/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 The maximum strength of the eart... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The coil needs at least 1361 turns.

Step by step solution

01

Understand the Problem

We need to find out how many turns are required in a coil to produce an RMS voltage of 120 V using a magnetic field of strength \(6.9 \times 10^{-5} \, \text{T}\) and an area per turn of \(0.022 \, \text{m}^2\). The frequency is given as 60 Hz.
02

Relevant Equation for the EMF

The RMS voltage (\(V_{rms}\)) for an alternating EMF in a coil is given by the formula \(V_{rms} = \frac{1}{\sqrt{2}} \times N \times A \times B \times \omega\), where \(N\) is the number of turns, \(A\) is the area per turn, \(B\) is the magnetic field strength, and \(\omega\) is the angular frequency.
03

Calculate Angular Frequency

The angular frequency \(\omega\) can be determined by the formula \(\omega = 2\pi f\), where \(f\) is the frequency. In this problem, \(f = 60 \, \text{Hz}\). Thus, \(\omega = 2\pi \times 60\).
04

Rearrange the Formula to Solve for N

Rearrange the formula to solve for \(N\): \[N = \frac{V_{rms} \times \sqrt{2}}{A \times B \times \omega}\]
05

Plug in Values and Compute N

Substitute the known values into the equation: \[N = \frac{120 \times \sqrt{2}}{0.022 \times 6.9 \times 10^{-5} \times 2 \pi \times 60}\]Calculate \(N\) to find the minimum number of turns needed.

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Key Concepts

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

Earth's Magnetic Field
Earth's magnetic field, often called the geomagnetic field, is an invisible force field that protects the Earth from solar wind and cosmic radiation. It extends from the Earth's interior out into space. This magnetic field is what makes a compass point north and helps birds migrate along geomagnetic lines. Near the south magnetic pole, the field strength can reach up to \(6.9 \times 10^{-5} \, \text{T}\) (Tesla), which might not seem much, but is significant when it comes to electromagnetic applications. In many educational settings, students explore how it can be used to generate electricity by utilizing principles related to electromagnetic induction. The uniform nature of this field allows scientists and engineers to predict and manipulate its effects reliably to develop innovative technologies, including renewable electric power systems.
AC Electricity Generation
Alternating Current (AC) electricity generation is a core concept in physics and electrical engineering. It involves converting mechanical energy into electrical energy through the rotation of coils within a magnetic field. As these coils spin within the magnetic field, they cut through magnetic lines of force, inducing an electric current. The principle governing this is known as electromagnetic induction. The frequency of AC is typically measured in hertz (Hz), with a common value being 60 Hz in some regions, including the United States. This means the current changes direction 60 times per second. This alternation is beneficial due to its ease of transformation to different voltages, which makes it ideal for long-distance power transmission. Students often learn to calculate the various parameters of AC electricity by applying formulas that describe the voltage or current in terms of mechanical rotation and coil characteristics.
RMS Voltage
RMS voltage, or Root Mean Square voltage, is a way of expressing the effective value of an alternating voltage. For an AC system, it provides a measure that is comparable to a constant DC voltage in terms of the power delivered. Mathematically, the RMS voltage is calculated by taking the average of the squared values of the instantaneous voltage over a cycle, then taking the square root of this average. This complex-sounding calculation results in a simple formula in practical scenarios:
  • \(V_{rms} = \frac{V_{peak}}{\sqrt{2}}\)
  • In some systems, known parameters can be used directly to solve for or apply \(V_{rms}\) in electric power equations.
RMS voltage allows engineers and students to evaluate AC circuits more effectively by understanding how much work a particular voltage can perform over time, similar to DC systems.
Coil Turns Calculation
The number of turns in a coil is a crucial factor in determining the induced electromotive force (EMF) in AC generation. Each turn of the coil can be thought of as a loop that interacts with the magnetic field. Depending on the number of loops, the effective cutting of magnetic lines of force can be modulated, affecting the voltage generated by the system. The core formula for calculating the required coil turns (\(N\)) to achieve a desired RMS voltage involves substituting known variables such as area per turn (\(A\)), magnetic field strength (\(B\)), and angular frequency (\(\omega\)):
  • \(N = \frac{V_{rms} \times \sqrt{2}}{A \times B \times \omega}\)
This formula demonstrates how adjusting the coil specifications impacts the overall generated voltage, allowing design flexibility for various electrical applications. By manipulating these factors, engineers can tailor coil systems for specific voltage outputs, addressing the efficiency and efficacy needs based on available magnetic field strengths like that of Earth's.

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Most popular questions from this chapter

A generator has a square coil consisting of 248 turns. The coil rotates at \(79.1 \mathrm{rad} / \mathrm{s}\) in a \(0.170\) -T magnetic field. The peak out put of the generator is \(75.0 \mathrm{~V}\). What is the length of one side of the coil?

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 \(0.15 \mathrm{~V}\). What is its magnitude in part \(b\) ?

A generator is connected across the primary coil \(\left(N_{\mathrm{p}}\right.\) 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 \(R_{1}=\left(N_{\mathrm{p}} / N_{\mathrm{s}}\right)^{2} R_{2},\) by starting with Ohm's law as applied to the secondary coil.

The rechargeable batteries for a laptop computer need a much smaller voltage than what a wall socket provides. Therefore, a transformer is plugged into the wall socket and produces the necessary voltage for charging the batteries. (a) Is the transformer a step-up or a step-down transformer? (b) Is the current that goes through the batteries greater than, equal to, or smaller than the current coming from the wall socket? (c) If the transformer has a negligible resistance, is the electric power delivered to the batteries greater than, equal to, or less than the power coming from the wall socket? In all cases, provide a reason for your answer. the batteries of a laptop computer are rated at \(9.0 \mathrm{~V}\), and a current of \(225 \mathrm{~mA}\) is used to charge them. The wall socket provides a voltage of \(120 \mathrm{~V}\). (a) Determine the turns ratio of the transformer, (b) What is the current coming from the wall socket? (c) Find the power delivered by the wall socket and the power sent to the batteries. Be sure your answers are consistent with your answers to the Concept Questions.

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?

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