00 Chapter 1 Active-Recall Diagnostic Quiz

00 Chapter 1 Active-Recall Diagnostic Quiz (Vector Calculus & Fundamentals)

How to use this

Answer before reading the notes to find your gaps, then again after. Every solution is collapsed — write the full answer on paper first, then expand. Questions marked 🔥 have appeared in real exams; years are given.

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Question 1 — Differential Area in Cylindrical Coordinates

Write the vector expression for the differential surface area element $d{\vec{s}}_r$ on a cylindrical surface of constant radius $r$.

Solution

$d{\vec{s}}_r={\hat{a}}_r(rd\varphi dz)$ The surface normal is ${\hat{a}}_r$. The patch is bounded by the differential arc length $rd\varphi$ (note the scale factor $h_{\varphi }=r$) and the height $dz$.

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Question 2 — Metric Coefficients 🔥 (2019)

Define “metric coefficient” and state the three metric coefficients for each coordinate system.

Solution

A metric coefficient (scale factor) $h_i$ converts a differential change in a coordinate variable into a physical differential length: $d{l}_i=h_{i}d{u}_i$. It is needed because coordinates such as $\theta$ and $\varphi$ are angles, not distances.

System	$(h_1,h_2,h_3)$
Cartesian $(x,y,z)$	$(1,1,1)$
Cylindrical $(r,\varphi ,z)$	$(1,r,1)$
Spherical $(R,\theta ,\varphi )$	$(1,R,R\sin \theta )$

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Question 3 — Physical Meaning of Divergence 🔥 (2018)

What does $\nabla \cdot \vec{A}$ physically represent, and what is its significance in electromagnetics?

Solution

Divergence is the net outward flux per unit volume at a point, in the limit as the volume shrinks to zero: $\nabla \cdot \vec{A}={\lim }_{\Delta v\to 0}\frac{{\oint }_S\vec{A}\cdot d\vec{s}}{\Delta v}$

• $\nabla \cdot \vec{A}>0$ — a source (flux originates here)
• $\nabla \cdot \vec{A}<0$ — a sink (flux terminates here)
• $\nabla \cdot \vec{A}=0$ — solenoidal (no sources or sinks)

In EM: $\nabla \cdot \vec{D}={\rho }_v$ says electric flux lines begin and end on electric charges. $\nabla \cdot \vec{B}=0$ says magnetic flux lines have no sources at all — no monopoles — so they must close on themselves.

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Question 4 — Physical Meaning of Curl 🔥 (2022)

What does $\nabla \times \vec{A}$ represent, and what are the consequences of a field being curl-free?

Solution

Curl is the maximum circulation per unit area at a point, directed normal to the plane of that maximum circulation (right-hand rule): $\nabla \times \vec{A}={\lim }_{\Delta s\to 0}{\hat{a}}_n\frac{[{\oint }_C\vec{A}\cdot d\vec{l}{]}_{\max }}{\Delta s}$ If $\nabla \times \vec{A}=0$ (irrotational):

1. ${\oint }_C\vec{A}\cdot d\vec{l}=0$ on every closed path (Stokes’s theorem)
2. The field is conservative — line integrals are path-independent
3. A scalar potential exists: $\vec{A}=-\nabla V$
4. Zero net work moving a charge around a closed loop — the field form of KVL
5. Example: the static electric field, $\nabla \times \vec{E}=0$

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Question 5 — Divergence Theorem

State the Divergence Theorem and specify exactly which integrals it converts.

Solution

${\oint }_S\vec{A}\cdot d\vec{s}={\int }_V(\nabla \cdot \vec{A})dv$ It converts a volume integral of the divergence over $V$ into a closed surface flux integral over the closed surface $S$ bounding $V$. The surface must be closed.

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Question 6 — Stokes’s Theorem

State Stokes’s Theorem and specify which integrals it converts.

Solution

${\oint }_C\vec{A}\cdot d\vec{l}={\int }_S(\nabla \times \vec{A})\cdot d\vec{s}$ It converts an open surface integral of the curl into a closed line integral (circulation) around the contour $C$ bounding that surface. The surface must be open, and $d\vec{s}$ must follow the right-hand rule relative to the traversal of $C$.

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Question 7 — Null Identity I: Curl of a Gradient

Prove that $\nabla \times (\nabla V)\equiv 0$.

