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

Short Answer

Expert verified
The side length of the coil is approximately 0.15 meters.

Step by step solution

01

Understand the Problem

We need to find the length of one side of the square coil. We have been given the number of turns (248), the angular velocity (79.1 rad/s), the magnetic field strength (0.170 T), and the peak voltage output (75.0 V). The formula for the peak voltage ( E_{peak}) of a rotating loop in a magnetic field is given by: e_{peak} = NABω where N is the number of turns, A is the area of the coil, B is the magnetic field, and ω is the angular velocity. Since the coil is square, A = l², where l is the length of one side of the coil.
02

Plug in Known Values

We will substitute the known values into the equation E_{peak} = NABω to solve for the area first.\(75.0 = 248 \times (l^2) \times 0.170 \times 79.1\)
03

Simplify the Equation

First, rearrange the equation to solve for l²:\(l^2 = \frac{75.0}{248 \times 0.170 \times 79.1}\)Now calculate the denominator:\(248 \times 0.170 \times 79.1 = 3335.656\)So,\(l^2 = \frac{75.0}{3335.656}\)
04

Solve for l² and then for l

Calculate the value of l²:\(l^2 = \frac{75.0}{3335.656} \approx 0.022482\)Now, solve for l:\(l = \sqrt{0.022482} \approx 0.14994\) meters.
05

Conclusion

The length of one side of the coil, given the peak output voltage and other parameters, is approximately 0.15 meters.

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Key Concepts

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

Understanding Magnetic Field
The concept of a magnetic field is fundamental when analyzing how generators work. In simple terms, a magnetic field is a region where a magnetic force is exerted on charges or another magnet. We often visualize this as the area around a magnet where magnetic forces can be experienced. In the case of an electric generator, the magnetic field plays a crucial role in inducing voltage.

Typically, the strength of a magnetic field is measured in Tesla (T), which represents how intensely the field can interact with electric charges. In our example, a magnetic field strength of 0.170 T is used, which affects how the coil interacts with the movement and generates electricity.

Key points to remember about magnetic fields in generators:
  • They provide the necessary force to move electrons within the wire coils.
  • The strength of the field directly influences the voltage generated.
Exploring Angular Velocity
Angular velocity ( ω ) is a measure of how fast an object rotates or spins around a particular axis and is usually given in radians per second (rad/s). In generators, understanding angular velocity is crucial since it impacts the amount of voltage induced in the coil.

Angular velocity is not only about the speed of rotation but also indicates the rate of change of angular position. For a generator coil rotating in a magnetic field, high angular velocity can mean more frequent cutting of magnetic field lines, thereby inducing more voltage.

In our exercise, the angular velocity of the coil is given as 79.1 rad/s. With this value, we calculated the peak voltage by determining how effectively the coil cuts through magnetic field lines at this rate.

Benefits of knowing angular velocity:
  • Allows prediction of the generated voltage for different rotational speeds.
  • Helps fine-tune generator performance by adjusting rotation rates.
Understanding Peak Voltage
Peak voltage (E_{peak}) is the maximum voltage output from a generator, observed when conditions are optimal for electrical generation. It can be thought of as the highest voltage point achieved during a cycle of the generator's operation.

For alternating current (AC) generators like the one described in our example, this is a crucial measurement because it signifies the generator's capacity to produce electricity under perfect conditions. Here, the peak voltage is given as 75.0 V.

Calculating peak voltage involves using the formula:
\[ e_{peak} = NABω \]
where:
  • N: Total number of coil turns
  • A: Area of the coil
  • B: Magnetic field strength
  • ω: Angular velocity
By knowing these values, one can determine the maximum power output capability of the generator.
Calculating Coil Area
The area of the coil closely associates with the overall output and efficiency of a generator. For a square coil, the area can be expressed as the side length squared (A = l^2). Determining the coil's area is vital as it affects how much magnetic field each section of the wire looped inside the coil can occupy, thus directly influencing the generator's output.

In our solved problem, to find the side length of the coil, we rearranged the peak voltage equation:
\[ e_{peak} = NABω \]
Plugging in all other known values allowed us to solve for the coil area (A), subsequently helping us find the side:\[ l^2 = \frac{75}{3335.656} \approx 0.022482 \]
\[ l \approx 0.14994 \] meters.

Understanding coil area:
  • It's a crucial factor in the induced voltage's strength and efficiency.
  • Helps determine physical dimensions needed for specific voltage outputs.

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Most popular questions from this chapter

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?

A \(120.0-\mathrm{V}\) motor draws a current of \(7.00 \mathrm{A}\) when running at normal speed. The resistance of the armature wire is \(0.720 \Omega .\) (a) Determine the back emf generated by the motor. (b) What is the current at the instant when the motor is just turned on and has not begun to rotate? (c) What series resistance must be added to limit the starting current to \(15.0 \mathrm{A} ?\)

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 vacuum cleaner is plugged into a \(120.0-\mathrm{V}\) socket and uses \(3.0 \mathrm{A}\) of current in normal operation when the back emf generated by the electric motor is \(72.0 \mathrm{V}\). Find the coil resistance of the motor.

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.

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