Magnetic field of straight wire, circular loop, solenoid, toroid: Visual walkthrough

1.8.24 · D2 · Physics › Electromagnetism › Magnetic field of straight wire, circular loop, solenoid, to

Step 1 — Wire ko tiny straight pieces mein kaato

KYA. Hum ek flat circular loop of wire lete hain, radius $a$ , jisme current $I$ beh raha hai (charge ring ke around flow kar raha hai). Hum wire ko anginat tiny segments mein kaatate hain. Aisa hi ek segment ek chhota straight arrow hai jise hum $d l$ kehte hain.

KYUN. Ek curved wire ke baare mein ek saath sochna mushkil hai. Lekin itna chhota tukda ki woh straight lage — woh easy hai: Biot–Savart law hume ek straight current piece ka field bata deta hai. Agar hum ek piece ka field nikal sakein, toh bas sab pieces ko jod do. Yeh "anginat tiny cheezein jodo" wali baat exactly wahi hai jo integral sign $\oint$ ka matlab hoga.

PICTURE. Green loop dekho. Thick green arrow ek segment $d l$ hai — yeh current ke saath point karta hai (us jagah charge jis direction mein travel kar raha hai). Hum field chahte hain red point $P$ par, jo axis par baitha hai — loop ke centre se hoti hui, loop ke plane ke perpendicular line par — centre se $x$ door.

Figure — Magnetic field of straight wire, circular loop, solenoid, toroid

Step 2 — Biot–Savart kyun, Ampère kyun nahi

KYA. Hum Biot–Savart law yaad karte hain: ek tiny piece se tiny field hoti hai $d B = \frac{μ _{0}}{4 π} \frac{I d l \times r ^}{r ^{2}}$

KYUN. Formula ko dheere dheere padho, term by term, jahan har symbol exist karta hai:

$d B$ — is ek segment se tiny field contribution.
$\frac{μ _{0}}{4 π}$ — nature ka ek fixed constant jo strength set karta hai ( $μ_{0} = 4 π \times 1 0^{- 7} T\cdotm/A$ ).
$I d l$ — "kitna current, times piece kitna lamba hai": bada, lamba, strong piece bada field push karta hai.
$\overset{r}{^}$ — ek unit-length arrow jo segment se field point $P$ ki taraf point karta hai. Yeh batata hai ki $P$ kis direction mein hai.
$r^{2}$ bottom mein — field distance ke square ke saath kamzor hoti hai, bilkul gravity ya Coulomb's law ki tarah.
$d l \times \overset{r}{^}$ — cross product. Yahi key hai: yeh kehta hai $d B$ dono wire piece aur $P$ ki taraf jaane wali line ke perpendicular point karta hai. Magnetic field current ke around wrap hoti hai, seedha $P$ ki taraf nahi point karti.

Ampere's Law kyun nahi? Ampère tab hi help karta hai jab aap koi aisa loop dhundh sako jis par $B$ ki value har jagah same ho, taaki use factor out kar sako. Ek coil ke axis par aisa koi loop nahi hota — field strength point to point change hoti hai, exploit karne ke liye koi clean symmetry nahi hai. Toh hum mazboor hain pieces ko haath se jodne ke liye. Biot–Savart isi kaam ke liye hai.

PICTURE. Figure mein ek segment dikhta hai, vector $\overset{r}{^}$ jo $P$ tak pahuncha hai, aur resulting $d B$ jo $d l$ aur $\overset{r}{^}$ wale plane ke perpendicular bahar nikli hui hai. Dhyan do $d B$ tilted hai — yeh axis ke along nahi hai.

Step 3 — Distance $r$ har piece ke liye same hai

KYA. Ring ka har segment $P$ se same slant-distance par hai: $r = a^{2} + x^{2}$

KYUN. Ek right triangle imagine karo. Ek leg $a$ hai (centre se wire tak baahir), doosri leg $x$ hai (centre se axis ke along $P$ tak). Hypotenuse — wire se $P$ tak seedhi line — slant distance $r$ hai. Pythagoras deta hai $r = a^{2} + x^{2}$ . Kyunki point $P$ axis par hai (bilkul centre mein), ring ka har piece exactly itni hi door hai. Toh $r$ ek constant hai jab hum loop ke around jaate hain — yeh ek badi simplification hai.

$a^{2}$ — "outward" leg squared.
$x^{2}$ — "along-axis" leg squared.
inke sum ka square root — seedha honest straight-line distance.

PICTURE. Blue right triangle: legs $a$ aur $x$ , hypotenuse $r$ . Segment ko ring ke around spin karo aur triangle bas rotate hota hai — $r$ kabhi nahi badlta.

