22.7

The Electric Generator

| How a Generator Produces an Emf

Electric generators, such as those in Figure 22-20, produce virtually all of the world’s electrical energy. A generator produces electrical energy from mechanical work, just the opposite of what a motor does. In a motor, an input electric current causes a coil to rotate, thereby doing mechanical work on any object attached to the shaft of the motor. In a generator, the shaft is rotated by some mechanical means, such as an engine or a turbine, and an emf is induced in a coil. If the generator is connected to an external circuit, an electric current is the output of the generator.

Figure 22-20 Electric generators such as these supply electric power by producing an induced emf according to Faraday’s law of electromagnetic induction. (Michael Melford/The Image Bank/Getty Images)

The physics of an electric generator.

In its simplest form, an ac generator consists of a coil of wire that is rotated in a uniform magnetic field, as Figure 22-21a indicates. Although not shown in the picture, the wire is usually wound around an iron core. As in an electric motor, the coil/core combination is called the armature. Each end of the wire forming the coil is connected to the external circuit by means of a metal ring that rotates with the coil. Each ring slides against a stationary carbon brush, to which the external circuit (the lamp in the drawing) is connected.

Figure 22-21 (a) This electric generator consists of a coil (only one loop is shown) of wire that is rotated in a magnetic field by some mechanical means. (b) The current I arises because of the magnetic force exerted on the charges in the moving wire. (c) The dimensions of the coil.

To see how current is produced by the generator, consider the two vertical sides of the coil in Figure 22-21b. Since each is moving in a magnetic field , the magnetic force exerted on the charges in the wire causes them to flow, thus creating a current. With the aid of RHR-1 (fingers of extended right hand point along , thumb along the velocity , palm pushes in the direction of the force on a positive charge), it can be seen that the direction of the current is from bottom to top in the left side and from top to bottom in the right side. Thus, charge flows around the loop. The upper and lower segments of the loop are also moving. However, these segments can be ignored because the magnetic force on the charges within them points toward the sides of the wire and not along the length.
The magnitude of the motional emf developed in a conductor moving through a magnetic field is given by Equation 22.1. To apply this expression to the left side of the coil, whose length is L (see Figure 22-21c), we need to use the velocity component that is perpendicular to . Letting be the angle between and , it follows that , and, with the aid of Equation 22.1, the emf can be written as

The emf induced in the right side has the same magnitude as that in the left side. Since the emfs from both sides drive current in the same direction around the loop, the emf for the complete loop is . If the coil consists of N loops, the net emf is N times as great as that of one loop, so

It is convenient to express the variables v and in terms of the angular speed at which the coil rotates. Equation 8.2 shows that the angle is the product of the angular speed and the time, , if it is assumed that rad when t = 0 s. Furthermore, any point on each vertical side moves on a circular path of radius r = W/2, where W is the width of the coil (see Figure 22-21c). Thus, the tangential speed v of each side is related to the angular speed via Equation 8.9 as . Substituting these expressions for and v in the previous equation for E, and recognizing that the product LW is the area A of the coil, we can write the induced emf as

| (22.4)

In this result, the angular speed is in radians per second and is related to the frequency f [in cycles per second or hertz (Hz)] according to (Equation 10.6).

Squeezing this hand-held generator produces electric power for portable devices such as cell phones, CD players, and pagers. (Mario Tama/Getty Images News and Sport Services)

Although Equation 22.4 was derived for a rectangular coil, the result is valid for any planar shape of area A (e.g., circular) and shows that the emf varies sinusoidally with time. The peak, or maximum, emf occurs when and has the value . Figure 22-22 shows a plot of Equation 22.4 and reveals that the emf changes polarity as the coil rotates. This changing polarity is exactly the same as that discussed for an ac voltage in Section 20.5 and illustrated in Figure 20-11. If the external circuit connected to the generator is a closed circuit, an alternating current results that changes direction at the same frequency f as the emf changes polarity. Therefore, this electric generator is also called an alternating current (ac) generator. The next two examples show how Equation 22.4 is applied.

