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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 0.1146 Tesla.

Step by step solution

01

Understand the Problem

We need to find the magnitude of the magnetic field required for an AC generator that yields a maximum electromotive force (emf) of 5500 V, with a coil of 150 turns and an area per turn of 0.85 m², operating at a frequency of 60.0 Hz.
02

Review the Formula for Maximum EMF

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

Calculate the Angular Frequency

Angular frequency \(\omega\) can be calculated using the formula:\[\omega = 2\pi f\]where \(f\) is the frequency. For our generator:\[\omega = 2\pi \times 60.0 \, \text{Hz} = 120\pi \, \text{rad/s}\]
04

Rearrange Formula to Solve for B

To find the magnetic field \(B\), rearrange the maximum emf formula to:\[B = \frac{\mathcal{E}_{\text{max}}}{N\cdot A\cdot \omega}\]
05

Plug in the Known Values

Substitute the known values into the formula:\[B = \frac{5500}{150 \times 0.85 \times 120\pi}\]Calculate the result to find \(B\).
06

Calculate the Result

Perform the calculation:\[B = \frac{5500}{150 \times 0.85 \times 120\pi} = \frac{5500}{15300 \pi} = \frac{5500}{48042}\approx 0.1146 \, \text{T}\]
07

Finalize the Answer

The magnetic field \(B\) required for the generator to achieve the given parameters is approximately 0.1146 Tesla.

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Key Concepts

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

Electromotive Force
Electromotive force (EMF) is the engine that powers the movement of electrons, generating a current within a circuit. Think of it as the voltage that gives the initial push to electrons, helping them to flow. In an AC (alternating current) generator, the EMF is produced as a result of the interaction between the magnetic field, the coil, and how fast the coil rotates. The maximum EMF, often referred to as peak EMF, measures how strong this push can be at its greatest moment. This is determined by the formula:
  • \( \mathcal{E}_{\text{max}} = NAB\omega \)
Here, \(N\) is the number of turns in the coil, \(A\) is the area of each turn, \(B\) is the magnetic field, and \(\omega\) is the angular frequency. In our exercise, the generator's EMF is set to reach a peak of 5500 V. Understanding how each component contributes to this value helps in designing generators that meet specific output needs.
Magnetic Field
The magnetic field in an AC generator is essential for inducing electromotive force. Magnetic fields are invisible forces surrounding magnets or electric currents, and they influence other nearby magnetic and conductive materials. In the context of an AC generator, the magnetic field and the coil work together; as the coil spins through this field, it cuts through magnetic lines, inducing an EMF.
The strength of the magnetic field (denoted as \(B\)) is critical. A stronger magnetic field will induce a higher EMF. The generator's design requires finding the right magnetic field strength based on desired EMF, coil specifications (like turns and area), and the angular frequency.
Using the rearranged formula:
  • \( B = \frac{\mathcal{E}_{\text{max}}}{N\cdot A\cdot \omega} \)
We can find the necessary \(B\) to produce 5500 V, as calculated to be approximately 0.1146 Tesla.
Angular Frequency
Angular frequency is a measure of how fast something rotates or oscillates within a circle. It’s particularly important in alternating current circuits and generators. Represented by \(\omega\), it is linked to the linear frequency \(f\) through the equation:
  • \(\omega = 2\pi f\)
Here, \(f\) represents the frequency of the AC, which, for this exercise, is 60.0 Hz (common in many everyday devices).
Angular frequency can be visualized as how many times per second the coil completes a full cycle, converted into radians (since a full circle is \(2\pi\) radians). Calculating \(\omega\) for our generator gives 120\(\pi\) rad/s, critical for finding how the coil’s rotation speed influences the EMF. Understanding angular frequency helps in grasping how rapidly changing magnetic environments affect electron movement, and thus, how effective the generator will be.

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Most popular questions from this chapter

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 'I. 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?

A flat circular coil with 105 turns, a radius of \(4.00 \times 10^{-2} \mathrm{m}\), and a resistance of \(0.480 \Omega\) is exposed to an external magnetic field that is directed perpendicular to the plane of the coil. The magnitude of the external magnetic field is changing at a rate of \(\Delta B / \Delta t=0.783 \mathrm{T} / \mathrm{s}\), thereby inducing a current in the coil. Find the magnitude of the magnetic field at the center of the coil that is produced by the induced current.

A step-down transformer (turns ratio \(=1: 8\) ) is used with an electric train to reduce the voltage from the wall receptacle to a value needed to operate the train. When the train is running, the current in the secondary coil is \(1.6 \mathrm{A}\). What is the current in the primary coil?

The drawing shows a type of flow meter that can be used to measure the speed of blood in situations when a blood vessel is sufficiently exposed (e.g., during surgery). Blood is conductive enough that it can be treated as a moving conductor. When it flows perpendicularly with respect to a magnetic field, as in the drawing, electrodes can be used to measure the small voltage that develops across the vessel. Suppose that the speed of the blood is \(0.30 \mathrm{m} / \mathrm{s}\) and the diameter of the vessel is \(5.6 \mathrm{mm} .\) In a \(0.60-\mathrm{T}\) magnetic field what is the magnitude of the voltage that is measured with the electrodes in the drawing?

A generating station is producing \(1.2 \times 10^{6} \mathrm{W}\) of power that is to be sent to a small town located \(7.0 \mathrm{km}\) away. Each of the two wires that comprise the transmission line has a resistance per kilometer of \(5.0 \times 10^{-2} \Omega / \mathrm{km}\) (a) Find the power used to heat the wires if the power is transmitted at \(1200 \mathrm{V}\). (b) A 100: 1 step-up transformer is used to raise the voltage before the power is transmitted. How much power is now used to heat the wires?

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