Worked examples — Maxwell's equations — integral form, all four

This is the worked-examples deep dive for the parent note on Maxwell's four integral laws. We will not re-derive the four laws — we will stress-test them against every kind of input they can face: every sign, the zero cases, the degenerate (broken-symmetry) cases, the limiting values, one real-world word problem, and one exam-style twist.

If any symbol here feels unfamiliar, the prerequisites live at Gauss's Law, Faraday's Law and Lenz's Law, Displacement Current, Divergence and Curl, and Capacitors. The Hinglish companion is here.

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The scenario matrix

Every problem in this topic is one of the cells below. The four laws only ever ask one of two questions — "how much flux pokes through?" or "how much does the field loop?" — but the answer changes drastically depending on symmetry, sign, and whether something is changing in time.

Cell	What makes it distinct	Law used	Example
C1 Positive source	field points outward, flux $>0$	Gauss-E	(A)
C2 Negative source	field points inward, flux $<0$	Gauss-E	(B)
C3 Zero enclosed / external charge	net flux $=0$ even with field present	Gauss-E	(C)
C4 No-monopole case	closed surface, $\vec{B}$ flux always $0$	Gauss-B	(D)
C5 Sign of induced EMF (growing vs shrinking flux)	the minus sign of Lenz	Faraday	(E)
C6 Rotating loop → limiting/peak behaviour	flux through an angle, all quadrants	Faraday	(F)
C7 Displacement current, no moving charge	Ampère–Maxwell in a gap	Ampère–Maxwell	(G)
C8 Real-world word problem	transformer-style number crunch	Faraday	(H)
C9 Exam twist: which surface do you bound?	the trap that forced Maxwell's fix	Ampère–Maxwell	(I)

We will hit every cell. Let's go.

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C1 — Positive source (flux out)

1. Write the law. ${\oint }_S\vec{E}\cdot d\vec{A}=\frac{Q_{enc}}{{\epsilon }_0}$. Why this step? We want flux through a closed surface with a charge inside — that is exactly Gauss's job, no integration needed.
2. Plug the enclosed charge. $Q_{enc}=+3\times 10^{-6}\text{C}$, ${\epsilon }_0=8.854\times 10^{-12}\text{F/m}$. ${\Phi }_E=\frac{3\times 10^{-6}}{8.854\times 10^{-12}}\approx 3.39\times 10^5{\text{N⋅m}}^2/\text{C}.$ Why this step? The right-hand side of Gauss is literally "charge over ${\epsilon }_0$" — no geometry enters.
3. Field on the surface needs the geometry. By spherical symmetry $E$ is constant over the sphere, so ${\Phi }_E=E(4\pi r^2)$: $E=\frac{{\Phi }_E}{4\pi r^2}=\frac{3.39\times 10^5}{4\pi (0.20{)}^2}\approx 6.74\times 10^5\text{N/C}.$ Why this step? Flux is field × area only when the field is uniform and perpendicular — the sphere is chosen precisely so this holds.

Verify: Units of RHS: $\text{C}/(\text{F/m})=\text{C}\cdot \text{m}/\text{F}=\text{V⋅m}={\text{N⋅m}}^2/\text{C}$ ✓. And ${\Phi }_E=E\cdot 4\pi r^2=6.74\times 10^5\times 4\pi (0.2{)}^2\approx 3.39\times 10^5$ ✓ — closes the loop.

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C2 — Negative source (flux in)

1. Same law, new sign. ${\Phi }_E=\frac{Q_{enc}}{{\epsilon }_0}=\frac{-3\times 10^{-6}}{8.854\times 10^{-12}}\approx -3.39\times 10^5{\text{N⋅m}}^2/\text{C}.$ Why this step? Nothing about the surface changed — only the sign of the source. Gauss is linear in $Q$.
2. Interpret the sign. Negative flux means net lines enter the surface: a negative charge is a sink, arrows point inward. Why this step? The whole value of the sign is physical, not cosmetic — it tells you the direction of the field.

Verify: $|{\Phi }_E|$ identical to (A), sign flipped. Ratio ${\Phi }_B/{\Phi }_A=-1$ exactly.

