Inductors and self-inductance

What an inductor does

• An inductor is a coil of wire, often wound on a core. Its purpose is to oppose changes in the current flowing through it.
• The mechanism is self-induction:
  • a current through the coil creates a magnetic field and hence a flux through the coil itself,
  • if the current changes, that flux changes,
  • by Faraday's law a back emf is induced in the coil,
  • by Lenz's law that back emf opposes the change in current that produced it.

$$\varepsilon = -L\frac{\Delta I}{\Delta t}$$

• $\varepsilon$ — induced (back) emf (V)

• $L$ — self-inductance (henries, H)

• $\dfrac{\Delta I}{\Delta t}$ — rate of change of current (A s⁻¹)

• Self-inductance measures how much back emf a coil produces per unit rate of change of current. A coil of $1$ H produces $1$ V of back emf when the current through it changes at $1$ A s⁻¹.

• Inductances in practice are usually in millihenries ($1$ mH $= 10^{-3}$ H).

The capacitor–inductor duality

Learning these as a pair is the most efficient way to master this section:

	Capacitor	Inductor
Opposes a change in	voltage	current
Stores energy in	an electric field	a magnetic field
Energy stored	$E = \tfrac{1}{2}CV^2$	$E = \tfrac{1}{2}LI^2$
Time constant	$\tau = RC$	$\tau = L/R$
At the instant of switching on	acts like a wire (no voltage, max current)	acts like a break (no current, max voltage)
After a long time	acts like a break (no current)	acts like a wire (no voltage across it)
Cannot change instantly	its voltage	its current

• Those last two rows are the key to every switching question. A capacitor's voltage and an inductor's current can never jump instantaneously — that would require infinite current or infinite emf respectively.

Energy stored in an inductor

$$E = \tfrac{1}{2}LI^2$$

• $E$ — energy stored in the magnetic field (J)
• The $\tfrac{1}{2}$ appears for the same reason as with a capacitor: the current builds from zero, so the average is half the final value.
• The energy is stored in the magnetic field, and is returned to the circuit when the current falls.

Switching an inductive circuit on and off

Switching on (an inductor in series with a resistor and a supply):

• at $t = 0$ the current is zero and the back emf is at its largest, opposing the rise,
• the current then grows toward its final value $I_0 = V/R$,
• the growth curve has the same shape as a charging capacitor's voltage curve.

Switching off:

• the current is forced to fall rapidly, so $\Delta I/\Delta t$ is very large and negative,
• the back emf is correspondingly very large, and now acts to maintain the current,
• this can produce a voltage far higher than the supply, often enough to arc across the opening switch contacts.
• This is why switching off an inductive load (a motor, a relay, an electromagnet) produces sparks, and why such circuits include a protective diode to give the decaying current a safe path.

The time constant

$$\tau = \frac{L}{R}$$

• Check the units: (H)/(Ω) = seconds.
• After $1\tau$ the current has reached $63\%$ of its final value when switching on, or fallen to $37\%$ when switching off — the same landmarks as an RC circuit.
• Larger $L$ → slower change, because more back emf opposes it.
• Larger $R$ → faster change, because the final current is smaller and reached sooner. Note this is the opposite of an RC circuit, where a larger $R$ slows things down.