Step 4 — Har piece ka field: direction se pehle size

KYA. Hamare segment ke liye, wire piece $d l$ $\overset{r}{^}$ ke perpendicular hai (wire ring ke around chalti hai, jabki $\overset{r}{^}$ seedha $P$ ki taraf jaata hai). Do perpendicular directions ke cross product ka size lengths ke product ke barabar hota hai, toh $∣ d l \times \overset{r}{^} ∣ = d l \times 1 = d l$ . Isliye tiny field ka magnitude hai $d B = \frac{μ _{0} I d l}{4 π ( a ^{2} + x ^{2} )}$

KYUN. Humne Step 3 ka $r^{2} = a^{2} + x^{2}$ bottom mein substitute kiya, aur use kiya ki cross product ka size bas $d l$ hai (kyunki $d l$ aur $\overset{r}{^}$ ke beech ka angle $9 0^{\circ}$ hai, jiska sine $1$ hai).

top $μ_{0} I d l$ — strength times current times piece-length.
bottom $4 π (a^{2} + x^{2})$ — constant aur distance-squared.

Yeh tilted arrow ki length hai. Ab hum uski direction deal karte hain.

PICTURE. Green wire arrow $d l$ aur yellow $\overset{r}{^}$ arrow ke beech right angle ek small square se mark kiya gaya hai. Resulting $d B$ ki length $d B$ hai aur yeh axis aur $\overset{r}{^}$ wale plane mein hai.

Step 5 — Badi cancellation: sirf axial part bachta hai

KYA. Har tilted $d B$ ko do pieces mein split karo: ek axis ke along (ise $d B_{∥}$ kaho) aur ek axis ke perpendicular (ise $d B_{⊥}$ kaho). Ab us segment ko ring ke bilkul doosri taraf wale segment se pair karo. Unke perpendicular parts exactly opposite directions mein point karte hain aur zero mein cancel ho jaate hain. Unke axial parts same direction mein point karte hain aur add up ho jaate hain.

KYUN. Yahi poora trick hai, aur yeh pure symmetry hai. Har us piece ke liye jo apna $d B_{⊥}$ "upar" push karta hai, ring ke doosre side ka piece equal $d B_{⊥}$ "neeche" push karta hai. Puri ring par sum karo: saare sideways parts ek doosre ko khatam kar dete hain. Sirf along-axis parts pile up hote hain. Toh hum $d B_{⊥}$ bilkul chod sakte hain aur sirf yeh rakh sakte hain: $d B_{∥} = d B cos α,$ jahan $α$ $d B$ ka axis se tilt angle hai.

Angle $α$ hamare triangle mein $P$ wala angle hi hai, aur triangle se $cos α = \frac{a}{a ^{2} + x ^{2}} .$

$cos α$ — "tilted arrow ka kitna hissa axis ke along lie karta hai" ( $1$ = fully axial, $0$ = fully sideways).
$a$ top par — outward leg; $a^{2} + x^{2}$ hypotenuse. Yeh ratio $P$ ke corner se opposite-over-hypotenuse hai, jo precisely axially point karne wala fraction hai.

PICTURE. Do opposite segments, unke do tilted $d B$ arrows, aur dashed sideways components jo opposite directions mein point karte hain (woh annihilate hote hain). Solid axial components dono axis ke along same direction mein point karte hain.

Step 6 — Saare pieces add karo (integral)

KYA. Har axial contribution ka size same hai (ring par har jagah same $r$ , same $α$ ), toh summing easy hai: $B = \oint d B cos α = \frac{μ _{0} I}{4 π} \frac{a}{( a ^{2} + x ^{2} ) ^{3/2}} \oint d l .$

KYUN. Humne constant factors — $d l$ ke alawa sab kuch — sum ke baahir nikaal diye, kyunki jab hum loop ke around chalte hain toh woh change nahi hote. Phir $\oint d l$ ka matlab hai "ring ke around saari tiny lengths jodo" = total circumference $2 π a$ .

$\frac{1}{( a ^{2} + x ^{2} ) ^{3/2}}$ isliye aaya kyunki $d B$ ne ek power $\frac{1}{a ^{2} + x ^{2}}$ di aur $cos α$ ne factor $\frac{a}{a ^{2} + x ^{2}}$ diya — multiply karo: $\frac{a}{( a ^{2} + x ^{2} ) ^{3/2}}$ .

$\oint d l = 2 π a$ substitute karo: $B = \frac{μ _{0} I}{4 π} \cdot \frac{a}{( a ^{2} + x ^{2} ) ^{3/2}} \cdot 2 π a = \frac{μ _{0} I a ^{2}}{2 ( a ^{2} + x ^{2} ) ^{3/2}}$

$μ_{0} I$ — strength × current.
$a^{2}$ top par — ek $a$ $cos α$ se, ek circumference se.
$2$ bottom mein — $4 π$ kyunki circumference ka $2 π$ cancel hua.
$(a^{2} + x^{2})^{3/2}$ — distance dependence: field axis ke along fade hoti hai.

PICTURE. Puri ring ke around saare axial arrows, head-to-tail stack hokar $P$ par ek bada red axial arrow bante hain — total field.

