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

1.8.25 · D4 · Physics › Electromagnetism › Magnetic flux Φ = ∫B·dA

Yeh hai tumhara training ground Magnetic Flux ke liye. Har problem ko pehle khud solve karo, tab uska solution kholo. Levels seedhi stairs ki tarah chadhte hain: pehle tum formula pehchante ho, phir apply karte ho, phir thode tricky setups analyse karte ho, phir kai ideas ko synthesise karte ho, aur aakhir mein mastery problems tak pahunchte ho jo flux ko Faraday's Law of Induction, Lenz's Law aur Gauss's Law for Magnetism ke saath combine karte hain.

Sab kuch ek formula aur ek caution par tika hai:

Chalte hain pehle us vocabulary ko pakka karte hain jo har problem mein milegi, taaki koi bhi symbol anjaan na lage.

Is picture mein, blue arrows field $B$ hain, green disc surface hai, yellow arrow uska normal $\hat{n}$ hai, aur unke beech ka red angle $θ$ hai. Har problem ke liye yeh image apne dimag mein rakho.

Level 1 — Recognition

Kya tum $B$ , $A$ , $θ$ identify karke plug in kar sakte ho?

L1.1 — Straight-on square

Ek square loop jiska side $0.30 m$ hai, apne normal ke saath ek uniform field $B = 0.80 T$ ke saath rakha hua hai. $Φ$ nikalo.

Recall Solution L1.1

WHAT: har quantity identify karo. WHY: normal ka field ke saath hona matlab $θ = 0^{\circ}$ hai, woh head-on case jahan har line pass karti hai.

Area: $A = (0.30)^{2} = 0.090 m^{2}$ (square area = side²).
$cos 0^{\circ} = 1$ .
$Φ = B A cos θ = 0.80 \times 0.090 \times 1 = 0.072 Wb$ .

L1.2 — Edge-on ring

Ek circular ring jiska radius $0.10 m$ hai, edge-on pakda hua hai taaki field ring ke plane mein flat ho, $B = 1.2 T$ . $Φ$ nikalo.

Recall Solution L1.2

WHAT: edge-on matlab field surface ke andar hai, toh normal $B$ se $⊥$ hai, yaani $θ = 9 0^{\circ}$ .

$cos 9 0^{\circ} = 0$ .
$Φ = B A cos 9 0^{\circ} = 0 Wb$ . WHAT IT LOOKS LIKE: lines ring ke saath seedhi nikal jaati hain jaise baarish sideways pakde page ke saath slide kare — koi bhi pierce nahi karti.

Level 2 — Application

Ab angle $0$ ya $9 0^{\circ}$ nahi hai — tumhe $cos θ$ honestly use karna hoga, aur kabhi kabhi picture se $θ$ padhna hoga.

L2.1 — Tilted loop

Ek rectangular loop $0.40 m \times 0.25 m$ ek uniform field $B = 0.60 T$ mein rakhi hai. Uska normal $B$ se $3 0^{\circ}$ banata hai. $Φ$ nikalo.

Recall Solution L2.1

$A = 0.40 \times 0.25 = 0.10 m^{2}$ .
$θ = 3 0^{\circ}$ (pehle se normal se diya gaya hai — achha).
$cos 3 0^{\circ} = \frac{3}{2} \approx 0.8660$ .
$Φ = 0.60 \times 0.10 \times 0.8660 = 0.0520 Wb$ .

L2.2 — Angle given from the surface, not the normal

Ek flat coil jiska area $0.050 m^{2}$ hai, field aur coil ke plane ke beech $4 0^{\circ}$ ka angle banata hai. Field $B = 0.90 T$ . $Φ$ nikalo.

Recall Solution L2.2

WHAT: problem plane se angle deta hai, ise $α = 4 0^{\circ}$ kaho. WHY convert: formula ko normal se angle chahiye. Normal plane se $9 0^{\circ}$ par hai, toh $θ = 9 0^{\circ} - α = 9 0^{\circ} - 4 0^{\circ} = 5 0^{\circ} .$ WHAT IT LOOKS LIKE: figure mein, $α$ (green) surface ke saath chipka hai, $θ$ (red) yellow normal tak pahunchta hai; dono milkar right angle banate hain.

