Visual walkthrough — Understand inductance and the henry

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Step 0 — Fix an orientation first (so signs mean something)

WHAT. Before any signs can be trusted, we must pin down which way is "positive." We adopt the passive-sign convention: pick a direction for the current $I$ through the coil, then label the coil's two terminals $+$ and $-$ so that $I$ enters the $+$ terminal. The voltage $V$ is measured from $+$ to $-$.

WHY. Lenz's "minus sign" is meaningless until "plus" is defined. With this one choice, every polarity later is unambiguous and you can drop it straight into a circuit diagram.

PICTURE. The coil below has a chosen current arrow, a labelled $+$ terminal it enters, and the measured voltage $V$ across it.

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Step 1 — A current draws a magnetic field

WHAT. With the current direction now fixed, push current through a single loop. The moving charge wraps the loop in an invisible magnetic field. We draw the field as arrows curling through the loop.

WHY. This is the very first physical fact — everything downstream is bookkeeping on top of it. If current made no field, a coil would be just a resistor and there would be nothing to store or oppose.

PICTURE. Look at the loop below. The teal arrows are the magnetic field lines threading through the hole of the loop.

The key experimental fact, drawn as "more current → more arrows": $\Phi \propto I$

• $\Phi$ ::: the field poking through one loop (count the arrows)
• $\propto$ ::: "grows in step with" — double one, double the other
• $I$ ::: the current you are pushing (direction fixed in Step 0)

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Step 2 — Winding turns multiplies the field's grip

WHAT. Wind the same wire into $N$ stacked loops. Now the same flux $\Phi$ passes through each of the $N$ turns. We add up the flux "caught" by every turn.

WHY. Stacking turns does two things at once, and this doubling-up is the secret behind the famous $N^2$ later. Watch: (a) each turn carries the current, so the field itself gets stronger; (b) there are now more turns for that field to thread.

PICTURE. The stack below shows $N=3$ turns; the same bundle of field lines threads all three, so the field is "linked" three times.

$\lambda =N\Phi$

• $\lambda$ ::: total flux caught by the whole coil
• $N$ ::: how many turns are stacked (a plain count)
• $\Phi$ ::: flux through one single turn (from Step 1)

Since $\Phi \propto I$, the whole linkage grows with current: $\lambda \propto I$.

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Step 3 — Bundle all the geometry into one number, $L$

WHAT. Because $\lambda$ is proportional to $I$ (double the current, double the linkage), their ratio is a constant for a fixed coil. We name that constant $L$.

WHY. We never want to recompute field lines again. A single number $L$ hides all the messy geometry — turns, area, core material — so the coil's behaviour becomes a one-liner. This is exactly the trick of "let the slope be a name."

PICTURE. Plot $\lambda$ up, $I$ across. A coil made of ordinary (linear) material is a straight line through the origin; the steepness of that line is $L$. A fat, stubborn coil is a steep line; a feeble one is shallow.

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Step 4 — Faraday: changing the flux creates a voltage

WHAT. Now shake the current up and down in time. As $I$ changes, $\lambda$ changes with it, and a changing $\lambda$ conjures an EMF around the loop.

WHY. We need a tool that turns "flux is changing" into "a voltage appears." That tool is the derivative $\frac{d}{dt}$ — it means the instantaneous rate of change, the slope of the graph at one instant. We pick the derivative and not a plain ratio because we care about how fast right now, not the total. This law is Faraday's Law of Induction.

PICTURE. Below, current is a wobbling curve. Where it climbs steeply the induced EMF spikes; where it is flat the EMF is zero. It tracks the steepness, not the height.

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Step 5 — Substitute, reconcile $\epsilon$ and $V$, read the sign

WHAT. Replace $\lambda$ by $LI$ from Step 3. Since $L$ is a fixed constant (linear regime), only $I$ varies with time, so the rate of change of $LI$ is $L$ times the rate of change of $I$. Then convert the coil's own EMF $\epsilon$ into the terminal voltage $V$ using Step 0's convention $V=-\epsilon$.

WHY. Steps 0 and 4 defined $\epsilon$ (what the coil generates) and $V$ (what you measure at the terminals). Here we finally link them so the sign is unambiguous in a circuit. Lenz lives in the minus sign of $\epsilon$; flipping to $V$ turns it positive, which is why the driving source "feels" the opposition as a real voltage it must overcome.

PICTURE. With current entering the $+$ terminal: when $I$ rises, the coil makes $\epsilon <0$ (opposing), so the terminal voltage $V=-\epsilon >0$ pushes back — drawn as the teal arrow bracing against the orange rise.

$\epsilon =-\frac{d(LI)}{dt}=-L\frac{dI}{dt}\stackrel{V=-\epsilon }{\to }\boxed{V=L\frac{dI}{dt}}$

• $L$ ::: pulled out front because it is constant in the linear regime
• $\frac{dI}{dt}$ ::: how fast the current is changing — the whole driver of the voltage
• $\epsilon$ ::: coil's self-generated EMF (carries Lenz's minus)
• $V=-\epsilon$ ::: terminal voltage in the passive-sign convention — positive when current entering $+$ is rising

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Step 6 — The degenerate case: steady DC

WHAT. Hold the current perfectly constant. Then its slope is zero: $\frac{dI}{dt}=0$, so $V=0$ and $\epsilon =0$.

