Worked examples — Ampere's circuital law — magnetostatic form

1.8.23 · D3 · Physics › Electromagnetism › Ampere's circuital law — magnetostatic form

Yeh parent note ka ek companion hai. Yahan hum dobara kya hai law — yeh nahi seekhte — hum ise stress-test karte hain. Hum har tarah ki situation se guzarte hain jo ek problem de sakti hai: currents ek taraf point karti hain, phir doosri taraf; loops jo current ko enclose karti hain, phir loops jo nahi karti; field bilkul centre par ($s\to 0$), door ($s\to \infty$), aur wire ki exact boundary par; ek real-world cable; aur ek exam trap jisme do nested currents hain.

Koi bhi symbol aane se pehle: mujhe un kuch symbols ke naam bata dene do jo hum har jagah reuse karte hain, seedhe words mein.

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Scenario matrix

Har Ampère's-law problem inhi cells mein se ek mein rehti hai. Is page ka goal hai ki koi bhi cell untouched na rahe.

Cell	Kya khas hai ismein	Covered by
A. Sign / direction	current upar vs. neeche; kya circulation sign flip karta hai?	Ex 1
B. Loop encloses current	full turn, $\oint d\varphi =2\pi$	Ex 1, 2
C. Loop enclose NAHI karta	angle returns, $\oint d\varphi =0$, circulation $=0$	Ex 3
D. Degenerate: axis par point	$s\to 0$ limit, kya $B$ finite hai?	Ex 4
E. Boundary case	thick wire exactly at $s=R$, inside-formula meets outside-formula	Ex 4
F. Limiting: far field	$s\to \infty$, $B\to 0$	Ex 4
G. Cancellation trap	do opposite currents, net $I_{\text{enc}}=0$ lekin $B\ne 0$	Ex 5
H. Real-world word problem	coaxial cable, current inside + return outside	Ex 6
I. Exam twist: nested sheets/solenoid	do enclosed currents ka superposition	Ex 7

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Example 1 — Sign & direction (Cell A, B)

Recall Forecast

Guess karo: kya current ko reverse karne se circulation ka sign flip hoga, ya magnitude, ya dono?

Steps (a):

1. Right-hand rule: apna thumb page se bahar point karo (current direction), fingers counter-clockwise curl karti hain. Yeh step kyun? Yeh batata hai ki $\vec{B}$ hamare chosen walk direction ke saath point karta hai, isliye har $\vec{B}\cdot d\vec{l}$ term positive hai.
2. Law directly apply karo: $\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_{\text{enc}}={\mu }_0(5)$. Yeh step kyun? Loop poora current enclose karta hai (Cell B), isliye $I_{\text{enc}}=I=5\text{A}$.
3. Compute karo: $\oint \vec{B}\cdot d\vec{l}=(4\pi \times 10^{-7})(5)=2\pi \times 10^{-6}\text{T⋅m}\approx 6.28\times 10^{-6}\text{T⋅m}$.

Steps (b): 4. Current ab into the page. Sign convention ke hisaab se, current jo surface ko loop ke right-hand normal ke against pierce kare, woh negative count hoti hai: $I_{\text{enc}}=-5\text{A}$. Yeh step kyun? Loop ke liye right-hand rule (counter-clockwise chalo ⇒ normal page se bahar point karta hai) ab current ki direction se disagree karta hai. 5. Isliye $\oint \vec{B}\cdot d\vec{l}=-2\pi \times 10^{-6}\text{T⋅m}$ — same magnitude, opposite sign.

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Example 2 — Loop encloses current, non-circular shape (Cell B)

Recall Forecast

Kya strange shape answer badal deta hai? Padhne se pehle guess karo.

