Biot-Savart law — magnetic field from current element: Worked examples

1.8.22 · D3 · Physics › Electromagnetism › Biot-Savart law — magnetic field from current element

Yahan sirf teen physical objects hain, toh symbols se pehle ek baar plain words mein naam kar lete hain:

Do master formulas jinpar hum rely karenge (dono parent note mein derive ki gayi hain):

Arc formula ko pehli baar use karte waqt justify karenge (Example E). Aur $μ_{0} = 4 π \times 1 0^{- 7} T\cdotm/A$ throughout.

The scenario matrix

Har straight-ya-circular Biot–Savart problem in cells mein se kisi ek mein aata hai. Last column us worked example ka naam deta hai jo ise cover karta hai.

#	Case class	The tricky bit	Worked in
C1	Infinite wire, dono ends door	$θ_{1} = θ_{2} = 9 0^{\circ}$	A
C2	Semi-infinite wire, ek end AT foot par	one angle $= 0^{\circ}$	B
C3	Finite wire, $P$ off to the side (asymmetric)	do alag non- $9 0^{\circ}$ angles, angle ka sign	C
C4	Point ON the line of the wire (degenerate)	$sin θ = 0$ har jagah ⇒ $B = 0$	D
C5	Circular arc (partial loop)	fraction $ϕ /2 π$ of a full loop	E
C6	Superposition — do straight pieces + ek arc (bent conductor)	vectors add karo, directions dekho	F
C7	On-axis point of a loop, limiting behaviour	far-field $\to 1/ x^{3}$ dipole	G
C8	Real-world word problem numbers aur units ke saath	unit bookkeeping, order of magnitude	H

Ab har cell clear karte hain.

Worked example Example A — Cell C1: Infinite straight wire

Ek bahut lamba straight wire $I = 10 A$ carry karta hai. $a = 5 cm = 0.05 m$ par $B$ find karo.

Forecast: Compute karne se pehle — $B$ $1 0^{- 5} T$ ke kareeb hoga ya $1 0^{- 2} T$ ke? (Earth ka field $\sim 5 \times 1 0^{- 5} T$ hai; abhi guess karo.)

Angles identify karo. Dono ends infinity tak jaate hain, toh $P$ se har end perpendicular se $9 0^{\circ}$ par dikh raha hai. Yeh step kyun? Workhorse formula ko $θ_{1}, θ_{2}$ chahiye; infinite ends dono extreme angle $9 0^{\circ}$ dete hain.
Plug in $sin 9 0^{\circ} = 1$ : $B = \frac{μ _{0} I}{4 π a} (1 + 1) = \frac{μ _{0} I}{2 π a}$ . Kyun? Do $1$ add hokar $2$ bante hain, finite formula famous infinite wali mein collapse ho jaati hai.
Numbers: $B = \frac{( 4 π \times 1 0 ^{- 7} ) ( 10 )}{2 π ( 0.05 )} = 4 \times 1 0^{- 5} T$ .

Verify: Units: $\frac{( T\cdotm/A ) ( A )}{m} = T$ . ✔ Magnitude $4 \times 1 0^{- 5} T$ — Earth ke field ke comparable, toh tumhara compass notice karta. $μ_{0} = 4 π \times 1 0^{- 7}$ ne $2 π$ ko neatly cancel kiya, ek accha sign hai.

Worked example Example B — Cell C2: Semi-infinite wire

Ek straight wire perpendicular foot $O$ se start hoti hai aur sirf ek direction mein $+ \infty$ tak jaati hai (socho jaise wall se nikali wire). Field point $P$ , $O$ se $a$ distance par hai. Dikhao $B$ infinite-wire value ka exactly half hai.

Forecast: Aadhi wire ⇒ aadha field? Algebra se pehle yes/no guess karo.

Near end par angle. Wire ka near end foot $O$ par hi baitha hai, $P$ ke seedha neeche. $P$ se $O$ ki taraf dekhne par, line perpendicular ke saath hi hai, toh $θ_{1} = 0^{\circ}$ . Yeh step kyun? Figure dekho: perpendicular foot se near end tak ka angle zero hai kyunki end hai hi foot.
Far end par angle. Far end infinity tak jaata hai, toh $θ_{2} = 9 0^{\circ}$ . Kyun? Us side ke liye infinite case wali reasoning same hai.
Plug in: $B = \frac{μ _{0} I}{4 π a} (sin 0^{\circ} + sin 9 0^{\circ}) = \frac{μ _{0} I}{4 π a} (0 + 1) = \frac{μ _{0} I}{4 π a}$ .

