A coil that answers back
A coil carrying a current \(I\) sets up a magnetic field, and that field passes through the coil's own turns. So the coil holds a flux linkage made by its own current. The field is proportional to the current, so the flux linkage is too. The constant of proportion is the coil's self-inductance \(L\).
Now let the current change. The coil's own flux linkage changes with it, and by Faraday's law an emf is induced in the coil itself. By Lenz's law this emf opposes the change of current, not the current. While the current grows, the emf pushes against it. While the current falls, the emf pushes to keep it going. This self-induced emf is called the back emf.
The self-inductance \(L\) of a coil is its flux linkage per unit current, \(L = N\Phi/I\). Equally, it is the back emf induced per unit rate of change of current, \(\varepsilon = -L\,dI/dt\). Its unit is the henry: \(1\,\text{H} = 1\,\text{Wb/A} = 1\,\text{V s/A}\).
A coil made for its inductance is called an inductor. In a circuit diagram it is drawn as a row of loops, with two bars beside it when it has an iron core. A coil of \(1\,\text{H}\) gives a back emf of \(1\,\text{V}\) when its current changes at \(1\,\text{A/s}\). A steady current, of any size, gives no back emf at all.
An inductor treats current the way mass treats velocity. A mass resists a change of its velocity, not the velocity; an inductor resists a change of its current, not the current. A large \(L\) is like a heavy body. It takes a long, firm push to get its current going, and a long time to stop it.
The current in a coil flows clockwise and is decreasing. Which way does the back emf push?
The inductance of a long solenoid
A long solenoid is the one coil whose inductance follows in a few lines. Its field inside is uniform, \(B = \mu_0 nI\), from the chapter Sources of the Magnetic Field. Each of its \(N = nl\) turns holds the same flux \(BA\). So the flux linkage is simple to write down, and dividing by the current gives \(L\).
\[ B = \mu_0 nI \]
Why. the field inside a long solenoid with \(n\) turns per metre.
\[ \Phi = BA = \mu_0 nIA \]
Why. the flux through one turn of cross-section \(A\).
\[ N\Phi = (nl)(\mu_0 nIA) = \mu_0 n^2 A l\,I \]
Why. the solenoid has \(N = nl\) turns, and each holds the same flux.
\[ L = \dfrac{N\Phi}{I} = \mu_0 n^2 A l \]
Read it. the definition of self-inductance; the current cancels.
A solenoid has \(1000\) turns per metre, a cross-section of \(10^{-3}\,\text{m}^2\) and a length of \(0.5\,\text{m}\). Find (a) its self-inductance, and (b) its inductance when a soft iron core of \(\mu_r = 1000\) fills it.
(a) \(L = \mu_0 n^2 A l = 4\pi\times10^{-7}\times10^6\times10^{-3}\times0.5 = 6.3\times10^{-4}\,\text{H}\), about \(0.63\,\text{mH}\).
(b) The core multiplies the field, and so the flux linkage, by \(\mu_r\): \(L = 1000\times6.3\times10^{-4} = 0.63\,\text{H}\).
Values used: \(\mu_0 = 4\pi\times10^{-7}\,\text{T m/A}\).
The square of \(n\) matters. Wind twice the turns on the same length, and the field inside doubles. There are also twice as many turns to link it. So the inductance becomes four times as large. The inductance does not depend on the current: it is fixed by the size, the turns and what fills the coil.
A solenoid \(25\,\text{cm}\) long has \(500\) turns and a cross-section of \(2\,\text{cm}^2\). What is its self-inductance, in millihenry?
A long solenoid is rewound with twice as many turns on the same length and the same tube. What happens to its self-inductance?
Reading the back emf from a graph
The back emf is \(-L\) times the slope of the current-time graph. So a straight rise gives a steady emf, a flat stretch gives none, and a straight fall gives a steady emf of the other sign. The size of the current does not matter, only how fast it changes.
In stage I of Fig 3.1 the slope is \(4/0.1 = 40\,\text{A/s}\), so the back emf is \(0.2\times40 = 8\,\text{V}\), against the rising current. In stage II the current is a steady \(4\,\text{A}\) and the emf is zero. In stage III the slope is \(-4/0.2 = -20\,\text{A/s}\), so the emf is \(4\,\text{V}\). It now points along the current, trying to keep it going.
The current in a \(50\,\text{mH}\) inductor falls steadily from \(2\,\text{A}\) to zero in \(5\,\text{ms}\). What is the size of the back emf, in volts?
Which of these are equal to one henry?
The spark at the switch
Open a switch in a circuit with a large inductor. The current must fall from its full value to zero in the instant the contacts part. That is a very large \(dI/dt\), so the back emf \(L\,dI/dt\) is very large. It appears across the gap between the contacts and can break down the air into a spark. Suppose a \(1\,\text{H}\) coil carries \(0.4\,\text{A}\), and the current is broken in \(0.5\,\text{ms}\). The emf is about \(1\times0.4/0.0005 = 800\,\text{V}\).
The choke of a tube light puts this spark to work. It is an inductor of many turns on an iron core, joined in series with the tube. A small switch called the starter sits across the ends of the tube.
A tube light will not glow on \(220\,\text{V}\) alone; its gas needs several hundred volts to strike. The choke supplies them. At switch-on the starter closes and a current flows through the choke and the two filaments, warming them. Then the starter's contacts spring apart and break the current suddenly. The choke's large \(L\,dI/dt\) appears across the tube and strikes it. Once the tube glows, its gas conducts, and the choke keeps the current in check, as the next chapter shows.
A spark often jumps at the switch when a circuit with a large coil is opened, but not when it is closed. Explain why.
On opening, the current must fall from its full value to zero in the instant the contacts part, so \(dI/dt\) is very large. The coil's back emf \(L\,dI/dt\) is then large enough to break down the air in the gap, and the energy stored in the coil goes into the spark. On closing, the current only grows gradually from zero, so the back emf never exceeds the battery's emf.
Joseph Henry taught at the Albany Academy in New York and built some of the strongest electromagnets of his time. In 1832 he noticed a bright spark when he broke the circuit of a long coil of wire. A short wire in the same circuit gave almost none. He reasoned that the coil's own changing field induced an emf in the coil itself, and he published this self-induction that year. The SI unit of inductance, the henry, is named after him.
- Thinking the back emf opposes the current.It opposes the change of current. While the current falls, the back emf points along it.
- Expecting a large steady current to give a large back emf.The back emf is \(L\,dI/dt\). A steady current, however large, gives none.
- Doubling L when the turns are doubled.\(L = \mu_0 n^2Al\). Twice the turns on the same length give four times the inductance.
- Thinking L changes with the current.L is fixed by the coil's size, turns and core. The current only sets how much flux the coil holds.