1.8.28Electromagnetism

Self-inductance L, mutual inductance M

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WHY does inductance exist at all?

The chain of logic, never skipped: I  Ampeˋre  B  integrate  Φ  Faraday  εI \;\xrightarrow{\text{Ampère}}\; B \;\xrightarrow{\text{integrate}}\; \Phi \;\xrightarrow{\text{Faraday}}\; \varepsilon


WHAT are LL and MM (definitions from flux)


HOW to derive LL for a long solenoid (from scratch)

A solenoid: length \ell, NN turns, cross-section AA, n=N/n=N/\ell turns per metre.

Step 1 — Field inside (Ampère's law). B=μ0nI=μ0NIB = \mu_0 n I = \mu_0\frac{N}{\ell}I Why this step? Inside a long solenoid the field is uniform and axial; an Amperian rectangle gives Bpath=μ0(npath)IB\ell_{\text{path}} = \mu_0 (n\ell_{\text{path}})I.

Step 2 — Flux through ONE turn. Φ=BA=μ0NIA\Phi = B A = \mu_0\frac{N}{\ell}I\,A Why this step? Flux =BdA=BA=\int \vec B\cdot d\vec A = BA because BB is uniform and perpendicular to the cross-section.

Step 3 — Flux LINKAGE (all NN turns). NΦ=μ0N2AIN\Phi = \mu_0\frac{N^2}{\ell}A\,I Why this step? Each of the NN turns is threaded by the same Φ\Phi, so total linkage is NΦN\Phi.

Step 4 — Read off L=NΦ/IL = N\Phi/I. L=μ0N2A=μ0n2A\boxed{\,L = \mu_0\frac{N^2 A}{\ell} = \mu_0 n^2 A\,\ell\,} Why this matters: LN2L\propto N^2 — doubling turns quadruples inductance, because each turn both makes more flux and links more flux.


HOW to derive MM for two coaxial solenoids

Inner long solenoid (the source) with n1=N1/n_1=N_1/\ell; outer/overlying coil of N2N_2 turns on the same area AA, length \ell.

Step 1. Inner current I1I_1 makes B1=μ0n1I1B_1=\mu_0 n_1 I_1 inside. Step 2. Flux through one turn of coil 2: Φ21=B1A=μ0n1I1A\Phi_{21}=B_1 A=\mu_0 n_1 I_1 A (only the inner-solenoid area carries field). Step 3. Linkage in coil 2: N2Φ21=μ0N2n1AI1N_2\Phi_{21}=\mu_0 N_2 n_1 A\,I_1. Step 4. M=N2Φ21I1=μ0N1N2A=μ0n1N2A\boxed{\,M = \frac{N_2\Phi_{21}}{I_1}=\mu_0\frac{N_1 N_2}{\ell}A = \mu_0 n_1 N_2 A\,}

Figure — Self-inductance L, mutual inductance M

Energy stored (why an inductor is a flywheel for current)


Worked examples


Common mistakes (steel-manned)


Recall Feynman: explain to a 12-year-old

Imagine pushing a heavy swing. A coil of wire is like a swing for electric current: once the current is moving, it doesn't want to suddenly stop or change — it "swings back" with a little electrical push (the back-EMF). Self-inductance is how stubborn the swing is about its own motion. Mutual inductance is when your swing is tied by a rope to a friend's swing: when you change yours, theirs feels a tug too. The more loops in the wire and the tighter they're packed, the more stubborn the swing.


