Self-inductance L, mutual inductance M: Worked examples

This page is a drill. The parent note built the formulas; here we throw every kind of input at them — every sign, zero cases, degenerate geometry, limiting values, a word problem, and an exam twist — so no scenario can surprise you.

Every symbol used here is defined in the parent, but we restate each one the moment we use it, so you can read from line one.

The scenario matrix

Below is the full list of "case classes" this topic can throw at you. Each later example is tagged with the cell(s) it covers.

Cell	What makes it distinct	Covered by
A. Pure geometry	Compute $L$ from turns/area/length	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	$d I / d t$ itself oscillates; peak EMF	Ex 3
F. Mutual $M$ from geometry + coupling $k$	Two coils, read off $M$ , check $k$	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	Translate English → symbols	Ex 7
J. Exam twist — series/combined coils	$L$ of two coils in series (aiding/opposing)	Ex 8

Let us knock them out.

The symbols, restated once

Definition The coupling coefficient

$k$ (defined before we use it) When two coils sit near each other, coil 1's flux does not, in general, fully reach coil 2 — some field leaks into the air. The coupling coefficient $k$ is the plain fraction "how much of coil 1's flux actually threads coil 2": $M = k L_{1} L_{2}, - 1 \leq k \leq 1.$

$∣ k ∣ = 1$ : every field line of one coil links the other (perfect linkage — the ideal transformer).
$k = 0$ : the coils share no flux (pulled far apart, or perpendicular).
Sign of $k$ : positive if the two coils are wound so their fluxes reinforce when carrying current the same way; negative if one coil is wound the opposite sense (or reversed in the circuit) so its flux opposes the other's. A negative $k$ (equivalently, a negative $M$ ) flips the sign of the cross-linkage term — this is exactly what turns "series-aiding" into "series-opposing" in Example 8. Textbooks often quote $0 \leq k \leq 1$ and carry the winding sense separately as a $\pm$ on $M$ ; both bookkeepings give the same physics.

Example 1 — Cell A: pure geometry

Worked example Compute

$L$ of a solenoid A solenoid has $N = 800$ turns, area $A = 2 \times 1 0^{- 4} m^{2}$ , length $ℓ = 0.40 m$ . Find $L$ .

Forecast: Guess the order of magnitude first — will it be henries, millihenries, or microhenries? (Coils this size are usually sub-millihenry.)

Step 1 — Write the box formula. $L = μ_{0} \frac{N ^{2} A}{ℓ}$ . Why this step? This is the only self-inductance formula for a long solenoid — it came from Ampère → flux → linkage in the parent. Everything is a plug-in.

Step 2 — Insert numbers. $L = (4 π \times 1 0^{- 7}) \frac{( 800 ) ^{2} ( 2 \times 1 0 ^{- 4} )}{0.40}$ Why this step? Keep $μ_{0}$ symbolic until the end so we don't round early.

Step 3 — Crunch. $(800)^{2} = 6.4 \times 1 0^{5}$ . Times $2 \times 1 0^{- 4} = 128$ . Divide by $0.40 = 320$ . Times $4 π \times 1 0^{- 7}$ : $L \approx 4.02 \times 1 0^{- 4} H = 0.402 mH .$ Why this step? We collapse the numerator and denominator before touching $μ_{0}$ so a single final multiply by $4 π \times 1 0^{- 7}$ carries all the rounding — this avoids compounding round-off across several intermediate products and makes the arithmetic auditable term by term.

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: the three signs of self-EMF

We use the coil from Ex 1 ( $L = 0.402 mH$ ) and change how the current moves. The formula is always $ε = - L \frac{d I}{d t},$ with the sign convention fixed above ( $ε > 0$ = pushes with the current).

The figure below plots the same current through three phases — a rising ramp, a flat top, and a falling ramp — so you can read the slope $d I / d t$ straight off the curve and see which sign of $ε$ it forces. The red segment is the rising phase (Cell B); the flat and falling phases are the black segments.

