Biot-Savart law — magnetic field from current element

1.8.22 · Physics › Electromagnetism

YEH HAI KYA?

Is formula mein bake hue key facts:

Direction: $d B$ , $d l$ aur $\overset{r}{^}$ dono ke $⊥$ hai (cross product ki wajah se). Yeh current element ko wrap karne wale circles ka tangent hai.
Magnitude: $d B = \frac{μ _{0}}{4 π} \frac{I d l sin θ}{r ^{2}}$ , jahan $θ$ , $d l$ aur $\overset{r}{^}$ ke beech ka angle hai.
$1/ r^{2}$ ke saath kam hota hai — Coulomb's law ki tarah, lekin point charge ki jagah current ke tukde ke liye.

Figure — Biot-Savart law — magnetic field from current element

HAR PIECE KYUN MATTER KARTA HAI? (Har factor ko steel-man karo)

EK REAL FIELD KAISE DERIVE KARTE HAIN: STRAIGHT WIRE

Bare formula tab tak kaam ka nahi jab tak hum ise integrate na karein. Chaliye classic case scratch se karte hain.

Worked example Long straight wire se

$a$ doori par field Setup. Wire $y$ -axis ke along hai aur $I$ current upar ki taraf carry kar raha hai. Field point $P$ , perpendicular distance $a$ par hai. Ek segment $d l = d y \overset{y}{^}$ height $y$ par hai.

Step 1 — geometry. Maano $θ$ , $d l$ aur $\overset{r}{^}$ ke beech ka angle hai. Figure se, $r = a^{2} + y^{2}, sin θ = \frac{a}{a ^{2} + y ^{2}} .$ Yeh step kyun? Biot–Savart magnitude ko $r$ aur $sin θ$ ek hi variable mein chahiye taaki hum integrate kar sakein.

Step 2 — $d B$ likho.

= \frac{\mu_0 I}{4\pi}\frac{a\,dy}{(a^2+y^2)^{3/2}}.$$ *Yeh step kyun?* Geometry substitute karo taaki sab kuch $y$ mein ho jaye. Saare segments $\vec B$ ko $P$ par **page se bahar** push karte hain (same direction), isliye hum magnitudes add karte hain — koi vector cancellation nahi. **Step 3 — poori wire par integrate karo** ($y$, $-\infty$ se $+\infty$ tak): $$B = \frac{\mu_0 I a}{4\pi}\int_{-\infty}^{\infty}\frac{dy}{(a^2+y^2)^{3/2}}.$$ Standard integral $\displaystyle\int\frac{dy}{(a^2+y^2)^{3/2}}=\frac{y}{a^2\sqrt{a^2+y^2}}$ hai. $-\infty$ se $\infty$ tak evaluate karne par $\dfrac{2}{a^2}$ milta hai. *Yeh step kyun?* Limits daalne par: $y\to+\infty$ par bracket $\to 1/a^2$ hoti hai; $-\infty$ par $\to -1/a^2$; difference $=2/a^2$. **Step 4 — result.** $$\boxed{B = \frac{\mu_0 I}{2\pi a}}$$ *Yeh kyun matter karta hai:* Yahi famous **infinite straight-wire field** hai — yeh $1/a$ se ghatta hai (na ki $1/r^2$ se!) kyunki humne infinitely many elements ko sum kiya.

Common mistake Common mistakes (Steel-manned)

1. " $r$ origin se use karo, element se nahi." Kyun sahi lagta hai: Bahut se problems mein hum ek fixed origin se measure karte hain. Fix: Biot–Savart mein $r$ hamesha current element $d l$ se field point $P$ tak jaata hai, aur har element ke liye yeh change hota hai.

2. " $B = μ_{0} I / (2 π a)$ kisi bhi wire ke liye kaam karta hai." Kyun sahi lagta hai: Yahi formula sabko yaad hota hai. Fix: Yeh sirf infinitely long straight wire ke liye hai. Finite wires ke liye $sin θ_{1} + sin θ_{2}$ form use karo; loops ke liye, $μ_{0} I /2 R$ .

3. $sin θ$ bhool jaana, raw $d l / r^{2}$ chodna. Kyun sahi lagta hai: Coulomb's $1/ r^{2}$ jaisa lagta hai. Fix: Magnetism ek cross product se aata hai; directional $sin θ$ factor zaroori hai — seedhe aage ka contribution zero hota hai.

4. $d B$ ko scalars ki tarah add karna jab woh alag-alag directions mein point karte hain. Fix: Magnitudes tab hi add karo jab saare $d B$ parallel hoon (straight wire at $P$ , loop centre). Warna pehle components mein resolve karo (jaise loop ke axis par, sirf axial components bachte hain).

Forecast-then-Verify

Flashcards

Biot–Savart law (vector form)

d B = \frac{μ _{0}}{4 π} \frac{I d l \times r ^}{r ^{2}}

Biot–Savart mein

r

kahan se kahan point karta hai?

Current element

d l

se field point

P

tak.

d B

ki magnitude angle ke saath

d B = \frac{μ _{0}}{4 π} \frac{I d l sin θ}{r ^{2}}

Current element ke seedhe aage field zero kyun hoti hai?

θ = 0 \Rightarrow sin θ = 0

(cross product zero ho jaata hai).

Infinite straight wire ka field

B = \frac{μ _{0} I}{2 π a}

Circular loop ke centre par field

B = \frac{μ _{0} I}{2 R}

Finite straight wire ka field

B = \frac{μ _{0} I}{4 π a} (sin θ_{1} + sin θ_{2})

μ_{0}

ki value

4 π \times 1 0^{- 7} T\cdotm/A

d B

ki direction

d l

aur

\overset{r}{^}

ke relative

Dono ke perpendicular (right-hand rule), element ke chaaro taraf circles ka tangent.

Loop ka on-axis field

x

doori par

B = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}}

Recall Feynman: 12-saal ke bachche ko samjhao

Socho ek hose hai jo paani ki jagah invisible "swirl" spray karta hai. Wire ke har chhote tukde mein se electricity flow karti hai, aur woh swirl sideways spray karta hai — na seedha aage, na seedha peeche. Swirl wire ke chaaro taraf chhote rings banata hai. Door jaane par swirl weak hota jaata hai. Ek jagah par total swirl paane ke liye, wire ke har chhote tukde ka swirl add karte hain. Yahi adding-up Biot–Savart law hai.

Connections

Ampère's Law — symmetric cases ke liye aasaan shortcut; wire ke liye same $μ_{0} I /2 π a$ deta hai.
Magnetic Field of a Solenoid — loop fields ko integrate karke banta hai.
Magnetic Dipole Moment — loop ka far-field ek dipole hota hai ( $\propto 1/ x^{3}$ ).
Coulomb's Law — electric $1/ r^{2}$ analogue; contrast: magnetism mein cross product hai aur koi monopoles nahi.
Lorentz Force — yeh $B$ phir moving charges par kya karta hai.
Right-Hand Rule — yahan saari directions fix karta hai.