Verify: Infinite value $\frac{μ _{0} I}{2 π a}$ se compare karo: ratio exactly $\frac{1}{2}$ hai. ✔ Toh semi-infinite wire half deti hai — forecast "half" sahi tha, lekin kyun note karo: far half poora $9 0^{\circ}$ side contribute karta hai aur near end $0$ deta hai, iska matlab yeh nahi ki tumne literally wire ko do equal field-halves mein kaat diya.

Worked example Example C — Cell C3: Finite wire,

$P$ ek taraf (asymmetric) Ek straight wire $y$ -axis ke saath $y = 0$ se $y = L$ tak jaati hai, $I = 8 A$ . Field point $P$ perpendicular distance $a = 3 m$ par hai, bottom end ( $y = 0$ ) ke level par, $L = 4 m$ . $B$ find karo.

Forecast: Kya do end-angles equal honge ya alag? Kaun sa sign issue tumhe kaatega?

Perpendicular foot kahan hai? $P$ bottom end ke level par hai, toh foot $O$ bottom end par hi hai $(y = 0)$ . Yeh step kyun? Formula ke angles foot se measure hote hain; pehle use locate karna padega. Yahan yeh ek end ke saath coincide karta hai.
Bottom end tak angle: yeh foot par hi hai, toh $θ_{1} = 0^{\circ}$ , $sin θ_{1} = 0$ . Kyun? Example B ke near end wali degeneracy same hai.
Top end tak angle: top end height $L = 4$ par hai, horizontal distance $a = 3$ . Perpendicular se $P \to top$ line tak ka angle satisfy karta hai $sin θ_{2} = \frac{L}{a ^{2} + L ^{2}} = \frac{4}{9 + 16} = \frac{4}{5} = 0.8$ . Yeh step kyun? Right triangle par (legs $a$ aur $L$ ), angle-at- $P$ ke opposite side vertical leg $L$ hai; isliye opposite/hypotenuse $= L / a^{2} + L^{2}$ . Yahi wajah hai $sin$ use karte hain: yeh exactly is triangle par opposite-over-hypotenuse hai.
Combine: $B = \frac{μ _{0} I}{4 π a} (0 + 0.8) = \frac{( 1 0 ^{- 7} ) ( 8 ) ( 0.8 )}{3}$ ( $μ_{0} /4 π = 1 0^{- 7}$ use karke).
Number: $B = \frac{8 \times 0.8 \times 1 0 ^{- 7}}{3} = 2.13 \times 1 0^{- 7} T$ .

Verify: Sanity — yeh $a = 3$ par infinite-wire value se kam hona chahiye, jo hai $\frac{μ _{0} I}{2 π a} = \frac{( 2 \times 1 0 ^{- 7} ) ( 8 )}{3} = 5.33 \times 1 0^{- 7} T$ . Hamara $2.13 \times 1 0^{- 7}$ chhota hai ✔ (wire ka ek tukda infinite se kam deta hai). Sign trap: agar $P$ dono ends ke beech level par hota, toh ek angle foot ke doosri taraf measure hota aur dono $sin$ values positives ki tarah add hote — har angle ko perpendicular se positive amount lo aur add karo.

Worked example Example D — Cell C4: Wire ki line par point (degenerate)

Ek straight wire $x$ -axis par hai. Field point $P$ $x$ -axis par bhi baitha hai, wire ke end se 2 m aage. $P$ par $B$ kya hai?

Forecast: Non-zero ya exactly zero? Padhne se pehle commit karo.

Har element ke liye $θ$ dekho. Har element $d l$ ke liye ( $x$ ke saath point karta hua) $P$ ki taraf vector $\overset{r}{^}$ bhi $x$ ke saath point karta hai. Toh unke beech angle $θ = 0^{\circ}$ hai. Yeh step kyun? Biot–Savart ki magnitude mein $sin θ$ hai; integrate karne se pehle $θ$ inspect karna zaroori hai.
$sin θ$ apply karo: $sin 0^{\circ} = 0$ , toh $d B = 0$ har element ke liye. Kyun? Ek current element field sirf sideways spray karta hai — seedha aage ya seedha peeche nahi. $P$ puri wire ke seedha aage hai.
Integrate karo: $B = \int 0 = 0$ .

Verify: $B = 0$ exactly. ✔ Yeh parent note ka "straight-ahead zero" fact literally saamne hai: wire ke apne axis par baitha point us wire se koi field feel nahi karta. Koi bhi thodi si sideways offset is tie ko todti hai aur finite-wire formula deti hai.