Solution

Apply Stokes’s theorem to $\nabla V$ over any open surface $S$ bounded by $C$: ${\int }_S[\nabla \times (\nabla V)]\cdot d\vec{s}={\oint }_C\nabla V\cdot d\vec{l}$ Since $\nabla V\cdot d\vec{l}=dV$, we have ${\int }_A^B\nabla V\cdot d\vec{l}=V(B)-V(A)$. On a closed contour $A=B$: ${\oint }_C\nabla V\cdot d\vec{l}=V(A)-V(A)=0$ This holds for every surface $S$, so the integrand vanishes identically: $\nabla \times (\nabla V)\equiv 0■$ Consequence: any irrotational field can be written as $-\nabla V$.

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Question 8 — Null Identity II: Divergence of a Curl

Prove that $\nabla \cdot (\nabla \times \vec{A})\equiv 0$.

Solution

Apply the Divergence Theorem to $\nabla \times \vec{A}$ over volume $V$ bounded by closed surface $S$: ${\int }_V\nabla \cdot (\nabla \times \vec{A})dv={\oint }_S(\nabla \times \vec{A})\cdot d\vec{s}$ Split $S$ into two open halves $S_1$, $S_2$ sharing the same rim $C$. By Stokes’s theorem each contributes ${\oint }_C\vec{A}\cdot d\vec{l}$, but their outward normals are opposed, so the two contours are traversed in opposite senses and cancel: ${\oint }_S(\nabla \times \vec{A})\cdot d\vec{s}=0$ True for any $V$, so the integrand is identically zero: $\nabla \cdot (\nabla \times \vec{A})\equiv 0■$ Consequence: any solenoidal field can be written as $\nabla \times \vec{A}$ — this licenses the vector magnetic potential.

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Question 9 — Field Concept vs Action-at-a-Distance 🔥 (2024)

Why does the classical “action-at-a-distance” picture fail in dynamic electromagnetics?

Solution

Action-at-a-distance assumes force is transmitted instantaneously ($u=\infty$). In reality electromagnetic disturbances propagate at the finite speed $u=1/\sqrt{\mu ϵ}\le c\approx 3\times 10^8$ m/s. Consequences:

1. If a charge moves, a distant charge does not feel the change until a retardation time $R/u$ has elapsed.
2. During that delay, energy and momentum must reside somewhere — in the field itself, not in the charges.
3. Newton’s third law is not satisfied instantaneously between the two charges; the field carries the difference.

The field concept resolves this: a charge alters the state of the space around it, and that field is the physical agent that exerts force on other charges.

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Question 10 — Inadequacy of Circuit Theory 🔥 (2018, 2019, 2021)

Under what physical conditions does circuit theory break down? Give two examples.

Solution

Inadequacy 1 — Finite propagation delay. KVL/KCL assume instantaneous signal propagation. Real signals travel at $u=1/\sqrt{\mu ϵ}$. When the circuit dimension $d$ becomes comparable to $\lambda =u/f$, different parts of the same conductor sit at different phases and KVL/KCL fail.

• Example: at 50 Hz, $\lambda =6000$ km, so a 0.3 m circuit has $d/\lambda \approx 5\times 10^{-8}$ — circuit theory is exact. At 3 GHz, $\lambda =10$ cm, so a 10 cm PCB trace spans a full wavelength and its two ends can be $180^{\circ }$ out of phase.

Inadequacy 2 — Radiation and parasitic coupling. Circuit theory assumes energy stays inside the wires. At high frequency the fields detach and radiate, and stray capacitive/inductive coupling appears between traces.

• Example: a half-wave dipole at 900 MHz radiates nearly all its input power, yet its circuit model — a short wire — predicts almost no loss. The missing power is the radiation resistance, a purely field-theoretic quantity.

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Question 11 — Quasi-Static Conditions 🔥 (2020)

What is implied by “quasi-static conditions”?

Solution

A system is quasi-static when the frequency is low enough (or dimensions small enough) that time-variation of the fields is negligible, so static formulas remain accurate approximations even though the sources vary with time.

• Condition: $d\ll \lambda$, equivalently $f\ll u/d$.
• $\partial \vec{B}/\partial t$ is negligible, so $\nabla \times \vec{E}\approx 0$ and a single-valued potential still exists, $\vec{E}\approx -\nabla V$.
• Phase delays across conductors are ignored, so KVL/KCL remain valid.
• Not the same as static: the sources are time-varying; we merely neglect retardation.

