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The armature of a simple ac generator has 20 circular loops of wire, each with a radius of \(10 \mathrm{~cm} .\) It is rotated with a frequency of \(60 \mathrm{~Hz}\) in a uniform magnetic field of \(800 \mathrm{mT}\). (a) What is the maximum emf induced in the loops? (b) How often is this value attained? (c) If the time period in part (b) were cut in half, what would be the new maximum emf value?

Short Answer

Expert verified
(a) 189 V; (b) 120 times/sec; (c) 378 V.

Step by step solution

01

Identify Known Variables

We have several known quantities: Number of loops, \(N = 20\); Radius of loops, \(r = 10\) cm; Frequency, \(f = 60\) Hz; Magnetic field strength, \(B = 800\) mT. Convert the radius to meters: \(r = 0.1\) m and convert magnetic field to Tesla: \(B = 0.8\) T.
02

Calculate the Area of One Loop

The area \(A\) of one loop is given by the formula \(A = \pi r^2\). Substituting the known value for radius gives \(A = \pi (0.1)^2 = 0.0314\) square meters.
03

Determine Maximum Emf Induced

The maximum emf \(\varepsilon_{max}\) induced in the armature can be calculated with the formula \(\varepsilon_{max} = NAB \cdot 2\pi f\). Substituting the known values gives \(\varepsilon_{max} = 20 \cdot 0.0314 \cdot 0.8 \cdot 2\pi \cdot 60 \approx 189\) V.
04

Calculate Frequency of Maximum Emf

The maximum emf value is reached twice every cycle of rotation, since the ac wave reaches its positive and negative peaks in each cycle. Hence, the frequency of the maximum emf is \(2f = 2 \times 60 = 120\) times per second.
05

Adjust for Halved Time Period

Halving the time period doubles the frequency. The new frequency is \(120\) Hz. Using \(\varepsilon_{max} = NAB \cdot 2\pi (2f)\) with the new frequency, the maximum emf becomes twice the previous value, which is \(2 \times 189 = 378\) V.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Maximum EMF
The maximum electromotive force (emf) in an ac generator is a crucial concept to grasp. An emf is the electrical action produced by a non-electrical source, like a rotating coil in a magnetic field. For our ac generator task, we calculate the maximum emf using the formula: \(\varepsilon_{max} = NAB \cdot 2\pi f\). Here, \(N\) is the number of loops, \(A\) is the area of one loop, \(B\) is the magnetic field strength, and \(f\) is the frequency of rotation.

- **Number of Loops (N):** This defines how many circles of wire coil are in the generator. More loops mean a greater emf.
- **Area of One Loop (A):** For each loop of wire, consider the size. The larger the loop, the larger the area, and in turn, the more emf can be generated. Calculating area involves using \(A = \pi r^2\), where \(r\) is the radius of the loop.
- **Emf Calculation:** By plugging all known values into our formula, we can predict the maximum emf produced by the generator. In this example, the maximum calculated is approximately 189 volts. Understanding and calculating the maximum emf reflects how much "push" the generator can give to the electric charge.
Magnetic Field Strength
Magnetic field strength, often represented by \(B\), plays an important role as a part of the generator’s effectiveness. A magnetic field is a field produced around a magnet, and its strength manifests in how it influences nearby currents and magnetic materials. In this exercise, it is measured in Tesla (T).

- **What It Means:** A stronger magnetic field results in a higher emf when interacting with the rotating loops. - **Unit Conversion:** If the magnetic field strength is given in milliTesla (mT), converting it to Tesla involves dividing by 1000. So, 800 mT converts to 0.8 T.
- **Interaction with Coils:** When the coils of the generator cut through the magnetic field lines, an emf is induced, proportional to both the strength of the field and the rate at which lines are cut.
The magnetic field provides the necessary force that, along with the mechanical motion of the coil, helps produce the current. Its strength directly impacts the output of the ac generator.
Frequency of Rotation
Frequency of rotation, denoted as \(f\), tells us how fast the generator's armature spins. It's a critical factor in determining the output of an ac generator.

- **Understanding Frequency:** The frequency is measured in Hertz (Hz), which counts the number of cycle completions per second. A higher frequency generally results in a higher induced emf.
- **Impact on EMF:** Since the maximum emf is proportional to the frequency, doubling the frequency, for instance, leads to doubling the maximum emf (assuming other conditions remain constant).
- **Cycle Completion:** Each cycle involves the armature reaching its positive peak and negative peak of emf. Therefore, if the frequency is 60 Hz, the maximum values of emf occur 120 times a second.

By understanding and manipulating frequency, we can adjust the generator's performance. This showcases the dynamic nature of how ac generators can be optimized for various uses by merely changing the speed of rotation.

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Most popular questions from this chapter

The armature of an ac generator has 100 turns. Each turn is a rectangular loop measuring \(8.0 \mathrm{~cm}\) by \(12 \mathrm{~cm} .\) The generator has a sinusoidal voltage output with an amplitude of \(24 \mathrm{~V}\). (a) If the magnetic field of the generator is \(250 \mathrm{mT}\), with what frequency does the armature turn? (b) If the magnetic field was doubled and the frequency cut in half, what would be the amplitude of the output?

Find the frequencies of electromagnetic waves with wavelengths of (a) \(3.0 \mathrm{~cm},\) (b) \(650 \mathrm{nm}\), and (c) \(1.2 \mathrm{fm}\). (d) Classify the type of light in each case.

An ideal transformer has 840 turns in its primary coil and 120 turns in its secondary coil. If the primary coil draws 2.50 A at \(110 \mathrm{~V}\), what are (a) the current and (b) the output voltage of the secondary coil?

An ideal solenoid with a current of 1.5 A has a radius of \(3.0 \mathrm{~cm}\) and a turn density of 250 turns \(/ \mathrm{m}\). (a) What is the magnetic flux (due to its own field) through only one of its loops at its center? (b) What current would be required to double the flux value in part (a)?

A student makes a simple ac generator by using a single square wire loop \(10 \mathrm{~cm}\) on a side. The loop is then rotated at a frequency of \(60 \mathrm{~Hz}\) in a magnetic field of \(0.015 \mathrm{~T}\). (a) What is the maximum emf output? (b) If she wanted to make the maximum emf output ten times larger by adding loops, how many should she use in total?

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