Self-inductance L, mutual inductance M: Worked examples

1.8.28 · D3 · Physics › Electromagnetism › Self-inductance L, mutual inductance M

Yeh page ek drill hai. Parent note ne formulas banaye; yahan hum unpe har tarah ka input daalte hain — har sign, zero cases, degenerate geometry, limiting values, ek word problem, aur ek exam twist — taaki koi bhi scenario tumhe surprise na kar sake.

Yahan use kiya gaya har symbol parent mein define hai, lekin hum usse wahi restate karte hain jab use karte hain, taaki tum line one se padh sako.

Scenario matrix

Neeche un "case classes" ki poori list hai jo yeh topic tumse pooch sakta hai. Baad ke har example mein us cell ka tag lagaa hai jo use cover karta hai.

Cell	Kya cheez ise distinct banati hai	Covered by
A. Pure geometry	$L$ compute karo turns/area/length se	Ex 1
B. Sign of EMF — current rising	$d I / d t > 0 \Rightarrow$ EMF opposes (drops sign)	Ex 2a
C. Sign of EMF — current falling	$d I / d t < 0 \Rightarrow$ EMF aids current	Ex 2b
D. Zero/degenerate — steady current	$d I / d t = 0 \Rightarrow ε = 0$	Ex 2c
E. Oscillating (AC) current	$d I / d t$ khud oscillate karta hai; peak EMF	Ex 3
F. Mutual $M$ from geometry + coupling $k$	Do coils, $M$ read karo, $k$ check karo	Ex 4
G. Limiting behaviour	$N \to 2 N$ , $ℓ \to 0$ , $k \to 0$	Ex 5
H. Energy & flywheel	$U = \frac{1}{2} L I^{2}$ , energy release	Ex 6
I. Real-world word problem	English → symbols mein translate karo	Ex 7
J. Exam twist — series/combined coils	$L$ of do coils in series (aiding/opposing)	Ex 8

Chalte hain, inhe ek-ek karke solve karte hain.

Symbols, ek baar restate kiye

Definition Coupling coefficient

$k$ (use se pehle define kiya gaya) Jab do coils ek-doosre ke paas hoti hain, coil 1 ka flux generally poora coil 2 tak nahi pahunchta — kuch field hawa mein leak ho jaata hai. Coupling coefficient $k$ woh seedha fraction hai "coil 1 ka kitna flux actually coil 2 se thread hota hai": $M = k L_{1} L_{2}, - 1 \leq k \leq 1.$

$∣ k ∣ = 1$ : ek coil ki har field line doosri se link hoti hai (perfect linkage — ideal transformer).
$k = 0$ : coils koi flux share nahi karti (door khich gayi, ya perpendicular).
$k$ ka sign: positive agar do coils aise wound hain ki unke fluxes same direction mein current lete waqt reinforce karte hain; negative agar ek coil opposite sense mein wound hai (ya circuit mein reverse ki gayi hai) taaki uska flux doosre ka oppose kare. Negative $k$ (equivalently, negative $M$ ) cross-linkage term ka sign flip kar deta hai — yahi cheez Example 8 mein "series-aiding" ko "series-opposing" banati hai. Textbooks often $0 \leq k \leq 1$ quote karte hain aur winding sense ko alag se $M$ pe $\pm$ ke roop mein rakhte hain; dono bookkeepings same physics deti hain.

Definition Induced EMF

$ε$ aur uski polarity convention $ε$ (Greek "epsilon") induced electromotive force hai — woh voltage jo changing flux ek loop ke around "dhakelta" hai, volts (V) mein measure kiya jaata hai. Yeh koi force nahi hai; yeh induced field ka work per unit charge hai jo ek loop ke around karta hai. Polarity convention jo poori page pe use hogi: hum $ε$ ki positive direction ko driving current $I$ ki same direction lete hain. Tab:

$ε > 0$ matlab induced EMF current ke saath push karta hai (use sustain karne ki koshish karta hai);
$ε < 0$ matlab induced EMF current ke against push karta hai (use ghatane ki koshish karta hai). Is choice ke saath self-EMF hai $ε = - L \frac{d I}{d t}$ , aur leading minus sign hi Lenz's law hai: rising current ( $d I / d t > 0$ ) $ε < 0$ deta hai (opposes), falling current $ε > 0$ deta hai (aids).

