Inductance & LR/LC Circuits

A coil that fights its own change

Pull the plug on a running vacuum cleaner in a dark room and you sometimes see a blue spark jump inside the socket. Nothing was pushing that spark a moment earlier: the appliance was running quietly on a steady current. The spark appears at the instant the current is cut off, and it can carry a voltage far larger than the 120 V the wall was supplying.

In Motional EMF you moved a conductor through a field and got an emf out of it. Nothing had to move here. The coil inside the motor made its own magnetic field, that field threaded its own turns, and when the current died the field died with it. A changing flux through a loop induces an emf, whether the flux came from a magnet across the room or from the very current flowing in the loop. That second case is what follows, together with the two circuits (LR and LC) where it dominates everything.

Flux linkage, and the number called L

Start with a single coil of $N$ turns carrying current $I$. That current makes a magnetic field everywhere, and everywhere in space the field is proportional to $I$: double the current and you double $B$ at every point. So the flux $\Phi$ through one turn is proportional to $I$ too, and so is the flux linkage $N\Phi$, the total flux counted once per turn.

Proportional means there's a constant of proportionality. Name it $L$:

$N\Phi = LI$

$L$ is the self-inductance, measured in henries (1 H = 1 Wb/A). Because both sides scale with $I$ in the same way, $L$ itself does not depend on $I$ at all. It depends only on the shape of the coil: how many turns, how big, how long. Inductance is geometry, exactly as capacitance is geometry.

Now put that into Faraday's law. The emf around the coil is $-d(N\Phi)/dt$, and since $L$ is a constant,

$\mathcal{E} = -L\frac{dI}{dt}$

Read it in words. The coil produces a voltage proportional to the rate of change of its current. A steady current, however large, produces none. The minus sign is Lenz's law: the emf opposes the change. Push the current up and the coil pushes back like a headwind. Cut the current and the coil pushes forward, trying to keep it going: that is the spark in the socket.

Getting L from the geometry

The recipe is always the same three steps: assume a current $I$, find the field it makes, integrate the field to get the flux linkage, then divide by $I$.

Do it for a long solenoid of length $\ell$, cross-sectional area $A$, with $N$ turns, so $n = N/\ell$ turns per metre. Ampère's law gave you $B = \mu_0 n I$ inside and essentially zero outside. Each turn is threaded by $\Phi = BA = \mu_0 n I A$, and there are $N = n\ell$ of them, so

$N\Phi = n\ell \cdot \mu_0 n I A = \mu_0 n^2 A \ell \, I$

Divide by $I$ and the current vanishes, as it must:

$L = \mu_0 n^2 A \ell$

Worked example 1. A solenoid has 500 turns over a length of 0.25 m with cross-section $4.0\times10^{-4}$ m². Find $L$, and find the emf when the current in it rises at a steady 40 A/s.

Turn density first: $n = 500/0.25 = 2000$ turns per metre. Then

$L = (4\pi\times10^{-7})(2000)^2(4.0\times10^{-4})(0.25)$

Take it piece by piece. $(2000)^2 = 4.0\times10^6$. Times $4\pi\times10^{-7}$ gives $5.027$. Times $4.0\times10^{-4}$ gives $2.011\times10^{-3}$. Times $0.25$ gives

$L = 5.03\times10^{-4}\ \text{H}$

The emf is $|\mathcal{E}| = L\,|dI/dt| = (5.03\times10^{-4})(40) = 2.0\times10^{-2}$ V. Small, because the coil is small. A 0.15 H coil with the same 40 A/s would give 6.0 V.

A toroid works the same way but its field isn't uniform. For a toroid of $N$ turns, rectangular cross-section of height $h$, inner radius $a$ and outer radius $b$, Ampère's law on a circle of radius $r$ inside gives $B = \mu_0 N I/(2\pi r)$. The flux through one turn needs an integral because $B$ falls off across the cross-section:

$\Phi = \int_a^b \frac{\mu_0 N I}{2\pi r} h\,dr = \frac{\mu_0 N I h}{2\pi}\ln\frac{b}{a}$

so

$L = \frac{\mu_0 N^2 h}{2\pi}\ln\frac{b}{a}$

Inductors in series and in parallel combine like resistors, not like capacitors, as long as they are far enough apart that neither one's field threads the other. In series the same $dI/dt$ runs through both and the emfs add, so $L = L_1 + L_2$. In parallel the voltages match and the currents add, so $1/L = 1/L_1 + 1/L_2$. Two coils of 0.30 H and 0.60 H in parallel give $L = (0.30)(0.60)/0.90 = 0.20$ H.