$cos 5 0^{\circ} \approx 0.6428$ .
$Φ = 0.90 \times 0.050 \times 0.6428 = 0.0289 Wb$ .

Level 3 — Analysis

Field ya surface vary karti hai — integral apna kaam karta hai. Aur bhi: normal ka sign aur direction.

L3.1 — Field jo surface par badhti hai

Ek field $\hat{z}$ direction mein point karti hai jiska magnitude $B = (0.50 x) \hat{z} T$ hai ( $x$ metres mein). Yeh $x y$ -plane mein ek rectangle se pass hoti hai jo $0 \leq x \leq 3$ , $0 \leq y \leq 2$ span karta hai. Normal $\hat{z}$ hai. $Φ$ nikalo.

Recall Solution L3.1

WHY integrate: $B$ ka $x$ par depend karna matlab koi single $B A$ kaam nahi karega — patli strips mein slice karo jahan $B$ nearly constant ho, sum karo, limit lo.

Kyunki $B ∥ \hat{n}$ (dono $\hat{z}$ ke saath hain), $B \cdot d A = 0.50 x d A$ . $Φ = \int_{0}^{2} \int_{0}^{3} 0.50 x d x d y = 0.50 [\frac{x ^{2}}{2}]_{0}^{3} [y]_{0}^{2} = 0.50 \times \frac{9}{2} \times 2.$
$Φ = 0.50 \times 4.5 \times 2 = 4.5 Wb$ .

L3.2 — Field perpendicular direction mein badhti hai

Same rectangle $x y$ -plane mein ( $0 \leq x \leq 3$ , $0 \leq y \leq 2$ ), lekin ab $B = (0.50 z) \hat{z} T$ . $Φ$ nikalo.

Recall Solution L3.2

WHAT's the catch: poora rectangle $z = 0$ par rakha hai. Toh surface par $z = 0$ har jagah hai, matlab $B = 0.50 \times 0 = 0$ har patch par. $Φ = \int_{S} 0 d A = 0 Wb .$ WHY it matters: field $z$ ke saath vary karti hai, lekin surface kabhi $z \neq = 0$ ko sample hi nahi karti. Varying field tabhi flux badhati hai jab surface actually wahan ho jahan field nonzero ho.

L3.3 — Normal ko reverse karna

L3.1 ke setup ke liye, maan lo hum normal ko $- \hat{z}$ ki taraf choose karte hain. Ab $Φ$ kya hai?

Recall Solution L3.3

WHY: $\hat{n}$ ko flip karna $B \cdot d A$ ka sign har jagah flip kar deta hai, kyunki ab field aur normal anti-parallel hain ( $θ = 18 0^{\circ}$ , $cos 18 0^{\circ} = - 1$ ). $Φ = - 4.5 Wb .$ Flux ka sign is par depend karta hai ki tum "out" kis taraf mante ho — yeh ek bookkeeping choice hai. Magnitude fixed hai; sign tumhare chosen normal ke hisaab se aata hai. Yahi choice Faraday's Law of Induction mein sign fix karti hai.

Level 4 — Synthesis

Flux ko turns ke saath, closed surfaces ke saath, aur rate of change ke saath combine karo.

L4.1 — Coil with many turns and a rate of change

Ek coil jiske $N = 200$ turns aur area $A = 0.020 m^{2}$ hai, normal field ke parallel rakhi hai, aur field $0.30 s$ mein uniformly $0$ se $0.60 T$ tak badhti hai. (a) Final flux linkage $N Φ$ nikalo, aur (b) induced EMF ki magnitude $∣ E ∣ = N \frac{d Φ}{d t}$ nikalo.

Recall Solution L4.1

(a) WHAT: ek turn se flux nikalo, phir $N$ se multiply karo kyunki har turn same flux link karta hai.

$Φ = B A cos 0 = 0.60 \times 0.020 \times 1 = 0.012 Wb$ .
Flux linkage $= N Φ = 200 \times 0.012 = 2.4 Wb$ .