Steps:

1. Parent ke Step 3 se recall karo: polar coordinates mein $\oint \vec{B}\cdot d\vec{l}=\frac{{\mu }_{0}I}{2\pi }\oint d\varphi$. Yeh step kyun? $\vec{B}$ ka koi radial component nahi hai, isliye sirf angular sweep bachti hai — path ki shape irrelevant hai, sirf yeh mayne rakhta hai ki woh kitna angle sweep karta hai.
2. Ek loop jo wire ke around ek baar wrap kare woh $\oint d\varphi =2\pi$ sweep karta hai. Yeh step kyun? Tum exactly ek full turn ke baad apne starting angle par wapas aate ho.
3. Isliye $\oint \vec{B}\cdot d\vec{l}=\frac{{\mu }_0(5)}{2\pi }(2\pi )={\mu }_0(5)=2\pi \times 10^{-6}\text{T⋅m}$.

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Example 3 — Loop wire ko enclose NAHI karta (Cell C)

Recall Forecast

Do cheezein guess karo: (1) circulation ki value, (2) kya loop par $\vec{B}=0$ hai.

Steps:

1. $I_{\text{enc}}=0$: koi current $C$ ke bound kisi bhi surface ko pierce nahi karti. Yeh step kyun? Wire bahar hai — tum $C$ ke across ek aisi surface stretch kar sakte ho jise wire kabhi puncture nahi karta.
2. Isliye ${\oint }_C\vec{B}\cdot d\vec{l}={\mu }_0(0)=0$.
3. Lekin loop par $\vec{B}\ne 0$! Figure dekho: wire ke paas $\vec{B}$ strong hai, aur yeh loop ke near edge par andar thread karta hai aur far edge par bahar. Yeh step kyun? Positive contributions (jahan $\vec{B}$ walk ke saath point karta hai) exactly negative walo ko cancel karte hain — yeh Cell G ka flavor hai lekin ek single external wire ke saath.

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Example 4 — Degenerate, boundary & limiting behaviour (Cells D, E, F)

Recall Forecast

Apne dimag mein $B$ vs $s$ sketch karo. $B$ sabse bada kahan hai? Kya axis par finite hai?

Steps (D) — axis, $s\to 0$:

1. Inside formula (parent Ex 2): $B_{\text{in}}=\frac{{\mu }_{0}Is}{2\pi R^2}$. Yeh step kyun? Sirf radius $s$ ke andar ki current loop ko thread karti hai, aur woh fraction $s^2/R^2$ hai.
2. $s=0$ set karo: $B=0$. Yeh step kyun? Zero-radius loop se koi current enclosed nahi hoti — isliye axis par field exactly zero hai, finite aur well-behaved. (Contrast: ek thin wire blow up kar jaata, lekin real wires nahi karte.)

Steps (E) — surface, $s=R$ dono sides se: 3. Inside se: $B_{\text{in}}(R)=\frac{{\mu }_{0}IR}{2\pi R^2}=\frac{{\mu }_{0}I}{2\pi R}$. 4. Outside se: $B_{\text{out}}(R)=\frac{{\mu }_{0}I}{2\pi R}$ (poora current enclosed). Yeh step kyun? Boundary par dono formulas agree karne chahiye — field continuous hai, koi sudden jump nahi. $R=0.01$ plug karo: $B(R)=\frac{(4\pi \times 10^{-7})(5)}{2\pi (0.01)}=1.0\times 10^{-4}\text{T}$. Yeh maximum hai — graph ki peak.

Steps (F) — far field, $s\to \infty$: 5. Outside: $B_{\text{out}}=\frac{{\mu }_{0}I}{2\pi s}$. $s=0.10$ par: $B=\frac{(4\pi \times 10^{-7})(5)}{2\pi (0.10)}=1.0\times 10^{-5}\text{T}$. 6. Jab $s\to \infty$, $B\to 0$. Yeh step kyun? $1/s$ decay ka matlab hai ki field fade hoti hai lekin kabhi abruptly nahi rukti — ek infinite straight wire ke saath consistent.

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Example 5 — The cancellation trap (Cell G)

Recall Forecast

Net enclosed current zero hai — toh kya field bhi zero hai? Dhyan rakhna.

