \(1/R\); phase rotates every \(\lambda\). A very small region of a giant sphere looks planar — far-zone field locally behaves as a plane wave.
The radiation pattern is \(|\sin\theta|\): doughnut-shaped, maximum in the equatorial plane, zero along the dipole axis.
The condition \(\beta R \gg 1\) is equivalent to \(R \gg \lambda/2\pi\) — at lower frequencies the far zone is farther away.
11-2.2 The Elemental Magnetic Dipole (Small Current Loop)
A small filamentary loop of area \(S = \pi b^2\) carrying current \(I\) is a magnetic dipole with phasor moment \(\mathbf{m} = \mathbf{a}_z I\pi b^2\). Its far-field solution is the dual of the electric dipole — obtained by the substitution \(I\,d\ell \leftrightarrow j\beta m\):
Same \(|\sin\theta|\) pattern, same \(1/R\) decay. \(E_\theta\) and \(E_\phi\) are in space quadrature, so combining an electric and a magnetic dipole (with proper phase) produces circular polarization.
11-3 Antenna Pattern and Parameters
Radiation intensity U
The radiation intensity is the time-average power per unit solid angle radiated in the direction \((\theta, \phi)\):
\(U\) is independent of \(R\). Total radiated power:
Directive gain, directivity, beamwidth
The directive gain measures how the antenna concentrates radiation compared to an isotropic radiator with the same \(P_r\):
Its maximum value is the directivity:
The (half-power) beamwidth is the angular width between the directions where \(U\) drops to half its maximum, i.e. the \(-3\) dB points on the power pattern. Narrower beam ⇒ higher directivity.
For the Hertzian dipole: \(G_D = 1.5\sin^2\theta\), \(D = 1.5\) (1.76 dB), beamwidth 90°.
Radiation resistance
A hypothetical resistance that would dissipate the same power as the antenna radiates, at the antenna's peak current:
For a Hertzian dipole (carrying the standard 80π² integral through):
Note the \((d\ell/\lambda)^2\) dependence — a \(d\ell = 0.01\lambda\) Hertzian dipole has only \(R_r \cong 0.08\ \Omega\). Short dipoles are very poor radiators. The input impedance of a short dipole also has a large capacitive reactance, making it hard to feed efficiently.
Power gain and radiation efficiency
Not all input power is radiated; some is lost as ohmic heat in the wire (loss resistance \(R_\ell\)) or ground:
Well-constructed antennas achieve \(\eta_r\) close to 100%.
11-4 Thin Linear Antennas
A center-fed dipole of total length \(2h\) carrying a sinusoidal standing-wave current:
(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
\(|F(\theta)|\) is the \(E\)-plane pattern function of a thin dipole; the \(H\)-plane pattern is a circle (azimuthally symmetric). Pattern shape depends strongly on \(\beta h = \pi(2h/\lambda)\): for \(2h = \lambda/2\) or \(\lambda\), a single lobe dominates; for \(2h \ge 3\lambda/2\), multiple lobes appear and the main beam tilts away from the broadside direction.
Half-wave dipole (2h = λ/2)
The workhorse of practical antenna work. Setting \(\beta h = \pi/2\):
Integrating over a great sphere gives \(P_r = 36.54\,I_m^2\) watts, from which
A half-wave dipole's input impedance is approximately \(73.1\ \Omega\) — conveniently close to standard 75-Ω coaxial cable, so matching is simple. A small length trim below λ/2 makes the input reactance vanish.
Quarter-wave monopole over a conducting ground
A vertical wire of length \(\lambda/4\) driven against a large conducting ground plane. By the method of images, the ground plane plus monopole is equivalent to a half-wave dipole radiating into the upper half-space only. Consequences:
- Same pattern shape as the half-wave dipole (in upper hemisphere).
- Same \(U_{\max}\); half the radiated power (radiation only upward).
- Radiation resistance is half: \(R_r = 36.5\ \Omega\).
- Directivity is double: \(D = 3.28 \approx 5.16\) dB.
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:
This is the principle of pattern multiplication.
Two-element array
Two identical elements separated by \(d\) along the x-axis, with currents equal in magnitude but with element 2 phase-shifted by \(\xi\) relative to element 1. The array factor is
Two important configurations:
- Broadside array (\(\xi = 0\)): all elements in phase. Maximum radiation is perpendicular (broadside) to the array axis. Nulls along the axis.
