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Chapter 22: Problem 42

You need to design a \(60.0-\mathrm{Hz}\) ac generator that has a maximum emf of 5500 \(\mathrm{V}\) . The generator is to contain a 150 -turn coil that has an area per turn of 0.85 \(\mathrm{m}^{2}\) . What should be the magnitude of the magnetic field in which the coil rotates?

Short Answer

Expert verified
The magnetic field should be approximately 0.29 T.

Step by step solution

01

Understand the Problem

We are tasked with designing an AC generator with a specific maximum electromotive force (emf), operating frequency, number of coil turns, and area per turn. We need to find the magnetic field's magnitude that will allow the generator to achieve the specified maximum emf.
02

Use the Formula for Maximum EMF

The formula for the maximum emf (\( \varepsilon_{max} \)) of an AC generator is \( \varepsilon_{max} = NAB\omega \), where \( N \) is the number of turns, \( A \) is the area per turn, \( B \) is the magnetic field's magnitude, and \( \omega \) is the angular frequency. Our job is to solve this equation for \( B \).
03

Calculate Angular Frequency

The angular frequency \( \omega \) is related to the frequency \( f \) by the equation \( \omega = 2\pi f \). Given \( f = 60.0 \, \mathrm{Hz} \), we calculate \( \omega = 2 \pi \times 60.0 \, \mathrm{rad/s} \).
04

Insert Known Values

We know \( \varepsilon_{max} = 5500 \, \mathrm{V} \), \( N = 150 \), \( A = 0.85 \, \mathrm{m}^2 \), and \( \omega = 120\pi \, \mathrm{rad/s} \). Substituting these into the emf formula: \( 5500 = 150 \times 0.85 \times B \times 120\pi \).
05

Solve for Magnetic Field (B)

Rearrange the equation to solve for \( B \): \( B = \frac{5500}{150 \times 0.85 \times 120\pi} \). Calculate this to find \( B \approx 0.29 \, \mathrm{T} \) (Tesla).

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Key Concepts

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

Maximum EMF
The concept of maximum electromotive force (EMF) is pivotal in the design of an AC generator. EMF is the electrical action produced by a non-electrical source, and for an AC generator, it represents the peak voltage generated. Maximum EMF is crucial as it determines the potential energy transfer from the generator to the circuit. In this exercise, the AC generator is required to have a maximum EMF of 5500 volts. This value is the peak voltage that the generator can supply, and achieving this requires precise control of various factors such as the number of coil turns, coil area, and the magnetic field strength.
Magnetic Field Strength
Magnetic field strength refers to the intensity of the magnetic field in which the coil of a generator rotates. This field is denoted by the symbol \( B \) and is a crucial factor for inducing EMF. In the formula for maximum EMF, \( B \) represents the magnitude of the magnetic field. Stronger magnetic fields increase the induced EMF, making them critical in designing efficient generators. In our scenario, the calculation of the magnetic field strength is essential to achieve the target maximum EMF of 5500 V, producing a value of approximately 0.29 Tesla. This balance ensures the generator's design meets the set requirements, providing the necessary energy output efficiently.
Angular Frequency
Angular frequency, denoted as \( \omega \), links the linear frequency of an AC system to the rotational speed of the loop and magnetic interactions. It is calculated with the formula \( \omega = 2\pi f \), where \( f \) is the frequency in hertz. For the given generator design problem, the frequency is 60 Hz. Consequently, the angular frequency calculation provides \( \omega = 120\pi \) rad/s. This frequency affects how quickly the AC voltage cycles, impacting the generator's output characteristics. Understanding angular frequency is fundamental to grasping how quickly alternating current changes its direction, which is a central feature of AC systems.
Electromotive Force Formula
The electromotive force (EMF) formula helps dictate how generators function by describing how EMF is induced based on several parameters. The specific formula used is \( \varepsilon_{max} = NAB\omega \), where \( N \) is the number of turns in the coil, \( A \) is the area per turn, \( B \) is the magnetic field strength, and \( \omega \) is the angular frequency. Each variable has a direct impact on the potential difference generated. In this generator problem, the formula is rearranged to solve for the magnetic field required to produce the specified maximum EMF of 5500 V. By substituting the known values of other factors into the equation, we determine the necessary magnetic field strength, ensuring proper generator function and design.

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Most popular questions from this chapter

Parts \(a\) and \(b\) of the drawing show the same uniform and constant (in time) magnetic field \(\overrightarrow{\mathbf{B}}\) directed perpendicularly into the paper over a rectangular region. Outside this region, there is no field. Also shown is a rectangular coil (one turn), which lies in the plane of the paper. In part \(a\) the long side of the coil (length \(=L )\) is just at the edge of the field region, while in part \(b\) the short side (width \(=W\) is just at the edge. It is known that \(L / W=3.0 .\) In both parts of the drawing the coil is pushed into the field with the same velocity \(\overrightarrow{\mathbf{v}}\) until it is completely within the field region. The magnitude of the average emf induced in the coil in part \(a\) is 0.15 V. What is its magnitude in part \(b\) ?

ssm The drawing shows three identical rods (A, B, and C) moving in different planes. A constant magnetic field of magnitude 0.45 T is directed along the \(+y\) axis. The length of each rod is \(L=1.3 \mathrm{m},\) and the rods each have the same speed, \(v_{\mathrm{A}}=v_{\mathrm{B}}=v_{\mathrm{C}}=2.7 \mathrm{m} / \mathrm{s}\) . For each rod, find the magnitude of the motional emf, and indicate which end (1 or 2 ) of the rod is positive.

ssm A piece of copper wire is formed into a single circular loop of radius 12 \(\mathrm{cm} .\) A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.60 \(\mathrm{T}\) in a time of 0.45 s. The wire has a resistance per unit length of \(3.3 \times 10^{-2} \Omega / \mathrm{m} .\) What is the average electrical energy dissipated in the resistance of the wire?

ssm A \(3.0-\mu \mathrm{F}\) capacitor has a voltage of 35 \(\mathrm{V}\) between its plates. What must be the current in a \(5.0-\mathrm{mH}\) inductor so that the energy stored in the inductor equals the energy stored in the capacitor?

A planar coil of wire has a single turn. The normal to this coil is parallel to a uniform and constant (in time) magnetic field of 1.7 \(\mathrm{T}\) . An emf that has a magnitude of 2.6 \(\mathrm{V}\) is induced in this coil because the coil's area \(A\) is shrinking. What is the magnitude of \(\Delta A / \Delta t,\) which is the rate (in \(\mathrm{m}^{2} / \mathrm{s} )\) at which the area changes?
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