6.5 Electric Generators

Sections

Learning Objectives

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

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 6.3 Calculating the Emf Induced in a Generator Coil

The generator coil shown in Figure 6.20 is rotated through one-fourth of a revolution (from $θ = 0 º to θ = 90 º)$ 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?

The figure shows a schematic diagram of an electric generator. It consists of a rotating rectangular coil placed between the two poles of a permanent magnet shown as two rectangular blocks curved on side facing the coil. The magnetic field B is shown pointing from the North to the South Pole. The two ends of this coil are connected to the two small rings. The two conducting carbon brushes are kept pressed separately on both the rings. The coil is attached to an axle with a handle at the other end. Outer e

Figure 6.20 When this generator coil is rotated through one-fourth of a revolution, the magnetic flux

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 $Δ t.$

6.11

emf = - N \frac{Δ Φ}{Δ t}

We know that $N = 200$ and $Δ t = 15.0 m s,$ and so we must determine the change in flux $Δ Φ$ to find emf.

Solution

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

6.12

Δ Φ = Δ (BA cos θ) = AB Δ (cos θ) .

Now, $Δ (cos θ) = - 1 . 0,$ since it was given that $θ$ goes from $0º$ to $90º.$ Thus, $Δ Φ = - AB,$ and

6.13

emf = N \frac{AB}{Δ t} .

The area of the loop is $A = π r^{2} = (3.14. . .) {(0.0500 m)}^{2} = 7.85 \times 10^{- 3} m^{2} .$ Entering this value gives

6.14

e m f = 200 \frac{(7.85 \times 10^{- 3} m^{2}) (1.25 T)}{15.0 \times 10^{- 3} s} = 131 V .

Discussion

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

The emf calculated in Example 6.3 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 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 $w$ and height $ℓ$ in a uniform magnetic field, as illustrated in Figure 6.21.

The figure shows a schematic diagram of an electric generator with a single rectangular coil. The rotating rectangular coil is placed between the two poles of a permanent magnet shown as two rectangular blocks curved on side facing the coil. The magnetic field B is shown pointing from the North to the South Pole. The North Pole is on the left and the South Pole is to the right and hence the direction of field is from left to right. The angular velocity of the coil is given as omega. The velocity vector v

Figure 6.21 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 $e m f = B ℓ v,$ where the velocity v is perpendicular to the magnetic field $B .$ Here the velocity is at an angle $θ$ with $B,$ so that its component perpendicular to $B$ is $v sin θ$ (see Figure 6.21). Thus, in this case, the emf induced on each side is $emf = Bℓv sin θ,$ and they are in the same direction. The total emf around the loop is then

6.15

emf = 2 Bℓv sin θ .

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 $ω .$ The angle $θ$ is related to angular velocity by $θ = ωt,$ so that

6.16

emf = 2 Bℓv sin ωt .

Now, linear velocity $v$ is related to angular velocity $ω$ by $v = r ω .$ Here $r = w / 2$ , so that $v = (w / 2) ω,$ and

6.17

emf = 2 Bℓ \frac{w}{2} ω sin ωt = (ℓ w) Bω sin ωt .

Noting that the area of the loop is $A = ℓ w,$ and allowing for $N$ loops, we find that

6.18

emf = NAB ω sin ωt

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

6.19

emf = {emf}_{0} sin ωt,

where

6.20

{emf}_{0} = NAB ω

is the maximum (peak) emf. Note that the frequency of the oscillation is $f = ω / 2 π,$ and the period is $T = 1 / f = 2 π / ω .$ Figure 6.22 shows a graph of emf as a function of time, and it now seems reasonable that AC voltage is sinusoidal.

The first part of the figure shows a schematic diagram of a single coil electric generator. It consists of a rotating rectangular loop placed between the two poles of a permanent magnet shown as two rectangular blocks curved on side facing the loop. The magnetic field B is shown pointing from the North to the South Pole. The two ends of this loop are connected to the two small rings. The two conducting carbon brushes are kept pressed separately on both the rings. The loop is rotated in the field with an a

Figure 6.22 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.

{emf}_{0}

is the peak emf. The period is

T = 1 / f = 2 π / ω,

where

f

is the frequency. Note that the script E stands for emf.

The fact that the peak emf, ${emf}_{0} = NAB ω,$ 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 $ω)$ , 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 6.23 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.

The first part of the figure shows a schematic diagram of a single coil D C electric generator. It consists of a rotating rectangular loop placed between the two poles of a permanent magnet shown as two rectangular blocks curved on side facing the loop. The magnetic field B is shown pointing from the North to the South Pole. The two ends of this loop are connected to the two sides of a split ring. The two conducting carbon brushes are kept pressed separately on both sides of the split rings. The loop is r

Figure 6.23 Split rings, called commutators, produce a pulsed DC emf output in this configuration.

Example 6.4 Calculating the Maximum Emf of a Generator

Calculate the maximum emf, emf₀, of the generator that was the subject of Example 6.3.

Strategy

Once $ω,$ the angular velocity, is determined, ${emf}_{0} = NAB ω$ can be used to find $e m f_{0} .$ All other quantities are known.

Solution

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

6.21

ω = \frac{Δ θ}{Δ t}

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

6.22

ω = \frac{π /2 rad}{0.0150 s} .

104.7 rad/s is exactly 1,000 rpm. We substitute this value for $ω$ and the information from the previous example into $e m f_{0} = N A B ω,$ yielding

6.23

\begin{array}{l} {emf}_{0} & = & NAB ω \\ = & 200 (7 . 85 \times 10^{- 3} m^{2}) (1.25 T) (104.7 rad/s) \\ = & 206 V . \end{array}

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 6.24 shows a cutaway view of a steam turbine; steam moves over the blades connected to the shaft, which rotates the coil within the generator.

Photograph of a steam turbine connected to a generator.

Figure 6.24 Steam turbine/generator. The steam produced by burning coal impacts the turbine blades, turning the shaft that is connected to the generator. (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.

6.5 Electric Generators

6.5 Electric Generators

Learning Objectives

Learning Objectives

Example 6.3 Calculating the Emf Induced in a Generator Coil

Example 6.4 Calculating the Maximum Emf of a Generator

Related Items

Resources

Videos

Documents

Links