Example 1 — Cell A: pure geometry

Worked example Solenoid ka

$L$ compute karo Ek solenoid mein $N = 800$ turns hain, area $A = 2 \times 1 0^{- 4} m^{2}$ , length $ℓ = 0.40 m$ hai. $L$ find karo.

Forecast: Pehle order of magnitude guess karo — kya yeh henries, millihenries, ya microhenries mein hoga? (Is size ke coils usually sub-millihenry hote hain.)

Step 1 — Box formula likho. $L = μ_{0} \frac{N ^{2} A}{ℓ}$ . Yeh step kyun? Yeh long solenoid ke liye ek hi self-inductance formula hai — yeh parent mein Ampère → flux → linkage se aaya tha. Sab kuch plug-in hai.

Step 2 — Numbers daalo. $L = (4 π \times 1 0^{- 7}) \frac{( 800 ) ^{2} ( 2 \times 1 0 ^{- 4} )}{0.40}$ Yeh step kyun? $μ_{0}$ ko end tak symbolic rakho taaki hum jaldi round na karein.

Step 3 — Calculate karo. $(800)^{2} = 6.4 \times 1 0^{5}$ . Times $2 \times 1 0^{- 4} = 128$ . Divide by $0.40 = 320$ . Times $4 π \times 1 0^{- 7}$ : $L \approx 4.02 \times 1 0^{- 4} H = 0.402 mH .$ Yeh step kyun? Hum numerator aur denominator ko pehle collapse karte hain $μ_{0}$ ko touch karne se pehle — taaki ek final multiply by $4 π \times 1 0^{- 7}$ saara rounding carry kare — yeh kai intermediate products mein round-off compounding se bachata hai aur arithmetic ko term by term auditable banata hai.

Verify: Units: $(T\cdotm/A) \cdot m^{2} / m = T\cdotm^{2} / A = Wb/A = H$ . ✓ Magnitude sub-millihenry as forecast. ✓

Example 2 — Cells B, C, D: self-EMF ke teen signs

Hum Ex 1 wali coil use karte hain ( $L = 0.402 mH$ ) aur change karte hain kaise current move karta hai. Formula hamesha $ε = - L \frac{d I}{d t},$ hai, upar fix ki gayi sign convention ke saath ( $ε > 0$ = current ke saath push karta hai).

Neeche ka figure teen phases mein wahi current plot karta hai — ek rising ramp, ek flat top, aur ek falling ramp — taaki tum seedha curve se slope $d I / d t$ padhh sako aur dekh sako ki woh $ε$ ka kaunsa sign force karta hai. Red segment rising phase hai (Cell B); flat aur falling phases black segments hain.

Figure — Self-inductance L, mutual inductance M

Recall 2c kyun poora point hai

$I$ ki value $ε = - L d I / d t$ mein kabhi appear nahi hoti — sirf uska slope karta hai. ::: Sahi; ek constant current zero self-EMF deta hai chahe kitna bhi bada ho.

Example 3 — Cell E: oscillating (AC) current

Worked example Sinusoidal drive

Current $I (t) = 3 sin (200 t) A$ ek coil mein flow karta hai jiska $L = 0.402 mH$ hai. Induced EMF aur uski peak value find karo. (Yahan $200$ $ω$ hai rad/s mein.)

Forecast: EMF ek cosine hoga (sine ka derivative) — 90° out of phase. Kya uski peak $3$ se badi hogi ya choti?

Step 1 — Differentiate karo. $\frac{d I}{d t} = 3 \cdot 200 cos (200 t) = 600 cos (200 t) A/s$ . Differentiate kyun? Kyunki EMF rate of change hai; sine ke liye, woh rate derivative hai, jo ek cosine hai $ω \times$ amplitude se scale kiya hua.

Step 2 — Formula apply karo. $ε = - L \frac{d I}{d t} = - (4.02 \times 1 0^{- 4}) (600) cos (200 t) = - 0.241 cos (200 t) V .$

Step 3 — Peak. Cosine $\pm 1$ ke beech swing karta hai, isliye $ε_{peak} = 0.241 V$ . Yeh step kyun? Exams peak (amplitude) poochte hain — yeh cosine ke aage ka coefficient hai.

