236. Electric GeneratorsLearning Objectives
Example 1: Calculating the Emf Induced in a Generator CoilThe generator coil shown in Figure 1 is rotated through onefourth of a revolution (from $\theta =\mathrm{0\xba}$ to $\theta =\text{90\xba}$ ) in 15.0 ms. The 200turn circular coil has a 5.00 cm radius and is in a uniform 1.25 T magnetic field. What is the average emf induced? Strategy We use Faraday’s law of induction to find the average emf induced over a time $\mathrm{\Delta}t$: $$\text{emf}=N\frac{\mathrm{\Delta}\Phi}{\mathrm{\Delta}t}\text{.}$$We know that $N=\text{200}$ and $\mathrm{\Delta}t=\text{15}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{ms}$, and so we must determine the change in flux $\mathrm{\Delta}\Phi $ to find emf. Solution Since the area of the loop and the magnetic field strength are constant, we see that $$\mathrm{\Delta}\Phi =\mathrm{\Delta}(\text{BA}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta )=\text{AB}\mathrm{\Delta}(\text{cos}\phantom{\rule{0.25em}{0ex}}\theta )\text{.}$$Now, $\mathrm{\Delta}(\text{cos}\phantom{\rule{0.25em}{0ex}}\theta )=1\text{.}0$, since it was given that $\theta $ goes from $\text{0\xba}$ to $\text{90\xba}$ . Thus $\mathrm{\Delta}\Phi =\text{AB}$, and $$\text{emf}=N\frac{\text{AB}}{\mathrm{\Delta}t}.$$The area of the loop is $A={\mathrm{\pi r}}^{2}=(3\text{.}\text{14}\text{.}\text{.}\text{.})(0\text{.}\text{0500}\phantom{\rule{0.25em}{0ex}}\text{m}{)}^{2}=7\text{.}\text{85}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$. Entering this value gives $$\text{emf}=\text{200}\frac{(7\text{.}\text{85}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2})(1\text{.}\text{25}\phantom{\rule{0.25em}{0ex}}\text{T})}{\text{15}\text{.}0\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{s}}=\text{131}\phantom{\rule{0.25em}{0ex}}\text{V.}$$Discussion This is a practical average value, similar to the 120 V used in household power. The emf calculated in Example 1 is the average over onefourth 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 $w$ and height $\ell $ in a uniform magnetic field, as illustrated in Figure 2. 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 $\text{emf}=\mathrm{B\ell v}$, where the velocity v is perpendicular to the magnetic field $B$. Here the velocity is at an angle $\theta $ with $B$, so that its component perpendicular to $B$ is $v\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $ (see Figure 2). Thus in this case the emf induced on each side is $\text{emf}=\mathrm{B\ell v}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $, and they are in the same direction. The total emf around the loop is then $$\text{emf}=2\mathrm{B\ell v}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \text{.}$$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 $\omega $. The angle $\theta $ is related to angular velocity by $\theta =\mathrm{\omega t}$, so that $$\text{emf}=\text{2}\mathrm{B\ell v}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\mathrm{\omega t}\text{.}$$Now, linear velocity $v$ is related to angular velocity $\omega $ by $v=\mathrm{r\omega}$. Here $r=w/2$, so that $v=(w/2)\omega $, and $$\text{emf}=2\mathrm{B\ell}\frac{w}{2}\omega \phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\mathrm{\omega t}=(\ell w)\mathrm{B\omega}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\mathrm{\omega t}\text{.}$$Noting that the area of the loop is $A=\ell w$, and allowing for $N$ loops, we find that $$\text{emf}=\text{NAB}\omega \phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\mathrm{\omega t}$$is the where $${\text{emf}}_{0}=\text{NAB}\omega $$is the maximum The fact that the peak emf, ${\text{emf}}_{0}=\text{NAB}\omega $, 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 $\omega $), 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 4 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 ripplefree DC. Example 2: Calculating the Maximum Emf of a GeneratorCalculate the maximum emf, ${\text{emf}}_{0}$, of the generator that was the subject of Example 1. Strategy Once $\omega $, the angular velocity, is determined, ${\text{emf}}_{0}=\text{NAB}\omega $ can be used to find ${\text{emf}}_{0}$. All other quantities are known. Solution Angular velocity is defined to be the change in angle per unit time: $$\omega =\frac{\mathrm{\Delta}\theta}{\mathrm{\Delta}t}\text{.}$$Onefourth of a revolution is $\mathrm{\pi /2}$ radians, and the time is 0.0150 s; thus, $$\begin{array}{lll}\omega & =& \frac{\pi /2\phantom{\rule{0.25em}{0ex}}\text{rad}}{\text{0.0150 s}}\\ & =& \text{104}.\text{7 rad/s}\text{.}\end{array}$$104.7 rad/s is exactly 1000 rpm. We substitute this value for $\omega $ and the information from the previous example into ${\text{emf}}_{0}=\text{NAB}\omega $, yielding $$\begin{array}{lll}{\text{emf}}_{0}& =& \text{NAB}\omega \\ & =& \text{200}(7\text{.}\text{85}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2})(1\text{.}\text{25}\phantom{\rule{0.25em}{0ex}}\text{T})(\text{104}\text{.}7\phantom{\rule{0.25em}{0ex}}\text{rad/s})\\ & =& \text{206}\phantom{\rule{0.25em}{0ex}}\text{V}\end{array}\text{.}$$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 5 shows a cutaway view of a steam turbine; steam moves over the blades connected to the shaft, which rotates the coil within the generator. 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. Section Summary
