Inductance and magnetic energy

Faraday's law says a circuit responds to any change in the flux through it, and nothing in it excludes the flux the circuit produces itself.

That loophole is the whole of this lesson. It gives a circuit element with no static counterpart, an energy stored in a magnetic field, the transformer that made electrical distribution possible, and, at the end, a circuit that oscillates at a frequency assembled from $ε_{0}$ and $μ_{0}$ . It assumes Faraday's law and Lenz's law from the previous lesson, the solenoid field from the one before, and the technique of solving a first order differential equation by separating variables.

Self-inductance

Send a current $I$ through a coil and it produces a magnetic field, and that field threads the coil's own turns. Every contribution to the field is proportional to $I$ , so the total flux linkage is proportional to $I$ too, and the constant of proportionality is a property of the geometry:

N Φ_{B} = L I

$L$ is the self-inductance, in webers per ampere, called henries after Joseph Henry, who discovered self-induction in 1832. Combining this with Faraday's law,

E = - L \frac{d I}{d t}

The coil opposes changes in its own current. Increase the current and it pushes back; decrease it and it pushes forward, trying to keep the current going. This is not a metaphorical inertia. Put $L$ beside $m$ , $I$ beside $v$ , and $E$ beside force, and the equation is Newton's second law.

For a long solenoid the inductance is computable. The field inside is $B = μ_{0} n I$ from the eighth lesson, the flux through one turn is $μ_{0} n I A$ , and there are $n l$ turns in a length $l$ , so $N Φ = μ_{0} n^{2} A l I$ and

L = μ_{0} n^{2} A l = \frac{μ_{0} N^{2} A}{l}

The square on the turn count is why inductors are wound with many turns: doubling $N$ doubles both the field produced and the flux caught.

Example. A solenoid 25 cm long is wound with 500 turns over a cross section of 4.0 cm². What is its inductance, and what EMF appears across it if its current is changing at 200 A/s?

$L = μ_{0} N^{2} A / l = (4 π \times 10^{- 7}) (500)^{2} (4.0 \times 10^{- 4}) / (0.25) = 5.03 \times 10^{- 4}$ H, about half a millihenry. The induced EMF is $L d I / d t = (5.03 \times 10^{- 4}) (200) = 0.101$ V. A tenth of a volt from a rate of change that would take a large current to sustain, which is why inductors of any usefulness are wound on iron: the core multiplies $L$ by the relative permeability of the material, often by a factor of a thousand.

Now you. A solenoid 30 cm long with 800 turns over 6.0 cm² carries a current changing at 150 A/s. Find its inductance and the EMF across it.

Answer

$L = (4 π \times 10^{- 7}) (800)^{2} (6.0 \times 10^{- 4}) / (0.30) = 1.61 \times 10^{- 3}$ H, and the EMF is $(1.61 \times 10^{- 3}) (150) = 0.241$ V.

The LR circuit

Connect an inductor $L$ in series with a resistor $R$ across a source of EMF $E$ . The loop rule, in the amended form the previous lesson insisted on, gives

E - I R - L \frac{d I}{d t} = 0

which has exactly the structure of the RC circuit's equation. Separating variables and imposing $I = 0$ at the instant of connection,

I (t) = \frac{E}{R} (1 - e^{- t / τ}), τ = \frac{L}{R}

At the first instant the current is zero, because the inductor's back EMF exactly cancels the source: an inductor with no current in it behaves momentarily like a break in the circuit. After several time constants the current settles at $E / R$ and the inductor, with $d I / d t$ now zero, behaves like a plain piece of wire. The inductor's whole existence is in the transition.

Opening the switch is the dangerous case. The current is forced to fall abruptly, $d I / d t$ is enormous, and $L d I / d t$ can reach thousands of volts across the coil, which is where the spark at a switch comes from and how the ignition coil in a petrol engine makes its spark from a 12 V supply. Circuits driving inductive loads carry a diode across the coil for exactly this reason, giving the current somewhere to go while it dies away.

Example. A 0.50 H inductor in series with 25 Ω is connected to 12 V. Find the time constant, the final current, the current after 10 ms, and the energy finally stored.

$τ = L / R = 0.50 / 25 = 0.020$ s. The final current is $E / R = 12 / 25 = 0.48$ A. At $t = 0.010$ s, which is half a time constant, $I = 0.48 (1 - e^{- 0.5}) = 0.48 \times 0.3935 = 0.189$ A. The energy stored at full current is $\frac{1}{2} L I^{2} = \frac{1}{2} (0.50) (0.48)^{2} = 0.0576$ J, a result derived in the next section.