Energy in the magnetic field

Building up a current in an inductor costs work, because the whole time you are pushing against the induced emf. The external source must supply a voltage $L\,dI/dt$ just to overcome it, so the power it delivers is

$P = I \cdot L\frac{dI}{dt}$

Energy is the time integral of power, and $I\,dI/dt\,dt = I\,dI$, so pushing the current from 0 up to some final $I$ costs

$U = \int_0^{I} LI'\,dI' = \tfrac12 L I^2$

That energy is not lost. It sits in the magnetic field and comes back out when the current falls. A 0.20 H coil carrying 3.0 A holds $\tfrac12(0.20)(9.0) = 0.90$ J.

You can also count the energy field-point by field-point. Put $\tfrac12 LI^2$ for a solenoid next to its volume $A\ell$ and its field $B = \mu_0 n I$:

$\frac{U}{A\ell} = \frac{\tfrac12 \mu_0 n^2 A\ell I^2}{A\ell} = \tfrac12 \mu_0 n^2 I^2 = \frac{B^2}{2\mu_0}$

The geometry dropped out entirely, which is the sign that this is a general truth about fields, not a fact about solenoids. The magnetic energy density anywhere is

$u_B = \frac{B^2}{2\mu_0}$

A field of 0.80 T carries $0.64/(2\times4\pi\times10^{-7}) = 2.55\times10^{5}$ J per cubic metre.

Worked example 2. A coaxial cable carries 10 A out along an inner conductor of radius 1.0 mm and back along an outer conductor at radius 5.0 mm. How much magnetic energy is stored per metre?

There's no coil here, so $\tfrac12 LI^2$ has nothing obvious to plug into. Use the density instead. Between the conductors, Ampère's law on a circle of radius $r$ encloses the full 10 A, giving $B = \mu_0 I/(2\pi r)$. Outside the outer conductor the enclosed current is zero, so there is no field and no energy there.

Take a thin cylindrical shell of radius $r$, thickness $dr$, one metre long. Its volume is $2\pi r\,dr$, and

$dU = \frac{B^2}{2\mu_0}\,2\pi r\,dr = \frac{\mu_0 I^2}{8\pi^2 r^2}\,2\pi r\,dr = \frac{\mu_0 I^2}{4\pi}\frac{dr}{r}$

Integrate from $a$ to $b$:

$U = \frac{\mu_0 I^2}{4\pi}\ln\frac{b}{a} = (10^{-7})(100)\ln 5$

$\ln 5 = 1.609$, so $U = (10^{-5})(1.609) = 1.61\times10^{-5}$ J per metre. Thin shells and a logarithm: that combination shows up every time the field goes like $1/r$.

Mutual inductance

Now two coils. Run current $I_1$ through the first; some of its flux threads the second. That flux linkage is again proportional to $I_1$, so define

$N_2\Phi_2 = M I_1, \qquad \mathcal{E}_2 = -M\frac{dI_1}{dt}$

$M$ is the mutual inductance, also in henries. A remarkable fact, which you can take on trust here: $M$ comes out the same whichever coil you drive. Two coils with $M = 0.020$ H, one of them carrying a current changing at 150 A/s, induce $(0.020)(150) = 3.0$ V in the other.

The practical recipe is to drive whichever coil has the field you can compute. For a small 100-turn coil of radius 1.0 cm sitting at the centre of a large 20-turn coil of radius 0.20 m, do not try to find the little coil's field at the big one. Drive the big coil instead: at its centre $B = \mu_0 N_1 I/(2R) = (4\pi\times10^{-7})(20)I/(0.40) = 6.28\times10^{-5}I$ T, essentially uniform over the tiny coil. Its area is $\pi(0.010)^2 = 3.14\times10^{-4}$ m², so one turn catches $1.97\times10^{-8} I$ Wb and 100 turns catch $1.97\times10^{-6}I$. Hence $M = 1.97\times10^{-6}$ H.