Figure 22-22 An ac generator produces this alternating emf E according to .

| Example 10 | | | An AC Generator

In Figure 22-21 the coil of the ac generator rotates at a frequency of f = 60.0 Hz and develops an emf of 120 V (rms). The coil has an area of A = 3.0 × 10−3 m2 and consists of N = 500 turns. Find the magnitude of the magnetic field in which the coil rotates.
Reasoning The magnetic field can be found from the relation . However, in using this equation we must remember that is the peak emf, whereas the given value of 120 V is not a peak value but an rms value. The peak emf is related to the rms emf by , according to Equation 20.13.
Solution Solving for B and using the facts that and , we find that the magnitude of the magnetic field is

Problem solving insight When using the equation , remember that the angular frequency must be in rad/s and is related to the frequency f (in Hz) according to (Equation 10.6).

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 11 | | | A Bike Generator

A bicyclist is traveling at night, and a generator mounted on the bike powers a headlight. A small rubber wheel on the shaft of the generator presses against the bike tire and turns the coil of the generator at an angular speed that is 44 times as great as the angular speed of the tire itself. The tire has a radius of 0.33 m. The coil consists of 75 turns, has an area of 2.6 × 10−3 m2, and rotates in a 0.10-T magnetic field. When the peak emf being generated is 6.0 V, what is the linear speed of the bike?

The physics of a bike generator.

Reasoning Since the tires are rolling, the linear speed v of the bike is related to the angular speed of its tires by (see Section 8.6), where r is the radius of a tire. We are given that the angular speed of the coil is 44 times as great as that of the tire. Thus, and the linear speed of the bike can be related to the angular speed of the coil. However, according to the discussion in this section on electric generators, the angular speed of the coil is related to the peak emf developed by the rotating coil, the number of turns in the coil, the area of the coil, and the magnetic field, all of which are known.
Knowns and Unknowns The data for this problem are:

Description | Symbol | Value | Comment
Radius of tire | r | 0.33 m |
Number of turns in coil | N | 75 |
Angular speed of coil |  |  | Angular speed of coil is 44 times as great as that of tire.
Area of coil | A | 2.6 × 10−3 m2 |
Magnitude of magnetic field | B | 0.10 T |
Peak emf produced by generator |  | 6.0 V |
Unknown Variable |  |  |
Linear speed of bike | v | ? |

Modeling the Problem

| Step 1 Rolling Motion
| When a tire rolls without slipping on the ground, the linear speed v of the tire (the speed at which its axle is moving forward) is related to the angular speed of the tire about the axle. This relationship is given by

| (8.12)

where r is the radius of the tire. We are given that the angular speed of the coil is 44 times as great as the angular speed of the tire, so . Solving this equation for and substituting the result into , we obtain Equation 22.10 at the right. The radius of the tire is known, but the angular speed of the coil is not; we will evaluate it in Step 2.

| (22.10)

| Step 2 Peak Emf Induced in a Rotating Planar Coil
| A generator produces an emf when a coil of wire rotates in a magnetic field. According to Equation 22.4, the peak emf is given by , where N is the number of turns in the coil, A is the area of the coil, B is the magnitude of the magnetic field, and is the angular speed of the rotating coil. Solving this relation for gives

Note from the data table that all the variables on the right side of this equation are known. We substitute this result into Equation 22.10, as indicated at the right.

Solution Algebraically combining the results of the two steps, we have

The linear speed of the bicycle is

Related Homework: Problem 42

| The Electrical Energy Delivered by a Generator and the Countertorque

Some power-generating stations burn fossil fuel (coal, gas, or oil) to heat water and produce pressurized steam for turning the blades of a turbine whose shaft is linked to the generator. Others use nuclear fuel or falling water as a source of energy. As the turbine rotates, the generator coil also rotates and mechanical work is transformed into electrical energy.

Need more practice?

Interactive LearningWare22.3
An armature of an electric generator consists of a 275-turn circular coil of radius 15 cm. The armature is rotated in a 0.80-T magnetic field, and the polarity changes every 4.2 × 10−3 s. What is the peak emf produced by the generator?

