Self Inductance

The phenomenon

• Consider a circuit (coil) in which the current is changing — for example a coil in series with a battery, a rheostat and a key K. The changing current varies the magnitude of the magnetic flux linked with the coil itself, and consequently an emf is induced in that same circuit.
• The production of an induced emf in the circuit (coil) itself, on account of a change in the current in it, is termed the phenomenon of self-induction, and the coefficient involved is the self-inductance.
• Diagram in words: a single coil connected through a key K and a rheostat to a cell; arrows show the flux threading the coil produced by its own current $i$.

Definition of the coefficient

• Let the magnetic flux linked with the circuit itself at any instant be $\phi$, corresponding to a current $i$ in it. Since the field, and hence the flux, is produced by that current,

$$\phi \propto i \qquad\Rightarrow\qquad \phi = Li \qquad\text{or}\qquad L = \dfrac{\phi}{i} \qquad \ldots (1)$$

• where $L$ is the constant of proportionality, called the self-inductance (or coefficient of self induction) of the coil.
• For a closely wound coil of $N$ turns, the same magnetic flux is linked with all the turns, and each turn contributes to the induced emf when the flux changes. The quantity used is then the flux linkage $N\phi_B$:

$$N\phi_B\propto i,\qquad N\phi_B = Li \qquad\Rightarrow\qquad L = \dfrac{N\phi_B}{i} \qquad \ldots (2)$$

• The inductance $L$ depends only on the geometry of the coil and the material properties of its core.

Self-induced emf and the unit

• According to Faraday's law, $e = -\dfrac{d\phi}{dt}$. Using $\phi = Li$ with $L$ constant,

$$e = -\dfrac{d}{dt}(Li) = -L\dfrac{di}{dt} \qquad \ldots (3)$$

• Hence the unit of $L$ is

$$\text{unit of } L = \dfrac{\lvert e\rvert}{\lvert di/dt\rvert} = \dfrac{\text{volt}}{\text{ampere/second}} = \text{henry (H)}$$

• Also $1\ \text{H} = 1\ \text{Wb A}^{-1} = 1\ \text{T m}^2\text{A}^{-1} = 1\ \text{ohm}\cdot\text{second}$.

The three definitions of $L$ the board accepts

• From equation (1), $\phi = Li$: the self-inductance of a circuit is the ratio of the magnetic flux (produced by the current in the circuit) linked with the circuit to the current flowing in it. In words, the magnetic flux produced per unit current in the circuit.
• From equation (3), $L = \dfrac{\lvert e\rvert}{\lvert di/dt\rvert}$: the self-inductance is the ratio of the induced emf produced around the circuit to the rate of change of current in it — that is, the induced emf produced per unit rate of change of current.
• From the work done. When the current increases, the induced emf acts against it, so work has to be done to establish the flux associated with a steady current $i_0$. Work done in time $dt$ is

$$dW = e\cdot i\,dt = \left(L\dfrac{di}{dt}\right)i\,dt = L\,i\,di$$

$$W = \int dW = \int_0^{i_0} L\,i\,di = \dfrac{1}{2}Li_0^2 \qquad\text{(magnitude)}$$

• Putting $i_0 = 1$ gives $W = \dfrac{1}{2}L$, so $L = 2W$ numerically. Hence the self-inductance of a circuit is numerically equal to twice the work done in establishing the magnetic flux associated with unit current in the circuit.
• This work represents the energy of the circuit: energy $=\dfrac{1}{2}Li_0^2$.

Electrical inertia

• Comparing $\dfrac{1}{2}Li^2$ with the kinetic energy $\dfrac{1}{2}mv^2$ shows that the self-inductance $L$ of an electrical circuit plays the same role — electrical inertia — that mass (inertia) plays in mechanical motion. A large $L$ resists any change of current, just as a large mass resists any change of velocity.

Inductance of a solenoid

• If a current $i$ is established in the windings of a long solenoid, it produces a magnetic flux $\phi_B$ through the central region. The inductance is $L = \dfrac{N\phi_B}{i}$, where $N\phi_B$ is the flux linkage.
• For a length $l$ near the middle of the solenoid, with $n$ turns per unit length and cross-sectional area $A$, and with $\theta = 0^\circ$ between $\vec{B}$ and $\vec{A}$,

$$N\phi_B = (nl)\left(\vec{B}\cdot\vec{A}\right) = nlBA$$

• The field inside a solenoid is $B = \mu_0 n i$, so

$$L = \dfrac{N\phi_B}{i} = \dfrac{nl(\mu_0 n i)A}{i} = \mu_0 n^2 A l$$

• where $Al$ is the interior volume of the solenoid. Therefore the inductance per unit length near the middle of a long solenoid is

$$\dfrac{L}{l} = \mu_0 n^2 A = \dfrac{\mu_0 n^2\pi d^2}{4}$$

• $d$ being the diameter of the solenoid. This implies $L\propto n^2$ and $L\propto d^2$.
• In terms of the total number of turns $N = nl$, the same result is $L = \dfrac{\mu_0 N^2 A}{l}$.
• Since $n$ is a number per unit length, the inductance is a product of $\mu_0$ and a quantity with the dimension of length — which is why $\mu_0$ can be expressed in henry per metre.

Inductance of a toroid

• For a toroid of $N$ turns, circular cross-section of radius $r$ and major radius $R$, the field inside is $B = \dfrac{\mu_0 N i}{2\pi r_0}$ where $r_0$ is the distance from the toroidal axis. If $r\ll R$ we may take $r_0\approx R$, so $B = \dfrac{\mu_0 N i}{2\pi R}$.
• The flux through the cavity, which links each turn, is

$$\phi = \pi r^2 B = \dfrac{\mu_0 N i r^2}{2R}$$

• By Faraday's law $e = -N\dfrac{d\phi}{dt} = -\dfrac{\mu_0 N^2 r^2}{2R}\dfrac{di}{dt}$; comparing with $e = -L\dfrac{di}{dt}$,

$$L = \dfrac{\mu_0 N^2 r^2}{2R}\qquad (r\ll R)$$

Combination of inductances

• If several inductances are connected in series or in parallel, the total inductance is given by

$$L_{\text{total}} = L_1+L_2+L_3+\cdots \qquad\text{(series)}$$

$$\dfrac{1}{L_{\text{total}}} = \dfrac{1}{L_1}+\dfrac{1}{L_2}+\dfrac{1}{L_3}+\cdots \qquad\text{(parallel)}$$

• The parallel combination is always smaller than the smallest of the individual coils.

Standard results

• $\phi = Li$; $L = \dfrac{N\phi_B}{i}$; $e = -L\dfrac{di}{dt}$; $L = \dfrac{\lvert e\rvert}{\lvert di/dt\rvert}$.
• $W = \dfrac{1}{2}Li^2$, and $L = 2W$ numerically for unit current.
• Long solenoid: $L = \mu_0 n^2 Al = \dfrac{\mu_0 N^2A}{l}$; toroid: $L = \dfrac{\mu_0 N^2r^2}{2R}$.
• Coaxial cable of length $l$, inner radius $a$, outer radius $b$: $L = \dfrac{\mu_0 l}{2\pi}\ln\left(\dfrac{b}{a}\right)$.
• Two parallel wires of length $l$, radius $a$, separation $d$: $L = \dfrac{\mu_0 l}{\pi}\ln\left(\dfrac{d}{a}\right)$.
• Unit: henry; $1\ \text{H} = 1\ \text{Wb A}^{-1} = 1\ \text{ohm}\cdot\text{second}$.