Verify: Peak $= L ω I_{0} = (4.02 \times 1 0^{- 4}) (200) (3) = 0.241 V$ . ✓ Note karo $L ω$ inductive reactance hai — wahi object jo RL circuits aur Transformers ko govern karta hai.

Example 4 — Cell F: mutual inductance & coupling

Worked example Do coaxial solenoids

Do solenoids ek hi core share karte hain, length $ℓ = 0.30 m$ , area $A = 5 \times 1 0^{- 4} m^{2}$ . Inner: $N_{1} = 600$ . Outer: $N_{2} = 300$ . Assume karo perfect overlap taaki saara inner flux outer coil se thread ho. $M$ , phir $L_{1}$ , $L_{2}$ find karo, aur $k$ check karo.

Forecast: Perfect overlap ke saath, coupling coefficient $k$ ki kya value honi chahiye?

Step 1 — Mutual inductance. $M = μ_{0} \frac{N _{1} N _{2}}{ℓ} A$ . Yeh step kyun? Inner current $B_{1} = μ_{0} (N_{1} / ℓ) I_{1}$ banata hai; outer ke $N_{2}$ turns mein se har ek flux $B_{1} A$ catch karta hai; linkage ko $I_{1}$ se divide karo. $M = (4 π \times 1 0^{- 7}) \frac{( 600 ) ( 300 )}{0.30} (5 \times 1 0^{- 4}) = 3.77 \times 1 0^{- 4} H .$

Step 2 — Do self-inductances. $L_{1} = μ_{0} \frac{N _{1}^{2} A}{ℓ} = (4 π \times 1 0^{- 7}) \frac{60 0 ^{2} ( 5 \times 1 0 ^{- 4} )}{0.30} = 7.54 \times 1 0^{- 4} H,$ $L_{2} = μ_{0} \frac{N _{2}^{2} A}{ℓ} = (4 π \times 1 0^{- 7}) \frac{30 0 ^{2} ( 5 \times 1 0 ^{- 4} )}{0.30} = 1.88 \times 1 0^{- 4} H .$ Yeh step kyun? Har coil ki self-inductance uski apni $N^{2}$ ko same box formula $L = μ_{0} N^{2} A / ℓ$ mein daalne se milti hai — geometry ( $A$ , $ℓ$ ) shared hai, sirf turn count badlta hai. Hume dono $L_{1}$ aur $L_{2}$ chahiye kyunki coupling check $M$ ko unke geometric mean $L_{1} L_{2}$ se compare karta hai.

Step 3 — Coupling coefficient. $k = \frac{M}{L _{1} L _{2}}$ (uske definition se rearrange kiya gaya). $L_{1} L_{2} = (7.54 \times 1 0^{- 4}) (1.88 \times 1 0^{- 4}) = 3.77 \times 1 0^{- 4}$ . $k = \frac{3.77 \times 1 0 ^{- 4}}{3.77 \times 1 0 ^{- 4}} = 1.00.$ Yeh kyun matter karta hai: $k = 1$ "perfect linkage" idealisation confirm karta hai — transformer limit. (Agar outer coil opposite sense mein wound hoti, to hum $k = - 1$ padhte: $M$ ka same magnitude, flipped sign.)

Verify: $M = k L_{1} L_{2}$ with $k = 1$ Step 1 ka $M$ exactly reproduce karta hai. ✓ Aur $M_{12} = M_{21}$ reciprocity se — swap karo kaunsa coil source hai toh same $M$ milega. ✓

Example 5 — Cell G: limiting behaviour

Worked example Edges par kya hota hai?

Ek base coil $L_{0} = μ_{0} \frac{N ^{2} A}{ℓ}$ se shuru karo. Teen limit questions answer karo.

Forecast: Kaunsa change $L$ ko char se multiply karta hai? Kaunsa use infinity tak bheja deta hai?

