Exercises — Displacement current — Maxwell's addition to Ampere's law

Sirf (b), ek changing electric field. Maxwell ki insight yeh hai ki $I_d={\epsilon }_{0}d{\Phi }_E/dt$ — flux ka rate of change, field khud nahi. Ek steady field mein $d{\Phi }_E/dt=0$ hota hai, isliye $I_d=0$; ek stationary charge mein bilkul bhi current nahi hota.

Jhooth. Koi bhi charge gap cross nahi karta — electrons ek plate par pile ho jaate hain aur doosri plate chhod dete hain. Jo conceptually cross hota hai woh growing field ka effect hai, jise $I_d$ capture karta hai. Yahi poora reason hai ki Maxwell ko sirf "more current" ki jagah ek nayi term ki zaroorat thi.

Sirf $I_d$. Conduction current $I_c$ wahan zero hai (gap mein se koi charge flow nahi karta), lekin us surface se electric flux grow ho raha hai, isliye $I_d={\epsilon }_{0}d{\Phi }_E/dt\ne 0$. Us surface par poora magnetic field $I_d$ se aata hai.

Kyunki $Q=CV$ aur $I_d=I_c=\frac{dQ}{dt}=C\frac{dV}{dt}$: $I_d=(2\times 10^{-6})(10^6)=2\text{A}.$ Dhyaan do ki humne field ko kabhi touch nahi kiya — kyunki $I_d$ hamesha capacitor ko feed karne wale conduction current ke barabar hota hai.

Seedha definition se: $I_d={\epsilon }_0\frac{d{\Phi }_E}{dt}=(8.85\times 10^{-12})(4\times 10^6)=3.5\times 10^{-5}\text{A}.$ Kahin bhi koi charges appear nahi hote — yeh pure-vacuum form hai jo ek light wave ke andar bhi kaam karta rehta hai.

Flux hai ${\Phi }_E=EA$ (uniform field ⟂ area ke), isliye $\frac{d{\Phi }_E}{dt}=A\frac{dE}{dt}$. Tab $I_d={\epsilon }_{0}A\frac{dE}{dt}=(8.85\times 10^{-12})(0.02)(5\times 10^9)=8.85\times 10^{-4}\text{A}.$

Symmetry (yeh KAISA dikhta hai): setup ki circular symmetry se (figure dekho), $\vec{B}$ tangential hai aur radius $r$ ke circle par constant hai, isliye $\oint \vec{B}\cdot d\vec{l}=B(2\pi r)$. Enclosed $I_d$: field plate ke across uniform hai, isliye radius $r$ ke loop se guzarne wala flux total ka $\frac{\pi r^2}{\pi R^2}$ fraction hai. Isliye $I_{d,\text{enc}}=I\frac{r^2}{R^2}.$ Law apply karo: $B(2\pi r)={\mu }_{0}I\frac{r^2}{R^2}$, isliye $B=\frac{{\mu }_{0}Ir}{2\pi R^2}=\frac{(4\pi \times 10^{-7})(2)(0.03)}{2\pi (0.05{)}^2}=4.8\times 10^{-6}\text{T}.$ Form mein bilkul waise hi jaise $B$ uniform current carry karne wale wire ke andar hota hai — woh promised symmetry.

Andar ($r<R$): sirf flux ka fraction $r^2/R^2$ enclosed hota hai, jisse $B=\frac{{\mu }_{0}Ir}{2\pi R^2}$ milta hai — $r$ ke saath linearly badhta hai. Bahar ($r>R$): loop ab poora flux enclose karta hai, isliye $I_{d,\text{enc}}=I$ aur $B(2\pi r)={\mu }_{0}I\Rightarrow B=\frac{{\mu }_{0}I}{2\pi r}$ — $1/r$ ki tarah girta hai. Maximum edge par hai $r=R$. Dono formulas wahan check karo: $B_{\text{in}}(R)=\frac{{\mu }_{0}IR}{2\pi R^2}=\frac{{\mu }_{0}I}{2\pi R},B_{\text{out}}(R)=\frac{{\mu }_{0}I}{2\pi R}.$ Dono match karte hain — field continuous hai. Numerically $B_{\max }=\frac{(4\pi \times 10^{-7})(2)}{2\pi (0.05)}=8.0\times 10^{-6}\text{T}$.

