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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 magnitude should be approximately \(0.29\) T.

Step by step solution

01

Identify Given Values

We are given the frequency \( f = 60.0 \text{ Hz} \), maximum emf \( \mathcal{E}_{\text{max}} = 5500 \text{ V} \), number of turns \( N = 150 \), and area per turn \( A = 0.85 \text{ m}^2 \).
02

Use the EMF Formula

The formula for maximum emf in an AC generator is \( \mathcal{E}_{\text{max}} = NAB\omega \), where \( N \) is the number of turns, \( A \) is the area per turn, \( B \) is the magnetic field, and \( \omega \) is the angular velocity.
03

Find Angular Velocity

The angular velocity \( \omega \) is related to frequency \( f \) by the formula \( \omega = 2\pi f \). So, \( \omega = 2\pi \times 60 = 120\pi \text{ rad/s} \).
04

Rearrange the EMF Formula

Solve the formula \( \mathcal{E}_{\text{max}} = NAB\omega \) for \( B \). Rearranging gives \( B = \frac{\mathcal{E}_{\text{max}}}{NA\omega} \).
05

Substitute Values

Substitute the known values into the rearranged equation: \( B = \frac{5500}{150 \times 0.85 \times 120\pi} \approx 0.29 \text{ T} \).
06

Calculate the Result

Perform the calculation to find \( B \). The final answer is \( B \approx 0.29 \text{ T} \), rounded to two decimal places.

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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 represents the peak voltage that the generator can produce. It's an essential parameter in designing generators as it determines how much voltage is available for pushing current through a circuit. This maximum emf occurs when the coil within the generator cuts through the magnetic field lines at the most optimal angle, resulting in the highest possible change in magnetic flux. The formula for calculating the maximum emf is \( \mathcal{E}_{\text{max}} = NAB\omega \). Each factor in this equation鈥攏umber of turns \( N \), area \( A \), magnetic field \( B \), and angular velocity \( \omega \)鈥攊nfluences the outcome, making it vital to understand how they contribute to achieving the desired emf.
magnetic field
The magnetic field strength, represented by \( B \), is crucial in the functioning of an AC generator. It is the invisible force that interacts with the coil to produce the emf. When the generator's coil rotates through this magnetic field, an electric current is induced due to the change in magnetic flux. Increasing the strength of the magnetic field enhances the ability to generate a higher emf, which is beneficial for a stronger electrical output. Key considerations on the magnetic field include:
  • Magnitude: The strength of the magnetic field directly influences the emf produced. A stronger magnetic field results in a greater change in flux per unit time, leading to higher emf.
  • Configuration: The placement and structure of the magnetic field affect how the coil interacts with it and thus the efficiency of the generator.
Maintaining an optimal magnetic field strength is vital for efficient generator operation.
coil turns
The number of turns in the coil, indicated by \( N \), significantly affects the emf produced by an AC generator. Each loop or turn of the coil contributes to the total emf generated because it allows for more interactions with the magnetic field. The basic operation follows Faraday's Law of Induction, which states that the induced emf in a coil is proportional to the rate of change of magnetic flux linkage. Consider the following:
  • More turns mean more interaction with the magnetic field, resulting in a larger induced emf.
  • Increasing the number of turns also increases the coil's resistance, which can impact the generator's efficiency.
Thus, balancing the number of coil turns is crucial for optimizing both the voltage output and the efficiency of the generator.
angular velocity
Angular velocity, denoted by \( \omega \), describes how fast the coil spins within the magnetic field of an AC generator. It is a key factor influencing the maximum emf, as indicated by the equation \( \mathcal{E}_{\text{max}} = NAB\omega \). Angular velocity is connected to the generator's frequency and is calculated using \( \omega = 2\pi f \), where \( f \) is the frequency. Important points about angular velocity include:
  • Direct Impact: Higher angular velocity results in a faster change in magnetic flux, increasing the induced emf.
  • Frequency Dependency: Angular velocity is proportional to the frequency, meaning that generators set to higher frequencies will also have higher angular velocities and thus potentially higher emfs.
Adjusting angular velocity is a common method of controlling the generator's output to meet specific power requirements.

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Most popular questions from this chapter

A long solenoid of length \(8.0 \times 10^{-2} \mathrm{m}\) and cross-sectional area \(5.0 \times 10^{-5} \mathrm{m}^{2}\) contains 6500 turns per meter of length. Determine the emf induced in the solenoid when the current in the solenoid changes from 0 to 1.5 A during the time interval from 0 to 0.20 s.

Magnetic resonance imaging (MRI) is a medical technique for producing pictures of the interior of the body. The patient is placed within a strong magnetic field. One safety concern is what would happen to the positively and negatively charged particles in the body fluids if an equipment failure caused the magnetic field to be shut off suddenly. An induced emf could cause these particles to flow, producing an electric current within the body. Suppose the largest surface of the body through which flux passes has an area of \(0.032 \mathrm{m}^{2}\) and a normal that is parallel to a magnetic field of \(1.5 \mathrm{T}\). Determine the smallest time period during which the field can be allowed to vanish if the magnitude of the average induced emf is to be kept less than \(0.010 \mathrm{V}\).

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 \(b ?\)

The resistances of the primary and secondary coils of a transformer are 56 and \(14 \Omega,\) respectively. Both coils are made from lengths of the same copper wire. The circular turns of each coil have the same diameter. Find the turns ratio \(N_{J} / N_{\mathrm{p}}\).

The maximum strength of the earth's magnetic field is about \(6.9 \times 10^{-5} \mathrm{T}\) near the south magnetic pole. In principle, this field could be used with a rotating coil to generate \(60.0-\mathrm{Hz}\) ac electricity. What is the minimum number of turns (area per turn \(=0.022 \mathrm{m}^{2}\) ) that the coil must have to produce an rms voltage of \(120 \mathrm{V} ?\)
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