College Physics for AP® Courses

# 23.5Electric Generators

College Physics for AP® Courses23.5 Electric Generators

### Learning Objectives

By the end of this section, you will be able to:

• Calculate the emf induced in a generator.
• Calculate the peak emf that can be induced in a particular generator system.

The information presented in this section supports the following AP® learning objectives and science practices:

• 4.E.2.1 The student is able to construct an explanation of the function of a simple electromagnetic device in which an induced emf is produced by a changing magnetic flux through an area defined by a current loop (i.e., a simple microphone or generator) or of the effect on behavior of a device in which an induced emf is produced by a constant magnetic field through a changing area. (SP.6.4)

Electric generators induce an emf by rotating a coil in a magnetic field, as briefly discussed in Induced Emf and Magnetic Flux. We will now explore generators in more detail. Consider the following example.

### Example 23.3Calculating the Emf Induced in a Generator Coil

The generator coil shown in Figure 23.20 is rotated through one-fourth of a revolution (from $θ = 0º θ = 0º$ to $θ = 90º θ = 90º$ ) in 15.0 ms. The 200-turn circular coil has a 5.00 cm radius and is in a uniform 1.25 T magnetic field. What is the average emf induced?

Figure 23.20 When this generator coil is rotated through one-fourth of a revolution, the magnetic flux $Φ Φ$ changes from its maximum to zero, inducing an emf.

Strategy

We use Faraday’s law of induction to find the average emf induced over a time $ΔtΔt size 12{Δt} {}$:

$emf=−NΔΦΔt.emf=−NΔΦΔt. size 12{"emf"= - N { {ΔΦ} over {Δt} } } {}$
23.11

We know that $N=200N=200 size 12{N="200"} {}$ and $Δt=15.0msΔt=15.0ms size 12{Δt="15" "." 0"ms"} {}$, and so we must determine the change in flux $ΔΦΔΦ size 12{ΔΦ} {}$ to find emf.

Solution

Since the area of the loop and the magnetic field strength are constant, we see that

$ΔΦ=Δ(BAcosθ)=ABΔ(cosθ).ΔΦ=Δ(BAcosθ)=ABΔ(cosθ). size 12{ΔΦ=Δ $$ital "BA""cos"θ$$ = ital "AB"Δ $$"cos"θ$$ } {}$
23.12

Now, $Δ(cosθ)=−1.0Δ(cosθ)=−1.0 size 12{Δ $$"cos"θ$$ = - 1 "." 0} {}$, since it was given that $θ θ$ goes from $0º 0º$ to $90º 90º$ . Thus $ΔΦ=−ABΔΦ=−AB size 12{ΔΦ= - ital "AB"} {}$, and

$emf=NABΔt.emf=NABΔt. size 12{"emf"=N { { ital "AB"} over {Δt} } } {}$
23.13

The area of the loop is $A=πr2=(3.14...)(0.0500m)2=7.85×10−3m2A=πr2=(3.14...)(0.0500m)2=7.85×10−3m2 size 12{A=πr rSup { size 8{2} } = $$3 "." "14" "." "." "."$$ $$0 "." "0500"m$$ rSup { size 8{2} } =7 "." "85" times "10" rSup { size 8{ - 3} } `m rSup { size 8{2} } } {}$. Entering this value gives

$emf=200(7.85×10−3m2)(1.25 T)15.0×10−3 s=131V.emf=200(7.85×10−3m2)(1.25 T)15.0×10−3 s=131V. size 12{"emf"="200" { { $$7 "." "85" times "10" rSup { size 8{ - 3} } " m" rSup { size 8{2} }$$ $$1 "." "25"" T"$$ } over {"15" "." 0 times "10" rSup { size 8{ - 3} } " s"} } ="131"" V"} {}$
23.14

Discussion

This is a practical average value, similar to the 120 V used in household power.

