Understand inductance and the henry: Worked examples

Intuition What this page is

The parent note gave you three formulas. Here we drill them across every case class an exam or a real circuit can throw at you: rising current, falling current, zero rate of change, degenerate geometry, limiting behaviour, a word problem, and a nasty exam twist. Before each solution you forecast the answer — guessing first is how the idea sticks.

The three tools, one more time, so no symbol is unearned:

$L = \frac{N Φ}{I}$ — inductance = flux linkage per amp. $N$ = number of turns, $Φ$ = flux through one loop (webers), $I$ = current (amps).
$V = L \frac{d I}{d t}$ — voltage across the coil = inductance × how fast current changes. The symbol $\frac{d I}{d t}$ just means "the slope of the current-versus-time graph": how many amps it climbs per second.
$E = \frac{1}{2} L I^{2}$ — energy parked in the magnetic field.

The scenario matrix

Cell	Case class	What's special	Example
A	Current rising ( $d I / d t > 0$ )	$V > 0$ , coil opposes the rise	Ex 1
B	Current falling ( $d I / d t < 0$ )	$V < 0$ (sign flips), coil opposes the fall	Ex 2
C	Zero rate of change (steady DC)	$d I / d t = 0 \Rightarrow V = 0$ , coil = plain wire	Ex 3
D	Energy: charge vs discharge	$E$ depends on $I^{2}$ , so direction doesn't matter	Ex 4
E	Geometry / degenerate input	$L \propto N^{2}$ ; what if $N$ , $A$ , or $ℓ$ change	Ex 5
F	Limiting behaviour — the spark	$d I / d t \to$ huge as current is cut → huge $V$	Ex 6
G	Real-world word problem	Sizing a coil for a target voltage	Ex 7
H	Exam twist — read the definition backwards	Find $Φ$ or $N$ from $L$ and $I$	Ex 8

Eight examples, eight cells, no gaps.

Example 1 — Cell A: current rising

Worked example Rising current

A $2 H$ inductor carries a current increasing steadily at $3 A/s$ . Find the induced voltage magnitude and its sign relative to the current direction.

Forecast: Big inductance, brisk rise → a handful of volts. Guess a number before reading on.

Step 1. Identify what changes: the current slope is $\frac{d I}{d t} = + 3 A/s$ (positive = rising). Why this step? $V$ depends on the slope, not on the current value — so the first job is always to read the slope and its sign.

Step 2. Apply the behaviour law: $V = L \frac{d I}{d t} = 2 \times 3 = 6 V$ Why this step? This is the defining equation of the henry (1 V per 1 A/s per 1 H).

Step 3. Sign: by Lenz's Law the induced voltage opposes the rise, so it points to push back against the increasing current. Why this step? The magnitude tells you "how hard," Lenz tells you "which way."

Verify: Units: $H \times \frac{A}{s} = \frac{V \cdot s}{A} \times \frac{A}{s} = V$ . ✓ Answer $= 6 V$ .

Example 2 — Cell B: current falling (sign flips)

Worked example Falling current

The same $2 H$ inductor now has its current falling at $3 A/s$ (from $9 A$ toward $0$ ). Find the induced voltage, with sign.

Forecast: Same size as Ex 1, but does the sign change? Guess before reading.

Step 1. Read the slope: falling means $\frac{d I}{d t} = - 3 A/s$ . Why this step? "Falling" is the entire content of the sign — miss it and you get the direction backwards.

Step 2. Apply the law: $V = L \frac{d I}{d t} = 2 \times (- 3) = - 6 V$ Why this step? Same formula; the negative slope carries straight through to a negative voltage.

Step 3. Interpret: the sign flipped from Ex 1. Now the coil tries to keep the current going — it acts like a temporary battery pushing current forward. Why this step? This is exactly why cutting an inductor's current can spark: it fights the decrease.

Verify: Magnitude equals Ex 1 ( $6 V$ ) because $∣ d I / d t ∣$ is the same; only the sign differs. ✓

Example 3 — Cell C: zero rate of change (steady DC)

Worked example Constant current

A $5 H$ coil carries a constant $4 A$ from a battery. What voltage appears across the ideal inductor?

Forecast: Big $L$ , real current flowing — surely some voltage? Guess, then be surprised.

Step 1. Read the slope: current is constant, so $\frac{d I}{d t} = 0$ . Why this step? Steady means "no change over time," and no change means zero slope.

Step 2. Apply the law: $V = L \frac{d I}{d t} = 5 \times 0 = 0 V$ Why this step? The formula only cares about change. A constant current gives nothing.

Step 3. Interpret: an ideal inductor in steady DC is just a plain wire — zero volts across it. Why this step? This kills the most common misconception (" $V = I R$ -style thinking"). There is no $I$ in the inductor's voltage law.

Verify: $L \times 0 = 0$ regardless of how large $L$ or $I$ is. ✓ The $4 A$ value is a red herring.

Example 4 — Cell D: energy (direction-independent)

Example 5 — Cell E: geometry & the degenerate cases

Worked example Sizing and re-sizing a solenoid

An air-core solenoid: $N = 200$ turns, area $A = 1 \times 1 0^{- 4} m^{2}$ , length $ℓ = 0.05 m$ , $μ_{0} = 4 π \times 1 0^{- 7}$ . (a) Find $L$ . (b) What is $L$ if you double the turns to $400$ ? (c) Degenerate check: what happens to $L$ as $ℓ \to \infty$ (an infinitely stretched coil)?

