Exercises — Maxwell's equations — integral form, all four

(a) Gauss's Law for Electricity — closed surface, enclosed charge. Dekho Gauss's Law. (b) Gauss's Law for Magnetism ($\oint \vec{B}\cdot d\vec{A}=0$, no monopoles). (c) Faraday's Law — changing magnetic flux se circulating $\vec{E}$ banta hai. Dekho Faraday's Law and Lenz's Law. (d) Ampère–Maxwell Law — gap mein current term zero hai, toh changing-electric-flux term kaam karta hai (the Displacement Current).

Gauss: $\Phi =\frac{Q_{enc}}{{\epsilon }_0}=\frac{3.0\times 10^{-6}}{8.85\times 10^{-12}}=3.39\times 10^5\text{N}\cdot {\text{m}}^2/\text{C}$. Bahar wala charge kuch nahi badalta: uski lines closed surface mein enter karti hain aur exit karti hain, net contribution zero. Sirf enclosed charge count hota hai.

Cylinder kyun? Symmetry se $\vec{E}$ seedha bahar (radially) point karta hai aur curved wall par everywhere same size hai; flat caps par $\vec{E}$ surface ke parallel hai toh uska flux zero hai. Toh saara flux curved side se aata hai jiska area $2\pi rL$ hai. $\oint \vec{E}\cdot d\vec{A}=E(2\pi rL)=\frac{Q_{enc}}{{\epsilon }_0}=\frac{\lambda L}{{\epsilon }_0}.$ Length $L$ cancel ho jaata hai — theek hai, ek infinite line ki koi special length nahi hoti: $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}=\frac{2.0\times 10^{-8}}{2\pi (8.85\times 10^{-12})(0.050)}=7.19\times 10^3\text{N/C}.$

Flux ${\Phi }_B=B\cdot (\pi R^2)$ (field $\perp$ loop hai, toh pura area count hota hai). $|\text{EMF}|=\left|\frac{d{\Phi }_B}{dt}\right|=\pi R^2\frac{dB}{dt}=\pi (0.10{)}^2(0.30)=9.42\times 10^{-3}\text{V}.$ Yahan sirf $B$ change ho raha hai, area fixed hai — isliye $\pi R^2$ bahar aa jaata hai.

Plate area $A$ ke saath, field $E=Q/({\epsilon }_{0}A)$ hai, toh ${\Phi }_E=EA=Q/{\epsilon }_0$. Phir $I_d={\epsilon }_0\frac{d{\Phi }_E}{dt}={\epsilon }_0\cdot \frac{1}{{\epsilon }_0}\frac{dQ}{dt}=\frac{dQ}{dt}=4.0\text{A}.$ Displacement current bilkul wire current ke barabar hai — current gap se bhi continuous rahti hai.

Plates ke beech koi conduction current nahi hai, toh displacement-current term use karo. Displacement current plate area mein uniformly spread hai, toh radius $r$ ke loop ke andar jo fraction enclosed hai wo $\frac{\pi r^2}{\pi R^2}=\frac{r^2}{R^2}$ hai. Circular loop ke around Ampère–Maxwell (jahan $\vec{B}$ tangential aur constant hai): $B(2\pi r)={\mu }_0{I}_{d,\text{enc}}={\mu }_{0}I\frac{r^2}{R^2}.$ $B=\frac{{\mu }_{0}Ir}{2\pi R^2}=\frac{(4\pi \times 10^{-7})(4.0)(0.010)}{2\pi (0.020{)}^2}=2.0\times 10^{-5}\text{T}.$

${\Phi }_B=B{s}^2$. Sirf area change ho raha hai, toh chain rule se $\text{EMF}=-\frac{d}{dt}(B{s}^2)=-2Bs\frac{ds}{dt}=-2(0.50)(0.20)(-0.10)=+0.020\text{V}.$ Magnitude $=0.020\text{V}$. Kyunki $ds/dt<0$ (shrink ho raha hai), do minus signs milke EMF positive banate hain, matlab induced current decreasing out-of-page flux se lad rhi hai by trying to add out-of-page flux (counterclockwise current, Lenz ke hisaab se).