Figure — Self-inductance L, mutual inductance M

Recall Why 2c is the whole point

The value of $I$ never appears in $ε = - L d I / d t$ — only its slope does. ::: Correct; a constant current gives zero self-EMF no matter how large.

Example 3 — Cell E: oscillating (AC) current

Worked example Sinusoidal drive

Current $I (t) = 3 sin (200 t) A$ flows through a coil of $L = 0.402 mH$ . Find the induced EMF and its peak value. (Here $200$ is $ω$ in rad/s.)

Forecast: The EMF will be a cosine (the derivative of sine) — 90° out of phase. Will its peak be bigger or smaller than $3$ ?

Step 1 — Differentiate. $\frac{d I}{d t} = 3 \cdot 200 cos (200 t) = 600 cos (200 t) A/s$ . Why differentiate? Because EMF is the rate of change; for a sine, that rate is the derivative, which is a cosine scaled by $ω \times$ amplitude.

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

Step 3 — Peak. The cosine swings between $\pm 1$ , so $ε_{peak} = 0.241 V$ . Why this step? Exams ask for peak (amplitude) — it is the coefficient in front of the cosine.

Verify: Peak $= L ω I_{0} = (4.02 \times 1 0^{- 4}) (200) (3) = 0.241 V$ . ✓ Note $L ω$ is the inductive reactance — the same object that governs RL circuits and Transformers.

Example 4 — Cell F: mutual inductance & coupling

Worked example Two coaxial solenoids

Two solenoids share the same core, length $ℓ = 0.30 m$ , area $A = 5 \times 1 0^{- 4} m^{2}$ . Inner: $N_{1} = 600$ . Outer: $N_{2} = 300$ . Assume perfect overlap so all inner flux threads the outer coil. Find $M$ , then $L_{1}$ , $L_{2}$ , and check $k$ .

Forecast: With perfect overlap, what value must the coupling coefficient $k$ take?

Step 1 — Mutual inductance. $M = μ_{0} \frac{N _{1} N _{2}}{ℓ} A$ . Why this step? Inner current makes $B_{1} = μ_{0} (N_{1} / ℓ) I_{1}$ ; the outer's $N_{2}$ turns each catch flux $B_{1} A$ ; divide linkage by $I_{1}$ . $M = (4 π \times 1 0^{- 7}) \frac{( 600 ) ( 300 )}{0.30} (5 \times 1 0^{- 4}) = 3.77 \times 1 0^{- 4} H .$

Step 2 — The two self-inductances. $L_{1} = μ_{0} \frac{N _{1}^{2} A}{ℓ} = (4 π \times 1 0^{- 7}) \frac{60 0 ^{2} ( 5 \times 1 0 ^{- 4} )}{0.30} = 7.54 \times 1 0^{- 4} H,$ $L_{2} = μ_{0} \frac{N _{2}^{2} A}{ℓ} = (4 π \times 1 0^{- 7}) \frac{30 0 ^{2} ( 5 \times 1 0 ^{- 4} )}{0.30} = 1.88 \times 1 0^{- 4} H .$ Why this step? Each coil's self-inductance is its own $N^{2}$ fed into the same box formula $L = μ_{0} N^{2} A / ℓ$ from Ex 1 — the geometry ( $A$ , $ℓ$ ) is shared, only the turn count changes. We need both $L_{1}$ and $L_{2}$ because the coupling check compares $M$ against their geometric mean $L_{1} L_{2}$ .

Step 3 — Coupling coefficient. $k = \frac{M}{L _{1} L _{2}}$ (rearranged from its definition above). $L_{1} L_{2} = (7.54 \times 1 0^{- 4}) (1.88 \times 1 0^{- 4}) = 3.77 \times 1 0^{- 4}$ . $k = \frac{3.77 \times 1 0 ^{- 4}}{3.77 \times 1 0 ^{- 4}} = 1.00.$ Why this matters: $k = 1$ confirms the "perfect linkage" idealisation — the transformer limit. (If the outer coil had been wound the opposite sense, we would read $k = - 1$ : same magnitude of $M$ , flipped sign.)

