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A generator uses a coil that has 100 turns and a \(0.50-\mathrm{T}\) magnetic field. The frequency of this generator is \(60.0 \mathrm{Hz},\) and its emf has an rms value of \(120 \mathrm{V}\). Assuming that each turn of the coil is a square (an approximation), determine the length of the wire from which the coil is made.

Short Answer

Expert verified
The total wire length is 38.0 meters.

Step by step solution

01

Understand the relation between rms emf and peak emf

The root mean square (rms) value of the electromotive force (emf) is related to the peak emf by the equation: \( \text{emf}_{\mathrm{rms}} = \frac{\text{emf}_{\text{peak}}}{\sqrt{2}} \). Given that \( \text{emf}_{\mathrm{rms}} = 120\, \mathrm{V} \), we solve for the peak emf: \( \text{emf}_{\text{peak}} = 120\sqrt{2}\, \mathrm{V} \).
02

Calculate the peak emf using given values

Given the rms value \( \text{emf}_{\mathrm{rms}} = 120\, \mathrm{V} \), we can calculate the peak emf: \( \text{emf}_{\text{peak}} = 120\sqrt{2} \approx 169.7\, \mathrm{V} \).
03

Use Faraday’s Law of Induction

According to Faraday’s Law, the peak emf is given by: \( \text{emf}_{\text{peak}} = NAB\omega \sin(\omega t) \), where \( N \) is the number of turns, \( A \) is the area of the coil, \( B \) is the magnetic field, and \( \omega \) is the angular frequency. The maximum value is at \( \sin(\omega t) = 1 \), so \( \text{emf}_{\text{peak}} = NAB\omega \).
04

Calculate the angular frequency

The angular frequency \( \omega \) is related to the frequency \( f \) by \( \omega = 2\pi f \). Given \( f = 60.0\, \mathrm{Hz} \), we find \( \omega = 2\pi \times 60.0 = 120\pi\, \mathrm{rad/s} \).
05

Solve for the area of one turn of the coil

Using \( \text{emf}_{\text{peak}} = NAB\omega \), and substituting \( N = 100 \), \( B = 0.50\, \mathrm{T} \), \( \omega = 120\pi \), and \( \text{emf}_{\text{peak}} = 169.7 \), we can calculate the area of one turn \( A \) as follows: \[ A = \frac{169.7}{100 \times 0.50 \times 120\pi} \approx 0.009 \mathrm{m}^2. \]
06

Determine the side length of each square turn

Since each turn is a square, if \( A = 0.009\, \mathrm{m}^2 \), then each side of the square has a length \( s \), where \( s^2 = A \). Thus, \( s = \sqrt{0.009} \approx 0.095\, \mathrm{m} \).
07

Calculate the total length of the wire

Each square turn has a perimeter, which is four times the side length \( s \). Thus, the total length of the wire is: \[ \text{Total length} = 100 \times 4 \times 0.095 = 38.0\, \mathrm{m}. \]

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Key Concepts

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

Faraday's Law
Faraday's Law of electromagnetic induction is fundamental to understanding how electric generators work. It states that an electromotive force (emf) is induced in a conductor when the magnetic flux through the conductor changes. The induced emf is directly proportional to the rate of change of magnetic flux. In the context of the provided exercise, Faraday's Law is applied to calculate the peak emf produced by the generator. The formula is given by: \[ \text{emf}_{\text{peak}} = NAB\omega \sin(\omega t) \]where:
  • N is the number of turns in the coil (100 turns in this case)
  • A is the area of one loop in the coil
  • B is the magnetic field strength (0.50 T here)
  • \(\omega\) is the angular frequency in radians per second
Faraday's Law shows how a rotating coil in a magnetic field can produce an AC voltage. When the position of the coil changes, the amount of magnetic field lines passing through it changes, inducing an emf.
rms Voltage
The root mean square (rms) voltage of an alternating current system, like the one produced by the generator in the exercise, measures the effective value of an alternating voltage. It is useful because it represents the DC equivalent of an AC voltage in terms of its ability to perform work in a resistive circuit. The relationship between the rms voltage and the peak voltage in an AC circuit is given by:\[ \text{emf}_{\text{rms}} = \frac{\text{emf}_{\text{peak}}}{\sqrt{2}} \] In this exercise, the generator has an rms voltage of 120 V. By rearranging the formula, we can calculate the peak voltage:\[ \text{emf}_{\text{peak}} = 120\sqrt{2} \approx 169.7 \text{ V} \]Understanding rms values is crucial in electricity and electronics because they allow us to compare AC and DC circuits directly and work with real power considerations.
Angular Frequency
Angular frequency, often symbolized by \( \omega \), is a measure of how quickly the phase of a wave progresses. It is particularly relevant in systems involving rotating objects or periodic changes such as AC currents.In this context, angular frequency is found using the equation:\[ \omega = 2\pi f \]where \( f \) is the frequency of the generator. Given that \( f = 60.0 \text{ Hz} \), we calculate:\[ \omega = 2\pi \times 60.0 = 120\pi \text{ rad/s} \]Utilizing angular frequency allows us to explore the timing-related aspects of oscillating systems and address issues involving rotational speed and periodic waveforms, which are essential for a generator's operation.
Magnetic Field
A magnetic field, denoted typically by \( B \), is a vector field that describes the magnetic influence on charges, moving charges, and magnetic materials. In the scope of electromagnetic induction, the magnetic field plays a critical role in inducing electromotive force in the loop of wire within the generator coil.The strength of the magnetic field in this problem is given as 0.50 Tesla (T). This value reflects how 'strong' or 'weak' the magnetic effect is on the generator's coil. The interaction between the rotating coil and the stable magnetic field generates an electromagnetic force thanks to Faraday's Law.Magnetic fields form the basis of many technologies, from simple magnets to complex electronic systems like transformers, motors, and generators. They are critical in numerous applications such as data storage, navigation, and even medical technologies.

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Most popular questions from this chapter

A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is \(7.0 \times 10^{-3} \mathrm{Wb}\). What is the flux that passes through the circular loop?

Two coils of wire are placed close together. Initially, a current of 2.5 A exists in one of the coils, but there is no current in the other. The current is then switched off in a time of \(3.7 \times 10^{-2}\) s. During this time, the average emf induced in the other coil is \(1.7 \mathrm{V}\). What is the mutual inductance of the two-coil system?

A generator is connected across the primary coil \(\left(N_{p}\right.\) turns) of a transformer, while a resistance \(R_{2}\) is connected across the secondary coil \(\left(N_{\mathrm{s}}\right.\) turns). This circuit is equivalent to a circuit in which a single resistance \(R_{1}\) is connected directly across the generator, without the transformer. Show that \(R_{1}=\left(N_{p} / N_{\mathrm{s}}\right)^{2} R_{2},\) by starting with Ohm's law as applied to the secondary coil.

A magnetic field has a magnitude of \(0.078 \mathrm{T}\) and is uniform over a circular surface whose radius is \(0.10 \mathrm{m}\). The field is oriented at an angle of \(\phi=25^{\circ}\) with respect to the normal to the surface. What is the magnetic flux through the surface?

The secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter. The turns ratio of the transformer is \(50: 1 .\) The primary coil is plugged into a standard \(120-\mathrm{V}\) outlet. The current in the secondary coil is \(1.7 \times 10^{-3} \mathrm{A}\). Find the power consumed by the air filter.

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