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C3 — Zero enclosed charge (field present, flux zero)

1. Apply Gauss with $Q_{enc}=0$. ${\oint }_S\vec{E}\cdot d\vec{A}=\frac{0}{{\epsilon }_0}=0.$ Why this step? Gauss cares only about charge inside the surface. The outside charge encloses nothing here.
2. Reconcile with the obvious field. Every field line from the outside charge that enters one face must exit another — the ins and outs cancel exactly. Why this step? This is the geometric reason the law can ignore external charge (see the parent-note mistake box).

Recall Why "field ≠ 0" but "flux = 0" is not a contradiction

Question ::: How can the field be nonzero on every face yet the net flux be zero? Answer ::: Flux is a signed sum; entering contributes negative, leaving contributes positive, and for a source outside the two are always equal and opposite.

Verify: $\Phi =Q_{enc}/{\epsilon }_0=0$ for any nonzero ${\epsilon }_0$ ✓.

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C4 — No-monopole case

1. Write Gauss-B. ${\oint }_S\vec{B}\cdot d\vec{A}=0$ — always, no exceptions. Why this step? There are no magnetic monopoles, so no "enclosed magnetic charge" can ever appear on the RHS.
2. See it. $\vec{B}$ lines are closed loops: every line leaving through the top of the sphere must curve around and re-enter through the bottom. Why this step? Closed loops guarantee equal in-and-out flux for any surface, degenerate cases included.

Verify: RHS is identically $0$; nothing to plug — the law is unconditional.

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C5 — Sign of the induced EMF (Lenz)

1. Flux first. ${\Phi }_B=B\cdot A=B\cdot \pi R^2$ (field uniform, loop flat and perpendicular). Why this step? Faraday needs the flux through the open surface bounded by the loop; the disc is the simplest such surface.
2. Differentiate. Only $B$ changes, so $\text{EMF}=-\frac{d{\Phi }_B}{dt}=-\pi R^2\frac{dB}{dt}=-\pi (0.10{)}^2(0.50)\approx -1.57\times 10^{-2}\text{V}.$ Why this step? The chain rule picks out the one time-varying factor; the minus sign is Lenz's law and we keep it.
3. Read the direction. Flux out of page is growing, so the induced current must create flux into the page to oppose it → clockwise (viewed from the front). Why this step? The negative sign is not decoration — it dictates the physical current direction.

Verify: $\text{EMF}=-\pi (0.10{)}^2(0.50)\approx -0.0157\text{V}$; units ${\text{m}}^2\cdot \text{T/s}=\text{Wb/s}=\text{V}$ ✓. Sign negative for growing flux ✓.

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C6 — Rotating loop, all quadrants and the limiting peaks

1. Flux with an angle. ${\Phi }_B=BA\cos \theta =BA\cos (\omega t)$. Why this step? Only the component of $\vec{B}$ along the normal pierces the loop; that projection is a cosine (see s02).
2. Differentiate through the angle. $\text{EMF}=-\frac{d}{dt}\left(BA\cos \omega t\right)=BA\omega \sin (\omega t).$ Why this step? The chain rule brings down $-\omega \sin$, and the outer minus flips it to $+\sin$.
3. Peak value. $\sin$ maxes at $1$: ${\text{EMF}}_{\max }=BA\omega =0.4\times 0.02\times 100=0.80\text{V}.$ Why this step? Limiting behaviour: the extreme case tells you the amplitude of the AC generator's output.
4. All four quadrants of $\theta$:
   • $\theta$ near $0$ (loop faces field): flux maximal, $\sin \approx 0$ → EMF ≈ 0.
   • $\theta$ near $90^{\circ }$ (edge-on): flux zero, $\sin =1$ → EMF peak $+0.80$ V.
   • $\theta$ near $180^{\circ }$: flux minimal (negative), $\sin \approx 0$ → EMF ≈ 0.
   • $\theta$ near $270^{\circ }$: $\sin =-1$ → EMF peak $-0.80$ V. Why this step? Covering every quadrant shows the counter-intuitive truth: EMF is biggest when flux is changing fastest, i.e. when the loop is edge-on, not face-on.

Verify: ${\text{EMF}}_{\max }=BA\omega =0.4\cdot 0.02\cdot 100=0.80$ V; units ${\text{T⋅m}}^2/\text{s}=\text{Wb/s}=\text{V}$ ✓.