$I_d=I_c=\frac{dQ}{dt}=\frac{Q_0}{\tau }{e}^{-t/\tau }$.

• $t=0$ par: $I_d=\frac{Q_0}{\tau }=\frac{6\times 10^{-6}}{2\times 10^{-3}}=3\times 10^{-3}\text{A}=3\text{mA}$.
• Jab $t\to \infty$: $e^{-t/\tau }\to 0$, isliye $I_d\to 0$. Ek baar capacitor full ho jaane ke baad field change hona band ho jaata hai — no more displacement current, hence no more $B$.

(a) Plates ke beech field $E=\frac{Q}{{\epsilon }_{0}A}$ jahan $A=\pi R^2$. Differentiate karo: $\frac{dE}{dt}=\frac{1}{{\epsilon }_{0}A}\frac{dQ}{dt}=\frac{I}{{\epsilon }_0\pi R^2}=\frac{1.5}{(8.85\times 10^{-12})\pi (0.04{)}^2}=3.37\times 10^{13}\text{V/(m⋅s)}.$ (b) $I_d={\epsilon }_{0}A\frac{dE}{dt}={\epsilon }_0\pi R^2\cdot \frac{I}{{\epsilon }_0\pi R^2}=I=1.5\text{A}.$ Gap mein displacement current exactly wire mein conduction current ke barabar hai — paradox quantitatively dissolve ho gaya.

(i) Edge par poora current enclosed hai: $B=\frac{{\mu }_{0}I}{2\pi R}=\frac{(4\pi \times 10^{-7})(1.5)}{2\pi (0.04)}=7.5\times 10^{-6}\text{T}.$ (ii) $B(2\pi R)={\mu }_0{\epsilon }_0\frac{d{\Phi }_E}{dt}={\mu }_0{\epsilon }_0(\pi R^2)\frac{dE}{dt}$. L4.1 se $dE/dt$ use karte hue: $B=\frac{{\mu }_0{\epsilon }_{0}R}{2}\frac{dE}{dt}=\frac{(4\pi \times 10^{-7})(8.85\times 10^{-12})(0.04)}{2}(3.37\times 10^{13})=7.5\times 10^{-6}\text{T}.$ Same number — kyunki ${\epsilon }_{0}d{\Phi }_E/dt$ current $I$ hi hai.

Units: ${\mu }_0$ hai $\text{T⋅m/A}={\text{kg⋅m/A}}^2{\text{s}}^2$ aur ${\epsilon }_0$ hai ${\text{A}}^2{\text{s}}^4/({\text{kg⋅m}}^3)$. Unka product ${\mu }_0{\epsilon }_0$ ki units hain ${\text{s}}^2/{\text{m}}^2$, isliye $1/\sqrt{{\mu }_0{\epsilon }_0}$ ki units hain $\text{m/s}$ — ek speed. $\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=\frac{1}{\sqrt{(4\pi \times 10^{-7})(8.85\times 10^{-12})}}=3.00\times 10^8\text{m/s}.$ Yeh $c$ hai, Speed of Light. Wahi term jo Maxwell ne add ki ($I_d$) woh hai jo ${\epsilon }_0$ ko wave equation mein daalti hai — displacement current ke bina na $c$ hoga na light. Dekho Electromagnetic Waves.

Vacuum mein $I_c=0$ har jagah, isliye $\vec{B}$ ka ek maatra source displacement term ${\mu }_0{\epsilon }_0\partial \vec{E}/\partial t$ hai. Yeh bilkul Faraday's Law of Induction ka mirror hai (changing $\vec{B}$ $\vec{E}$ banata hai); saath milke yeh ek self-feeding wave banate hain. Numerically, $\vec{E}$ ko patch ke $\perp$ treat karte hue: $I_d={\epsilon }_{0}A\frac{dE}{dt}=(8.85\times 10^{-12})(1)(3\times 10^{11})=2.66\text{A}.$ Ek pure field change — koi electrons nahi — ek multi-amp "current" carry karta hai. Yahi displacement current hai jo bilkul apne dum par khada hai.