(b) WHY a derivative: EMF jawab deta hai "flux kitni tezi se change ho raha hai?" — yeh rate of change hai, Faraday's Law of Induction se derive hota hai.

$\frac{d Φ}{d t} = A cos 0 \cdot \frac{d B}{d t} = 0.020 \times \frac{0.60}{0.30} = 0.020 \times 2.0 = 0.040 Wb/s$ .
$∣ E ∣ = N \frac{d Φ}{d t} = 200 \times 0.040 = 8.0 V$ .

L4.2 — Closed surface (Gauss for magnetism)

Ek cube jiska side $0.10 m$ hai, ek uniform field $B = 2.0 T$ mein rakha hai jo $+ \hat{x}$ ki taraf point karta hai. Cube ki poori closed surface se total flux nikalo.

Recall Solution L4.2

WHY zero without computing six faces: Gauss's Law for Magnetism kehta hai $\oint B \cdot d A = 0$ kisi bhi closed surface ke liye — field lines ka koi start ya end nahi hota, toh har line jo cube mein enter karti hai woh bahar bhi nikalti hai.

Face at $x = 0$ : outward normal $- \hat{x}$ hai, flux $= - B \cdot (0.10)^{2} = - 0.020 Wb$ .
Face at $x = 0.10$ : outward normal $+ \hat{x}$ hai, flux $= + 0.020 Wb$ .
Baaki chaar faces: normal $\hat{x}$ se $⊥$ hai, toh $θ = 9 0^{\circ}$ , har ek $0$ contribute karta hai.
Total $= - 0.020 + 0.020 + 0 = 0 Wb$ . ✓

Level 5 — Mastery

Multi-step reasoning: rotating loops, aur Lenz's law se direction padhna.

L5.1 — Rotating loop, flux as a function of time

Ek loop jiska area $A = 0.040 m^{2}$ hai ek uniform field $B = 0.50 T$ mein rotate karta hai taaki uske normal aur field ke beech angle $θ (t) = ω t$ ho, jahan $ω = 100 rad/s$ . (a) $Φ (t)$ likho. (b) Ek turn ke liye peak EMF $∣ E ∣_{m a x}$ nikalo.

Recall Solution L5.1

(a) WHAT: $θ (t) = ω t$ ko $Φ = B A cos θ$ mein substitute karo. $Φ (t) = B A cos (ω t) = 0.50 \times 0.040 \times cos (100 t) = 0.020 cos (100 t) Wb .$

(b) WHY a derivative and why it peaks: EMF $= - \frac{d Φ}{d t}$ . $cos$ differentiate karne par $- sin$ milta hai: $E = - \frac{d Φ}{d t} = B A ω sin (ω t) .$

$sin$ $- 1$ aur $+ 1$ ke beech swing karta hai, toh peak magnitude $B A ω$ hai: $∣ E ∣_{m a x} = B A ω = 0.50 \times 0.040 \times 100 = 2.0 V .$ WHAT IT LOOKS LIKE: flux tab maximum hota hai jab loop field face kare ( $cos = 1$ ), lekin EMF maximum hota hai ek quarter-turn baad, jab loop edge-on ho aur flux sabse tezi se change ho raha ho ( $sin = 1$ ).

L5.2 — Direction of induced current (Lenz)

L5.1 mein, us instant par jab loop ka flux decrease ho raha ho (face-on se edge-on ki taraf ja raha ho), induced current kis taraf flow karti hai — girte hue flux ko reinforce karne ke liye ya oppose karne ke liye?

Recall Solution L5.2

WHY oppose: Lenz's Law kehta hai induced current hamesha flux ki change se ladti hai (energy conservation — nature free lunches se resist karta hai).

Flux gir raha hai, toh induced current us direction mein flow karti hai jiska apna magnetic field loop se $B$ ke same direction mein point kare — flux ko prop up karne ki koshish karta hai, yaani ise maintain karne ki.
Yeh change ko reinforce nahi karti; yeh change ko oppose karti hai (yahan, decrease ko oppose karti hai flux add karke).

Wrap-up

Related building blocks: Surface and Line Integrals, Magnetic Field B, Electric Flux, Inductance.