The emf calculated in Example 23.3 is the average over one-fourth of a revolution. What is the emf at any given instant? It varies with the angle between the magnetic field and a perpendicular to the coil. We can get an expression for emf as a function of time by considering the motional emf on a rotating rectangular coil of width $ww size 12{w} {}$ and height $ℓℓ size 12{l} {}$ in a uniform magnetic field, as illustrated in Figure 23.21.

Figure 23.21 A generator with a single rectangular coil rotated at constant angular velocity in a uniform magnetic field produces an emf that varies sinusoidally in time. Note the generator is similar to a motor, except the shaft is rotated to produce a current rather than the other way around.

Charges in the wires of the loop experience the magnetic force, because they are moving in a magnetic field. Charges in the vertical wires experience forces parallel to the wire, causing currents. But those in the top and bottom segments feel a force perpendicular to the wire, which does not cause a current. We can thus find the induced emf by considering only the side wires. Motional emf is given to be $emf=Bℓvemf=Bℓv size 12{"emf"=Bℓv} {}$, where the velocity v is perpendicular to the magnetic field $BB size 12{B} {}$. Here the velocity is at an angle $θθ size 12{θ} {}$ with $BB size 12{B} {}$, so that its component perpendicular to $BB size 12{B} {}$ is $vsin θvsin θ size 12{v"sin"θ} {}$ (see Figure 23.21). Thus in this case the emf induced on each side is $emf=Bℓvsinθemf=Bℓvsinθ size 12{"emf"=Bℓv"sin"θ} {}$, and they are in the same direction. The total emf around the loop is then

$emf=2Bℓvsinθ.emf=2Bℓvsinθ. size 12{"emf"=2Bℓv"sin"θ} {}$
23.15

This expression is valid, but it does not give emf as a function of time. To find the time dependence of emf, we assume the coil rotates at a constant angular velocity $ωω size 12{ω} {}$. The angle $θθ size 12{θ} {}$ is related to angular velocity by $θ=ωtθ=ωt size 12{θ=ωt} {}$, so that

$emf=2Bℓvsinωt.emf=2Bℓvsinωt. size 12{"emf"=Bℓv"sin"ωt} {}$
23.16

Now, linear velocity $v v$ is related to angular velocity $ω ω$ by $v=rωv=rω size 12{v=rω} {}$. Here $r=w/2r=w/2 size 12{r=w/2} {}$, so that $v=(w/2)ωv=(w/2)ω size 12{v= $$w/2$$ ω} {}$, and

$emf=2Bℓw2ωsinωt=(ℓw)Bωsinωt.emf=2Bℓw2ωsinωt=(ℓw)Bωsinωt. size 12{"emf"=2Bℓ { {w} over {2} } ω"sin"ωt= $$ℓw$$ Bω"sin"ωt} {}$
23.17

Noting that the area of the loop is $A=ℓwA=ℓw size 12{A=ℓw} {}$, and allowing for $NN size 12{N} {}$ loops, we find that

$emf = NAB ω sin ωt emf = NAB ω sin ωt size 12{"emf"= ital "NAB"ω"sin"ωt} {}$
23.18

is the emf induced in a generator coil of $NN size 12{N} {}$ turns and area $AA size 12{A} {}$ rotating at a constant angular velocity $ω ω$ in a uniform magnetic field $BB size 12{B} {}$. This can also be expressed as

$emf=emf0sinωt,emf=emf0sinωt, size 12{"emf"="emf" rSub { size 8{0} } "sin"ωt} {}$
23.19

where

$emf 0 = NAB ω emf 0 = NAB ω size 12{"emf" rSub { size 8{0} } = ital "NAB"ω} {}$
23.20

is the maximum maximum (peak) emf. Note that the frequency of the oscillation is $f=ω/2πf=ω/2π size 12{f=ω/2π} {}$, and the period is $T=1/f=2π/ωT=1/f=2π/ω size 12{T=1/f=2π/ω} {}$. Figure 23.22 shows a graph of emf as a function of time, and it now seems reasonable that AC voltage is sinusoidal.

