14.2 Self-inductance and inductors  (Page 3/6)

A good approach for calculating the self-inductance of an inductor consists of the following steps:

To demonstrate this procedure, we now calculate the self-inductances of two inductors.

Cylindrical solenoid

Consider a long, cylindrical solenoid with length l , cross-sectional area A , and N turns of wire. We assume that the length of the solenoid is so much larger than its diameter that we can take the magnetic field to be $B={\mu }_{0}nI$ throughout the interior of the solenoid, that is, we ignore end effects in the solenoid. With a current I flowing through the coils, the magnetic field produced within the solenoid is

so the magnetic flux through one turn is

Using [link] , we find for the self-inductance of the solenoid,

If $n=N\text{/}l$ is the number of turns per unit length of the solenoid, we may write [link] as

where $V=Al$ is the volume of the solenoid. Notice that the self-inductance of a long solenoid depends only on its physical properties (such as the number of turns of wire per unit length and the volume), and not on the magnetic field or the current. This is true for inductors in general.

Rectangular toroid

A toroid with a rectangular cross-section is shown in [link] . The inner and outer radii of the toroid are $R_1\text{and}{R}_2,\text{and}h$ is the height of the toroid. Applying Ampère’s law in the same manner as we did in [link] for a toroid with a circular cross-section, we find the magnetic field inside a rectangular toroid is also given by

where r is the distance from the central axis of the toroid. Because the field changes within the toroid, we must calculate the flux by integrating over the toroid’s cross-section. Using the infinitesimal cross-sectional area element $da=hdr$ shown in [link] , we obtain

Now from [link] , we obtain for the self-inductance of a rectangular toroid