- End-fire array (\(\xi = -\beta d\)): elements progressively phase-lagged so wavefronts add along the array axis. Main beam along the axis.
With \(N\) elements and graded amplitudes (uniform, binomial, Dolph–Chebyshev), arrays can achieve narrow main lobes and controlled sidelobe levels. \(N\)-fold increase in elements gives approximately \(N\)-fold gain.
11-6 Receiving Antennas and Reciprocity
By the reciprocity theorem, an antenna's pattern, directivity, and input impedance are identical whether used to transmit or receive. The open-circuit voltage induced by an incident field:
where \(\mathbf{\ell}_e\) is the antenna's vector effective length. \(V_{oc}\) is maximum when \(\mathbf{E}_i\) is parallel to the antenna (polarization matched) and zero when perpendicular — this is polarization mismatch.
Effective aperture and the Friis formula
A receiving antenna intercepts incoming wave power with an effective aperture:
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:
The \((\lambda/R)^2\) free-space path loss governs every communications link budget. For radar (target of radar cross section \(\sigma\) at distance \(R\)):
The \(R^4\) dependence (vs. \(R^2\) for one-way) is why radar is fundamentally hard.
11-7 Wave Propagation Near the Earth
A transmit antenna at height \(h_1\) and a receive antenna at height \(h_2\) separated by ground distance \(d\) see the direct ray plus a ground-reflected ray. Their superposition at the receiver gives a path-gain factor:
\(|F|\) oscillates between 0 and 2 as \(h_2/d\) changes — the basis for multipath fading in terrestrial radio and cellular systems. More careful treatment on a curved earth, with imperfect ground reflection, gives more complicated formulas; software-based ray tracing is used in practice.
11-8 Other Practical Antenna Types
Brief survey of common antennas beyond linear dipoles:
- Traveling-wave antennas — terminated long-wire antennas with a pure traveling-wave current; strong end-fire radiation. Wider bandwidth than standing-wave dipoles.
- Yagi–Uda antenna — a driven half-wave dipole plus parasitic elements (a reflector behind, several directors in front) that produce a sharp end-fire beam. Ubiquitous as rooftop TV antennas.
- Helical antennas — a wire wound into a helix over a ground plane. In normal mode (\(\text{circumference} \ll \lambda\)), radiates broadside with linear polarization; in axial mode (\(\text{circumference} \approx \lambda\)), radiates endfire with circular polarization — popular for satellite uplinks.
- Broadband antennas — log-periodic, biconical, spiral: geometries defined by angles rather than length scales, giving frequency-independent patterns over wide bands.
- Aperture antennas — horns, reflectors, and apertures. The field is computed via equivalence: the field distribution across the aperture is equivalent to an array of infinitesimal sources. Gain scales with aperture area \(G \propto A_{\text{phys}}/\lambda^2\); parabolic dish antennas at GHz frequencies commonly achieve 30–60 dB.
Chapter 11 at a Glance
- Radiation workflow: \(\mathbf{J} \to \mathbf{A}\) (retarded integral) \(\to \mathbf{H} = (\nabla\times\mathbf{A})/\mu \to \mathbf{E} = (\nabla\times\mathbf{H})/(j\omega\epsilon)\).
- Retarded potential: \(\mathbf{A} = (\mu/4\pi)\int(\mathbf{J}\,e^{-jkR}/R)\,dv'\), \(k = 2\pi/\lambda\).
- Hertzian dipole far field: \(E_\theta = j(I\,d\ell/4\pi R)\eta_0\beta\sin\theta\,e^{-j\beta R}\); \(|\sin\theta|\) pattern; \(D = 1.5\); \(R_r = 80\pi^2(d\ell/\lambda)^2\).
- Dual magnetic dipole: same pattern, swap \(I\,d\ell \leftrightarrow j\beta m\), \(\mathbf{E}_e = \eta_0\mathbf{H}_m\).
- Near field ≈ quasi-static; far field (\(\beta R \gg 1\)): locally plane wave, \(E/H = \eta_0\).
- Directivity \(D = 4\pi U_{\max}/P_r\); radiated in dB often.
- Radiation resistance \(R_r = 2P_r/I^2\); short dipoles radiate poorly (\(R_r \propto (d\ell/\lambda)^2\)).
- Efficiency: \(\eta_r = R_r/(R_r + R_\ell) = G_P/D\).
- Half-wave dipole: \(F(\theta) = \cos((\pi/2)\cos\theta)/\sin\theta\); \(R_r = 73.1\ \Omega\); \(D = 2.15\) dB; HPBW 78°.
