Exercises — Ampere's circuital law — magnetostatic form

Line integral sirf enclosed current pe depend karta hai, radius pe kabhi nahi: ${\oint }_C\vec{B}\cdot d\vec{l}={\mu }_0{I}_{\text{enc}}={\mu }_0(5.0)=(4\pi \times 10^{-7})(5.0)\approx 6.28\times 10^{-6}\text{T⋅m}.$ Radius ko $6.0\text{cm}$ tak double karne se integral ke liye kuch nahi badalta: wahi $5.0\text{A}$ abhi bhi enclosed hai. (Local field $B$ zaroor half hota hai, lekin woh ek alag quantity hai.)

(a) haan aur (c) haan — dono mein ek symmetry hai (wire ke liye cylindrical, solenoid ke liye translational + outside-is-zero trick) jo ek smart loop ke saath $\vec{B}$ ko constant aur parallel banati hai. (b) nahi aur (d) nahi — koi coordinate nahi jiske along $\vec{B}$ uniform ho, toh $B$ ko bahar nahi kheench sakte. Law abhi bhi sach hai, bas solve karne ke liye useful nahi. Wahan Biot–Savart law use karo.

Loop = radius $s$ ka circle, toh $\oint \vec{B}\cdot d\vec{l}=B(2\pi s)$ aur $I_{\text{enc}}=I$: $B=\frac{{\mu }_{0}I}{2\pi s}=\frac{(4\pi \times 10^{-7})(12)}{2\pi (0.040)}=6.0\times 10^{-5}\text{T}=60\mu \text{T}.$

Ek long solenoid ke andar field uniform hoti hai: $B={\mu }_{0}nI$. $B=(4\pi \times 10^{-7})(1500)(2.0)=3.77\times 10^{-3}\text{T}\approx 3.8\text{mT}.$ Note karo yeh axis se tumhari distance pe depend nahi karta (jab tak tum well inside ho). Solenoid and toroid fields dekho.

$B=\frac{{\mu }_{0}NI}{2\pi r}=\frac{(4\pi \times 10^{-7})(800)(3.0)}{2\pi (0.10)}=4.8\times 10^{-3}\text{T}\approx 4.8\text{mT}.$ Solenoid ke unlike, $B\propto 1/r$: inner rim pe zyada strong hai, outer rim pe zyada weak hai.

(a) Inside ($s<R$): sirf radius $s$ ke andar wali current enclosed hai. Uniform current density $J=I/(\pi R^2)$ ke saath, $I_{\text{enc}}=J\pi s^2=I\frac{s^2}{R^2},B(2\pi s)={\mu }_{0}I\frac{s^2}{R^2}\Rightarrow B=\frac{{\mu }_{0}Is}{2\pi R^2}.$ $B=\frac{(4\pi \times 10^{-7})(10)(0.0010)}{2\pi (0.0020{)}^2}=5.0\times 10^{-4}\text{T}=0.50\text{mT}.$ (b) Outside ($s>R$): saare $10\text{A}$ enclosed hain, wire ek thin wire ki tarah behave karti hai: $B=\frac{{\mu }_{0}I}{2\pi s}=\frac{(4\pi \times 10^{-7})(10)}{2\pi (0.0040)}=5.0\times 10^{-4}\text{T}=0.50\text{mT}.$ Dono yahan chosen radii ki coincidence se $0.50\text{mT}$ equal hain — lekin behaviour alag hai: inside linear rise (figure mein amber line dekho), outside $1/s$ fall.

$I_{\text{enc}}=0$ (koi current surface ko pierce nahi karti), toh $\oint \vec{B}\cdot d\vec{l}={\mu }_0(0)=0.$ Lekin loop pe $\vec{B}\ne 0$ — wire ka field har jagah present hai. Near side pe positive contributions door side ke negative contributions ko exactly cancel kar dete hain. Zero circulation ka matlab zero field nahi hota.