Conceptual QuestionsExercise 1Using RHR1, show that the emfs in the sides of the generator loop in Figure 4 are in the same sense and thus add. Exercise 2The source of a generator’s electrical energy output is the work done to turn its coils. How is the work needed to turn the generator related to Lenz’s law? Problems & ExercisesExercise 1Calculate the peak voltage of a generator that rotates its 200turn, 0.100 m diameter coil at 3600 rpm in a 0.800 T field. Show/Hide Solution Solution474 V Exercise 2At what angular velocity in rpm will the peak voltage of a generator be 480 V, if its 500turn, 8.00 cm diameter coil rotates in a 0.250 T field? Exercise 3What is the peak emf generated by rotating a 1000turn, 20.0 cm diameter coil in the Earth’s $5\text{.}\text{00}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{T}$ magnetic field, given the plane of the coil is originally perpendicular to the Earth’s field and is rotated to be parallel to the field in 10.0 ms? Show/Hide Solution Solution0.247 V Exercise 4What is the peak emf generated by a 0.250 m radius, 500turn coil is rotated onefourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field. (This is 60 rev/s.) Exercise 5(a) A bicycle generator rotates at 1875 rad/s, producing an 18.0 V peak emf. It has a 1.00 by 3.00 cm rectangular coil in a 0.640 T field. How many turns are in the coil? (b) Is this number of turns of wire practical for a 1.00 by 3.00 cm coil? Show/Hide Solution Solution(a) 50 (b) yes Exercise 6Integrated Concepts This problem refers to the bicycle generator considered in the previous problem. It is driven by a 1.60 cm diameter wheel that rolls on the outside rim of the bicycle tire. (a) What is the velocity of the bicycle if the generator’s angular velocity is 1875 rad/s? (b) What is the maximum emf of the generator when the bicycle moves at 10.0 m/s, noting that it was 18.0 V under the original conditions? (c) If the sophisticated generator can vary its own magnetic field, what field strength will it need at 5.00 m/s to produce a 9.00 V maximum emf? Exercise 7(a) A car generator turns at 400 rpm when the engine is idling. Its 300turn, 5.00 by 8.00 cm rectangular coil rotates in an adjustable magnetic field so that it can produce sufficient voltage even at low rpms. What is the field strength needed to produce a 24.0 V peak emf? (b) Discuss how this required field strength compares to those available in permanent and electromagnets. Show/Hide Solution Solution(a) 0.477 T (b) This field strength is small enough that it can be obtained using either a permanent magnet or an electromagnet. Exercise 8Show that if a coil rotates at an angular velocity $\omega $, the period of its AC output is $\mathrm{2\pi /\omega}$. Exercise 9A 75turn, 10.0 cm diameter coil rotates at an angular velocity of 8.00 rad/s in a 1.25 T field, starting with the plane of the coil parallel to the field. (a) What is the peak emf? (b) At what time is the peak emf first reached? (c) At what time is the emf first at its most negative? (d) What is the period of the AC voltage output? Show/Hide Solution Solution(a) 5.89 V (b) At t=0 (c) 0.393 s (d) 0.785 s Exercise 10(a) If the emf of a coil rotating in a magnetic field is zero at $t=0$, and increases to its first peak at $t=0\text{.}\text{100}\phantom{\rule{0.25em}{0ex}}\text{ms}$, what is the angular velocity of the coil? (b) At what time will its next maximum occur? (c) What is the period of the output? (d) When is the output first onefourth of its maximum? (e) When is it next onefourth of its maximum? Exercise 11Unreasonable Results A 500turn coil with a $0\text{.}\text{250}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$ area is spun in the Earth’s $5\text{.}\text{00}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{T}$ field, producing a 12.0 kV maximum emf. (a) At what angular velocity must the coil be spun? (b) What is unreasonable about this result? (c) Which assumption or premise is responsible? Show/Hide Solution Solution(a) $1\text{.}\text{92}\times {\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{rad/s}$ (b) This angular velocity is unreasonably high, higher than can be obtained for any mechanical system. (c) The assumption that a voltage as great as 12.0 kV could be obtained is unreasonable.