Verify: $M = k L_{1} L_{2}$ with $k = 1$ reproduces Step 1's $M$ exactly. ✓ And $M_{12} = M_{21}$ by reciprocity — swapping which coil is the source gives the same $M$ . ✓

Example 5 — Cell G: limiting behaviour

Worked example What happens at the edges?

Start from a base coil $L_{0} = μ_{0} \frac{N ^{2} A}{ℓ}$ . Answer three limit questions.

Forecast: Which change multiplies $L$ by four? Which blows it up to infinity?

(i) Double the turns, $N \to 2 N$ (length fixed, so turns packed tighter). Why: $L \propto N^{2}$ . $L \to (2 N)^{2} / N^{2} \cdot L_{0} = 4 L_{0}$ . Turns do double duty (make + link flux). ✓

(ii) Shrink the length, $ℓ \to 0$ (turns fixed, packed infinitely tight). Why: $L = μ_{0} N^{2} A / ℓ$ . As $ℓ \to 0$ , $L \to \infty$ . A physically impossible limit, but it shows tighter packing $\Rightarrow$ larger $L$ .

(iii) Pull the two coils of Ex 4 far apart, $k \to 0$ . Why: $M = k L_{1} L_{2} \to 0$ . No shared flux, no mutual effect. The coils "forget" each other.

Verify (numeric anchor for (i)): Take Ex 1's coil, $L_{0} = 0.402 mH$ . Doubling $N$ to $1600$ gives $L = μ_{0} (1600)^{2} (2 \times 1 0^{- 4}) /0.40 = 1.608 mH = 4 L_{0}$ . ✓

Example 6 — Cell H: stored energy & the flywheel

Worked example Energy and its release

An inductor with $L = 0.402 mH$ carries a steady $I = 5 A$ . (a) How much energy is stored? (b) If that current is switched off in $0.5 ms$ , what average power is dumped?

Forecast: The energy is small (sub-millijoule); but because it exits quickly, the power might be surprisingly large.

Step 1 — Energy. $U = \frac{1}{2} L I^{2}$ . Why this formula? It is the work done against the back-EMF while building the current from 0 to $I$ (integral of $L I d I$ ), stored in the magnetic field. $U = \frac{1}{2} (4.02 \times 1 0^{- 4}) (5)^{2} = 5.02 \times 1 0^{- 3} J = 5.02 mJ .$

Step 2 — Average power on switch-off. Energy leaves in $t = 0.5 \times 1 0^{- 3} s$ : $P_{avg} = \frac{U}{t} = \frac{5.02 \times 1 0 ^{- 3}}{0.5 \times 1 0 ^{- 3}} = 10.0 W .$ Why this step? Power = energy per unit time. The rapid switch-off is exactly why breaking an inductor's circuit sparks — the flywheel refuses to stop and forces energy out fast.

Verify: Units: J/s = W. ✓ Small stored energy but 10 W of transient power — the flywheel intuition confirmed. ✓

Example 7 — Cell I: real-world word problem

Worked example Doorbell / relay coil

"A relay coil has self-inductance $L = 0.20 H$ . When the switch opens, its $0.30 A$ current collapses to zero in $1.0 ms$ . Estimate the spike voltage across the opening contacts."

Forecast: The current is tiny (0.3 A) but the collapse is fast. Will the spike be a few volts or tens of volts?

Step 1 — Translate English to symbols. "Collapses to zero in 1 ms" $\Rightarrow \frac{d I}{d t} = \frac{0 - 0.30}{1.0 \times 1 0 ^{- 3}} = - 300 A/s$ . Why this step? The scary number is not $I$ , it's the slope — always identify $d I / d t$ from the words.