LR circuits: the switch closes, the switch opens

Put a battery $\mathcal{E}$, a resistor $R$ and an inductor $L$ in a loop and close the switch at $t = 0$. Going round the loop, the battery lifts the potential by $\mathcal{E}$, the resistor drops $IR$, and the inductor drops $L\,dI/dt$:

$\mathcal{E} = IR + L\frac{dI}{dt}$

At the first instant $I = 0$ (it was zero before, and current is continuous), so the whole battery voltage goes into $L\,dI/dt$: the current starts climbing at $\mathcal{E}/L$. A 15 V battery on a 50 mH coil starts at $15/0.050 = 300$ A/s whatever the resistance is. As $I$ grows, the resistor eats more of the voltage and the climb flattens, until at $dI/dt = 0$ the current sits at $I_\infty = \mathcal{E}/R$. Separating variables in the loop equation gives the curve between:

$I(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right), \qquad \tau = \frac{L}{R}$

Check that $\tau = L/R$ has units of seconds: henries are volt-seconds per amp, ohms are volts per amp, so the division leaves seconds. And note the direction: $\tau$ grows with $L$ but shrinks with $R$, opposite to the RC circuit. More resistance drives the current to its final value faster.

Open the switch on a decaying loop instead and the same algebra with no battery gives $I(t) = I_0 e^{-t/\tau}$. A 0.50 H coil dumping into 20 Ω has $\tau = 0.025$ s, and falling to a tenth of its starting current takes $t = \tau\ln 10 = (0.025)(2.303) = 0.0576$ s.

Worked example 3. A 10 V battery, a 5.0 Ω resistor and a 0.50 H inductor are connected in series at $t = 0$. How much heat does the resistor dissipate between $t = 0$ and $t = \tau$?

Heat is not $I^2R\tau$, because $I$ is changing the whole time. You have to integrate. First the constants: $I_\infty = 10/5.0 = 2.0$ A and $\tau = 0.50/5.0 = 0.10$ s. Then

$Q = \int_0^{\tau} I^2 R\,dt = I_\infty^2 R\int_0^{\tau}\left(1 - e^{-t/\tau}\right)^2 dt$

Expand the square into $1 - 2e^{-t/\tau} + e^{-2t/\tau}$ and integrate each piece over $0$ to $\tau$. The first gives $\tau$. The second gives $-2\tau(1 - e^{-1})$. The third gives $\tfrac{\tau}{2}(1 - e^{-2})$. With $e^{-1} = 0.3679$ and $e^{-2} = 0.1353$, the bracket is

$\tau\left[1 - 2(0.6321) + \tfrac12(0.8647)\right] = \tau\left[1 - 1.2642 + 0.4323\right] = 0.1681\,\tau$

So $Q = (2.0)^2(5.0)(0.10)(0.1681) = (2.0)(0.1681) = 0.336$ J. Sanity check: if the current had been at its full 2.0 A the whole time you'd get $I^2R\tau = 2.0$ J, so 0.336 J is comfortably smaller, as it must be.

Worked example 4. A 0.40 H inductor and an 8.0 Ω resistor are switched across a battery at $t = 0$. When does the inductor hold half of its eventual energy?

Half the energy, not half the current: that's the whole problem. Energy goes like $I^2$, so half the energy means $I = I_\infty/\sqrt{2} = 0.7071 I_\infty$. Set that equal to the rise curve:

$1 - e^{-t/\tau} = 0.7071 \;\Longrightarrow\; e^{-t/\tau} = 0.2929$

Take logs: $t/\tau = -\ln(0.2929) = 1.228$. With $\tau = 0.40/8.0 = 0.050$ s,

$t = (1.228)(0.050) = 0.0614\ \text{s}$

Notice it takes longer than one time constant, because at $t = \tau$ the current is only 63% of final and the energy only $0.63^2 = 40\%$.

That continuity rule also explains the spark. Suppose a 12 V battery sits in series with 4.0 Ω, and an ideal inductor is wired in parallel with a 6.0 Ω resistor. After a long time the inductor is a plain wire, so it shorts out the 6.0 Ω and carries the full $12/4.0 = 3.0$ A. Now open the switch. The inductor's 3.0 A cannot vanish, and the only path left is the 6.0 Ω resistor, so 3.0 A is forced through it: $V = (3.0)(6.0) = 18$ V, larger than the battery that started it. Real switches with air gaps and big coils are how you get thousands of volts, and a spark.