Forecast: Guess (a) in microhenries; for (b) will $L$ double or quadruple?

Step 1. Apply the solenoid formula: $L = μ_{0} \frac{N ^{2} A}{ℓ} = 4 π \times 1 0^{- 7} \cdot \frac{20 0 ^{2} \cdot 1 0 ^{- 4}}{0.05}$ Why this step? This bundles all geometry into $L$ ; the $N^{2}$ is the star of the show.

Step 2. Crunch: $20 0^{2} = 40000$ , so numerator $= 40000 \times 1 0^{- 4} = 4$ ; divide by $0.05$ → $80$ ; times $4 π \times 1 0^{- 7}$ : $L \approx 1.0 \times 1 0^{- 4} H = 100 μ H$ Why this step? Keeping powers of ten separate avoids calculator slips.

Step 3. Double the turns: $L \propto N^{2}$ , so $L \to 4 \times = 400 μ H$ . Why this step? More turns raise both the field and the loops linking it, so the effect squares — not doubles.

Step 4. Degenerate limit $ℓ \to \infty$ : $L = μ_{0} N^{2} A / ℓ \to 0$ . Why this step? Stretch the same wire into an endless coil and the field thins out — inductance collapses toward zero. This shows the formula behaves sensibly at its extreme.

Verify: (a) $\approx 1.005 \times 1 0^{- 4} H$ . (b) $= 4 \times$ that. (c) $\to 0$ . ✓

Example 6 — Cell F: limiting behaviour (the spark)

Worked example Cutting the current fast

A $0.5 H$ relay coil carries $2 A$ . A switch opens and forces the current to $0$ in just $1 ms$ ( $0.001 s$ ). Estimate the peak induced voltage.

Forecast: Only half a henry and two amps — but the time is tiny. Big or small voltage?

Step 1. Estimate the slope: current falls from $2$ to $0$ in $0.001 s$ : $\frac{d I}{d t} \approx \frac{0 - 2}{0.001} = - 2000 A/s$ Why this step? A sharp change means a huge slope — this is where the drama comes from.

Step 2. Apply the law: $V = L \frac{d I}{d t} = 0.5 \times (- 2000) = - 1000 V$ Why this step? The magnitude ( $1000 V$ ) dwarfs the supply — because $V$ scales with speed of change, not current size.

Step 3. Interpret the limit: as the cut time $\to 0$ , $∣ V ∣ \to \infty$ . In reality this spikes across the switch contacts as a spark. Why this step? This is the practical face of "inductors fight a decrease" — and why real circuits add a flyback diode.

Verify: Units $H \times A/s = V$ ; $0.5 \times 2000 = 1000 V$ . ✓ Faster cut → larger voltage, unbounded in the ideal limit. ✓

Example 7 — Cell G: real-world word problem

Example 8 — Cell H: exam twist (read the definition backwards)

Worked example Find the flux from L and I

A $150 mH$ coil has $N = 300$ turns and carries $0.4 A$ . Find the flux through one turn, $Φ$ .

Forecast: We normally compute $L$ from $Φ$ ; here we run it in reverse. Expect a tiny weber value.

Step 1. Start from the definition $L = \frac{N Φ}{I}$ and solve for $Φ$ : $Φ = \frac{L I}{N}$ Why this step? Every symbol here is known except $Φ$ ; rearranging the definition is the whole trick of the twist.

Step 2. Unit hygiene: $150 mH = 0.150 H$ . Substitute: $Φ = \frac{0.150 \times 0.4}{300} = \frac{0.06}{300} = 2 \times 1 0^{- 4} Wb$ Why this step? Plug in cleanly; the flux per turn is small because a single loop links only a slice of the total flux linkage.

Step 3. Sanity: total flux linkage $N Φ = 300 \times 2 \times 1 0^{- 4} = 0.06 Wb$ , and $L = N Φ/ I = 0.06/0.4 = 0.15 H$ . ✓ Why this step? Feeding the answer back through the original definition confirms consistency.

Verify: $Φ = 2 \times 1 0^{- 4} Wb$ , and $N Φ/ I = 0.15 H$ matches the given $L$ . ✓

Recall Quick self-test across the matrix

Rising current, 3 H, 4 A/s — what voltage? ::: $V = 3 \times 4 = 12 V$ (opposes the rise). Constant 10 A through a 7 H coil — voltage? ::: $0 V$ ; $d I / d t = 0$ so the value of current is irrelevant. Reverse the current direction — does stored energy change? ::: No; $E = \frac{1}{2} L I^{2}$ and $I^{2}$ ignores sign. Double the turns of a solenoid — new inductance? ::: $4 \times$ the old value, because $L \propto N^{2}$ . Cut current faster — bigger or smaller induced voltage? ::: Bigger; $V = L d I / d t$ grows with the steepness of the change.

Connections

Understand inductance and the henry (index 1.1.12) — the parent note these examples drill.
Faraday's Law of Induction — the source of $V = - d λ / d t$ behind Ex 1–2 & 6.
Lenz's Law — decides the sign in every rising/falling case.
RL Circuits and Time Constant — where the spark of Ex 6 is tamed by $τ = L / R$ .
Magnetic Field of a Solenoid — the field behind the geometry of Ex 5.
Energy Storage in Fields — the $\frac{1}{2} L I^{2}$ used in Ex 4.
Capacitance and the Farad — the dual: swap current↔voltage to get the capacitor versions.