WHY. Every derivation must survive its edge cases. Here the "stubborn" coil goes completely limp — in steady DC an ideal inductor is just a plain wire. Skip this and a reader is baffled when a real coil shows no voltage on a battery.

PICTURE. A flat current line (slope 0) sits above a dead-flat zero-voltage line.

Recall Check the sign in both directions (current enters the

$+$ terminal) Current rising → $\frac{dI}{dt}>0$ → sign of terminal $V$? ::: positive — the coil pushes back against the rise ($\epsilon <0$). Current falling → $\frac{dI}{dt}<0$ → sign of terminal $V$? ::: negative — it props the current up (the spark when you open a switch). Current steady → $\frac{dI}{dt}=0$ → $V$ and $\epsilon$? ::: exactly zero — the coil looks like bare wire.

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Step 7 — Why energy is $\frac{1}{2}L{I}^2$ (the stored bonus)

WHAT. To build the current up you must supply power against the coil's opposing voltage. Power is $P=VI=\left(L\frac{dI}{dt}\right)I$. Adding up (integrating) that power as current climbs from $0$ to a final value $I$ gives the stored energy.

WHY. The intro promised $E=\frac{1}{2}L{I}^2$; here we earn it. We swap the time integral for a current integral using $\frac{dI}{dt}dt=dI$ — this is the same "cancel the $dt$" move that turns an awkward time-history into a clean area under a straight line.

PICTURE. On the $V$-vs-$I$ line for the inductor ($V\propto I$ during a steady ramp), the energy is the triangle area under the line — base $I$, height $LI$, so area $\frac{1}{2}\cdot I\cdot LI$.

$E=\int Pdt={\int }_0^{I}LIdI=\frac{1}{2}L{I}^2$

• $P=VI$ ::: power fed in each instant
• ${\int }_0^I\dots dI$ ::: sum over every step of current from off to $I$
• $\frac{1}{2}L{I}^2$ ::: the triangle's area — energy now living in the magnetic field, ready to return

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Step 8 — Where the geometry hides: the solenoid and $N^2$

WHAT. Compute $L$ for a long coil: $N$ turns, length $\ell$, area $A$, core permeability $\mu$. The field inside is $B=\mu \frac{N}{\ell }I$, flux per turn is $\Phi =BA$, and linkage is $N\Phi$.

WHY. This shows why Step 2's double-counting matters: $N$ appears once in the field and once in the linkage, so it squares. See Magnetic Field of a Solenoid for the field itself.

PICTURE. Two coils side by side: doubling the turns makes the bars in the $L$ chart jump 4×, not 2×.

$L=\frac{N\Phi }{I}=\mu \frac{N^{2}A}{\ell }$

• $\mu$ ::: how eagerly the core carries field (iron ≫ air) — constant only while unsaturated (Step 3)
• $N^2$ ::: the squared turn-count — the double-counting from Step 2
• $A$ ::: cross-section area — a wider hole catches more flux
• $\ell$ ::: length — spreading turns out thins the field, so it divides

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The one-picture summary

Everything on this page is one chain: orientation → current → flux → linkage → (slope $L$) → Faraday's rate → Lenz's sign → voltage, with energy $\frac{1}{2}L{I}^2$ as the stored triangle bonus.

Recall Feynman: the whole walk in plain words

First we pick which way current flows and which terminal is "plus," so that later a minus sign actually means something (Step 0). Push current through a loop and it grows an invisible magnetic field (Step 1). Stack the loop into many turns and the field threads every one of them, so you count it once per turn — that total is the flux linkage (Step 2). For a fixed coil of ordinary material, that linkage is always the same multiple of the current, and we name that multiple $L$ — the steepness of a straight line, valid until an iron core saturates and the line bends (Step 3). Now wobble the current: a changing linkage makes an EMF, and the derivative measures "how fast right now" (Step 4). Swapping in $LI$, pulling the constant $L$ out front, and flipping the coil's EMF into the terminal voltage gives $V=L\frac{dI}{dt}$; the sign is nature stubbornly opposing your change (Step 5). Hold the current still and the whole thing vanishes to zero — a coil in steady DC is just wire (Step 6). Building the current up costs energy, and that energy is the triangle under the $V$–$I$ line, $\frac{1}{2}L{I}^2$, stored in the field to be handed back later (Step 7). Finally, because the turn-count sneaks in twice, $L$ grows with $N^2$ (Step 8).

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Connections

• Understand inductance and the henry — the parent this page unpacks.
• Faraday's Law of Induction — Step 4's engine.
• Lenz's Law — Step 5's minus sign.
• Magnetic Field of a Solenoid — Step 8's $B=\mu NI/\ell$.
• RL Circuits and Time Constant — what happens when you wire $L$ to a resistor.
• Energy Storage in Fields — the $\frac{1}{2}L{I}^2$ bonus of Step 7.
• Capacitance and the Farad — the dual that resists voltage change instead.