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

1.8.31 · D3 · Physics › Electromagnetism › Maxwell's equations — integral form, all four

Yeh parent note on Maxwell's four integral laws ka worked-examples deep dive hai. Hum yahan charon laws ko re-derive nahi karenge — balki unhe har tarah ke input ke against stress-test karenge: har sign, zero cases, degenerate (broken-symmetry) cases, limiting values, ek real-world word problem, aur ek exam-style twist.

Agar koi bhi symbol unfamiliar lage, toh prerequisites yahan hain: Gauss's Law, Faraday's Law and Lenz's Law, Displacement Current, Divergence and Curl, aur Capacitors. Hinglish companion yahan hai.

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The scenario matrix

Is topic ka har problem neeche diye gaye cells mein se ek hai. Charon laws sirf do questions poochhti hain — "kitna flux ander ghusta hai?" ya "field kitna loop karta hai?" — lekin jawab drastically badal jaata hai symmetry, sign, aur time mein change hone ya na hone ke basis par.

Cell	Kya cheez ise alag banati hai	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 $=0$ field hone ke bawajood	Gauss-E	(C)
C4 No-monopole case	closed surface, $\vec{B}$ flux hamesha $0$	Gauss-B	(D)
C5 Sign of induced EMF (growing vs shrinking flux)	Lenz ka minus sign	Faraday	(E)
C6 Rotating loop → limiting/peak behaviour	ek angle se flux, sabhi 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: kaunsa surface bound karte ho?	woh trap jisne Maxwell ka fix force kiya	Ampère–Maxwell	(I)

Hum har cell cover karenge. Chalo shuru karte hain.

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C1 — Positive source (flux out)

1. Law likho. ${\oint }_S\vec{E}\cdot d\vec{A}=\frac{Q_{enc}}{{\epsilon }_0}$. Yeh step kyun? Hum flux chahte hain ek closed surface ke through jiske andar charge hai — yeh exactly Gauss ka kaam hai, koi integration nahi chahiye.
2. Enclosed charge plug karo. $Q_{enc}=+3\times 10^{-6}\text{C}$, ${\epsilon }_0=8.854\times 10^{-12}\text{F/m}$. ${\Phi }_E=\frac{3\times 10^{-6}}{8.854\times 10^{-12}}\approx 3.39\times 10^5{\text{N⋅m}}^2/\text{C}.$ Yeh step kyun? Gauss ka right-hand side literally "charge over ${\epsilon }_0$" hai — koi geometry enter nahi karta.
3. Surface par field ke liye geometry chahiye. Spherical symmetry se $E$ sphere par constant hai, toh ${\Phi }_E=E(4\pi r^2)$: $E=\frac{{\Phi }_E}{4\pi r^2}=\frac{3.39\times 10^5}{4\pi (0.20{)}^2}\approx 6.74\times 10^5\text{N/C}.$ Yeh step kyun? Flux = field × area tabhi hota hai jab field uniform aur perpendicular ho — sphere precisely isliye choose kiya jaata hai.

Verify: RHS ke units: $\text{C}/(\text{F/m})=\text{C}\cdot \text{m}/\text{F}=\text{V⋅m}={\text{N⋅m}}^2/\text{C}$ ✓. Aur ${\Phi }_E=E\cdot 4\pi r^2=6.74\times 10^5\times 4\pi (0.2{)}^2\approx 3.39\times 10^5$ ✓ — loop close ho gaya.

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C2 — Negative source (flux in)

1. Same law, naya sign. ${\Phi }_E=\frac{Q_{enc}}{{\epsilon }_0}=\frac{-3\times 10^{-6}}{8.854\times 10^{-12}}\approx -3.39\times 10^5{\text{N⋅m}}^2/\text{C}.$ Yeh step kyun? Surface mein kuch nahi badla — sirf source ka sign badla. Gauss $Q$ mein linear hai.
2. Sign interpret karo. Negative flux matlab net lines surface mein ghus rahi hain: negative charge ek sink hai, arrows andar point karte hain. Yeh step kyun? Sign ki poori value physical hai, cosmetic nahi — yeh field ki direction batata hai.

