Self-inductance

A spark at the plug

Pull the plug of a running motor out of the socket. Often a small blue spark jumps between the pin and the socket. The mains voltage is only a few hundred volts at most. Yet the spark needs far more to cross the gap.

Where does the voltage come from? From the motor itself. A coil can make an emf by its own changing current. This is self-induction.

The flux of a coil's own current

A current $I$ in a coil makes a magnetic field. The lines of this field pass through the turns of the same coil. Say the coil has $N$ turns, and the flux through each turn is $\phi$. The flux linkage is $N\phi$.

The field is in proportion to the current. So the flux linkage is in proportion to the current too:

The constant $L$ is the self-inductance of the coil. It tells us how much flux linkage each ampere makes. The unit is the henry (H):

A worked example. A coil of $400$ turns carries $2.5$ A. The flux through each turn is $6.0 \times 10^{-5}$ Wb.

The emf of a changing current

Now let the current change. The flux linkage changes, so by Faraday's law there is an emf:

The minus sign is Lenz's law. The emf opposes the change of the current. It does not oppose the current itself.

• When the current grows, the emf pushes against the current and slows the growth.
• When the current falls, the emf pushes along the current and slows the fall.

Think of a heavy trolley. It does not resist its speed. It resists a change of its speed. A coil does the same with its current. Inductance is a kind of electrical inertia.

A worked example. A coil of $0.75$ H has a current that grows steadily from $0$ to $3.2$ A in $0.40$ s.

So a coil of $1$ H gives $1$ V when its current changes at $1$ A/s. That is another way to define the henry.

A steady current is a different case. Then $\dfrac{dI}{dt} = 0$, and the self-induced emf is zero.

The spark

Now break the circuit. The current falls to zero in a very short time, and $\dfrac{\Delta I}{\Delta t}$ is huge.

A worked example. An electromagnet of $1.5$ H carries $0.80$ A. A switch opens, and the current falls to zero in $2.0$ ms.

The supply may give only $12$ V. But the coil makes $600$ V across the opening switch, and a spark jumps. The energy of the spark came out of the magnetic field of the coil. A diode joined across the coil gives the current another path, and removes the spark.

The inductance of a long solenoid

Take a long solenoid with $N$ turns, a length $l$ and a cross-section area $A$. The field inside is $B = \mu_0 nI$, with $n = N/l$ turns in each metre. The flux through one turn is

The flux linkage is $N\phi = \dfrac{\mu_0 N^2 A}{l}\,I$. Compare it with $N\phi = LI$:

Notice what is in it. The inductance depends only on the shape of the coil, and on what fills it. It does not depend on the current. A core that fills the solenoid multiplies $\mu_0$ by the relative permeability $\mu_r$ of the core: $L = \dfrac{\mu_r\mu_0 N^2 A}{l}$.

The number of turns appears as $N^2$. Double the turns, and the field doubles. The flux through each turn doubles, and there are twice as many turns. So $L$ grows by $4$.

A worked example. A solenoid has $1200$ turns, a length of $0.50$ m and a radius of $2.0$ cm.

1. The area. $A = \pi r^2 = \pi \times (0.020\ \text{m})^2 = 1.26 \times 10^{-3}\ \text{m}^2$.
2. The inductance in air. $L = \dfrac{4\pi \times 10^{-7}\ \text{T m/A} \times 1200^2 \times 1.26 \times 10^{-3}\ \text{m}^2}{0.50\ \text{m}} = 4.5\ \text{mH}$.
3. With a soft iron core. Say $\mu_r = 600$ for the iron. Then $L = 600 \times 4.5\ \text{mH} = 2.7\ \text{H}$.

Iron raises the inductance a great deal. The value of $\mu_r$ in a real core is not fixed. It changes with the field, so $2.7$ H is a typical figure for a modest current.

Put in a core of $\mu_r = 200$. How many times larger does $L$ become?

Practise with new numbers

Inductors in a circuit

Two inductors far apart, so that the flux of one does not pass through the other, combine like resistors.

• In series: $L = L_1 + L_2$.
• In parallel: $\dfrac{1}{L} = \dfrac{1}{L_1} + \dfrac{1}{L_2}$.

Sometimes a coil must have almost no inductance, for a precise resistor. The wire is doubled back before it is wound, so the current goes round one way in a turn and the other way in the next turn. The two fields cancel, and $L$ is close to zero.

Try it in Python

The program finds the inductance of one solenoid with three different cores. It also finds the emf when a current of $2.0$ A is switched off.

Common mistakes

• Saying the emf opposes the current. It opposes the change of the current.
• Thinking a coil blocks a steady current. In a steady current $\dfrac{dI}{dt} = 0$, so there is no self-induced emf. Only the resistance of the wire matters.
• Taking $L$ as in proportion to $N$. It goes as $N^2$.
• Forgetting to change cm to m. The area is $\pi r^2$, with $r$ in metres.
• Thinking the emf is zero when the current is zero. The emf depends on how fast the current changes, not on its size.

Remember these

• Flux linkage: $N\phi = LI$. Unit of $L$: the henry, $1\ \text{H} = 1\ \text{V s/A}$.
• Self-induced emf: $\varepsilon = -L\dfrac{dI}{dt}$. It opposes the change of the current.
• A long solenoid: $L = \dfrac{\mu_0 N^2 A}{l}$. With a core, multiply by $\mu_r$.
• Breaking a current quickly gives a large emf, and a spark.
• Series: $L_1 + L_2$. Parallel: $\dfrac{1}{L} = \dfrac{1}{L_1} + \dfrac{1}{L_2}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A coil carries a steady current of $3.0$ A. What is its self-induced emf?

2. A current in a coil is falling. In which way does the self-induced emf drive current?

3. A solenoid is wound again with twice as many turns, over the same length and area. By what factor does its inductance change?

4. The current in a coil of $0.40$ H changes from $5.0$ A to $1.0$ A in $0.20$ s. What is the size of the emf?

5. Why is a precision resistor wound with the wire doubled back on itself?