Steps:

1. $I_{\text{enc}}=(+7)+(-7)=0$. Yeh step kyun? Out-of-page positive count hota hai (right-hand normal), into-page negative; woh sum mein cancel ho jaate hain.
2. $\oint \vec{B}\cdot d\vec{l}={\mu }_0(0)=0$.
3. Phir bhi loop par kahi bhi $\vec{B}\ne 0$: har wire apna circular field produce karta hai, aur ek general point par yeh do fields cancel nahi hote (woh alag-alag directions mein alag-alag magnitudes ke saath point karte hain). Yeh step kyun? Law integral ko constrain karta hai, pointwise field ko nahi.

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Example 6 — Real-world word problem: coaxial cable (Cell H)

Recall Forecast

Hum cables ko is tarah kyun shield karte hain? Compute karne se pehle cable ke bahar field guess karo.

Steps (i) — conductors ke beech, $s=1\text{mm}$:

1. Radius $s=1\text{mm}$ ki loop sirf inner conductor ko enclose karti hai: $I_{\text{enc}}=3\text{A}$. Yeh step kyun? $b=2\text{mm}$ par return current is loop ke bahar hai, isliye count nahi hoti.
2. $B=\frac{{\mu }_{0}I}{2\pi s}=\frac{(4\pi \times 10^{-7})(3)}{2\pi (0.001)}=6.0\times 10^{-4}\text{T}$.

Steps (ii) — sab kuch ke bahar, $s=3\text{mm}$: 3. Loop ab dono currents ko enclose karti hai: $I_{\text{enc}}=(+3)+(-3)=0$. Yeh step kyun? Jaane wali aur aane wali dono currents andar hain — woh cancel ho jaati hain. 4. $B=\frac{{\mu }_0(0)}{2\pi s}=0$.

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Example 7 — Exam twist: nested current sheets / solenoid superposition (Cell I)

Recall Forecast

Innermost region ke andar, dono solenoids contribute karte hain. Kya fields add hange ya subtract?

Steps (a) — inner solenoid ke bilkul andar:

1. Akele har solenoid uske andar $B={\mu }_{0}nI$ deta hai (parent Ex 3). Yeh step kyun? Rectangular Amperian loop ${\mu }_{0}nLI$ pick up karta hai aur $BL={\mu }_{0}nLI\Rightarrow B={\mu }_{0}nI$.
2. Superposition se, inner region mein dono fields present hain; woh oppositely point karte hain, isliye subtract hote hain: $B={\mu }_{0}I(n_1-n_2)$. Yeh step kyun? Poora field vector sum hai; opposite windings ka matlab opposite signs.
3. $B=(4\pi \times 10^{-7})(2)(1000-400)=(4\pi \times 10^{-7})(2)(600)=1.508\times 10^{-3}\text{T}$.

Steps (b) — gap mein (inner ke bahar, outer ke andar): 4. Yahan sirf outer solenoid ka field present hai (inner solenoid ka field apne bahar ~zero hota hai). $B={\mu }_0{n}_{2}I=(4\pi \times 10^{-7})(400)(2)=1.005\times 10^{-3}\text{T}$, outer solenoid ke taraf point karta hua. Yeh step kyun? Gap mein ek Amperian loop sirf outer-solenoid turns ko enclose karta hai.

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Wrap-up

Recall Har example ne kaunsi cell cover ki?

Ex 1 :::: A (sign) + B (encloses) Ex 2 :::: B (shape-independence) Ex 3 :::: C (no enclosure, $\oint =0$, $\vec{B}\ne 0$) Ex 4 :::: D (axis $s\to 0$) + E (boundary $s=R$) + F (far field) Ex 5 :::: G (cancellation trap) Ex 6 :::: H (real-world coax) Ex 7 :::: I (nested solenoid superposition)

Related: Biot–Savart law · Solenoid and toroid fields · Gauss's law for magnetism · Stokes' theorem · Displacement current · Maxwell's equations.