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A generator uses a coil that has 100 turns and a \(0.50-\) 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 length of wire used is 38 m.

Step by step solution

01

Identify Known Values

We know the following values from the problem:- Number of turns, \( N = 100 \)- Magnetic field, \( B = 0.50 \, \text{T} \)- Frequency, \( f = 60.0 \, \text{Hz} \)- rms voltage, \( V_{rms} = 120 \, \text{V} \)- Assume each turn is a square.
02

Calculate Maximum emf

The rms voltage \( V_{rms} \) is related to the maximum emf \( \varepsilon_{max} \) by the equation:\[ \varepsilon_{max} = V_{rms} \times \sqrt{2} \]Substituting the given value:\[ \varepsilon_{max} = 120 \, \text{V} \times \sqrt{2} = 120 \times 1.414 = 169.68 \, \text{V} \]
03

Use emf Equation for a Coil

The maximum emf for a coil rotating in a magnetic field is given by:\[ \varepsilon_{max} = NAB \times \omega \times \sin(\omega t) \]For maximum emf, \( \sin(\omega t) = 1 \), so:\[ \varepsilon_{max} = NAB \omega \]
04

Calculate Angular Velocity

The angular velocity \( \omega \) is related to the frequency \( f \) by the equation:\[ \omega = 2\pi f \]Substituting the frequency:\[ \omega = 2 \pi (60) = 120\pi \, \text{rad/s} \]
05

Calculate Area per Turn

Re-arrange the maximum emf equation to solve for area \( A \):\[ A = \frac{\varepsilon_{max}}{NB\omega} \]Substitute the known values:\[ A = \frac{169.68}{100 \times 0.50 \times 120\pi} = \frac{169.68}{6000\pi} \approx 0.0090 \, \text{m}^2 \]
06

Calculate Side Length of Square

Since each turn of the coil is a square, the area \( A \) is equal to the side length squared, \( A = s^2 \). Solve for \( s \):\[ s = \sqrt{0.0090} \approx 0.095 \, \text{m} \]
07

Calculate Total Length of Wire

The total length of wire \( L \) is the perimeter of one turn multiplied by the number of turns:\[ \text{Perimeter of one turn} = 4s = 4 \times 0.095 = 0.38 \, \text{m} \]\[ L = 100 \times 0.38 = 38 \, \text{m} \]

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Key Concepts

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

Understanding Magnetic Field in Generators
In the context of generator coil calculations, the magnetic field, often denoted as \( B \), is a crucial element. It represents the intensity of the magnetic field through which the coil moves. The unit of this magnetic field is the Tesla (T). In our problem setup, the given magnetic field strength is \( 0.50 \, \text{T} \).
This magnetic field is what induces the electromotive force (emf) in the coil as it rotates. As the coil moves through this magnetic environment, the interaction between the coil and magnetic field generates electricity. This process works on the principle of electromagnetic induction. Understanding the magnitude of the magnetic field helps us calculate the maximum possible induced voltage across the coil, which is essential for determining the generator's output. Here are some points to remember:
  • The stronger the magnetic field, the higher the potential for generated voltage.
  • Generators rely on the magnetic field to induce electricity in the coil.
RMS Voltage and Its Role
RMS voltage, denoted as \( V_{rms} \), stands for root mean square voltage. This is a measure of the effective value of the alternating current (AC) voltage produced by the generator. In terms of generator design, \( V_{rms} \) helps gauge the usable output voltage of the system. For our generator, \( V_{rms} \) is specified as \( 120 \, \text{V} \).
To find the peak or maximum voltage \( \varepsilon_{max} \), needed to solve the problems related to coil length, we use the relationship: \[ \varepsilon_{max} = V_{rms} \times \sqrt{2} \]This gives us the highest voltage output from the generator, which takes place at the peak of the alternating current cycle. Understanding \( V_{rms} \) and its conversion to maximum voltage is key in calculating parameters like the coil's area and length.
  • \( V_{rms} \) provides a way to express AC voltages as effective DC equivalent.
  • It's crucial for determining practical power outputs in AC circuits.
Angular Velocity Explained
Angular velocity, often marked by \( \omega \), describes how fast the generator's coil rotates in the magnetic field. It’s related to the frequency \( f \) with the formula:\[ \omega = 2\pi f \]For our exercise problem, this gives an angular velocity of \( 120\pi \, \text{rad/s} \). Angular velocity is a cornerstone in analyzing how effectively the generator converts mechanical rotation into electrical energy.
Applying this angular velocity in calculations allows us to determine the changes needed for obtaining the desired electrical output. It plays a direct role in the formula for maximum emf: \[ \varepsilon_{max} = NAB \omega \] Here, it directly influences the coil’s electrical output when combined with other factors like the number of turns \( N \) and the magnetic field \( B \).
  • The faster the angular velocity, the quicker the coil moves, increasing potential voltage production.
  • It directly impacts the frequency of the AC produced.
Coil Turns and Their Calculations
Coil turns, symbolized by \( N \), indicate how many loops or turns of wire are inside the generator coil. In this exercise, we have \( N = 100 \) turns. The number of turns is one of the primary factors in calculating the emf in a coil. When calculating generator components, the formula \[ \varepsilon_{max} = NAB \omega \] demonstrates how emf is proportionally affected by the number of turns.
More turns in a coil effectively mean more voltage can be generated, assuming other factors remain constant. Calculating the total length of the wire involves summing the perimeters of all the coil turns:
  • More turns can lead to a greater induced voltage.
  • It necessitates longer wire length and more material.
  • Each turn contributes to the overall length calculation, and by considering them as squares, we simplify the calculation.
Overall, the coil turns are critical to understanding and manipulating the output efficiency of a generator.

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Most popular questions from this chapter

A \(5.40 \times 10^{-5} \mathrm{H}\) solenoid is constructed by wrapping 65 turns of wire around a cylinder with a cross-sectional area of \(9.0 \times 10^{-4} \mathrm{m}^{2} .\) When the solenoid is shortened by squeezing the turns closer together, the inductance increases to \(8.60 \times 10^{-5}\) H. Determine the change in the length of the solenoid.

Near San Francisco, where the vertically downward component of the earth's magnetic field is \(4.8 \times 10^{-5} \mathrm{T}, \mathrm{a}\) car is traveling forward at \(25 \mathrm{m} / \mathrm{s}\) The width of the car is \(2.0 \mathrm{m}\). (a) Find the emf induced between the two sides of the car. (b) Which side of the car is positive - the driver's side or the passenger's side?

A flat coil of wire has an area \(A, N\) turns, and a resistance \(R\). It is situated in a magnetic field, such that the normal to the coil is parallel to the magnetic field. The coil is then rotated through an angle of \(90^{\circ},\) so that the normal becomes perpendicular to the magnetic field. The coil has an area of \(1.5 \times 10^{-3} \mathrm{m}^{2}, 50\) turns, and a resistance of \(140 \Omega .\) During the time while it is rotating, a charge of \(8.5 \times 10^{-5} \mathrm{C}\) flows in the coil. What is the magnitude of the magnetic field?

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?

Mutual induction can be used as the basis for a metal detector. A typical setup uses two large coils that are parallel to each other and have a common axis. Because of mutual induction, the ac generator connected to the primary coil causes an emf of \(0.46 \mathrm{V}\) to be induced in the secondary coil. When someone without metal objects walks through the coils, the mutual inductance and, thus, the induced emf do not change much. But when a person carrying a handgun walks through, the mutual inductance increases. The change in emf can be used to trigger an alarm. If the mutual inductance increases by a factor of three, find the new value of the induced emf.

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