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

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

1. Gauss apply karo $Q_{enc}=0$ ke saath. ${\oint }_S\vec{E}\cdot d\vec{A}=\frac{0}{{\epsilon }_0}=0.$ Yeh step kyun? Gauss sirf surface ke andar ke charge ki parwah karta hai. Baahir ka charge yahan kuch enclose nahi karta.
2. Obvious field se reconcile karo. Baahre charge se jo bhi field line ek face se enter karti hai, woh doosre face se exit karni chahiye — ins aur outs exactly cancel ho jaate hain. Yeh step kyun? Yahi geometric reason hai ki law external charge ko ignore kar sakta hai (parent-note mistake box dekho).

Recall "field ≠ 0" lekin "flux = 0" contradiction kyun nahi hai

Question ::: Har face par field nonzero ho sakta hai phir bhi net flux zero kyun hai? Answer ::: Flux ek signed sum hai; enter karna negative contribute karta hai, leave karna positive, aur baahri source ke liye dono hamesha equal aur opposite hote hain.

Verify: $\Phi =Q_{enc}/{\epsilon }_0=0$ kisi bhi nonzero ${\epsilon }_0$ ke liye ✓.

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C4 — No-monopole case

1. Gauss-B likho. ${\oint }_S\vec{B}\cdot d\vec{A}=0$ — hamesha, koi exception nahi. Yeh step kyun? Koi magnetic monopole nahi hain, toh RHS par kabhi koi "enclosed magnetic charge" appear nahi ho sakta.
2. Dekhte hain. $\vec{B}$ lines closed loops hain: sphere ke top se nikalne wali har line curve karke bottom se wapas enter karni chahiye. Yeh step kyun? Closed loops kisi bhi surface ke liye equal in-and-out flux guarantee karte hain, degenerate cases mein bhi.

Verify: RHS identically $0$ hai; kuch plug karna hi nahi — law unconditional hai.

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C5 — Induced EMF ka sign (Lenz)

1. Pehle flux. ${\Phi }_B=B\cdot A=B\cdot \pi R^2$ (field uniform, loop flat aur perpendicular). Yeh step kyun? Faraday ko loop se bounded open surface ke through flux chahiye; disc sabse simple aisi surface hai.
2. Differentiate karo. Sirf $B$ badal raha hai, toh $\text{EMF}=-\frac{d{\Phi }_B}{dt}=-\pi R^2\frac{dB}{dt}=-\pi (0.10{)}^2(0.50)\approx -1.57\times 10^{-2}\text{V}.$ Yeh step kyun? Chain rule ek time-varying factor pick karta hai; minus sign Lenz's law hai aur hum ise rakhte hain.
3. Direction padhte hain. Page se baahir flux badh raha hai, toh induced current ko page mein andar flux create karna chahiye taaki oppose kare → clockwise (front se dekha toh). Yeh step kyun? Negative sign decoration nahi hai — yeh physical current direction dictate karta hai.

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

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C6 — Rotating loop, sabhi quadrants aur limiting peaks

1. Angle ke saath flux. ${\Phi }_B=BA\cos \theta =BA\cos (\omega t)$. Yeh step kyun? Sirf $\vec{B}$ ka normal ke saath component loop ko pierce karta hai; woh projection ek cosine hai (s02 dekho).
2. Angle ke through differentiate karo. $\text{EMF}=-\frac{d}{dt}\left(BA\cos \omega t\right)=BA\omega \sin (\omega t).$ Yeh step kyun? Chain rule $-\omega \sin$ neeche laata hai, aur outer minus usse $+\sin$ mein flip karta hai.
3. Peak value. $\sin$ 1 par max hota hai: ${\text{EMF}}_{\max }=BA\omega =0.4\times 0.02\times 100=0.80\text{V}.$ Yeh step kyun? Limiting behaviour: extreme case tumhe AC generator ke output ka amplitude batata hai.
4. $\theta$ ke sabhi chaar quadrants:
   • $\theta$ near $0$ (loop field ka samna kare): 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. Yeh step kyun? Har quadrant cover karne se counter-intuitive sach saamne aata hai: EMF tab sabse bada hota hai jab flux sabse tezi se change ho raha ho, yaani jab loop edge-on ho, face-on nahi.