(i) Turns double karo, $N \to 2 N$ (length fixed, to turns zyada tight pack hue). Kyun: $L \propto N^{2}$ . $L \to (2 N)^{2} / N^{2} \cdot L_{0} = 4 L_{0}$ . Turns double duty karte hain (flux banao + link karo). ✓

(ii) Length shrink karo, $ℓ \to 0$ (turns fixed, infinitely tight packed). Kyun: $L = μ_{0} N^{2} A / ℓ$ . Jaise $ℓ \to 0$ , $L \to \infty$ . Ek physically impossible limit, lekin yeh dikhata hai ki tighter packing $\Rightarrow$ larger $L$ .

(iii) Ex 4 ki do coils ko door kheencho, $k \to 0$ . Kyun: $M = k L_{1} L_{2} \to 0$ . Koi shared flux nahi, koi mutual effect nahi. Coils ek-doosre ko "bhool" jaati hain.

Verify (numeric anchor for (i)): Ex 1 ki coil lo, $L_{0} = 0.402 mH$ . $N$ ko double karke $1600$ karne par $L = μ_{0} (1600)^{2} (2 \times 1 0^{- 4}) /0.40 = 1.608 mH = 4 L_{0}$ milta hai. ✓

Example 6 — Cell H: stored energy & flywheel

Worked example Energy aur uski release

Ek inductor jiska $L = 0.402 mH$ hai, steady $I = 5 A$ carry karta hai. (a) Kitni energy store hai? (b) Agar woh current $0.5 ms$ mein switch off ho jaaye, to average power kitni dump hogi?

Forecast: Energy choti hai (sub-millijoule); lekin kyunki yeh jaldi exit karta hai, power surprisingly large ho sakti hai.

Step 1 — Energy. $U = \frac{1}{2} L I^{2}$ . Yeh formula kyun? Yeh woh work hai jo back-EMF ke against current ko 0 se $I$ tak build karte waqt kiya jaata hai (integral of $L I d I$ ), magnetic field mein store hota hai. $U = \frac{1}{2} (4.02 \times 1 0^{- 4}) (5)^{2} = 5.02 \times 1 0^{- 3} J = 5.02 mJ .$

Step 2 — Switch-off par average power. Energy $t = 0.5 \times 1 0^{- 3} s$ mein nikalti hai: $P_{avg} = \frac{U}{t} = \frac{5.02 \times 1 0 ^{- 3}}{0.5 \times 1 0 ^{- 3}} = 10.0 W .$ Yeh step kyun? Power = energy per unit time. Rapid switch-off exactly isliye inductor ka circuit break karne par spark aata hai — flywheel rukne se mana kar deta hai aur energy fast bahar force karta hai.

Verify: Units: J/s = W. ✓ Choti stored energy lekin 10 W transient power — flywheel intuition confirmed. ✓

Example 7 — Cell I: real-world word problem

Worked example Doorbell / relay coil

"Ek relay coil ki self-inductance $L = 0.20 H$ hai. Jab switch khulta hai, uska $0.30 A$ current $1.0 ms$ mein zero ho jaata hai. Opening contacts par spike voltage estimate karo."

Forecast: Current tiny hai (0.3 A) lekin collapse fast hai. Kya spike kuch volts hogi ya tens of volts?

Step 1 — English ko symbols mein translate karo. "1 ms mein zero ho jaata hai" $\Rightarrow \frac{d I}{d t} = \frac{0 - 0.30}{1.0 \times 1 0 ^{- 3}} = - 300 A/s$ . Yeh step kyun? Scary number $I$ nahi hai, slope hai — hamesha words se $d I / d t$ identify karo.

Step 2 — Spike EMF. $ε = - L \frac{d I}{d t} = - (0.20) (- 300) = + 60 V$ . Positive/large kyun? Current falling → coil use sustain karne ki koshish karti hai → ek coil se $60 V$ spike jo ek volt se bhi kam par chal rahi thi. Isliye relays ko flyback diodes milte hain.