- Quarter-wave monopole over ground: image method → upper-half-space half-wave dipole; \(R_r = 36.5\ \Omega\), \(D = 5.16\) dB.
- Pattern multiplication: total = element factor × array factor.
- Two-element array factor: \(2\cos[(\beta d\sin\theta\cos\phi + \xi)/2]\); broadside (\(\xi=0\)) vs. endfire (\(\xi=-\beta d\)).
- Reciprocity: Tx/Rx patterns, directivity, impedance all identical.
- Effective aperture: \(A_e = (\lambda^2/4\pi)D\).
- Friis formula: \(P_r/P_t = G_t G_r(\lambda/4\pi R)^2\); radar equation: \(P_r = P_t G_t G_r\lambda^2\sigma/[(4\pi)^3 R^4]\) — range as \(R^4\).
- Two-ray ground path gain: \(|F| = 2|\cos(2\pi h_1 h_2/\lambda d)|\) — multipath fading.
Problems and Solutions
Hertzian dipole: l = λ/50 at f = 1 GHz (λ = 0.3 m). Find R_rad, P_rad for I₀ = 1 A.
→ SolutionFor a z-directed Hertzian dipole, state the E-plane and H-plane patterns and find HPBW.
→ SolutionAn antenna has pattern U = U₀ cos²θ for 0 ≤ θ ≤ π/2, zero in lower hemisphere. Find D.
→ SolutionHalf-wave dipole (D = 1.64) at f = 3 GHz (λ = 0.1 m). Find A_eff.
→ SolutionShort 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°.
→ SolutionFind HPBW for the half-wave dipole. Its E-plane pattern is F(θ) = [cos(π/2 cosθ)/sinθ]².
→ SolutionI₀l = 0.1 A·m at f = 300 MHz (λ = 1 m). Find |E_θ| at r = 1 km, θ = 90°.
→ SolutionTwo 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.
→ SolutionCalculate F(θ) for the half-wave dipole at θ = 90°, 60°, 30°, 0°. Express in dB.
→ SolutionAntenna with D = 8 dBi, radiation efficiency η_e = 85%. Find gain G.
→ SolutionDescribe E-plane and H-plane patterns for (a) Hertzian dipole (z-directed), (b) half-wave dipole.
→ SolutionFor a broadside linear array of N elements, state the optimal element spacing and explain the grating lobe condition.
→ SolutionLink budget: P_t = 10 W, G_t = 15 dBi, G_r = 10 dBi, f = 2.4 GHz, r = 1 km. Find P_r.
→ SolutionMonostatic 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.
→ SolutionTransmitter: P_t = 100 W, feeder loss = 1 dB, antenna gain G = 20 dBi. Find EIRP.
→ SolutionTextbook Practice Problems
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 \(r=10\,\text{m}\), \(\theta=90^{\circ}\).
Sample: Same dipole. Find radiated power and \(R_r\).
Sample: Hertzian dipole pattern.
Sample: Half-wave dipole pattern.
Sample: Hertzian dipole. Find directivity.
Sample: Half-wave dipole at \(f=1\,\text{GHz}\). Find effective aperture.
Sample: Half-wave dipole at \(f=300\,\text{MHz}\), \(I_0=1\,\text{A}\). Find \(R_r,P_{rad}\).
Sample: Same antenna. Find \(|E_{\theta}|\) at \(r=100\,\text{m}\), \(\theta=90^{\circ}\).
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 \(P_r\).
Sample: Required \(P_t\) for \(P_r=10^{-12}\,\text{W}\) in above link?
Sample: Sinusoidal current \(I(z')=I_0\sin(\beta(\ell/2-|z'|))\) on dipole length \(\ell\). Far field?
Sample: Full-wave dipole (\(\ell=\lambda\)). Pattern.
Sample: Two-element broadside array, spacing \(d=\lambda/2\), equal currents in phase. Find array factor.
Sample: \(N=4\) uniform array, spacing \(d=\lambda/4\), progressive phase \(-\pi/2\) (end-fire).
Sample: Receiving antenna with \(G_r=10\,\text{dBi}\) at \(f=2\,\text{GHz}\). Find \(A_e\).
Sample: Antenna with \(R_r=73\,\Omega\) matched to \(Z_L=R_r\). Incident plane wave \(|E|=1\,\text{mV/m}\). Find received power.
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 \(P_r\).