7.13: Inductance of a Straight Coil
- Last updated
- Jul 7, 2024
- Save as PDF
- Page ID
- 97167
( \newcommand{\kernel}{\mathrm{null}\,}\)
In this section, we determine the inductance of a straight coil, as shown in Figure 7.13.1. The coil is circular with radius a and length l and consists of N windings of wire wound with uniform winding density. Also, we assume the winding density N/l is large enough that magnetic field lines cannot enter or exit between windings but rather must traverse the entire length of the coil. Since the coil forms a cylinder, the problem is easiest to work in cylindrical coordinates with the axis of the coil aligned along the z axis.

Inductance L in this case is given by (Section 7.12) L=NΦI
First, let’s determine the magnetic field. The magnetic flux density deep inside the coil is (Section 7.6): B≈ˆzμNIl
Next, we determine Φ. In this case, a natural choice for S is the interior cross-section of the coil in a plane perpendicular to the axis. The direction of ds must be +ˆz since this is the direction in which the fingers of the right hand point when the current flows in the direction indicated in Figure 7.13.1. Thus, we have
Φ≈∫S(ˆzμNIl)⋅(zds)=μNIl∫Sds=μNIlA
where A is the cross-sectional area of the coil.
Finally from Equation ??? we obtain L≈μN2Al (l≫a)
The inductance of a long straight coil is given approximately by Equation ???.
Again, this result is approximate because it neglects the non-uniform fringing field near the ends of the coil and the possibility that magnetic field lines escape between windings due to inadequate winding density. Nevertheless, this result facilitates useful engineering analysis and design.