Step 7 — Edge case: bilkul centre par field

KYA. $x = 0$ rakho ( $P$ loop ke centre par baitha hai): $B = \frac{μ _{0} I a ^{2}}{2 ( a ^{2} + 0 ) ^{3/2}} = \frac{μ _{0} I a ^{2}}{2 a ^{3}} = \frac{μ _{0} I}{2 a}$

KYUN. Centre par koi tilt hi nahi hota — har piece ka $d B$ pehle se hi seedha axis ke along point karta hai ( $α = 0$ , $cos α = 1$ ). Formula mein yeh reflect hona chahiye, aur hota bhi hai: $x = 0$ set karne se $(a^{2} + x^{2})^{3/2}$ collapse hokar $a^{3}$ ho jaata hai, aur clean $μ_{0} I / (2 a)$ bacha rehta hai. Chhote loops (small $a$ ) zyada strong centre fields dete hain — current zyada paas hoti hai.

PICTURE. Centre par, saare tiny arrows baahir fan out karte hain lekin har ek pehle se hi axis ke along flat hai; yahan cancellation ki zaroorat bhi nahi.

Step 8 — Limiting case: bahut door

KYA. Jab $P$ axis par bahut door ho, $x ≫ a$ , toh $a^{2}$ $x^{2}$ ke aage negligible hai: $B \approx \frac{μ _{0} I a ^{2}}{2 x ^{3}} \propto \frac{1}{x ^{3}} .$

KYUN. Bracket ke andar tiny $a^{2}$ drop karo: $(a^{2} + x^{2})^{3/2} \approx (x^{2})^{3/2} = x^{3}$ . Field ab $1/ x^{3}$ ki tarah fade karti hai — yeh magnetic dipole ki pehchaan hai. Door se ek current loop ek tiny bar magnet se alag nahi dikhta. Quantity $I a^{2}$ (ek factor $π$ tak) loop ka magnetic moment hai — uski "dipole strength".

PICTURE. $B$ versus $x$ ka curve: centre par tall aur finite, sagging, aur $1/ x^{3}$ ki tarah far out trail karta hua — exact curve aur uska far-field approximation saath mein draw kiya gaya.

Ek-picture summary

Puri derivation ek canvas par: loop kaato (Step 1) → Biot–Savart se har piece ke liye ek tilted $d B$ (Step 2) → sab ke liye same slant distance $r$ (Step 3) → sideways parts cancel, axial parts bachte hain (Step 5) → axial parts ko circumference $2 π a$ par sum karo (Step 6) → boxed answer, apne centre aur far-field faces ke saath (Steps 7–8).

Recall Feynman retelling — kisi dost ko batao

Hum chahte the ki ek current ring ke middle se guzarne wali line par baithne wale point par magnetic "swirl" strength kya hai. Hum easy shortcut (Ampère) use nahi kar sakte the kyunki yahan koi achha symmetric path nahi hai, toh humne honest tarika kiya: ring ko ek million chhote current-arrows mein kaata. Har chhota arrow ek chhota magnetic arrow banata hai, aur — yahi neat part hai — kyunki point axis par bilkul centre mein hai, har chhota current-arrow same distance par hai. Har chhota magnetic arrow tilted hai, toh humne ise "axis-ke-along" part aur "sideways" part mein split kiya. Har arrow ko ring ke seedha doosri taraf wale ke saath pair karo: unke sideways parts opposite point karte hain aur ek doosre ko wipe out karte hain, jabki unke along-axis parts team up karte hain. Toh humhe sirf axis parts add karne the — aur kyunki woh sab identical hain, unhe add karna bas "ek piece times circumference $2 π a$ " hai. Bookkeeping karo aur nikal aata hai $μ_{0} I a^{2} / [2 (a^{2} + x^{2})^{3/2}]$ . Centre par khado toh yeh $μ_{0} I /2 a$ ban jaata hai; door khado toh ring ek tiny bar magnet ki tarah $1/ x^{3}$ ke saath fade karti dikhti hai.

Recall

Loop ke axis ke liye Biot–Savart kyun use karna chahiye, Ampère kyun nahi? ::: Koi closed path nahi hai jis par $B$ constant ho, toh $B$ ko Ampère ke integral se baahir nahi nikala ja sakta — hume directly contributions add karni padegi. Axis par, $r = a^{2} + x^{2}$ har wire segment ke liye same kyun hai? ::: Kyunki $P$ axis par hai: har segment ek leg $a$ baahir aur ek leg $x$ along baitha hai, congruent right triangles banaata hai. $d B$ ke kaunse components cancel hote hain aur kyun? ::: Axis-ke-perpendicular parts; diametrically opposite segment ek equal aur opposite perpendicular part provide karta hai. $\oint d l$ kya evaluate hota hai aur constants integral se baahir kyun ja sakte hain? ::: $2 π a$ (circumference); baaki factors har segment ke liye identical hain, toh woh loop ke around constant hain. Loop ke centre par field ::: $B = \frac{μ _{0} I}{2 a}$ Far-field fall-off of an axial loop field ::: $B \propto 1/ x^{3}$ (dipole behaviour)

Related tools aur yeh kahan le jaata hai: Biot-Savart Law · Ampere's Law · Bar magnet and magnetic dipole · Magnetic force on a current-carrying conductor · Magnetic flux · Faraday's law of induction.