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ssm www The drawing shows a plot of the output emf of a generator as a function of time \(t\). The coil of this device has a cross-sectional area per turn of \(0.020 \mathrm{~m}^{2}\) and contains 150 turns. Find (a) the frequency \(f\) of the generator in hertz, (b) the angular speed \(\omega\) in \(\mathrm{rad} / \mathrm{s}\), and \((\mathrm{c})\) the magnitude of the magnetic field

Short Answer

Expert verified
Determine period from graph for frequency. Use \(\omega = 2\pi f\) for angular speed. Solve \(B = \frac{\text{emf}_{\max}}{NA\omega}\) for magnetic field.

Step by step solution

01

Understand the Given Data

The problem provides us with the cross-sectional area per turn \( A = 0.020 \text{ m}^2 \),and the number of turns in the coil \( N = 150 \text{ turns} \). Since it's a generator, we typically have an alternating current output, and we need to determine the frequency and other parameters.
02

Determine Frequency from the Graph

If a graph is given showing emf as a function of time \( t \), locate the period \( T \), which is the time it takes for one complete cycle of the waveform.The frequency \( f \) is then calculated using the formula:\[ f = \frac{1}{T} \]
03

Calculate Angular Speed

The angular speed \( \omega \) is related to the frequency by the equation \[ \omega = 2\pi f \]Use the frequency found in Step 2 and substitute it into this formula to find \( \omega \).
04

Establish the Formula for Emf

The induced emf is given by the equation:\[ \text{emf}_{\text{max}} = NAB\omega \]where \( B \) is the magnitude of the magnetic field, and \( \text{emf}_{\text{max}} \) is the maximum emf value observed in the graph.
05

Solve for Magnetic Field Magnitude

Rearrange the formula for maximum emf to solve for the magnetic field \( B \):\[ B = \frac{\text{emf}_{\text{max}}}{NA\omega} \]Plug in the known values: \( \text{emf}_{\text{max}} \) from the maximum value on the graph, \( N = 150 \), \( A = 0.020 \text{ m}^2 \), and \( \omega \) from Step 3, to compute \( B \).

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Key Concepts

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

Generator Frequency
In the context of electromagnetic induction, generator frequency refers to how often the cyclical output of a generator repeats itself per second. A generator turns mechanical energy into electrical energy, typically producing an alternating current (AC). This alternating output can be visualized as a wave, making the frequency an important characteristic defining how swiftly this wave oscillates.

To find the frequency of a generator, observe the period of the cycle in the waveform graph. The period, denoted as \( T \), is the time taken for one complete cycle of the output waveform. The frequency \( f \), measured in hertz (Hz), is calculated through the simple relationship:
  • \( f = \frac{1}{T} \)
This formula states that frequency is the reciprocal of the period. Frequency informs us of the rapidity of the generator's cycle changes, impacting how devices relying on this electrical output must operate.
Angular Speed
Angular speed in a generator is a measure of how fast the generator rotates, specifically concerning its electromagnetic coil. We denote angular speed with \( \omega \), which is measured in radians per second (rad/s). It's crucial because it governs how quickly the magnetic field lines are cut by the rotating coil, ultimately inducing an electromotive force (EMF).

To derive angular speed from the frequency, use the relationship:
  • \( \omega = 2\pi f \)
Here, \( 2\pi \) accounts for the angular displacement of the coil per full rotation. Thus, by substituting the frequency obtained earlier, one can calculate the angular speed of the generator accurately.
Understanding \( \omega \) is vital, particularly as it is a crucial parameter influencing the efficiency and power output of rotating generators.
Magnetic Field Magnitude
The magnitude of the magnetic field, denoted as \( B \), is an integral factor in the performance of a generator. It affects the strength of the induced EMF. A generator coil moving through a magnetic field experiences a change in magnetic flux, resulting in the induction of EMF.

To deduce the magnetic field's strength, employ the formula derived from the maximum observed EMF:
  • \( B = \frac{\text{emf}_{\text{max}}}{NA\omega} \)
Where:
  • \( \text{emf}_{\text{max}} \) is the peak EMF from the waveform graph,
  • \( N \) is the number of turns in the coil,
  • \( A \) is the cross-sectional area per turn, and
  • \( \omega \) is the angular speed.
By rearranging the standard EMF equation, we solve for \( B \), shedding light on how influential the magnetic field is in the induction process within the generator.
Induced EMF
In a generator, induced EMF represents the driving force behind the movement of electrical charge in a circuit. It arises when the conductive coil moves within a magnetic field, cutting through magnetic field lines, thus generating current.

The relationship dictating the generated EMF can be defined as:
  • \( \text{emf} = NAB\omega \sin(\theta) \)
Where:
  • \( \text{emf} \) is the electromotive force,
  • \( N \) is the number of turns,
  • \( A \) is the area of the coil,
  • \( B \) is the magnetic field magnitude,
  • \( \omega \) is the angular speed, and
  • \( \theta \) is the angular position of the coil.
This formula highlights how variations in these parameters can influence the magnitude and direction of the induced EMF. Understanding how to maximize induced EMF is fundamental in optimizing generator design and function.

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Most popular questions from this chapter

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 length 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?

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 \(3.7 \times 10^{-2} \mathrm{~s}\). 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?

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 \(0.010 \mathrm{~V}\).

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 \(60.0\) 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 \(120 \mathrm{~V}\) ?

The drawing shows a coil of copper wire that consists of two semicircles joined by straight sections of wire. In part \(a\) the coil is lying flat on a horizontal surface. The dashed line also lies in the plane of the horizontal surface. Starting from the orientation in part \(a\), the smaller semicircle rotates at an angular frequency \(\omega\) about the dashed line, until its plane becomes perpendicular to the horizontal surface, as shown in part \(b\). A uniform magnetic field is constant in time and is directed upward, perpendicular to the horizontal surface. The field completely fills the region occupied by the coil in either part of the drawing. (a) In which part of the drawing, if either, does a greater magnetic flux pass through the coil? Account for your answer. (b) As the shape of the coil changes from that in part \(a\) of the drawing to that in part \(b\), does an induced current flow in the coil, and, if so, in which direction does it flow? Give your reasoning. To describe the flow, imagine that you are above the coil looking down at it. (c) How is the period \(T\) of the rotational motion related to the angular frequency \(\omega\), and in terms of the period, what is the shortest time interval that elapses between parts \(a\) and \(b\) of the drawing? the magnitude of the magnetic field is \(0.35 \mathrm{~T}\). The resistance of the coil is \(0.025 \Omega\), and the smaller semicircle has a radius of \(0.20 \mathrm{~m}\). The angular frequency at which the small semicircle rotates is \(1.5 \mathrm{rad} / \mathrm{s}\). Determine the average current, if any, induced in the coil as the coil changes shape from that in part \(a\) of the drawing to that in \(\operatorname{part} b\).

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