Now you. A 0.20 H inductor in series with 40 Ω is connected to 24 V. Find the time constant, the final current, the current after 2.5 ms, and the stored energy.

Answer

$τ = 0.20 / 40 = 5.0$ ms, and the final current is $24 / 40 = 0.60$ A. At 2.5 ms, half a time constant, $I = 0.60 (1 - e^{- 0.5}) = 0.236$ A. The stored energy is $\frac{1}{2} (0.20) (0.60)^{2} = 0.036$ J.

Energy in the magnetic field

Establishing a current in an inductor takes work, because the back EMF opposes the increase the whole way up. The source delivers power $E I$ against that back EMF, which is $L I d I / d t$ , so the total work done in raising the current from zero to $I$ is

U = \int_{0}^{I} L I' d I' = \frac{1}{2} L I^{2}

exactly parallel to the capacitor's $\frac{1}{2} C V^{2}$ , and for the same reason: the opposition grows as the quantity being established grows, so the average is half the final value.

Where is it? Run the same substitution as for the capacitor. For a solenoid, $L = μ_{0} n^{2} A l$ and $B = μ_{0} n I$ , so $I = B / (μ_{0} n)$ and

U = \frac{1}{2} μ_{0} n^{2} A l \frac{B^{2}}{μ_{0}^{2} n^{2}} = \frac{B^{2}}{2 μ_{0}} \times (A l)

with $A l$ the volume inside the solenoid, where the field is. Dividing,

u = \frac{B^{2}}{2 μ_{0}}

joules per cubic metre. The electric and magnetic energy densities are now both in hand, $\frac{1}{2} ε_{0} E^{2}$ and $B^{2} / 2 μ_{0}$ , and in the twelfth lesson a light wave will carry both, in equal amounts.

Comparing them says something practical. At 1 T, the magnetic density is $1 / (2 μ_{0}) = 3.98 \times 10^{5}$ J/m³, four orders of magnitude above the 39.8 J/m³ an air capacitor can reach at the breakdown field. Magnetic storage is the better of the two, and superconducting magnetic energy storage exists commercially on that basis, though it still loses to chemical fuel by a factor of $10^{5}$ .

The same quantity is also a pressure, since joules per cubic metre are newtons per square metre, and it is the pressure the field exerts on whatever carries the current making it. In the 45 T of the strongest steady laboratory magnet, $B^{2} / 2 μ_{0} = 8.06 \times 10^{8}$ Pa, about 8000 atmospheres, which is comparable to the yield strength of good steel. That is the real limit on magnetic fields: not the current, not the heat, but the fact that a strong enough magnet tears itself apart.

Mutual inductance and the transformer

Two coils sharing flux induce in each other, with $E_{2} = - M d I_{1} / d t$ , and the mutual inductance $M$ is the same in both directions, which is a small theorem worth believing rather than deriving here. That is Faraday's iron ring, and it is a transformer.

Wind $N_{p}$ turns and $N_{s}$ turns on a common core so that essentially all the flux is shared. Each turn on either side sees the same $d Φ / d t$ , so

\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}}

and if the transformer is lossless, power in equals power out, so the currents go the other way: $I_{s} / I_{p} = N_{p} / N_{s}$ . A transformer trades voltage against current at fixed power, and it works only on alternating current, because a steady current produces no $d Φ / d t$ at all.

That constraint decided the war between direct and alternating current in the 1880s, and the reason is $I^{2} R$ . Send 500 MW down a line of 5 Ω resistance at 400 kV and the current is 1250 A, so the loss is $I^{2} R = 7.8$ MW, or 1.6 per cent. Send the same power at 132 kV and the current is 3788 A, so the loss is 71.7 MW, or 14.3 per cent. Nine times the loss for a third of the voltage, because the loss goes as the square. Long distance transmission demands high voltage, safe use demands low voltage, and only a transformer can convert between them. Edison's direct current system had no way to do it and lost on that point alone.

Real transformers fall short of the ideal in three named ways, and each is worth recognising because each is a term already met in this course. Not all the flux from one winding reaches the other, which is leakage and behaves as an unwanted series inductance. The windings have resistance, so the $I^{2} R$ that transmission avoids reappears inside the transformer, as copper loss. And the iron core is itself a conductor sitting in a changing flux, so eddy currents circulate in it and dissipate, which is why cores are built from thin laminations insulated from each other rather than from solid iron, and why the laminations are stacked along the flux direction so the circulating paths are broken. Add hysteresis loss, the energy spent driving the iron round its magnetisation loop on every cycle, and a large power transformer still manages better than 99 per cent efficiency, which is among the highest of any machine ever built.