Active-recall flashcards

What is the defining relation for self-inductance from flux?
NΦ=LIN\Phi = LI, so L=NΦ/IL=N\Phi/I (flux linkage per amp), unit henry.
Self-induced EMF formula and its sign reason?
ε=LdI/dt\varepsilon=-L\,dI/dt; minus from Lenz's law (opposes the change).
Why does a steady (constant) current produce no self-EMF?
Because dI/dt=0dI/dt=0; EMF depends on rate of change of current, not its value.
Inductance of a long solenoid?
L=μ0N2A/=μ0n2AL=\mu_0 N^2 A/\ell=\mu_0 n^2 A\ell.
Why is LN2L\propto N^2?
Each turn both produces extra flux (N\propto N) and links the flux (×N\times N), compounding to N2N^2.
Mutual inductance definition?
M=N2Φ21/I1M=N_2\Phi_{21}/I_1; EMF in coil 2 is ε2=MdI1/dt\varepsilon_2=-M\,dI_1/dt.
State the reciprocity result for MM.
M12=M21=MM_{12}=M_{21}=M always, regardless of coil sizes.
Coupling coefficient relation?
M=kL1L2M=k\sqrt{L_1L_2} with 0k10\le k\le1; k=1k=1 = perfect flux linkage.
Energy stored in an inductor and its derivation idea?
U=12LI2U=\tfrac12 LI^2, from LIdI\int LI\,dI = work done against back-EMF.
SI unit of inductance in base/derived terms?
Henry == Wb/A == V·s/A.

Connections

  • Faraday's law of induction — the parent law ε=d(NΦ)/dt\varepsilon=-d(N\Phi)/dt.
  • Lenz's law — explains the minus sign.
  • Ampère's law — gives BB inside the solenoid.
  • Magnetic flux — the quantity being linked.
  • Energy stored in magnetic fieldU=12LI2U=\tfrac12 LI^2 and u=B2/2μ0u=B^2/2\mu_0.
  • Transformers — application of MM and k1k\approx1.
  • RL circuits — where LdI/dtL\,dI/dt shapes current growth/decay.

Concept Map

Ampere's law

integrate over area

Faraday's law

flux linkage per amp

differentiate in time

minus sign

flux into neighbour coil

differentiate in time

reciprocity theorem

solenoid derivation

scales as N squared

Current I

Magnetic field B

Flux Phi

Induced EMF

Self-inductance L

EMF = -L dI/dt

Lenz's law opposes change

Mutual inductance M

EMF2 = -M dI1/dt

M12 = M21

L = mu0 N^2 A / l

Double turns quadruples L

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, self-inductance LL ka matlab simple hai: jab kisi coil mein current behta hai, woh apna magnetic field banata hai aur flux coil ke through pass hota hai. Ab agar current change hota hai, to flux change hota hai, aur Faraday's law kehta hai ki badalta flux EMF paida karta hai. Yeh EMF change ko oppose karta hai (Lenz's law, isliye minus sign): ε=LdI/dt\varepsilon=-L\,dI/dt. Isliye coil ek "current ka flywheel" jaisa behave karti hai — current ko achanak badalne nahi deti. Yaad rakho: EMF current par nahi, current ke rate of change par depend karta hai. Constant current = zero self-EMF.

Solenoid ke liye derive karna easy hai: andar field B=μ0nIB=\mu_0 nI, ek turn ka flux Φ=BA\Phi=BA, aur total linkage NΦN\Phi, to L=μ0N2A/L=\mu_0 N^2 A/\ell. Yahan dhyan do — LL, N2N^2 ke proportional hai, sirf NN ke nahi, kyunki har turn flux banata bhi hai aur link bhi karta hai (double duty). Yeh ek classic mistake hai jo students karte hain.

Mutual inductance MM tab aata hai jab do coils paas-paas hoti hain. Coil 1 ka current coil 2 mein flux bhejta hai; agar I1I_1 change ho to coil 2 mein EMF aata hai: ε2=MdI1/dt\varepsilon_2=-M\,dI_1/dt. Ek important fact: M12=M21M_{12}=M_{21}, chahe coils chhoti-badi kuch bhi ho (reciprocity). Aur M=kL1L2M=k\sqrt{L_1L_2}, jahan kk coupling batata hai — k=1k=1 matlab perfect coupling, jaise ideal transformer mein.

Practical importance: transformers, ignition coils, induction cooktops, aur RL circuits sab inhi ideas par chalte hain. Exam mein zyada marks isi se aate hain: L=μ0N2A/L=\mu_0 N^2A/\ell, ε=LdI/dt\varepsilon=-L\,dI/dt, U=12LI2U=\tfrac12 LI^2, aur M=kL1L2M=k\sqrt{L_1L_2} — yeh chaar formule pakke kar lo, baaki sab inhi se nikal aata hai.

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Connections