Worked example Example E — Cell C5: Circular arc (partial loop)

Ek wire quarter circle (arc angle $9 0^{\circ}$ ) mein bend ki gayi hai, radius $R = 0.2 m$ , current $I = 6 A$ . Circle ke centre par $B$ find karo.

Forecast: Full loop $μ_{0} I /2 R$ deta hai. Quarter ko uska kya fraction dena chahiye?

Har element apne $\overset{r}{^}$ ke perpendicular hai. Circle par tangent (direction of $d l$ ) hamesha radius ke $9 0^{\circ}$ par hoti hai (direction of $\overset{r}{^}$ centre ki taraf). Toh $sin θ = 1$ aur $r = R$ sab elements ke liye. Yeh step kyun? Yahi arc integral ko trivial banata hai — na varying $sin θ$ , na varying $r$ .
$d B$ likho aur arc length integrate karo. $d B = \frac{μ _{0} I}{4 π R ^{2}} d l$ , aur sab ek hi direction mein point karte hain (plane se bahar, right-hand rule), toh magnitudes add karo: $B = \frac{μ _{0} I}{4 π R ^{2}} \int d l$ . Kyun? Same direction ⇒ scalar sum, koi cancellation nahi.
$ϕ$ -radian arc ki arc length $R ϕ$ hai. Quarter ke liye, $ϕ = π /2$ , toh $\int d l = R \cdot \frac{π}{2}$ . Phir $B = \frac{μ _{0} I}{4 π R ^{2}} \cdot R \frac{π}{2} = \frac{μ _{0} I}{8 R}$ . Yeh step kyun? Yeh exactly general arc formula $B = \frac{μ _{0} I}{4 π R} ϕ$ hai $ϕ = π /2$ par evaluate kiya gaya.
Number: $B = \frac{( 4 π \times 1 0 ^{- 7} ) ( 6 )}{8 ( 0.2 )} = 4.71 \times 1 0^{- 6} T$ .

Verify: Fraction check: quarter arc $=$ full loop ka $\frac{1}{4}$ , aur $\frac{μ _{0} I}{8 R} = \frac{1}{4} \cdot \frac{μ _{0} I}{2 R}$ . ✔ Forecast "one quarter" confirm hua. Full-loop value hoti $\frac{μ _{0} I}{2 R} = 1.885 \times 1 0^{- 5} T$ ; hamara uska quarter hai.

Worked example Example F — Cell C6: Superposition (bent conductor)

Ek wire door se ek radius ke saath aati hai, ek point $O$ ke around half circle (radius $R = 0.1 m$ ) mein bend hoti hai, phir doosri radius se infinity tak chali jaati hai. Current $I = 5 A$ . $O$ par $B$ find karo.

Forecast: Straight pieces — kya yeh $O$ par contribute karte hain, ya zero?

Do straight radial pieces. Har straight piece seedha $O$ ki taraf ya us se door point karta hai, toh unke har element ke liye $\overset{r}{^}$ ( $O$ ki taraf) $d l$ ke parallel/antiparallel hai: $θ = 0^{\circ}$ ya $18 0^{\circ}$ , $sin θ = 0$ . Contribution $= 0$ . Yeh step kyun? Same degeneracy jaise Example D — field point ki taraf aimed wire wahan kuch nahi deti.
Half-circle arc. Arc formula use karo $ϕ = π$ ke saath: $B_{arc} = \frac{μ _{0} I}{4 π R} π = \frac{μ _{0} I}{4 R}$ . Kyun? Straights ne zero diya, toh total sirf semicircle hai.
Number: $B = \frac{( 4 π \times 1 0 ^{- 7} ) ( 5 )}{4 ( 0.1 )} = 1.571 \times 1 0^{- 5} T$ .

Verify: Half loop $\frac{μ _{0} I}{2 R}$ ka half hona chahiye: $\frac{1}{2} \cdot \frac{( 4 π \times 1 0 ^{- 7} ) ( 5 )}{2 ( 0.1 )} = \frac{1}{2} (3.14 \times 1 0^{- 5}) = 1.571 \times 1 0^{- 5}$ . ✔ Match karta hai. Lesson: bent conductor mein, radial-in / radial-out legs free hain — yeh arc ke centre par kabhi contribute nahi karte.

Worked example Example G — Cell C7: On-axis loop, limiting behaviour

Loop radius $R = 0.05 m$ , $I = 3 A$ . $B$ compute karo (a) centre par aur (b) axis par $x = 0.20 m$ par, aur confirm karo ki far-field $1/ x^{3}$ ki tarah fall karta hai.

