30.8 Inductors

Capacitors are useful circuit elements because they store potential energy UC in the electric field. Similarly, a coil of wire can be a useful circuit element because it stores energy in the magnetic field. In circuits, a coil is called an inductor because, as you’ll see, the potential difference across an inductor is an induced emf. An ideal inductor is one for which the wire forming the coil has no electric resistance. The circuit symbol for an inductor is The figure illustrates the circuit symbol of an inductor..

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We define the inductance L of a coil to be its flux-to-current ratio:

L=ΦmI (30.31)

Strictly speaking, this is called self-inductance because the flux we’re considering is the magnetic flux the solenoid creates in itself when there is a current. The SI unit of inductance is the henry, named in honor of Joseph Henry, defined as

1 henry=1 H≡1 Wb/A=1 T m2/A

Practical inductances are typically millihenries (mH) or microhenries (μH).

It’s not hard to find the inductance of a solenoid. In Chapter 29 we found that the magnetic field inside an ideal solenoid having N turns and length l is

B=μ0NIl

The magnetic flux through one turn of the coil is Φper turn=AB, where A is the cross-section area of the solenoid. The total magnetic flux through all N turns is

Φm=NΦper turn=μ0N2AlI (30.32)

Thus the inductance of a solenoid, using the definition of Equation 30.31, is

Lsolenoid=ΦmI=μ0N2Al (30.33)

The inductance of a solenoid depends only on its geometry, not at all on the current. You may recall that the capacitance of two parallel plates depends only on their geometry, not at all on their potential difference.

EXAMPLE 30.12 The length of an inductor

An inductor is made by tightly wrapping 0.30-mm-diameter wire around a 4.0-mm-diameter cylinder. What length cylinder has an inductance of 10 μH?

SOLVE The cross-section area of the solenoid is A=πr2. If the wire diameter is d, the number of turns of wire on a cylinder of length l is N=l/d. Thus the inductance is

L=μ0N2Al=μ0(l/d)2πr2l=μ0πr2ld2

The length needed to give inductance L=1.0×10−5 H is

l=d2Lμ0πr2=(0.00030 m)2 (1.0×10−5 H)(1.26×10−6 T m/A)π(0.0020 m)2=0.057 m=5.7 cm