(a) Total displacement current ke liye poora field aur poora area use hoga: $I_d={\epsilon }_0\frac{d{\Phi }_E}{dt}={\epsilon }_0(\pi R^2)\frac{dE}{dt}=(8.85\times 10^{-12})\pi (0.020{)}^2(1.0\times 10^{12})=1.11\times 10^{-2}\text{A}.$ (b) $r=R$ par loop saara $I_d$ enclose karta hai: $B=\frac{{\mu }_0{I}_d}{2\pi R}=\frac{(4\pi \times 10^{-7})(1.11\times 10^{-2})}{2\pi (0.020)}=1.11\times 10^{-7}\text{T}.$ Ye chota sa $\vec{B}$, jo purely changing $\vec{E}$ se bana hai bina kisi moving charge ke, bilkul wahi ingredient hai jo Electromagnetic Waves ko exist karne deta hai.

Ampère–Maxwell: ek changing $\vec{E}$ ($y$ along) $\vec{B}$ ko apne around curl karta hai. Wave ko $+x$ mein energy carry karni hai, toh field pattern ke liye $\vec{B}$ $\pm z$ along hona chahiye. Faraday: wo changing $\vec{B}$ ($z$ along) phir apni taraf ek $\vec{E}$ ko $y$ along curl karta hai. Dono laws ek self-consistent loop tab close karte hain jab $\vec{E}\times \vec{B}$ $+x$ (travel direction) along point kare — toh $\vec{E}$ $+y$ along hone se $\vec{B}$ $+z$ along force hota hai. Consistency check wave condition se: dono laws simultaneously satisfy hote hain jab $E=cB$ aur $\vec{E}\perp \vec{B}\perp \hat{x}$ ho. Koi contradiction nahi — ye same wave ke do aadhe hain.

Faraday kehta hai wiggling $\vec{B}$ se circulating $\vec{E}$ banta hai; Ampère–Maxwell kehta hai wiggling $\vec{E}$ se circulating $\vec{B}$ banta hai. Ek ko doosre mein feed karo ek plane wave ke liye aur ek wave equation milti hai jiska speed sirf coupling constants se set hoti hai: $c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}.$ Numerically: $c=\frac{1}{\sqrt{(4\pi \times 10^{-7})(8.85\times 10^{-12})}}=3.00\times 10^8\text{m/s}.$ Ye measured speed of light se match karta hai — wahi punchline jo Maxwell ko batayi ki light ek electromagnetic wave hai. Dekho Electromagnetic Waves.

Interior field: $B={\mu }_{0}nI$, toh $\frac{dB}{dt}={\mu }_{0}n\frac{dI}{dt}$. Loop ke through flux: ${\Phi }_B=B\pi r^2$ (andar uniform field hai). $|\text{EMF}|=\pi r^2\frac{dB}{dt}=\pi r^2{\mu }_{0}n\frac{dI}{dt}.$ $=\pi (0.030{)}^2(4\pi \times 10^{-7})(1000)(50)=1.78\times 10^{-4}\text{V}.$

$B=\frac{E}{c}=\frac{300}{3.00\times 10^8}=1.0\times 10^{-6}\text{T}.$ Note karo ki SI units mein $\vec{B}$, $\vec{E}$ ke comparison mein bahut chota hai — ye sirf factor $c$ ki wajah se hai, ye iska sign nahi ki magnetism "weaker" hai. Wave mein dono halves equal energy carry karte hain.

Kaun se law ki derivation se $c=1/\sqrt{{\mu }_0{\epsilon }_0}$ milta hai? ::: Faraday aur Ampère–Maxwell ka combination — do "changing-field" laws jo ek doosre ko feed karte hain.