College Physics

Figure 23.42Through rapid switching of an inductor, 1.5 V batteries can be used to induce emfs of several thousand volts. This voltage can be used to store charge in a

capacitor for later use, such as in a camera flash attachment.

It is possible to calculateLfor an inductor given its geometry (size and shape) and knowing the magnetic field that it produces. This is difficult in

most cases, because of the complexity of the field created. So in this text the inductanceLis usually a given quantity. One exception is the solenoid,

because it has a very uniform field inside, a nearly zero field outside, and a simple shape. It is instructive to derive an equation for its inductance. We

start by noting that the induced emf is given by Faraday’s law of induction asemf = −N(ΔΦ/ Δt)and, by the definition of self-inductance, as

emf = −L(ΔI/ Δt). Equating these yields

emf = −NΔΦ (23.37)

Δt

= −LΔI

Δt

.

Solving forLgives

(23.38)

L=NΔΦ

ΔI

.

This equation for the self-inductanceLof a device is always valid. It means that self-inductanceLdepends on how effective the current is in

creating flux; the more effective, the greaterΔΦ/ΔIis.

Let us use this last equation to find an expression for the inductance of a solenoid. Since the areaAof a solenoid is fixed, the change in flux is

ΔΦ= Δ(BA)=AΔB. To findΔB, we note that the magnetic field of a solenoid is given byB=μ 0 nI=μ 0 NI

ℓ

. (Heren=N/ℓ, whereNis

the number of coils andℓis the solenoid’s length.) Only the current changes, so thatΔΦ=AΔB=μ 0 NAΔI

ℓ

. SubstitutingΔΦinto

L=NΔΦ

ΔI

gives

(23.39)

L=NΔΦ

ΔI

=N

μ 0 NAΔℓI

ΔI

.

This simplifies to

(23.40)

L=

μ 0 N^2 A

ℓ

(solenoid).

This is the self-inductance of a solenoid of cross-sectional areaAand lengthℓ. Note that the inductance depends only on the physical

characteristics of the solenoid, consistent with its definition.

Example 23.7 Calculating the Self-inductance of a Moderate Size Solenoid

Calculate the self-inductance of a 10.0 cm long, 4.00 cm diameter solenoid that has 200 coils.

Strategy

This is a straightforward application ofL=

μ 0 N^2 A

ℓ

, since all quantities in the equation exceptLare known.

Solution

Use the following expression for the self-inductance of a solenoid:

(23.41)

L=

μ 0 N^2 A

ℓ

.

The cross-sectional area in this example isA=πr^2 = (3.14...)(0 .0200 m)^2 = 1.26×10

−3

m^2 ,Nis given to be 200, and the lengthℓis

0.100 m. We know the permeability of free space isμ 0 = 4π×10−7T ⋅ m/A. Substituting these into the expression forLgives

838 CHAPTER 23 | ELECTROMAGNETIC INDUCTION, AC CIRCUITS, AND ELECTRICAL TECHNOLOGIES

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