# Self Inductance

The ability of a coil to oppose any change in current is a measure of the self-inductance L of the coil.
For brevity, the prefix self is usually dropped. Inductance is measured in henries (H), after the American physicist Joseph Henry. Inductors are coils of various dimensions designed to introduce specified amounts of inductance into a circuit. The inductance of a coil varies directly with the magnetic properties of the coil. Ferromagnetic materials, therefore, are frequently employed to increase the inductance by increasing the flux linking the coil.
Fig. 1: Inductor configurations for which Equation (12.2) is appropriate
A close approximation, in terms of physical dimensions, for the inductance of the coils of Fig. 1 can be found using the following equation: $$L = {N^2 \mu A \over l} \, \text{(henries, H)} \tag{1}$$ where N represents the number of turns; $\mu$, the permeability of the core (recall that $\mu$ is not a constant but depends on the level of B and H since $\mu = B/H$); $A$, the area of the core in square meters; and $l$, the mean length of the core in meters.
Substituting $\mu = \mu_r \mu_o$ into Eq. (1) yields $$L = {N^2 \mu_r \mu_o A \over l} = \mu_r {N^2 \mu_o A \over l}$$ and $$L = \mu_r L_o \tag{2}$$ where $L_o$ is the inductance of the coil with an air core. In other words, the inductance of a coil with a ferromagnetic core is the relative permeability of the core times the inductance achieved with an air core. Equations for the inductance of coils different from those shown above can be found in reference handbooks. Most of the equations are more complex than those just described.
Example 1: Find the inductance of the air-core coil of Fig. 2.
Fig. 2: For Example 1.
Solution: For air core coil: $$L_o= {N^2 \mu_o A \over l} \tag{i}$$ Where $$\begin{split} A &= {\pi d^2 \over 4} \\ &= {\pi \times (4 \times 10^{-3} m)^2 \over 4}\\ &=12.57 \times10^{-6} m^2\\ \end{split}$$ Therefore Eq. (i) becomes $$\begin{split} L_o &= {(100)^2 (4 \pi \times 10^{-7})(12.57 \times10^{-6} \over 0.1} \\ &= 1.58 \mu H \end{split}$$