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Question 12 — Solenoidal vs Irrotational Fields

Define both, with their conditions, representations and EM examples.

Solution

	Solenoidal	Irrotational
Condition	$\nabla \cdot \vec{F}=0$	$\nabla \times \vec{F}=0$
Meaning	No sources/sinks; lines close on themselves	Zero circulation; conservative
Representation	$\vec{F}=\nabla \times \vec{A}$	$\vec{F}=-\nabla V$
EM example	Magnetic flux density $\vec{B}$	Static electric field $\vec{E}$

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Question 13 — Homogeneous, Linear, Isotropic 🔥 (2015, 2017, 2019, 2023, 2024)

Define homogeneous, linear and isotropic media with their mathematical conditions.

Solution

• Homogeneous: properties do not vary with position. $ϵ,\mu ,\sigma \ne f(x,y,z)$. Counter-example: the ionosphere.
• Linear: response proportional to excitation. $\vec{D}=ϵ\vec{E}$, $\vec{B}=\mu \vec{H}$ with $ϵ,\mu$ independent of field magnitude. Counter-example: a saturating ferromagnetic core.
• Isotropic: properties identical in all directions; $ϵ$ is a scalar and $\vec{D}\parallel \vec{E}$. Counter-example: quartz, where $ϵ$ is a $3\times 3$ tensor.

A medium that is all three is called a simple medium.

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Question 14 — Conductor or Insulator? 🔥 (2020)

State the condition under which the same medium acts as a good conductor versus a good insulator.

Solution

Everything follows from the loss tangent: $\tan {\delta }_c=\frac{|{\vec{J}}_c|}{|{\vec{J}}_d|}=\frac{\sigma }{\omega ϵ}$

• Good conductor: $\sigma \gg \omega ϵ$, so $\tan {\delta }_c\gg 1$; conduction current dominates.
• Good insulator: $\sigma \ll \omega ϵ$, so $\tan {\delta }_c\ll 1$; displacement current dominates.

Because $\omega$ appears, no material is permanently either — sea water is a good conductor at 1 kHz and a lossy dielectric at 10 GHz.

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Question 15 — Master Numerical 🔥 (2022, 2025)

For $\vec{A}=(3y-z){\hat{a}}_x+2xz{\hat{a}}_y$ (Wb/m) in free space, find $\nabla \cdot \vec{A}$, and $\vec{A},\vec{B},\vec{H},\vec{J}$ at $P(3,-1,2)$.

Solution

Divergence: $\nabla \cdot \vec{A}=\frac{\partial (3y-z)}{\partial x}+\frac{\partial (2xz)}{\partial y}+0=0$ — solenoidal, so $\vec{A}$ is a valid vector magnetic potential (Coulomb gauge).

$\vec{A}$ at $P$: ${\vec{A}}_P=(3(-1)-2){\hat{a}}_x+2(3)(2){\hat{a}}_y=-5{\hat{a}}_x+12{\hat{a}}_y$ Wb/m.

$\vec{B}=\nabla \times \vec{A}$: $\vec{B}=-2x{\hat{a}}_x-{\hat{a}}_y+(2z-3){\hat{a}}_z⟹{\vec{B}}_P=-6{\hat{a}}_x-{\hat{a}}_y+{\hat{a}}_z\text{T}$

$\vec{H}$: free space, so ${\vec{H}}_P={\vec{B}}_P/{\mu }_0=(-4.77{\hat{a}}_x-0.796{\hat{a}}_y+0.796{\hat{a}}_z)\times 10^6$ A/m.

$\vec{J}=\nabla \times \vec{H}=(\nabla \times \vec{B})/{\mu }_0=0$ — $P$ lies in a current-free region.

Check: $\nabla \cdot \vec{B}=-2+0+2=0$ ✓

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Score Yourself

Score	Verdict
13–15	Chapter 1 is exam-ready. Move to 02 Chapter Map - Electrostatics & Boundary Conditions.
9–12	Solid. Re-read the notes for the questions you missed.
5–8	Read 1.02 Vector Operators (Gradient, Divergence, Curl & Laplacian) and 1.03 Integral Theorems (Divergence Theorem & Stokes_s Theorem) & Identity Proofs properly before proceeding.
0–4	Start from 1.01 Spatial Coordinates & Differential Elements (Cartesian, Cylindrical, Spherical) and work through in order.

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Back to: 01 Chapter Map - Vector Calculus & Fundamentals