The LC circuit

Now connect a charged capacitor across an inductor, with no resistance. The loop rule gives $q / C + L d I / d t = 0$ , and with $I = d q / d t$ ,

L \frac{d^{2} q}{d t^{2}} + \frac{q}{C} = 0

which is the equation of simple harmonic motion, identical in form to a mass on a spring with $L$ playing the mass and $1 / C$ the spring constant. Its solution oscillates at

ω = \frac{1}{\sqrt{L C}}

Physically the energy sloshes. It starts entirely electric, $Q^{2} / 2 C$ in the capacitor. As the capacitor discharges the current grows and the energy moves into the magnetic field of the inductor, and at the moment the capacitor is empty the current is maximal and all the energy is magnetic, $\frac{1}{2} L I^{2}$ . The inductor's inertia then keeps the current going, recharging the capacitor with the opposite polarity, and the cycle repeats. The total is constant, and equating the two extremes gives the peak current directly: $\frac{1}{2} L I_{0}^{2} = Q_{0}^{2} / 2 C$ , so $I_{0} = Q_{0} / \sqrt{L C} = ω Q_{0}$ , which is the same relation between amplitude and peak speed that a mass on a spring obeys.

No real circuit does this forever. Every inductor has resistance, so the equation acquires a damping term and becomes $L d^{2} q / d t^{2} + R d q / d t + q / C = 0$ , which is the damped oscillator that Differential Equations solves in full: oscillation with an amplitude decaying as $e^{- R t / 2 L}$ while $R$ is small, and no oscillation at all once $R$ exceeds $2 \sqrt{L / C}$ . The ratio of stored energy to energy lost per radian is the quality factor $Q = ω L / R$ , and a good radio tuning circuit reaches a few hundred, which is what lets it pick one station out of a band. That is the practical use of the whole section: an LC circuit responds strongly at one frequency and weakly at others, so it selects.

Example. An inductor of 10 mH is connected to a capacitor of 100 nF. At what frequency does the circuit oscillate?

$L C = (10^{- 2}) (10^{- 7}) = 10^{- 9}$ , whose square root is $3.162 \times 10^{- 5}$ . Then $ω = 1 / (3.162 \times 10^{- 5}) = 3.162 \times 10^{4}$ rad/s, and $f = ω / 2 π = 5.03 \times 10^{3}$ Hz, an audible tone.

Now you. An inductor of 2.5 mH is connected to a capacitor of 40 nF. What is the oscillation frequency?

Answer

$L C = (2.5 \times 10^{- 3}) (4.0 \times 10^{- 8}) = 1.0 \times 10^{- 10}$ , so $\sqrt{L C} = 1.0 \times 10^{- 5}$ and $ω = 10^{5}$ rad/s. Then $f = 10^{5} / 2 π = 1.59 \times 10^{4}$ Hz.

A hint worth taking seriously

Look at what $ω = 1 / \sqrt{L C}$ is made of.

The inductance of a solenoid is $μ_{0} n^{2} A l$ , proportional to $μ_{0}$ and otherwise pure geometry. The capacitance of a parallel plate capacitor is $ε_{0} A / d$ , proportional to $ε_{0}$ and otherwise pure geometry. So for any circuit built out of such elements, $L C$ is $ε_{0} μ_{0}$ multiplied by something with the dimensions of length squared, and

\frac{1}{\sqrt{ε_{0} μ_{0}}}

has the dimensions of a speed. Putting in the numbers, $ε_{0} μ_{0} = (8.854 \times 10^{- 12}) (1.2566 \times 10^{- 6}) = 1.113 \times 10^{- 17}$ , whose reciprocal square root is $3.00 \times 10^{8}$ m/s.

That number should be recognisable. It is the speed of light, and it has just fallen out of a capacitor and a coil, apparatus with no light in it anywhere. Wilhelm Weber and Rudolf Kohlrausch obtained it in 1856 by charging a capacitor, measuring its charge electrostatically with a torsion balance and then magnetically by discharging it through a galvanometer, and taking the ratio. They got $3.107 \times 10^{8}$ m/s, and Fizeau's 1849 measurement of the speed of light gave $3.15 \times 10^{8}$ m/s. Weber and Kohlrausch noted the closeness and did not know what to make of it.

Maxwell did. But before the connection can be made, one of the four field equations has to be repaired, because as it stands Ampère's law contradicts the conservation of charge. Finding the contradiction is the next lesson.