"Page se bahar = positive" choose karo (right-hand rule, fingers loop traverse karne ke direction mein curl karti hain). Currents sign ke saath add hoti hain: Dono enclosed: $\oint \vec{B}\cdot d\vec{l}={\mu }_0(I_A-I_B)={\mu }_0(6.0-4.0)={\mu }_0(2.0)=2.51\times 10^{-6}\text{T⋅m}.$ Sirf A enclosed: $\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_A={\mu }_0(6.0)=7.54\times 10^{-6}\text{T⋅m}.$ Individual fields abhi bhi har jagah superpose hote hain; law simply Stokes' theorem ke through unka net threaded current bookkeep karta hai.

(a) Beech mein ($I_{\text{enc}}$ = sirf inner conductor $=+5.0\text{A}$): radius $s$ pe, $B=\frac{{\mu }_{0}I}{2\pi s}(\text{normal }1/s\text{ field}).$ (b) Bahar ($I_{\text{enc}}=+5.0-5.0=0$): $\oint \vec{B}\cdot d\vec{l}={\mu }_0(0)=0\Rightarrow B=0.$ Yahi wajah hai ki coaxial cables apna field confine karti hain — return current enclosed total ko cancel kar deta hai. Same logic toroid ke outside field ke zero hone jaisi hai, aur Gauss's law for magnetism ka mirror hai (koi net magnetic source nahi).

Dono surfaces ek hi loop share karti hain, toh $\oint \vec{B}\cdot d\vec{l}$ ek hi number hona chahiye. Magnetostatic form charging ke dauran incomplete hai kyunki plates ke beech $\partial \vec{E}/\partial t\ne 0$ hai. Maxwell displacement current add karta hai: $\oint \vec{B}\cdot d\vec{l}={\mu }_0\left(I_{\text{enc}}+{\epsilon }_0\frac{d{\Phi }_E}{dt}\right).$ S1 ke across $I_{\text{enc}}=I$ term dominate karta hai; S2 ke across koi wire nahi hai lekin rising $\vec{E}$ exactly ${\epsilon }_{0}d{\Phi }_E/dt=I$ deta hai. Dono surfaces ab ${\mu }_{0}I$ yield karti hain — consistent. Yahi woh fix hai jisne Maxwell's equations ko EM waves predict karne layak banaya. Displacement current dekho.

Symmetry se $\vec{B}$ $\mp y$ direction mein hai (sheet ke parallel, dono sides pe opposite), magnitude mein constant hai, aur loop ke do vertical legs $0$ contribute karte hain (wahan $\vec{B}\perp d\vec{l}$). Width $w$ ka ek loop lo jo sheet ko stradle kare, ek long side sheet se $d$ upar, doosri $d$ neeche. Dono long sides same sense mein $Bw$ contribute karti hain: $\oint \vec{B}\cdot d\vec{l}=2Bw.$ Loop se guzarne wali current: width $w$ ki strip $I_{\text{enc}}=Kw$ carry karti hai. Isliye $2Bw={\mu }_{0}Kw\Rightarrow \boxed{B=\frac{1}{2}{\mu }_{0}K}.$ Field sheet ke upar $-y$ mein point karta hai aur neeche $+y$ mein (sheet ke across reverse ho jaata hai). Jump $B_{\text{below}}-B_{\text{above}}={\mu }_{0}K$ current sheet ke liye general boundary condition hai.

Ampère's law ke right-hand side ko sirf ==$I_{\text{enc}}$== — loop ko thread karne wali net signed current — se matlab hai. Shape, size, aur external currents drop out ho jaate hain. Kya cheez ek loop ko useful banati hai (sirf valid nahi)? ::: Ek symmetry jo $\vec{B}$ ko uske along constant-and-parallel ya zero rakhti hai, taaki $B$ integral se bahar aa sake. Magnetostatic form kab break hoti hai? ::: Jab $\partial \vec{E}/\partial t\ne 0$ ho; tumhe displacement current ${\epsilon }_{0}d{\Phi }_E/dt$ add karna padta hai.