Question bank — Maxwell's equations — integral form, all four

False — uski field lines surface mein enter aur exit karti hain, isliye unka net flux contribution exactly zero hai; sirf enclosed charge $Q_{enc}$ hi Gauss's Law mein count hota hai.

False — zero net flux ka sirf matlab hai $Q_{enc}=0$; andar $+$ aur $-$ charge ki equal amounts, ya ek external field, surface par strong $\vec{E}$ de sakti hai jisme lines andar jaati hain aur wapas baahir aati hain.

False — yeh hamesha zero hota hai kyunki koi magnetic monopoles nahi hote; $\vec{B}$ lines kisi bhi current ke bawajood closed loops hoti hain.

False — ek steady $\vec{B}$ kuch nahi banata; sirf ek changing flux $d{\Phi }_B/dt$ hi loop $C$ ke around ek circulating $\vec{E}$ drive karta hai.

False — minus sign hi direction rule hai (Lenz's law), aur yeh sirf tab meaningful hota hai jab $d\vec{\ell }$ aur $d\vec{A}$ right-hand rule se paired hon; ise drop karne par induced currents apni khud ki cause ko reinforce karne lagte hain, jo energy conservation violate karta hai.

False — koi charge gap cross nahi karta; $I_d={\epsilon }_{0}d{\Phi }_E/dt$ ek changing electric flux hai jo ek current ki tarah magnetic field produce karta hai.

False — Gauss ek closed surface (ek bag) use karta hai jo $Q_{enc}$ trap karta hai; Ampère–Maxwell ek open surface use karta hai jo loop $C$ se bounded hoti hai.

True — $I_d={\epsilon }_{0}d{\Phi }_E/dt=dQ/dt=I$, isliye "current" gap ke through continuous rehta hai chahe wahan koi charge move na kare. Dekho Capacitors.

True — $Q_{enc}=0$, isliye ${\oint }_S\vec{E}\cdot d\vec{A}=0$; field zero nahi hai, lekin $+$ se har line bag ke andar $-$ par khatam hoti hai.

Cube symmetry destroy kar deta hai — $\vec{E}$ na to constant hai na hi $d\vec{A}$ ke perpendicular hai apni faces par, isliye ${\oint }_S\vec{E}\cdot d\vec{A}$ unsolvable ho jaata hai. Ek sphere choose karo taaki $E$ surface par constant aur radial (i.e., $d\vec{A}$ ke parallel) ho.

Ek sphere cylindrical symmetry se match nahi karta, isliye $\vec{E}\cdot d\vec{A}$ uske upar unpredictably vary karta hai; ek coaxial cylinder use karo jahan flux curved side par uniform ho aur flat ends par zero ho.

EMF hai $-d{\Phi }_B/dt$, flux ka rate of change, flux khud nahi. Ek still loop se steady field $d{\Phi }_B/dt=0$ deta hai, isliye EMF zero hai.

Yeh charging capacitor ke liye fail karta hai: wire ke through ek flat surface $I_{enc}=I$ deti hai, lekin usi loop $C$ se bounded ek bulged surface jo plates ke beech se jaati hai koi current nahi pakadti — contradiction, sirf ${\mu }_0{\epsilon }_{0}d{\Phi }_E/dt$ add karke fix hota hai.

Integral ek closed surface par hai aur sirf yeh kehta hai ki lines in = lines out; field khud bahut nonzero hai — flux cancel hota hai, field nahi.

Ek baar launch hone ke baad, changing $\vec{E}$ ek circulating $\vec{B}$ ko feed karta hai aur vice versa (Faraday + Ampère–Maxwell), isliye ek EM wave khud ko empty space mein sustain karti hai bina kisi charge ke.

Field strength $1/r^2$ ki tarah girta hai jabki sphere area $r^2$ ki tarah badhta hai, isliye unka product (flux ${\Phi }_E$) $r$-independent hai — yahi wajah hai ki Gauss's law kisi bhi radius ke liye hold karta hai.

Woh circulation ${\oint }_C$ ko flux ke barabar set karte hain us surface ke through jo loop $C$ bound karta hai, right-hand rule ke saath $d\vec{\ell }$ ko $d\vec{A}$ se pair karke; ek closed surface ka koi bounding loop nahi hota, isliye koi $C$ nahi hai aur koi consistent normal pair karne ke liye nahi — equation ke left side par kuch nahi hoga.

Real current capacitor plate par ruk jaata hai, lekin growing $\vec{E}$ exactly ${\epsilon }_{0}d{\Phi }_E/dt=dQ/dt$ supply karta hai, isliye loop $C$ se bounded kisi bhi surface ko thread karne wala total current same rehta hai.

Ek charge ka field baahir ki taraf point karta hai aur loop ke around zero net work karta hai; induced field circulate karta hai (${\oint }_C\vec{E}\cdot d\vec{\ell }\ne 0$), har charge par real work karta hai — yahi EMF hai.

Har law ka changing field doosre ko drive karta hai; unhe combine karne par ek wave equation force hoti hai jiska speed ${\mu }_0$ aur ${\epsilon }_0$ constants se fixed hoti hai, $c=1/\sqrt{{\mu }_0{\epsilon }_0}$ deta hai.

Koi isolated magnetic pole abhi tak nahi mila, isliye andar koi magnetic charge nahi hai; right side structurally zero hai.

${\Phi }_B=B\times \text{area}=0$ har time ke liye, isliye $d{\Phi }_B/dt=0$ aur EMF zero hai — koi area nahi, toh koi flux change karne ke liye nahi.

Nahi — $\vec{E}$ constant hai, isliye $d{\Phi }_E/dt=0$ aur $I_d=0$; displacement current vanish ho jaata hai aur koi $\vec{B}$ circulate nahi karta.

Zero — ek uniform field mein utni hi lines enter hoti hain jitni exit, isliye ${\Phi }_E=0$, jo Gauss's law se matlab hai koi enclosed charge nahi.

Surface integral ek mathematical point par ambiguous hai, isliye honest fix yeh hai ki surface ko infinitesimally nudge karo: agar tum ise deform karo ki charge just enclose ho toh $q/{\epsilon }_0$ milta hai, aur just exclude ho toh $0$. Ek smooth surface ke liye symmetric answer half hai, $q/(2{\epsilon }_0)$ — lekin hamesha ambiguity resolve karo yeh decide karke ki charge kis side hai, use kabhi surface par mat chhoddo.

Nahi — yeh ek magnetic monopole hoga; Gauss's law for magnetism ise forbid karta hai, isliye line khud par close honi chahiye.

Haan — ek changing $\vec{B}$ space mein khud ek real circulating $\vec{E}$ produce karta hai; wire sirf ise reveal karta hai current carry karke, yeh ise create nahi karta.

Haan — integral law us instant jo bhi charge enclosed hai uske liye exact hai. Subtlety yeh hai ki surface par field pattern instantaneous Coulomb field nahi hai: changes speed $c$ par propagate hote hain (retarded fields), isliye local $\vec{E}$ charge se lag karta hai — phir bhi woh retardation effects flux ko rearrange karte hain taaki total abhi bhi exactly $Q_{enc}/{\epsilon }_0$ ho.

Haan — wire current ko plates ke beech changing $\vec{E}$ se displacement current replace kar deta hai, same answer deta hai; yahi consistency Maxwell ka poora point tha.