# 5.7 Inductance

An inductor is the magnetic analogue of an electric capacitor. Just as a capacitor can store energy in the electric field in the medium between its conducting surfaces, an inductor can store energy in the magnetic field near its current-carrying conductors. A typical inductor consists of multiple turns of wire helically coiled around a cylindrical core (Fig. 5-26(a)). Such a structure is called a solenoid. Its core may be air filled or may contain a magnetic material with magnetic permeability μ. If the wire carries a current I and the turns are closely spaced, the solenoid will produce a relatively uniform magnetic field within its interior with magnetic field lines resembling those of the permanent magnet (Fig. 5-26(b)).

# Figure 5-26 Magnetic field lines of (a) a loosely wound solenoid and (b) a tightly wound solenoid.

(a) Loosely wound solenoid

Figure 5 26-a Full Alternative Text

(b) Tightly wound solenoid

Figure 5 26-b Full Alternative Text

# 5.7.1 Magnetic Field in a Solenoid

As a prelude to our discussion of inductance, we derive an expression for the magnetic flux density B in the interior region of a tightly wound solenoid. The solenoid is of length l and radius a and comprises N turns carrying current I. The number of turns per unit length is n=N/l, and the fact that the turns are tightly wound implies that the pitch of a single turn is small compared with the solenoid’s radius. Even though the turns are slightly helical in shape, we can treat them as circular loops (Fig. 5-27(a)). Let us start by considering the magnetic flux density B at point P on the axis of the solenoid. In Example 5-3, we derived the following expression for the magnetic field H along the axis of a circular loop of radius a, which is a distance z away from its center:

H=z ˆ I′a 2 2(a 2+z 2)3/2,(5.86)

where I′ is the current carried by the loop. If we treat an incremental length d z of the solenoid as an equivalent loop composed of n d z turns carrying a current I′=I n d z, then the induced field at point P is

d B=μ d H=z ˆ μ n I a 2 2(a 2+z 2)3/2 d z.(5.87)

# Figure 5-27 (a) Solenoid cross section showing geometry for calculating H at a point P on the solenoid axis and (b) solenoid inductance.

(a) Cross section

Figure 5 27-a Full Alternative Text

(b) Solenoid inductance

The total field B at P is obtained by integrating the contributions from the entire length of the solenoid. This is facilitated

by expressing the variable z in terms of the angle θ, as seen from P to a point on the solenoid rim. That is,

Upon substituting the last two expressions in Eq. (5.87) and integrating from θ 1 to θ 2, we have

B=z ˆ μ n I a 2 2∫θ 1 θ 2 a sec 2 θ d θ a 3 sec 3 θ=z ˆ μ n I 2(sin θ 2−sin θ 1).(5.89)

If the solenoid length l is much larger than its radius a, then for points P away from the solenoid’s ends θ 1≈−90° and θ 2≈90°, in which case Eq. (5.89) reduces to

Even though Eq. (5.90) was derived for the field B at the midpoint of the solenoid, it is approximately valid everywhere in the solenoid’s interior, except near the ends.

We now return to a discussion of inductance, which includes the notion of self-inductance, representing the magnetic flux linkage of a coil or circuit with itself, and mutual inductance, which involves the magnetic flux linkage in a circuit due to the magnetic field generated by a current in another one. Usually, when the term inductance is used, the intended reference is to self-inductance.

# 5.7.2 Self-Inductance of a Solenoid

From Eq. (5.66), the magnetic flux Φ linking a surface S is

In a solenoid characterized by an approximately uniform magnetic field throughout its cross-section and given by Eq. (5.90), the flux linking a single loop is

Φ=∫S z ˆ μ N l I⋅z ˆ d s=μ N l I S,(5.92)

where S is the cross-sectional area of the loop (Fig. 5-27(b)). Magnetic flux linkage Λ is defined as the total magnetic flux linking a given circuit or conducting structure. If the structure consists of a single conductor with multiple loops, as in the case of the solenoid, Λ equals the flux linking all loops of the structure. For a solenoid with N turns,

Λ=N Φ=μ N 2 l I S (Wb).(5.93)

The self-inductance of any conducting structure is defined as the ratio of the magnetic flux linkage Λ to the current I flowing through the structure:

The SI unit for inductance is the henry (H), which is equivalent to webers per ampere (Wb/A).