LC circuits: a circuit that oscillates

Take the resistor out. Charge a capacitor to $Q_0$, connect it across an inductor, and let go. The loop rule says the capacitor's voltage $Q/C$ equals the inductor's $L\,dI/dt$, and the current is the rate the capacitor discharges, $I = -dQ/dt$. Substituting:

$\frac{Q}{C} + L\frac{d^2Q}{dt^2} = 0 \quad\Longrightarrow\quad \frac{d^2Q}{dt^2} = -\frac{1}{LC}Q$

That is exactly the equation you solved for a mass on a spring, with $Q$ in place of $x$. So the answer is already known:

$Q(t) = Q_0\cos\omega t, \qquad \omega = \frac{1}{\sqrt{LC}}$

and the current $I = -dQ/dt = \omega Q_0 \sin\omega t$ has amplitude $I_{\max} = \omega Q_0$. The analogy is exact and worth keeping: charge plays position, current plays velocity, $L$ plays mass, and $1/C$ plays the spring constant. The current peaks a quarter period after the charge does, just as a mass moves fastest through equilibrium.

For $L = 2.0$ mH and $C = 5.0$ µF, $\omega = 1/\sqrt{(2.0\times10^{-3})(5.0\times10^{-6})} = 1/\sqrt{10^{-8}} = 10^{4}$ rad/s, so $f = \omega/2\pi = 1592$ Hz.

Energy makes most LC problems faster than solving for the time dependence. The total is constant:

$\frac{Q^2}{2C} + \tfrac12 LI^2 = \frac{Q_0^2}{2C} = \tfrac12 LI_{\max}^2$

Worked example 5. A 4.0 µF capacitor charged to 20 V is connected through a 2.0 mH inductor to an identical uncharged 4.0 µF capacitor. Resistance is negligible. What is the maximum current?

This one needs an idea, because there is no single capacitor with a fixed charge. Ask when the current is maximum. Current grows as long as there is a net voltage around the loop to push it, and the push vanishes when the two capacitor voltages are equal. Charge is conserved, so the 80 µC starts on one and ends split evenly: 10 V on each.

Energy before: $\tfrac12(4.0\times10^{-6})(20)^2 = 8.0\times10^{-4}$ J. Energy in the capacitors at that moment: two of them at 10 V, so $2\times\tfrac12(4.0\times10^{-6})(100) = 4.0\times10^{-4}$ J. The missing $4.0\times10^{-4}$ J is in the inductor:

$\tfrac12 L I^2 = 4.0\times10^{-4} \;\Longrightarrow\; I^2 = \frac{8.0\times10^{-4}}{2.0\times10^{-3}} = 0.40$

$I_{\max} = \sqrt{0.40} = 0.63$ A. Exactly half the stored energy ends up in the inductor at that instant: a result worth remembering for any two-capacitor sharing problem.

Recognising it

Reach for this lesson when you see any of these:

A number given in henries, or a component described as a coil, solenoid, toroid or choke. Then $L$ is either handed to you or has to be built from geometry with the three-step recipe.

The words "immediately after the switch is closed" or "a long time later". Those are the two limiting behaviours, and almost no algebra is needed: current continuous at the switch, $dI/dt = 0$ at the end.

An exponential asked for at a specific time, or a time asked for at a specific current. That's $\tau = L/R$ and one logarithm. If the question asks about energy rather than current, remember to take a square root before you take the log, as in example 4.

Heat, or total charge, over an interval rather than at an instant. That is an integral of $I^2R$ or $I$, never a product with $\tau$.

A capacitor and an inductor in a loop with no resistance mentioned. That's oscillation: get $\omega = 1/\sqrt{LC}$ first, and if the question asks for a current or a charge at some condition, try energy conservation before you try solving for time.

A surprising voltage across something, larger than any battery in the circuit. An inductor's current was interrupted and it forced that current through whatever path was left.

Next you leave fields behind for a while and pick up the other great conservation story in physics, starting with Temperature and the First Law, where the energy you have been shuttling between capacitor and coil starts leaking irreversibly into heat.