Magnetic flux Φ = ∫B·dA: Worked examples — Hinglish deep dive

1.8.25 · D3 · Physics › Electromagnetism › Magnetic flux Φ = ∫B·dA

Yeh page ek drill hai. Hum flux ka idea lete hain — $Φ = B A cos θ$ aur uska integral wala bada bhai $Φ = \int_{S} B \cdot d A$ — aur ise har tarah ke question se tackle karte hain jo yeh produce kar sakta hai. Shuru karne se pehle, ek reminder simple words mein:

Neeche sab kuch anhi do knobs ke baare mein hai, jo har possible setting par ghuma diye gaye hain.

The scenario matrix

Yeh un case-classes ki complete list hai jinmein flux problems aate hain. Har cell ko neeche kam se kam ek fully worked example milta hai.

#	Case class	Pehchaan karne wali khaasiyat	Example
C1	Face-on ( $θ = 0$ )	Field $∥$ normal, flux maximum $= B A$ hai	Ex 1
C2	Tilted ( $0 < θ < 9 0^{\circ}$ )	Partial projection, $cos θ$ 0 aur 1 ke beech	Ex 2
C3	Edge-on ( $θ = 9 0^{\circ}$ ) — degenerate	Field surface ke andar hai, flux $= 0$	Ex 3
C4	Obtuse angle ( $9 0^{\circ} < θ \leq 18 0^{\circ}$ ) — negative flux	Field surface ke peeche se pierce karta hai, $cos θ < 0$	Ex 4
C5	Non-uniform field	$B$ surface par vary karta hai → integrate karna padega	Ex 5
C6	Closed surface — degenerate	Jo bhi andar aata hai woh bahar bhi jaata hai, $\oint = 0$	Ex 6
C7	Multi-turn coil	Flux linkage $N Φ$ , bhoolna asaan hai	Ex 7
C8	Real-world word problem	Words ko $B, A, θ$ mein translate karo	Ex 8
C9	Exam twist: angle-with-surface trap	Woh angle plane se dete hain, normal se nahi	Ex 9
C10	Limiting behaviour	Kya hota hai jab $θ$ , $0 \to 18 0^{\circ}$ sweep karta hai	Ex 10

$cos θ$ ka sign is poori table ki reedh hai, isliye pehle ise visible karte hain.

Worked examples

C1 — Face-on loop (maximum flux)

Worked example Ex 1

Radius $r = 0.10 m$ ka ek circular loop ek uniform field $B = 0.80 T$ ke saath apna normal seedha field ki taraf karke rakha hua hai. $Φ$ nikalo.

Forecast: loop field ko seedha face kar raha hai. Guess karo: kya $cos θ$ yahan help karega ya hurt?

Circle ka area: $A = π r^{2} = π (0.10)^{2} = 0.0314 m^{2}$ . Yeh step kyun? Humein woh physical area chahiye jisse field guzarti hai; circle ka area $π r^{2}$ hota hai.
Normal $∥ B$ ka matlab hai $θ = 0$ , isliye $cos θ = 1$ . Yeh step kyun? $θ$ field aur normal ke beech ka angle hai; yahan dono ek hi direction mein hain — maximum-flux setting.
$Φ = B A cos θ = 0.80 \times 0.0314 \times 1 = 0.0251 Wb$ . Yeh step kyun? Face-on = $cos 0 = 1$ ke saath poora uniform-field formula.

Verify karo: Units $T \cdot m^{2} = Wb$ ✓. Yeh is loop ka is field mein sabse zyada flux hai — koi bhi tilt ise sirf kam karega. Sanity ✓.

C2 — Tilted loop (partial projection)

C3 — Edge-on loop (degenerate, zero flux)

C4 — Obtuse angle (NEGATIVE flux)

Worked example Ex 4

Wahi loop aur field, ab normal $B$ se $θ = 12 0^{\circ}$ par hai — yaani field loop ke back face se guzar rahi hai (chosen normal direction ke relative).

Forecast: flux number kuch aisa karega jo parent note ne kabhi nahi dikhaya. Uska sign guess karo.

$cos 12 0^{\circ} = - 0.5$ . Yeh step kyun? $9 0^{\circ}$ ke baad field surface ko normal ki opposite taraf se pierce karti hai, isliye normal ke saath uska component negative ho jaata hai.
$Φ = B A cos 12 0^{\circ} = 0.80 \times 0.0314 \times (- 0.5) = - 0.0126 Wb$ . Yeh step kyun? Dot product apne aap sign carry karta hai.

