Visual walkthrough — Ampere's circuital law — magnetostatic form

1.8.23 · D2 · Physics › Electromagnetism › Ampere's circuital law — magnetostatic form

Step 0 — Hum asal mein kya explain karne ki koshish kar rahe hain

Do symbols jo hum earn karenge, assume nahi karenge:

Jo quantity hum dhundh rahe hain woh hai circulation ${\oint }_C\underset{\text{field arrow yahan}}{\underbrace{\vec{B}}}\cdot \underset{\text{tiny step yahan}}{\underbrace{d\vec{l}}}$ Unke beech ka dot $\cdot$ dot product hai: yeh sirf $\vec{B}$ ka woh part rakhta hai jo $d\vec{l}$ ke saath lie karta hai. Hum ise aage define karte hain, kyunki is puri page ka ek hi maths piece yahi hai.

Step 1 — Dot product kyun? "Path ke saath kitna?"

Yeh tool kyun, koi aur kyun nahi? Humne poocha tha "field mujhe meri walk ke saath kitna push karta hai?" "Along" word ka matlab hi projection hai, aur projection hi $\cos \alpha$ compute karta hai. Koi aur operation us exact question ka jawab nahi deta. Isliye dot product — cross product nahi, plain multiplication nahi — yahan sahi instrument hai.

PICTURE: orange step $d\vec{l}$ aur blue field $\vec{B}$ angle $\alpha$ par milte hain; green dashed segment shadow $|\vec{B}|\cos \alpha$ hai — sirf yahi part count karta hai.

Step 2 — Ek input fact: straight wire ka field

Kuch bhi add karne se pehle, humein jaanna hai ki wire ke paas $\vec{B}$ actually kya hai. Yeh Biot–Savart law se aata hai (pichhla chapter), jise hum yahan given maante hain.

Sirf $s$ kyun? Symmetry ki wajah se wire har angle se aur apni length par har point se identical dikhti hai. Toh $\vec{B}$ kis angle par ya wire ke saath kahan par depend nahi kar sakta — sirf kitna door, $s$ par. Aur yeh circles mein wrap karna chahiye (right-hand rule: $I$ ke saath thumb, ungli jis taraf curl karein woh $\vec{B}$ ki direction).

PICTURE: wire page se bahar point kar raha hai (dot), field arrows concentric circles bana rahe hain; andar wale circles ke longer arrows (strong $B$), bahar wale circles ke shorter — $1/s$ falloff jo tum dekh sakte ho.

Step 3 — Friendly loop par jodo (ek circle)

Kyunki $\vec{B}\parallel d\vec{l}$ har jagah hai, $\alpha =0$ aur $\cos \alpha =1$, toh har tiny contribution sirf $B|d\vec{l}|$ hai: ${\oint }_C\vec{B}\cdot d\vec{l}={\oint }_{C}Bdl=B{\oint }_{C}dl$

• $B$ sum se bahar aa gaya kyunki fixed radius $s$ par yeh har point par same number hai (Step 2).
• ${\oint }_{C}dl$ sirf "circle ke aas-paas sab tiny lengths jodo" = circumference $=2\pi s$ hai.

${\oint }_C\vec{B}\cdot d\vec{l}=\underset{B}{\underbrace{\frac{{\mu }_{0}I}{2\pi s}}}\times \underset{\text{circumference}}{\underbrace{2\pi s}}={\mu }_{0}I$

PICTURE: blue circle loop orange $d\vec{l}$ steps ke saath, har ek blue $\vec{B}$ arrow ke upar — poori taraf perfectly aligned.

Step 4 — Koi bhi wiggly loop same answer kyun deta hai

Ab loop ko kisi bhi shape mein deform karo (wire ko phir bhi enclosing karte hue). Har step $d\vec{l}$ ko do independent moves se describe karo: ek radial part (wire se andar/bahar) aur ek angular part (wire ke aas-paas).

${\oint }_C\vec{B}\cdot d\vec{l}={\oint }_C\frac{{\mu }_{0}I}{2\pi s}(\underset{\text{sideways arc}}{\underbrace{sd\varphi }})=\frac{{\mu }_{0}I}{2\pi }{\oint }_{C}d\varphi$

• Radial moves $0$ contribute karte hain: $\vec{B}$ purely sideways (angular) point karta hai, toh yeh kisi bhi in/out step ke perpendicular hai — dot product zero (Step 1, $\alpha =90^{\circ }$).
• Radius $s$ par angle $d\varphi$ cover karte hue ek step ka sideways part ki length $sd\varphi$ hai. Yahan $s$ $B$ mein $1/s$ ko cancel kar deta hai — same secret, general loop.
• ${\oint }_{C}d\varphi$ = "ek baar ghoomne par sweep kiya gaya total angle."

PICTURE: ek lumpy loop; har step ek green radial piece mein split (dot product se kill) aur ek orange sideways arc mein (sirf yahi bachta hai).

Step 5 — Edge case: loop jo wire ko enclose nahi karta

PICTURE: wire loop ke bahar; near arc par $\vec{B}$ tumhari walk mein help karta hai (green, positive), far arc par oppose karta hai (red, negative). Unke sums cancel ho jaate hain: circulation $=0$ chahe $\vec{B}\ne 0$ kahin bhi ho.

Step 6 — Superpose karo: kisi bhi current bundle ke liye law

Real setups mein kai wires hote hain. Lekin circulation add hoti hai: agar kai wires se $\vec{B}={\vec{B}}_1+{\vec{B}}_2+\dots$, toh ${\oint }_C\vec{B}\cdot d\vec{l}={\oint }_C{\vec{B}}_1\cdot d\vec{l}+{\oint }_C{\vec{B}}_2\cdot d\vec{l}+\dots$ Har term ${\mu }_0{I}_k$ hai agar wire $k$ enclosed hai, aur $0$ agar nahi. Sirf enclosed wires ko jodne par:

PICTURE: teen wires; ek loop do ko enclose karta hai ($+I$ aur $-I$, direction se sign note karo) aur teesre ko miss karta hai. Sirf enclosed wale $I_{\text{enc}}$ mein aate hain; bahar wala grey out hai.

Ek-picture summary

Is page par sab kuch compressed: circular field ($B\propto 1/s$), ek wiggly loop jiske steps killed radial pieces aur surviving sideways pieces mein split hote hain, $s$-cancellation, aur verdict — ${\mu }_{0}I$ agar wire loop karo, $0$ agar nahi.

Flashcards