self-induction inductance and magnetic-field energy

When the current in a circuit changes, so does the magnetic flux it creates through that same circuit, and by Faraday's law a self-induced EMF appears: $\varepsilon_s = -\dfrac{\Delta \Phi}{\Delta t} = -L\dfrac{\Delta I}{\Delta t}$. The proportionality factor $L$ is called inductance; it is set by the geometry of the circuit and the magnetic properties of the medium, not by the current. The minus sign expresses Lenz's law: the self-induced EMF opposes the change of current, so the current in a coil cannot change instantaneously. For a long solenoid $L = \mu_0 \mu \dfrac{N^2 S}{l}$. A magnetic field carries energy: $W = \dfrac{L I^2}{2} = \dfrac{\Phi I}{2}$, with energy density $w = \dfrac{B^2}{2\mu_0\mu}$. Compare with a capacitor, where the energy $\dfrac{qU}{2}$ is stored in the electric field; here it is stored in the magnetic field.

A coil has inductance $L = 0.25$ H. The current in it changes uniformly from 2 A to 6 A in $0.02$ s. Find the magnitude of the self-induced EMF.

A solenoid has $N = 500$ turns, length $l = 0.4$ m, cross-sectional area $S = 4\cdot 10^{-4}$ m$^2$, and no core ($\mu = 1$). The current is $I = 2$ A. Find: 1) the inductance $L$; 2) the field strength $B$ inside the solenoid; 3) the magnetic-field energy $W$.

In a volume $V = 2\cdot10^{-3}$ m$^3$ a uniform magnetic field with strength $B = 0.5$ T is created (vacuum, $\mu = 1$). 1) Find the field energy in this volume. 2) What inductance must a coil have to store the same energy at a current $I = 3$ A? 3) Estimate whether this is realistic for a compact coreless coil.

A safe self-study plan in a simulator. Build an RL circuit in a virtual lab (for example, Falstad CircuitJS or PhET "Circuit Construction Kit"): a $6$ V source, a resistor $R = 100$ $\Omega$, a coil $L = 0.5$ H, a switch, an ammeter, and a current graph. Caution: do not reproduce switching-off of a real coil with an inductance of several henries — the inductive voltage spike produces a dangerous spark and can damage equipment; work only in the simulation.

1. Close the switch and record the $I(t)$ graph: verify that the current rises smoothly, not in a jump.
2. Determine from the graph the time $\tau$ in which the current reaches about $63\,\%$ of its steady value $I_0 = \dfrac{\mathcal{E}}{R}$, and compare it with the prediction $\tau = \dfrac{L}{R} = 5$ ms.
3. Double $L$ to $1$ H and check that the rise time doubles while the steady current stays the same.
4. Change $R$ and verify that $\tau$ is inversely proportional to $R$.
5. Write a two- or three-sentence conclusion in your notebook about inductance as "electrical inertia".

Show that the current in a coil rises smoothly, that the self-induced EMF opposes the change of current, and that the stored energy belongs to the magnetic field rather than to the wires as such.

A circuit diagram and a graph. (1) On the left, an RL circuit with a switch, a source $\mathcal{E}$, a resistor $R$, and a coil $L$ (the arrow for the self-induced EMF $\varepsilon_s$ points against the current while it grows). (2) On the right, the $I(t)$ graph after switching on: the curve rises smoothly towards $I_0 = \mathcal{E}/R$; the level $0.63\,I_0$ and the time $\tau = L/R$ are marked. (3) Below, a solenoid with field lines $B$ and the labels $W = \dfrac{LI^2}{2}$, $w = \dfrac{B^2}{2\mu_0\mu}$; the shaded volume $V = Sl$ shows where the energy is localized.

In a car ignition coil, the primary winding has inductance $L = 5\,\mathrm{mH}$. It carries a current $I = 8\,\mathrm{A}$. The breaker contacts open, and the current falls to zero in $\Delta t = 20\,\mathrm{\mu s}$. At that instant the coil's magnetic field stores energy $W = \frac12 L I^2 = \frac12 \times 5\times10^{-3}\,\mathrm{H} \times (8\,\mathrm{A})^2 = 0{,}16\,\mathrm{J}$. The average rate of change of current is $\frac{\Delta I}{\Delta t} = \frac{8\,\mathrm{A}}{20\times10^{-6}\,\mathrm{s}} = 4\times10^5\,\mathrm{A/s}$, so the self-induced EMF is $|\varepsilon| = L \frac{\Delta I}{\Delta t} = 5\times10^{-3} \times 4\times10^5 = 2\times10^3\,\mathrm{V}$. A secondary winding with turns ratio $100$ steps this up to about $200\,\mathrm{kV}$, and a spark jumps between the spark-plug electrodes.

Two key relations of this topic are at work: the self-induced EMF $\varepsilon = -L \frac{dI}{dt}$ when the circuit is broken quickly, and the magnetic-field energy $W = \frac12 L I^2$ stored in the coil.

In an induction cooktop, the coil under the glass has inductance $L = 50\,\mathrm{\mu H}$ and operates at a frequency $f = 25\,\mathrm{kHz}$ with current amplitude $I = 20\,\mathrm{A}$. At the current maximum, the magnetic field stores energy $W = \frac12 L I^2 = \frac12 \times 50\times10^{-6}\,\mathrm{H} \times (20\,\mathrm{A})^2 = 0{,}01\,\mathrm{J}$. The maximum rate of change of current is $\frac{dI}{dt} = 2\pi f I = 2\pi \times 25000\,\mathrm{Hz} \times 20\,\mathrm{A} \approx 3.14\times10^6\,\mathrm{A/s}$, so the amplitude of the self-induced EMF is $\varepsilon = L \frac{dI}{dt} \approx 50\times10^{-6} \times 3.14\times10^6 \approx 157\,\mathrm{V}$. This EMF limits the current and sets the operating regime of the circuit.

The scenario shows how an alternating current in a coil constantly experiences self-induction: $\varepsilon = -L \frac{dI}{dt}$ limits the rate of change of current, while the energy $W = \frac12 L I^2$ is periodically stored in the magnetic field and returned to the source.

In a wireless phone charger, the coil has inductance $L = 20\,\mathrm{\mu H}$ and operates at a frequency $f = 100\,\mathrm{kHz}$ with current amplitude $I = 1.5\,\mathrm{A}$. The energy stored in the magnetic field is $W = \frac12 L I^2 = \frac12 \times 20\times10^{-6}\,\mathrm{H} \times (1.5\,\mathrm{A})^2 = 2.25\times10^{-5}\,\mathrm{J}$. The maximum rate of change of current is $\frac{dI}{dt} = 2\pi f I = 2\pi \times 100000 \times 1.5 \approx 9.42\times10^5\,\mathrm{A/s}$, so the amplitude of the self-induced EMF is $\varepsilon = L \frac{dI}{dt} \approx 20\times10^{-6} \times 9.42\times10^5 \approx 18.8\,\mathrm{V}$. It is self-induction that prevents the current from changing instantaneously and enables energy transfer through the alternating magnetic field.

The link between inductance and magnetic-field energy is used here: $W = \frac12 L I^2$ determines how much energy the charging coil stores, and $\varepsilon = -L \frac{dI}{dt}$ explains why the current in the coil cannot change instantaneously and sets the required conversion frequency.