For a solenoid, use of Eq. (5.93) gives

# 5.7.3 Self-Inductance of Other Conductors

The solenoid consists of a single inductor of cross section S. If, on the other hand, the structure consists of two separate conductors, as in the case of the parallel-wire and coaxial transmission lines shown in Fig. 5-28, the flux linkage Λ associated with a length l of either line refers to the flux Φ through a closed surface between the two conductors, such as the shaded areas in Fig. 5-28. In reality, there is also some magnetic flux that passes through the conductors themselves, but it may be ignored by assuming that currents flow only on the surfaces of the conductors, in which case the magnetic field inside the conductors vanishes. This assumption is justified by the fact that our interest in calculating Λ is for the purpose of determining the inductance of a given structure, and inductance is of interest primarily in the ac case (i.e., time-varying currents, voltages, and fields). As we will see later in Section 7-5, the current flowing in a conductor under ac conditions is concentrated within a very thin layer on the skin of the conductor.

# Figure 5-28 To compute the inductance per unit length of a two-conductor transmission line, we need to determine the magnetic flux through the area S between the conductors.

(a) Parallel-wire transmission line

Figure 5 28-a Full Alternative Text

(b) Coaxial transmission line

Figure 5 28-b Full Alternative Text

For a two-conductor configuration similar to those of Fig. 5-28, the inductance is given by

# Example 5-8: Inductance of a Coaxial Transmission Line

Develop an expression for the inductance per unit length of a coaxial transmission line with inner and outer conductors of radii a and b (Fig. 5-29) and an insulating material of permeability μ.

# Figure 5-29 Cross-sectional view of coaxial transmission line (Example 5-8). and denote H field out of and into the page, respectively.

Figure 5 29 Full Alternative Text

Solution: The current I in the inner conductor generates a magnetic field B throughout the region between the two conductors. It is given by Eq. (5.30) as

B=ϕ ˆ μ I 2 π r,(5.97)

where r is the radial distance from the axis of the coaxial line. Consider a transmission-line segment of length l as shown in Fig. 5-29. Because B is perpendicular to the planar surface S between the conductors, the flux through S is

Φ=l∫a b B d r=l∫a b μ I 2 π r d r=μ I l 2 π ln b a.(5.98)

Using Eq. (5.96), the inductance per unit length of the coaxial transmission line is

# 5.7.4 Mutual Inductance

Magnetic coupling between two different conducting structures is described in terms of the mutual inductance between them. For simplicity, consider the case of two multiturn closed loops with surfaces S 1 and S 2. Current I 1 flows through the first loop (Fig. 5-30), and no current flows through the second one. The magnetic field B 1 generated by I 1 results in a flux Φ 12 through loop 2 given by

Φ 12=∫S 2 B 1⋅d s,(5.100)

and if loop 2 consists of N 2 turns all coupled by B 1 in exactly the same way, then the total magnetic flux linkage through loop 2 is

Λ 12=N 2 Φ 12=N 2∫S 2 B 1⋅d s.(5.101)

# Figure 5-30 Magnetic field lines generated by current I 1 in loop 1 linking surface S 2 of loop 2.

Figure 5 30 Full Alternative Text

The mutual inductance associated with this magnetic coupling is given by

Mutual inductance is important in transformers (as discussed in Chapter 6) wherein the windings of two or more circuits share a common magnetic core, as illustrated by the toroidal arrangement shown in Fig. 5-31.

# Figure 5-31 Toroidal coil with two windings used as a transformer.

Figure 5 31 Full Alternative Text

# Example 5-9: Mutual Inductance

The rectangular conducting loop shown in Fig. 5-32 is coplanar with the long, straight wire carrying current I. Determine the mutual inductance between the wire and the loop.

# Figure 5-32 Mutual inductance between two conductors (Example 5-9).

Figure 5 32 Full Alternative Text

Solution: According to Eq. (5.102), the mutual inductance is given by

L 12=N 2 I 1∫S 2 B 1⋅d s.

In the present case, I 1=I is the current carried by the first conductor (straight wire), S 2 is the cross-sectional area of the second conductor (rectangular loop), and B 1 is the magnetic field due to the first conductor (wire). Also, N 2=1 because the loop has a single turn. In the right-hand side of the y−z plane, the magnetic field due to current I is into the page (−x ˆ direction) and given by

B=−x ˆ μ 0 I 2 π y,

where y is the distance away from the wire. The expression is identical with that given by Eq. (5.97), after replacing ϕ ˆ with −x ˆ and r with y. To compute the magnetic flux passing through the loop (into the page), we need to define the differential area as d s=−x ˆ d y d z. Hence,

L 12=1 I∫y=0.05 0.20∫z=0.1 0.4(−x^μ 0 I 2 π y)⋅(−x^ d y d z)=μ 0 2 π(∫0.05 0.20 d y y)(∫0.1 0.4 d z)=0.3 μ 0 2 π(ln y)|0.05 0.20=0.3 μ 0 2 π ln(0.2 0.05)=83 nH.

Concept Question 5-15: What is the magnetic field like in the interior of a long solenoid?

Concept Question 5-16: What is the difference between self-inductance and mutual inductance?

Concept Question 5-17: How is the inductance of a solenoid related to its number of turns N?