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Photograph of a steam turbine connected to a generator.
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.

Test prep for ap courses

The emf induced in a coil that is rotating in a magnetic field will be at a maximum when

  1. the magnetic flux is at a maximum.
  2. the magnetic flux is at a minimum.
  3. the change in magnetic flux is at a maximum.
  4. the change in magnetic flux is at a minimum.

(c)

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A coil with circular cross section and 20 turns is rotating at a rate of 400 rpm between the poles of a magnet. If the magnetic field strength is 0.6 T and peak voltage is 0.2 V, what is the radius of the coil? If the emf of the coil is zero at t = 0 s, when will it reach its peak emf?

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Section summary

  • An electric generator rotates a coil in a magnetic field, inducing an emf given as a function of time by
    emf = NAB ω sin ωt , size 12{"emf"= ital "NAB"ω"sin"ωt} {}
    where A size 12{A} {} is the area of an N size 12{N} {} -turn coil rotated at a constant angular velocity ω size 12{ω} {} in a uniform magnetic field B size 12{B} {} .
  • The peak emf emf 0 size 12{"emf" rSub { size 8{0} } } {} of a generator is
    emf 0 = NAB ω . size 12{"emf" rSub { size 8{0} } = ital "NAB"ω} {}

Conceptual questions

Using RHR-1, show that the emfs in the sides of the generator loop in [link] are in the same sense and thus add.

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The 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?

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Problems&Exercises

Calculate the peak voltage of a generator that rotates its 200-turn, 0.100 m diameter coil at 3600 rpm in a 0.800 T field.

474 V

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At what angular velocity in rpm will the peak voltage of a generator be 480 V, if its 500-turn, 8.00 cm diameter coil rotates in a 0.250 T field?

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What is the peak emf generated by rotating a 1000-turn, 20.0 cm diameter coil in the Earth’s 5 . 00 × 10 5 T size 12{5 "." "00" times "10" rSup { size 8{ - 5} } `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?

0.247 V

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What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field. (This is 60 rev/s.)

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(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?

(a) 50

(b) yes

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Integrated 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?

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(a) A car generator turns at 400 rpm when the engine is idling. Its 300-turn, 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.

(a) 0.477 T

(b) This field strength is small enough that it can be obtained using either a permanent magnet or an electromagnet.

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Show that if a coil rotates at an angular velocity ω , the period of its AC output is 2π/ω .

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A 75-turn, 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?

(a) 5.89 V

(b) At t=0

(c) 0.393 s

(d) 0.785 s

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(a) If the emf of a coil rotating in a magnetic field is zero at t = 0 size 12{t=0} {} , and increases to its first peak at t = 0 . 100 ms size 12{t=0 "." "100"`"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 one-fourth of its maximum? (e) When is it next one-fourth of its maximum?

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Unreasonable Results

A 500-turn coil with a 0 . 250 m 2 size 12{0 "." "250"`m rSup { size 8{2} } } {} area is spun in the Earth’s 5 . 00 × 10 5 T size 12{5 "." "00" times "10" rSup { size 8{ - 5} } `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?

(a) 1 . 92 × 10 6 rad/s size 12{1 "." "92" times "10" rSup { size 8{6} } `"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.

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Questions & Answers

sound waves can be modeled as a change in pressure ,why is the change on in pressure used and not the actual pressure
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Roofs are sometimes pushed off vertically during a tropical cyclone, and buildings sometimes explode outward when hit by a tornado. Use Bernoulli’s principle to explain these phenomena.
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Explain why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both.
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Insulators (nonmetals) have a higher BE than metals, and it is more difficult for photons to eject electrons from insulators. Discuss how this relates to the free charges in metals that make them good conductors.
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P(pressure)=density ×depth×acceleration due to gravity Force =P×Area(28.0x8.5)
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Kyle
Practice Key Terms 3

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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