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C7 — Displacement current, no moving charge

1. Electric flux in the gap. ${\Phi }_E=EA$ (uniform field, plate as the open surface). Why this step? Ampère–Maxwell's new term needs $d{\Phi }_E/dt$; we must build ${\Phi }_E$ first.
2. Displacement-current formula. $I_d={\epsilon }_0\frac{d{\Phi }_E}{dt}={\epsilon }_{0}A\frac{dE}{dt}=8.854\times 10^{-12}\times 0.01\times 2\times 10^9\approx 1.77\times 10^{-4}\text{A}.$ Why this step? This is the term Maxwell added; it produces $\vec{B}$ exactly as a wire current would, with no charge in transit.

Verify: $I_d=8.854\times 10^{-12}\cdot 0.01\cdot 2\times 10^9\approx 1.77\times 10^{-4}$ A; units $\text{F/m}\cdot {\text{m}}^2\cdot \text{V/(m⋅s)}=\text{F⋅V/s}=\text{C/s}=\text{A}$ ✓.

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C8 — Real-world word problem

1. Faraday for $N$ turns. Each turn sees the same $d\Phi /dt$, and they add in series: $\text{EMF}=-N\frac{d{\Phi }_B}{dt}.$ Why this step? $N$ loops stacked means $N$ times the circulation — this is why transformers use many turns.
2. Rate of change (linear, so use $\Delta$). $\frac{d{\Phi }_B}{dt}=\frac{(2.0-8.0)\times 10^{-4}}{3.0\times 10^{-3}}=-0.20\text{Wb/s}.$ Why this step? Linear change means average rate = total change / total time; no calculus needed.
3. Combine. $\text{EMF}=-500\times (-0.20)=+100\text{V}.$ Why this step? The two minus signs (Lenz sign and falling flux) give a positive result — the coil pushes to maintain the vanishing flux.

Verify: $\text{EMF}=-500\cdot (-0.20)=100$ V; units $\text{Wb/s}=\text{V}$ ✓. It's volts, matching mains-scale transformers.

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C9 — Exam twist: which surface do you bound?

1. Surface (a): real current only. $\oint \vec{B}\cdot d\vec{\ell }={\mu }_0{I}_{enc}+{\mu }_0{\epsilon }_0\frac{d{\Phi }_E}{dt}={\mu }_0(2.0)+0={\mu }_0\cdot \mathrm{2.0.}$ Why this step? The wire pierces this disc, so $I_{enc}=2.0$ A; there's no changing $E$-flux through it.
2. Surface (b): displacement current only. No wire crosses it, so $I_{enc}=0$. But it lies between the plates where $E$ grows, and from example (C) of the parent note $I_d={\epsilon }_{0}d{\Phi }_E/dt=dQ/dt=I=2.0\text{A}$: $\oint \vec{B}\cdot d\vec{\ell }={\mu }_0(0)+{\mu }_0{\epsilon }_0\frac{d{\Phi }_E}{dt}={\mu }_0\cdot \mathrm{2.0.}$ Why this step? The displacement current picks up exactly what the wire dropped — the two surfaces agree.
3. Numeric value. ${\mu }_0=4\pi \times 10^{-7}\text{T⋅m/A}$: $\oint \vec{B}\cdot d\vec{\ell }=4\pi \times 10^{-7}\times 2.0\approx 2.51\times 10^{-6}\text{T⋅m}.$ Why this step? One number, reached two independent ways — that consistency is the whole reason Maxwell's fix is necessary, not optional.

Verify: both surfaces give ${\mu }_0\cdot 2.0=4\pi \times 10^{-7}\cdot 2.0\approx 2.513\times 10^{-6}$ T·m, and $I_d=I=2.0$ A ✓.

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Recall

Recall Match the scenario to the law (self-test)

Charge outside a closed surface — net flux? ::: Zero (C3): field lines enter and exit equally. Loop's flux out of page is growing — induced current direction? ::: Clockwise, to oppose the growth (C5, Lenz). When in a rotation is generator EMF largest? ::: When the loop is edge-on (flux zero, changing fastest) — C6. A capacitor gap carries no charge; is $\vec{B}$ there zero? ::: No — displacement current ${\epsilon }_{0}d{\Phi }_E/dt$ makes $\vec{B}$ (C7). Why do two different surfaces on one Amperian loop agree? ::: Real current + displacement current always sum to the same total (C9).