Step 2 — Spike EMF. $ε = - L \frac{d I}{d t} = - (0.20) (- 300) = + 60 V$ . Why positive/large? Current falling → coil fights to sustain it → a $60 V$ spike from a coil that ran on well under a volt. This is why relays get flyback diodes.

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

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

Worked example Two coupled coils in series

Two coils are wired in series so they carry the same current $I$ . Their self-inductances are $L_{1} = 3 mH$ and $L_{2} = 5 mH$ , and their mutual inductance has magnitude $∣ M ∣ = 2 mH$ . Find the total inductance for (a) series-aiding (both coils wound so their fluxes reinforce, $M > 0$ ) and (b) series-opposing (one coil wound the reverse sense, so its flux fights the other's, $M < 0$ ).

Forecast: In which connection is the pair more stubborn about current change — the one where the magnetic fields help each other, or the one where they fight?

Step 1 — Build the total flux linkage. Each coil links its own flux and the flux from its neighbour. With the same current $I$ in both, the flux linkages add: $N_{1} Φ_{1} = L_{1} I \pm M I, N_{2} Φ_{2} = L_{2} I \pm M I .$ Why this step? Coil 1 sees its self-linkage $L_{1} I$ plus the mutual linkage $\pm M I$ that coil 2 dumps onto it; coil 2 likewise. The sign is $+$ if the two fluxes point the same way (aiding) and $-$ if they oppose. This is where the coupling sign from the definitions box enters: $M > 0$ (positive $k$ ) $\Rightarrow +$ , $M < 0$ (negative $k$ ) $\Rightarrow -$ .

Step 2 — Sum into one equivalent inductance. The whole series pair behaves like a single inductor $L_{total}$ obeying $N Φ_{total} = L_{total} I$ . Add the two linkages: $L_{total} I = (L_{1} I \pm M I) + (L_{2} I \pm M I) \Rightarrow L_{total} = L_{1} + L_{2} \pm 2 M .$ Why the $2 M$ and not just $M$ ? The cross-linkage is counted twice — once as "coil 2's flux threading coil 1" and once as "coil 1's flux threading coil 2." Reciprocity ( $M_{12} = M_{21} = M$ ) makes those two equal, so they combine into a single $2 M$ .

Step 3 — Aiding ( $M > 0$ ). $L_{total} = 3 + 5 + 2 (2) = 12 mH$ . Why this step? Aiding takes the $+ 2 M$ branch; every contribution ( $3$ , $5$ , and $2 M = 4$ ) carries a plus sign because all fluxes reinforce — the pair is maximally stubborn.

Step 4 — Opposing ( $M < 0$ ). $L_{total} = 3 + 5 - 2 (2) = 4 mH$ . Why this step? Reversing one winding flips $k$ (and hence $M$ ) negative, so the cross term becomes $- 2 M = - 4$ ; the self terms $3 + 5$ stay positive but the opposing flux cancels part of the linkage, leaving a much smaller total.

Verify: Aiding (12) > opposing (4), as forecast — reinforcing fields make the stiffer inductor. Legality check: $∣ M ∣ \leq L_{1} L_{2} = 15 \approx 3.87 mH$ , and our $∣ M ∣ = 2$ obeys this ( $∣ k ∣ = 2/3.87 \approx 0.52 \leq 1$ ), so both numbers are physically admissible. ✓

Connections

Yeh note Hinglish mein — parent topic.
Faraday's law of induction — source of $ε = - d (N Φ) / d t$ used in every sign example.
Lenz's law — the minus sign in Ex 2a/2b/2c/7.
Ampère's law — gives $B$ for the geometry examples.
Magnetic flux — the linked quantity.
Energy stored in magnetic field — Ex 6.
RL circuits, Transformers — where $L ω$ and $k$ reappear.