The devices to which the generator supplies electricity are known collectively as the “load,” because they place a burden or load on the generator by taking electrical energy from it. If all the devices are switched off, the generator runs under a no-load condition, because there is no current in the external circuit and the generator does not supply electrical energy. Then, work needs to be done on the turbine only to overcome friction and other mechanical losses within the generator itself, and fuel consumption is at a minimum.
Figure 22-23 illustrates a situation in which a load is connected to a generator. Because there is now a current I = I1 + I2 in the coil of the generator and the coil is situated in a magnetic field, the current experiences a magnetic force . Figure 22-24 shows the magnetic force acting on the left side of the coil, the direction of being given by RHR-1. A force of equal magnitude but opposite direction also acts on the right side of the coil, although this force is not shown in the drawing. The magnetic force gives rise to a countertorque that opposes the rotational motion. The greater the current drawn from the generator, the greater the countertorque, and the harder it is for the turbine to turn the coil. To compensate for this countertorque and keep the coil rotating at a constant angular speed, work must be done by the turbine, which means more fuel must be burned. This is another example of the law of conservation of energy, since the electrical energy consumed by the load must ultimately come from the energy source used to drive the turbine.

Figure 22-23 The generator supplies a total current of I = I1 + I2 to the load.

Figure 22-24 (a) A current I exists in the rotating coil of the generator. (b) A top view of the coil, showing the magnetic force exerted on the left side of the coil.

| The Back Emf Generated by an Electric Motor

A generator converts mechanical work into electrical energy; in contrast, an electric motor converts electrical energy into mechanical work. Both devices are similar and consist of a coil of wire that rotates in a magnetic field. In fact, as the armature of a motor rotates, the magnetic flux passing through the coil changes and an emf is induced in the coil. Thus, when a motor is operating, two sources of emf are present: (1) the applied emf V that provides current to drive the motor (e.g., from a 120-V outlet), and (2) the emf E induced by the generator-like action of the rotating coil. The circuit diagram in Figure 22-25 shows these two emfs.

Figure 22-25 The applied emf V supplies the current I to drive the motor. The circuit on the right shows V along with the electrical equivalent of the motor, including the resistance R of its coil and the back emf E.

Consistent with Lenz’s law, the induced emf E acts to oppose the applied emf V and is called the back emf or the counter emf of the motor. The greater the speed of the motor, the greater is the flux change through the coil, and the greater is the back emf. Because V and E have opposite polarities, the net emf in the circuit is . In Figure 22-25, R is the resistance of the wire in the coil, and the current I drawn by the motor is determined from Ohm’s law as the net emf divided by the resistance:

| (22.5)

The next example uses this result to illustrate that the current in a motor depends on both the applied emf V and the back emf E.

| Example 12 | | | Operating a Motor

The coil of an ac motor has a resistance of . The motor is plugged into an outlet where V = 120.0 volts (rms), and the coil develops a back emf of volts (rms) when rotating at normal speed. The motor is turning a wheel. Find (a) the current when the motor first starts up and (b) the current when the motor is operating at normal speed.

The physics of operating a motor.

Reasoning Once normal operating speed is attained, the motor need only work to compensate for frictional losses. But in bringing the wheel up to speed from rest, the motor must also do work to increase the wheel’s rotational kinetic energy. Thus, bringing the wheel up to speed requires more work, and hence more current, than maintaining the normal operating speed. We expect our answers to parts (a) and (b) to reflect this fact.
Solution

a. | When the motor just starts up, the coil is not rotating, so there is no back emf induced in the coil and . The start-up current drawn by the motor is

| (22.5)

b. | At normal speed, the motor develops a back emf of volts, so the current is

Problem solving insight The current in an electric motor depends on both the applied emf V and any back emf E developed because the coil of the motor is rotating.

Example 12 illustrates that when a motor is just starting, there is little back emf, and, consequently, a relatively large current exists in the coil. As the motor speeds up, the back emf increases until it reaches a maximum value when the motor is rotating at normal speed. The back emf becomes almost equal to the applied emf, and the current is reduced to a relatively small value, which is sufficient to provide the torque on the coil needed to overcome frictional and other losses in the motor and to drive the load (e.g., a fan).

Check Your Understanding 4

You have a fixed length of wire and need to design a generator. You have only two options: use either a one-turn square coil or a two-turn square coil. Which option should you choose so that your generator will produce the greatest peak emf for a given frequency and magnetic field strength, or doesn’t it matter? (The answer is given at the end of the book).
Background: The peak emf for a generator is , where N is the number of turns in the coil, A is the area of the coil, B is the magnetic field strength, and is the angular frequency.
For similar questions (including calculational counterparts), consult Self-Assessment Test 22.2 , which is described at the end of Section 22.9.

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