Figure 23.22 The emf of a generator is sent to a light bulb with the system of rings and brushes shown. The graph gives the emf of the generator as a function of time. $emf0emf0 size 12{"emf" rSub { size 8{0} } } {}$ is the peak emf. The period is $T=1/f=2π/ωT=1/f=2π/ω size 12{T=1/f=2π/ω} {}$, where $ff size 12{f} {}$ is the frequency. Note that the script E stands for emf.

The fact that the peak emf, $emf0=NABωemf0=NABω size 12{"emf" rSub { size 8{0} } = ital "NAB"ω} {}$, makes good sense. The greater the number of coils, the larger their area, and the stronger the field, the greater the output voltage. It is interesting that the faster the generator is spun (greater $ωω size 12{ω} {}$), the greater the emf. This is noticeable on bicycle generators—at least the cheaper varieties. One of the authors as a juvenile found it amusing to ride his bicycle fast enough to burn out his lights, until he had to ride home lightless one dark night.

Figure 23.23 shows a scheme by which a generator can be made to produce pulsed DC. More elaborate arrangements of multiple coils and split rings can produce smoother DC, although electronic rather than mechanical means are usually used to make ripple-free DC.

Figure 23.23 Split rings, called commutators, produce a pulsed DC emf output in this configuration.

### Example 23.4Calculating the Maximum Emf of a Generator

Calculate the maximum emf, $emf0emf0 size 12{"emf" rSub { size 8{0} } } {}$, of the generator that was the subject of Example 23.3.

Strategy

Once $ωω size 12{ω} {}$, the angular velocity, is determined, $emf0=NABωemf0=NABω size 12{"emf" rSub { size 8{0} } = ital "NAB"ω} {}$ can be used to find $emf0emf0 size 12{"emf" rSub { size 8{0} } } {}$. All other quantities are known.

Solution

Angular velocity is defined to be the change in angle per unit time:

$ω=ΔθΔt.ω=ΔθΔt. size 12{ω= { {Δθ} over {Δt} } } {}$
23.21

One-fourth of a revolution is $π/2π/2 size 12{l} {}$ radians, and the time is 0.0150 s; thus,

$ω = π/2 rad0.0150 s = 104.7 rad/s. ω = π/2 rad0.0150 s = 104.7 rad/s.$
23.22

104.7 rad/s is exactly 1000 rpm. We substitute this value for $ωω size 12{ω} {}$ and the information from the previous example into $emf0=NABωemf0=NABω size 12{"emf" rSub { size 8{0} } = ital "NAB"ω} {}$, yielding

emf0 = NABω = 200(7.85×10−3m2)(1.25T)(104.7rad/s) = 206V. emf0 = NABω = 200(7.85×10−3m2)(1.25T)(104.7rad/s) = 206V.alignl { stack { size 12{"emf" rSub { size 8{0} } = ital "NAB"ω} {} # " "="200" $$7 "." "85" times "10" rSup { size 8{ - 3} } " m" rSup { size 8{2} }$$ $$1 "." "25"" T"$$ $$"104" "." 7" rad/s"$$ {} # " "="206"" V" {} } } {}
23.23

Discussion

The maximum emf is greater than the average emf of 131 V found in the previous example, as it should be.

In real life, electric generators look a lot different than the figures in this section, but the principles are the same. The source of mechanical energy that turns the coil can be falling water (hydropower), steam produced by the burning of fossil fuels, or the kinetic energy of wind. Figure 23.24 shows a cutaway view of a steam turbine; steam moves over the blades connected to the shaft, which rotates the coil within the generator.

Figure 23.24 Steam turbine/generator. The steam produced by burning coal impacts the turbine blades, turning the shaft which is connected to the generator. (credit: Nabonaco, Wikimedia Commons)

Generators illustrated in this section look very much like the motors illustrated previously. This is not coincidental. In fact, a motor becomes a generator when its shaft rotates. Certain early automobiles used their starter motor as a generator. In Back Emf, we shall further explore the action of a motor as a generator.