Verify: $∣ ε ∣ = L ∣ d I / d t ∣ = 0.20 \times 300 = 60 V$ . ✓ Tiny current, fast collapse ⇒ bada spike, as forecast. ✓

Example 8 — Cell J: exam twist, coils in series (aiding vs opposing)

Worked example Do coupled coils in series

Do coils series mein wire ki gayi hain taaki woh same current $I$ carry karein. Unki self-inductances $L_{1} = 3 mH$ aur $L_{2} = 5 mH$ hain, aur unki mutual inductance ka magnitude $∣ M ∣ = 2 mH$ hai. Total inductance find karo (a) series-aiding ke liye (dono coils aise wound ki unke fluxes reinforce karein, $M > 0$ ) aur (b) series-opposing ke liye (ek coil reverse sense mein wound, isliye uska flux doosre se fight kare, $M < 0$ ).

Forecast: Kaun se connection mein pair current change ke baare mein zyada stubborn hai — woh jahan magnetic fields ek-doosre ki help karein, ya woh jahan fight karein?

Step 1 — Total flux linkage banao. Har coil apna flux aur neighbour ka flux link karti hai. Dono mein same current $I$ hone par, flux linkages add hote hain: $N_{1} Φ_{1} = L_{1} I \pm M I, N_{2} Φ_{2} = L_{2} I \pm M I .$ Yeh step kyun? Coil 1 apni self-linkage $L_{1} I$ plus mutual linkage $\pm M I$ dekhti hai jo coil 2 uske upar daalta hai; coil 2 bhi similarly. Sign $+$ hai agar do fluxes same taraf point karein (aiding) aur $-$ agar oppose karein. Yahan coupling sign definitions box se enter hoti hai: $M > 0$ (positive $k$ ) $\Rightarrow +$ , $M < 0$ (negative $k$ ) $\Rightarrow -$ .

Step 2 — Ek equivalent inductance mein sum karo. Poora series pair ek single inductor $L_{total}$ ki tarah behave karta hai jo $N Φ_{total} = L_{total} I$ obey karta hai. Do linkages add karo: $L_{total} I = (L_{1} I \pm M I) + (L_{2} I \pm M I) \Rightarrow L_{total} = L_{1} + L_{2} \pm 2 M .$ $2 M$ kyun aur sirf $M$ nahi? Cross-linkage do baar count hoti hai — ek baar "coil 2 ka flux coil 1 se thread hona" ke roop mein aur ek baar "coil 1 ka flux coil 2 se thread hona" ke roop mein. Reciprocity ( $M_{12} = M_{21} = M$ ) un dono ko equal banata hai, isliye woh ek single $2 M$ mein combine hote hain.

Step 3 — Aiding ( $M > 0$ ). $L_{total} = 3 + 5 + 2 (2) = 12 mH$ . Yeh step kyun? Aiding $+ 2 M$ branch leta hai; har contribution ( $3$ , $5$ , aur $2 M = 4$ ) plus sign carry karta hai kyunki saare fluxes reinforce hote hain — pair maximally stubborn hai.

Step 4 — Opposing ( $M < 0$ ). $L_{total} = 3 + 5 - 2 (2) = 4 mH$ . Yeh step kyun? Ek winding reverse karna $k$ (aur hence $M$ ) ko negative flip karta hai, isliye cross term $- 2 M = - 4$ ban jaata hai; self terms $3 + 5$ positive rehte hain lekin opposing flux linkage ka kuch hissa cancel kar deta hai, bahut chota total chhod ke.

Verify: Aiding (12) > opposing (4), as forecast — reinforcing fields stiffer inductor banate hain. Legality check: $∣ M ∣ \leq L_{1} L_{2} = 15 \approx 3.87 mH$ , aur hamaara $∣ M ∣ = 2$ yeh obey karta hai ( $∣ k ∣ = 2/3.87 \approx 0.52 \leq 1$ ), isliye dono numbers physically admissible hain. ✓

Connections

Yeh note Hinglish mein — parent topic.
Faraday's law of induction — $ε = - d (N Φ) / d t$ ka source jo har sign example mein use hua.
Lenz's law — Ex 2a/2b/2c/7 mein minus sign.
Ampère's law — geometry examples ke liye $B$ deta hai.
Magnetic flux — linked quantity.
Energy stored in magnetic field — Ex 6.
RL circuits, Transformers — jahan $L ω$ aur $k$ phir se aate hain.