Self-inductance L, mutual inductance M: Question bank

1.8.28 · D5 · Physics › Electromagnetism › Self-inductance L, mutual inductance M

Spot the error

ε = - L I

, isliye 5 A wali coil mein 1 A wali coil se paanch guna zyada back-EMF hoga."

Formula galat hai: yeh

ε = - L d I / d t

hai, jisme current ka rate of change use hota hai, current khud nahi. Constant 5 A zero back-EMF deta hai, bilkul constant 1 A ki tarah.

L = Φ/ I

ek solenoid ke liye, isliye

L = μ_{0} N A / ℓ

Definition total flux linkage

N Φ

use karta hai, sirf

Φ

nahi:

L = N Φ/ I = μ_{0} N^{2} A / ℓ

. Extra

N

drop karne se poora

N^{2}

dependence kho jaata hai.

"Kyunki

B = μ_{0} n I

sirf turns-per-metre

n

par depend karta hai, same

n

wale do solenoids ka

L

same hoga."

B

toh

n

par depend karta hai, lekin

L = μ_{0} n^{2} A ℓ

cross-section

A

aur length

ℓ

par bhi depend karta hai. Same

n

lekin alag

A

ℓ

hone par alag

L

milega.

"Mutual inductance ke liye do coils ko electrically touch karna zaroori hai."

Koi electrical connection nahi chahiye — coupling shared magnetic flux (Magnetic flux) ke through hoti hai, shared current ke through nahi. Exactly isi tarah ek transformer kaam karta hai: do isolated windings.

L

find karne ke liye hum

N Φ = L I

differentiate karte hain aur paate hain

ε = - L d I / d t - I d L / d t

Rigid coil ke liye geometry fixed hoti hai, isliye

d L / d t = 0

aur doosra term vanish ho jaata hai. Extra term tab hi aata hai jab coil ki shape ya core physically change ho rahi ho.

"Ek single straight wire ki koi inductance nahi hoti kyunki yeh loop nahi hai."

Ek straight wire bhi apne aas-paas flux carry karta hai aur uski ek (chhoti) self-inductance hoti hai. Koi bhi current-carrying conductor magnetic energy store karta hai; loops sirf ise bahut bada aur calculate karna asaan banate hain.

Why questions

L

ko aksar "flux per amp" kyon kaha jaata hai?

Kyunki

L = N Φ/ I

literally measure karta hai ki current ka har ampere coil mein kitna flux linkage thread karta hai — ek bookkeeping constant taaki tumhe har baar field re-integrate na karna pade.

Induced EMF change ko aid karne ki jagah oppose kyon karta hai?

Lenz's law ke through energy conservation: agar yeh change ko aid karta, toh current bina kisi energy source ke badhta rehta — ek perpetual motion machine. Opposition ka matlab hai current change karne ke liye kaam karna padega.

Ek inductor ko "current ka flywheel" kyon kaha jaata hai?

Jaise ek spinning flywheel apni rotation mein change resist karta hai kyunki woh kinetic energy store karta hai, ek inductor current mein change resist karta hai kyunki woh magnetic energy

\frac{1}{2} L I^{2}

store karta hai; dono sudden changes ko smooth out karte hain.

L \propto N^{2}

kyun hota hai lekin

M \propto N_{1} N_{2}

Self-inductance mein ek hi coil "double duty" karta hai (uske apne turns dono flux banate hain aur link karte hain), jisse

N^{2}

milta hai. Mutual inductance coil 1 ka flux-making (

N_{1}

) aur coil 2 ka flux-linking (

N_{2}

) pair karta hai, jisse product milta hai.

Iron core daalne se

L

dramatically kyon badh jaata hai?

Iron ki high permeability

μ ≫ μ_{0}

same current ke liye

B

ko multiply kar deti hai, isliye

Φ

aur hence

L = μ n^{2} A ℓ

relative permeability factor se badh jaata hai — aksar hazaaron guna.

Hum ek long solenoid ke har turn se guzarne wale flux ko equal kyon maan sakte hain?

Ek long ideal solenoid ke andar field uniform aur axial hoti hai, isliye har turn (same area, same location) identical

Φ = B A

se threaded hota hai, jisse hum total linkage simply

N Φ

likh sakte hain.

RL circuit mein current apni final value tak gradually kyon pahunchta hai, ekdum jump kyon nahi karta?

Koi bhi sudden jump matlab infinite

d I / d t

, hence infinite back-EMF jo ise oppose kare — finite battery voltage ke saath yeh impossible hai. Isliye RL circuit mein current smoothly badhta hai timescale

τ = L / R

par.

Edge cases

Jab ek inductor jo steady current carry kar raha ho uska switch open hota hai, tab us instant par self-EMF kya hoga?

Bahut bada (ideally infinite): current almost instantly zero ki taraf force hota hai, jisse

d I / d t

huge aur negative ho jaata hai, isliye ek bada positive EMF appear hota hai — yahi wajah hai ki switches mein spark aata hai.

Agar ek coil ko 90° rotate kar diya jaaye taaki unke axes perpendicular ho jaayein, toh

M

ka kya hoga?

M \to 0

(aur

k \to 0

): coil 1 ka flux ab coil 2 ke area ko thread nahi karta, isliye koi linkage nahi aur koi induced EMF nahi, chahe dono coils exist karti rahein.

N \to 0

(koi turns nahi) ki limit mein coil ka

L

kya hoga?

L = μ_{0} N^{2} A / ℓ \to 0

. Koi turns nahi toh current carry karne ke liye koi loop nahi, koi flux nahi, koi linkage nahi — inductance vanish ho jaata hai, formula ke saath consistent.

Ek perfectly DC circuit jo bahut time se chal rahi ho, uska self-EMF kya hoga?

Zero. Jab current steady ho jaata hai,

d I / d t = 0

, isliye inductor ek ordinary wire (short) ki tarah behave karta hai — yeh sirf tab "react" karta hai jab current change ho raha ho.

Agar do coils identical hain (

L_{1} = L_{2} = L

) aur perfectly coupled hain, toh

M

kya hoga?

M = k L_{1} L_{2} = 1 \cdot L \cdot L = L

. Identical coils ka perfect coupling mutual inductance ko har coil ki self-inductance ke barabar bana deta hai.

Agar current linearly time ke saath badhta hai,

I = a t

, toh kya self-EMF arise hogi?

Ek constant EMF:

ε = - L d I / d t = - L a

. Current ka steady ramp ek steady (unchanging) back-EMF produce karta hai, growing nahi.

Connections

Faraday's law of induction — yahan har induced EMF ka source.
Lenz's law — minus sign aur saare "opposition" answers ki wajah.
Ampère's law — woh field deta hai jo ye traps assume karte hain.
Magnetic flux — woh quantity jo change honi chahiye.
Energy stored in magnetic field — jahan $\frac{1}{2} L I^{2}$ actually rehta hai.
RL circuits — gradual-rise edge case.
Transformers — no-electrical-contact coupling.

Question bank — Self-inductance L, mutual inductance M

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