Forecast: Dono points mein se kaun sa weaker field hai, aur roughly kitne factor se?

On-axis formula (parent note se) hai $B (x) = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}}$ .

Centre ( $x = 0$ ): $B (0) = \frac{μ _{0} I R ^{2}}{2 R ^{3}} = \frac{μ _{0} I}{2 R} = \frac{( 4 π \times 1 0 ^{- 7} ) ( 3 )}{2 ( 0.05 )} = 3.77 \times 1 0^{- 5} T$ . Yeh step kyun? $x = 0$ set karne par loop-centre result milna chahiye — ek built-in consistency check.
$x = 0.20$ par: $R^{2} + x^{2} = 0.0025 + 0.04 = 0.0425$ , $(0.0425)^{3/2} = 8.76 \times 1 0^{- 3}$ . Phir $B = \frac{( 4 π \times 1 0 ^{- 7} ) ( 3 ) ( 0.0025 )}{2 ( 8.76 \times 1 0 ^{- 3} )} = 5.38 \times 1 0^{- 7} T$ . Kyun? Direct substitution; $3/2$ power exact rakho.
Far-field check. Jab $x ≫ R$ , $(R^{2} + x^{2})^{3/2} \approx x^{3}$ , toh $B \approx \frac{μ _{0} I R ^{2}}{2 x ^{3}} \propto \frac{1}{x ^{3}}$ . Yeh step kyun? $x^{2}$ ke saath $R^{2}$ drop karne se dipole $1/ x^{3}$ tail expose hoti hai — dekho Magnetic Dipole Moment.

Verify: Ratio $B (0) / B (0.20) = 3.77 \times 1 0^{- 5} /5.38 \times 1 0^{- 7} \approx 70$ . Rough $1/ x^{3}$ scaling $x = 4 R$ ke liye bada drop predict karta, aur actual factor $\approx 70$ hai. ✔ $B (0.20)$ ke units: T (sab SI in, T out).

Worked example Example H — Cell C8: Real-world word problem

Ek household appliance cable $I = 15 A$ carry karta hai. Ek magnetic compass cable se $d = 2 cm$ door baitha hai. Kya field itni strong hai ki compass visibly deflect ho, yeh dekhte hue ki Earth ka horizontal field roughly $B_{E} = 2 \times 1 0^{- 5} T$ hai?

Forecast: Guess karo: Earth ke field ke comparable, das guna bada, ya das guna chhota?

Cable ko infinite straight wire model karo. $2 cm$ metre-long cable ke saath tiny hai, toh dono ends "infinite" lagte hain. Yeh step kyun? Finite formula ki jagah $B = μ_{0} I /2 π a$ use karna justify karta hai.
Compute karo: $a = 0.02 m$ , $B = \frac{( 4 π \times 1 0 ^{- 7} ) ( 15 )}{2 π ( 0.02 )} = 1.5 \times 1 0^{- 4} T$ .
Earth se compare karo: ratio $= \frac{1.5 \times 1 0 ^{- 4}}{2 \times 1 0 ^{- 5}} = 7.5$ . Kyun? Deflection wire field par relative to background depend karti hai jisse yeh compete karta hai.

Verify: $B_{wire} \approx 7.5 \times$ Earth's field ✔ — clearly enough to swing a compass noticeably. Units: T throughout. Order-of-magnitude sanity: 2 cm par 15 A wire se $1 0^{- 4} T$ textbook ballpark hai. Yahi wajah hai current-carrying cables nearby magnetometers disturb karte hain — is symmetric case ke liye Ampère's Law shortcut same number faster deta hai.

Recall Which cell is which? (self-test)

Point exactly wire ki apni line par baitha hai ::: $B = 0$ (Cell C4, $sin θ = 0$ ). Wire from the foot to infinity, sirf ek side ::: half the infinite value (Cell C2). Quarter arc at its centre ::: one quarter of $μ_{0} I /2 R$ (Cell C5). Radial straight leads into an arc ::: straights contribute nothing at the centre (Cell C6). Far away on a loop's axis ::: field $\propto 1/ x^{3}$ , a dipole (Cell C7).

Connections

Ampère's Law — symmetric cells (A, H) ke liye fast route.
Magnetic Field of a Solenoid — kai Example-G loops stack karo.
Magnetic Dipole Moment — Example G ka $1/ x^{3}$ tail.
Right-Hand Rule — upar use ki gayi har direction fix karta hai.
Lorentz Force — Example H mein compass needle par yeh fields kya karti hain.
Coulomb's Law — electric $1/ r^{2}$ contrast.