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, koi moving charge nahi

1. Gap mein electric flux. ${\Phi }_E=EA$ (uniform field, plate as the open surface). Yeh step kyun? Ampère–Maxwell ka naya term $d{\Phi }_E/dt$ chahta hai; pehle ${\Phi }_E$ banana hoga.
2. Displacement-current formula. $I_d={\epsilon }_0\frac{d{\Phi }_E}{dt}={\epsilon }_{0}A\frac{dE}{dt}=8.854\times 10^{-12}\times 0.01\times 2\times 10^9\approx 1.77\times 10^{-4}\text{A}.$ Yeh step kyun? Yahi woh term hai jo Maxwell ne add kiya; yeh $\vec{B}$ produce karta hai exactly waisa hi jaisa ek wire current karta, bina kisi charge ke transit mein.

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. $N$ turns ke liye Faraday. Har turn same $d\Phi /dt$ dekhta hai, aur woh series mein add hote hain: $\text{EMF}=-N\frac{d{\Phi }_B}{dt}.$ Yeh step kyun? $N$ loops stacked matlab $N$ times ki circulation — isliye transformers mein bahut saare turns use hote hain.
2. Rate of change (linear, toh $\Delta$ use karo). $\frac{d{\Phi }_B}{dt}=\frac{(2.0-8.0)\times 10^{-4}}{3.0\times 10^{-3}}=-0.20\text{Wb/s}.$ Yeh step kyun? Linear change matlab average rate = total change / total time; koi calculus nahi chahiye.
3. Combine karo. $\text{EMF}=-500\times (-0.20)=+100\text{V}.$ Yeh step kyun? Do minus signs (Lenz sign aur falling flux) ek positive result dete hain — coil push karta hai vanishing flux ko maintain karne ke liye.

Verify: $\text{EMF}=-500\cdot (-0.20)=100$ V; units $\text{Wb/s}=\text{V}$ ✓. Yeh volts hain, mains-scale transformers se match karta hai.

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C9 — Exam twist: kaunsa surface bound karte ho?

1. Surface (a): sirf real current. $\oint \vec{B}\cdot d\vec{\ell }={\mu }_0{I}_{enc}+{\mu }_0{\epsilon }_0\frac{d{\Phi }_E}{dt}={\mu }_0(2.0)+0={\mu }_0\cdot \mathrm{2.0.}$ Yeh step kyun? Wire is disc ko pierce karti hai, toh $I_{enc}=2.0$ A; isse koi changing $E$-flux nahi nikalta.
2. Surface (b): sirf displacement current. Isse koi wire cross nahi karti, toh $I_{enc}=0$. Lekin yeh plates ke beech lie karta hai jahan $E$ badh raha hai, aur parent note ke example (C) se $I_d={\epsilon }_{0}d{\Phi }_E/dt=dQ/dt=I=2.0\text{A}$: $\oint \vec{B}\cdot d\vec{\ell }={\mu }_0(0)+{\mu }_0{\epsilon }_0\frac{d{\Phi }_E}{dt}={\mu }_0\cdot \mathrm{2.0.}$ Yeh step kyun? Displacement current exactly woh pick up karta hai jo wire ne drop kiya — dono surfaces agree karte hain.
3. Numeric value. ${\mu }_0=4\pi \times 10^{-7}\text{T⋅m/A}$: $\oint \vec{B}\cdot d\vec{\ell }=4\pi \times 10^{-7}\times 2.0\approx 2.51\times 10^{-6}\text{T⋅m}.$ Yeh step kyun? Ek number, do independent tareekon se rakha gaya — yahi consistency ki wajah hai ki Maxwell ka fix zaroori hai, optional nahi.

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

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Recall

Recall Scenario ko law se match karo (self-test)

Charge ek closed surface ke baahir — net flux? ::: Zero (C3): field lines equally enter aur exit karti hain. Loop ka flux page se baahir badh raha hai — induced current direction? ::: Clockwise, growth ko oppose karne ke liye (C5, Lenz). Rotation mein generator EMF kab sabse bada hota hai? ::: Jab loop edge-on ho (flux zero, sabse tezi se change ho raha ho) — C6. Ek capacitor gap mein koi charge nahi; kya wahan $\vec{B}$ zero hai? ::: Nahi — displacement current ${\epsilon }_{0}d{\Phi }_E/dt$ $\vec{B}$ banata hai (C7). Ek Amperian loop par do alag surfaces kyun agree karte hain? ::: Real current + displacement current hamesha same total sum karte hain (C9).