Verify karo: Sign sirf ek bookkeeping label hai "lines chosen normal ke kis side se cross karti hain" ke liye. Iska magnitude $0.0126 Wb$ , $θ = 6 0^{\circ}$ par milne wale flux ke barabar hai (kyunki $∣ cos 12 0^{\circ} ∣ = ∣ cos 6 0^{\circ} ∣ = 0.5$ ) — $s 01$ curve par ek acha symmetry check. ✓

Yeh kyun matter karta hai: Faraday's law aur Lenz's Law ko $Φ$ ki change ki parwah hai, aur flux ka positive se negative ho jaana ek real change hai jo current drive karta hai.

C5 — Non-uniform field (integrate karna zaroori hai)

Worked example Ex 5

Ek field $\hat{z}$ ki taraf point karti hai lekin position ke saath badhti hai: $B = (0.20 + 0.10 x) \hat{z} T$ ( $x$ metres mein). $x y$ -plane mein rakhe square $0 \leq x \leq 3$ , $0 \leq y \leq 2$ se, normal $\hat{z}$ , flux nikalo.

Forecast: ek $B A$ use nahi kar sakte — kyun? Guess karo answer $x = 0$ par field use karne se bada hoga ya chota?

$B \cdot d A = (0.20 + 0.10 x) d A$ kyunki field $∥$ normal hai. Yeh step kyun? Jab dono vectors $\hat{z}$ ki taraf hon to dot product sirf magnitudes ka product hai — Surface and Line Integrals dekho.
Square par double integral set up karo: $Φ = \int_{0}^{2} \int_{0}^{3} (0.20 + 0.10 x) d x d y .$ Yeh step kyun? $B$ , $x$ ke saath change karta hai, isliye hum strips mein slice karte hain jahan $B$ locally constant ho, phir sum (integrate) karte hain.
Inner integral: $\int_{0}^{3} (0.20 + 0.10 x) d x = [0.20 x + 0.05 x^{2}]_{0}^{3} = 0.60 + 0.45 = 1.05$ . Yeh step kyun? Constant ka antiderivative linear hai; $x$ ka $x^{2} /2$ hai, aur $0.10/2 = 0.05$ .
Outer integral: $\int_{0}^{2} 1.05 d y = 1.05 \times 2 = 2.10 Wb$ . Yeh step kyun? Kuch bhi $y$ par depend nahi karta, isliye $y$ par integrate karna sirf width $2$ se multiply karna hai.

Verify karo: Agar field $0 \leq x \leq 3$ par apni average value par constant hoti, woh average $0.20 + 0.10 (1.5) = 0.35 T$ hai, jo $0.35 \times (3 \times 2) = 2.10 Wb$ deta hai — exact match, kyunki field linear hai isliye average trick exact hai. ✓

C6 — Closed surface (degenerate, hamesha zero)

Worked example Ex 6

Ek uniform field $B = 1.5 T$ , $\hat{x}$ ki taraf, side $0.20 m$ ke ek closed cube se guzarti hai. Cube se bahar jaane wala total flux nikalo.

Forecast: do faces field ke perpendicular hain. Sum karne se pehle total guess karo.

Sirf woh do faces flux rakhti hain jo $\hat{x}$ ke $⊥$ hain; baaki chaar edge-on hain ( $θ = 9 0^{\circ}$ , zero each). Yeh step kyun? Edge-on faces kuch contribute nahi karti ( $cos 9 0^{\circ} = 0$ ).
Face area $A = (0.20)^{2} = 0.040 m^{2}$ . Entry face par outward normal $- \hat{x}$ ki taraf point karta hai, isliye $θ = 18 0^{\circ}$ : $Φ_{in} = B A cos 18 0^{\circ} = 1.5 \times 0.040 \times (- 1) = - 0.060 Wb$ . Yeh step kyun? Outward normal field ko oppose karta hai jahan lines enter karti hain → negative flux.
Exit face par outward normal $+ \hat{x}$ hai, $θ = 0$ : $Φ_{out} = 1.5 \times 0.040 \times (+ 1) = + 0.060 Wb$ . Yeh step kyun? Lines outward normal ke along nikalti hain → positive flux.
Total: $\oint B \cdot d A = - 0.060 + 0.060 + 0 = 0 Wb$ . Yeh step kyun? Har face sum karo.

