Consider a coil of wire with current $i$ flowing through it that is changing at $\frac{di}{dt}$. The current $i$ in the circuit causes a magnetic field $\vec{B}$ in the coil with N turns of wire, and hence an average magnetic flux $\Phi_{B}$ through each turn of the coil.

From Faraday’s Law, the induced e.m.f. is given by:

$$\epsilon =-N \frac{d\Phi_{B}}{dt}$$

We introduce a quantity – self inductance to relate the magnetic flux and the current through the coil.

$$N \Phi_{B} = Li$$

Hence,

$$\epsilon =-L \frac{di}{dt}$$

When the current $i$ changes, the flux changes and a self-induced e.m.f. appears in the circuit.

The coil of wire is called an **inductor**. Inductor opposes changes in current in the circuit.

Energy cannot be delivered to the inductor infinitely fast, nor can it be dissipated instantaneously, in the form of heat or light by other circuit elements. Thus, power can never be infinite. This implies that the curve of current versus time must be continuous. A graph that is discontinuous means it contains a point at which the current jumps from one value to another without taking on all the values in between. When this happens, the slope of the curve at that location is infinite, which would imply infinite power.

### Example: Calculating Self Inductance

A toroidal solenoid with cross-sectional area A and mean radius r is closely wound with N turns of wire. The toroid is wound on a non-magnetic core. Determine its self-inductance L.

Assume that B is uniform across a cross section.

$$\begin{aligned} L &= \frac{N \Phi_{B}}{i} \\ &= \frac{N \left( BA \right)}{i} \\ &= \frac{N}{i} \left( \frac{ \mu_{0} Ni}{2 \pi r} \right) A \\ &= \frac{\mu_{0}N^{2}A}{2 \pi r} \end{aligned}$$

**Note:** The magnetic field of a toroidal solenoid can be found by applying Ampere’s Law.

The self-inductance of a circuit depends on its size, shape, and number of turns. For N turns close together, it is always proportional to $N^{2}$. It also depends on the magnetic properties of the materials enclosed by the circuit.