Self-Inductance

A coil's own current makes a magnetic field through the coil itself. Change the current and that flux changes, so the coil induces an emf in itself. A coil built to make use of this is an inductor, and the strength of the effect is its self-inductance.

Key idea

$$\varepsilon = -L\frac{dI}{dt}, \qquad L = \frac{N\Phi_B}{I}.$$ Unit: the henry, $1\ \text{H} = 1\ \text{V·s/A}$. For a long solenoid of $N$ turns, length $\ell$ and cross-section $A$, $L = \dfrac{\mu_0N^2A}{\ell}$.

Why it works

The flux a coil makes through itself is proportional to its current, so $N\Phi = LI$ for some constant $L$ fixed by the coil's shape. Faraday's law then gives $\varepsilon = -N\dfrac{d\Phi}{dt} = -L\dfrac{dI}{dt}$. For a solenoid, $B = \mu_0(N/\ell)I$, so $N\Phi = N\cdot\mu_0\dfrac{N}{\ell}I\cdot A$ and $L = \mu_0N^2A/\ell$. The $N^2$ appears because more turns make a stronger field and give that field more turns to thread. The minus sign means the emf fights the change: an inductor makes current sluggish, like mass makes motion sluggish.

An inductor on a battery: the back emf $L\,dI/dt$ is largest when the current is changing fastest, at the start, and fades as the current settles.

Worked examples

Worked example: A solenoid's inductance

A solenoid $25\ \text{cm}$ long has 500 turns of radius $2.0\ \text{cm}$. Find $L$.

$A = \pi(0.020)^2 = 1.26 \times 10^{-3}\ \text{m²}$.
$L = \dfrac{\mu_0N^2A}{\ell} = \dfrac{(4\pi \times 10^{-7})(500)^2(1.26 \times 10^{-3})}{0.25} = 1.58\ \text{mH}$.

Answer: $L \approx 1.58\ \text{mH}$

Worked example: The emf when the current changes

The current in that solenoid rises at $100\ \text{A/s}$. Find the emf.

$|\varepsilon| = L\dfrac{dI}{dt} = (1.58 \times 10^{-3})(100) = 0.158\ \text{V}$, opposing the rise.

Answer: $|\varepsilon| \approx 0.158\ \text{V}$

Worked example: An alternating current

A $50\ \text{mH}$ inductor carries $I = 2.0\sin(2\pi \cdot 60\,t)$ amperes. What is the largest emf across it?

$\dfrac{dI}{dt} = 2.0(2\pi \cdot 60)\cos(2\pi \cdot 60\,t)$, largest when the cosine is $\pm 1$: $754\ \text{A/s}$.
$|\varepsilon|_{\max} = (0.050)(754) = 37.7\ \text{V}$, when the current passes through zero.

Answer: $|\varepsilon|_{\max} \approx 37.7\ \text{V}$

Every circuit has some inductance, since every current makes a field around its own wire, but coils concentrate it. Winding the coil on an iron core multiplies $L$ by hundreds or thousands, because the iron's atoms line up with the coil's field and add their own. That is why transformers, motors and the chokes in power supplies are built around iron.

Common mistake

"An inductor opposes current." It opposes changes in current. A steady current, however large, has $dI/dt = 0$ and meets no back emf at all; the inductor then acts like a plain wire. Check with $\varepsilon = -L\,dI/dt$: the formula contains the rate of change, not the current itself.

Self-Inductance

Worked examples

Practice

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Sources and further reading