Verify karo: Yeh exactly Gauss's Law for Magnetism hai: koi monopole nahi, jo bhi line andar aati hai woh bahar bhi jaati hai, isliye closed surface hamesha $0$ deti hai — Electric Flux se contrast karo jahan enclosed charge ise nonzero banata hai. ✓

C7 — Multi-turn coil (flux linkage)

C8 — Real-world word problem

C9 — Exam twist: angle-to-the-surface trap

Worked example Ex 9

$A = 0.020 m^{2}$ area wala ek loop $B = 0.60 T$ mein is tarah tilted hai ki field loop ke plane ke saath $2 5^{\circ}$ banata hai. $Φ$ nikalo. (Wording dhyan se padho!)

Forecast: unhone angle plane se diya, normal se nahi. Guess karo kya $cos 2 5^{\circ}$ plug karoge ya kuch aur.

Convert karo: normal se angle $θ = 9 0^{\circ} - 2 5^{\circ} = 6 5^{\circ}$ . Yeh step kyun? Normal plane se $9 0^{\circ}$ dur hai, isliye plane se $2 5^{\circ}$ ka angle, normal se $6 5^{\circ}$ ka angle hai. Formula ka $θ$ hamesha normal se hota hai.
$Φ = B A cos 6 5^{\circ} = 0.60 \times 0.020 \times 0.4226 = 0.00507 Wb$ . Yeh step kyun? Ab $θ$ normal se theek se measure ki gayi hai.

Verify karo: Wrong angle $cos 2 5^{\circ} = 0.906$ use karne se $0.0109 Wb$ milta — sach se double se bhi zyada. Doosra sahi tarika hai $B A sin (angle-to-plane) = 0.60 \times 0.020 \times sin 2 5^{\circ} = 0.00507 Wb$ , jo match karta hai, confirm karta hai $cos θ = sin (plane angle)$ . ✓

C10 — Limiting behaviour (angle sweep karo)

Worked example Ex 10

Ek fixed loop lo, $B A = 0.10 Wb$ face-on ho toh. Flux tabulate karo jaise $θ$ (normal se angle) $0 \to 18 0^{\circ}$ sweep karta hai, aur trend describe karo.

Forecast: compute karne se pehle $0^{\circ}$ , $9 0^{\circ}$ , aur $18 0^{\circ}$ par value guess karo.

$θ = 0^{\circ}$ : $Φ = 0.10 cos 0 = + 0.10 Wb$ (maximum).
$θ = 6 0^{\circ}$ : $Φ = 0.10 cos 6 0^{\circ} = + 0.050 Wb$ (half).
$θ = 9 0^{\circ}$ : $Φ = 0.10 cos 9 0^{\circ} = 0 Wb$ (edge-on).
$θ = 12 0^{\circ}$ : $Φ = 0.10 cos 12 0^{\circ} = - 0.050 Wb$ (back-face, negative).
$θ = 18 0^{\circ}$ : $Φ = 0.10 cos 18 0^{\circ} = - 0.10 Wb$ (maximum reversed). Yeh steps kyun? Hum sirf $s 01$ cosine curve ko sample points par read kar rahe hain taaki poora behaviour range dekh sakein.

Verify karo: Flux continuously $[- 0.10, + 0.10] Wb$ range mein rehta hai, $θ = 9 0^{\circ}$ ke baare mein symmetric hai, aur exactly wahan zero hit karta hai. $\pm B A$ se bahar kabhi kuch bhi possible nahi — ek hard ceiling aur floor. ✓

Recall Self-test: yeh kaun sa cell hai?

A field lies at $7 0^{\circ}$ to the plane of a loop. Which matrix cell, and what's $θ$ ? ::: C9 (plane-angle trap); $θ = 9 0^{\circ} - 7 0^{\circ} = 2 0^{\circ}$ normal se. Flux through a sealed sphere in any field? ::: C6; magnetism ke Gauss's law se hamesha $0$ . A loop at $θ = 13 5^{\circ}$ from the normal — sign of flux? ::: C4; negative, kyunki $cos 13 5^{\circ} < 0$ . Why can't you use $B A$ when $B = 0.1 x \overset{z}{^}$ ? ::: C5; $B$ position ke saath vary karta hai, isliye strip by strip integrate karo.

Worked examples — Magnetic flux Φ = ∫B·dA