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Chapter 29: Problem 13

The armature of a small generator consists of a flat, square coil with 120 turns and sides with a length of 1.60 \(\mathrm{cm}\) . The coil rotates in a magnetic fleld of 0.0750 T. What is the angular speed of the coil if the maximum emf produced is 24.0 \(\mathrm{mV} ?\)

Short Answer

Expert verified
The angular speed \( \omega \) is approximately 10.42 rad/s.

Step by step solution

01

Understand the Formula for Maximum EMF

The maximum electromotive force (emf) in a rotating coil is given by the formula: \( \varepsilon_{max} = NAB\omega \), where \( N \) is the number of turns, \( A \) is the area of the coil, \( B \) is the magnetic field strength, and \( \omega \) is the angular speed. We need to solve for \( \omega \).
02

Calculate the Area of the Coil

The area \( A \) of the square coil is found by squaring the length of its side: \( A = (0.0160 \, \text{m})^2 = 0.000256 \, \text{m}^2 \).
03

Substitute Known Values into EMF Equation

Substitute the given values into the formula for maximum emf: \( 24.0 \, \mathrm{mV} = (120)(0.000256 \, \text{m}^2)(0.0750 \, \mathrm{T})\omega \). Convert the emf to volts: \( 24.0 \, \mathrm{mV} = 0.0240 \, \mathrm{V} \).
04

Solve for Angular Speed \( \omega \)

Rearrange the equation to solve for \( \omega \): \( \omega = \frac{0.0240}{(120)(0.000256)(0.0750)} \). Calculate \( \omega \): \( \omega \approx 10.42 \, \text{rad/s} \).

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Key Concepts

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

Angular Speed
Angular speed, often symbolized as \( \omega \), is a measure of how fast an object rotates about an axis. It is expressed in radians per second (rad/s). To visualize this concept, imagine a spinning top. The faster it spins, the greater its angular speed. For any rotating object, the angular speed signifies how quickly it completes a rotation.
Angular speed is crucial in many real-world applications, notably in electricity generation. In the context of generators, such as the one in our exercise, angular speed determines how swiftly the coil inside the generator turns.
To calculate angular speed in this exercise, we used the formula for maximum electromotive force (emf):
\[ \varepsilon_{max} = NAB\omega \]
Here:
  • \( N \) is the number of turns in the coil.
  • \( A \) is the area of the coil.
  • \( B \) represents the magnetic field strength.
By solving for \( \omega \), we determined how fast the coil must rotate to achieve a specific emf.
Maximum EMF
The maximum electromotive force (emf) is an important concept in electromagnetic induction. It represents the highest voltage that a generator can produce when operating at optimal conditions. In simple terms, emf is the electrical "push" or voltage generated by a changing magnetic field.
In our exercise, the maximum emf is given to be 24.0 mV, which we converted to volts for calculation as 0.0240 V. To find the maximum emf from a coil, you can use the equation \( \varepsilon_{max} = NAB\omega \). Here:
  • \( N \) is the number of turns in the coil.
  • \( A \) is the area of the coil, calculated here using \( A = (0.0160 \, \text{m})^2 \).
  • \( B \) is the magnetic field strength.
  • \( \omega \) is the angular speed of the coil.
This formula highlights the relationship between these factors and how they contribute to the production of emf. Understanding maximum emf is essential as it influences how effective a generator is at converting mechanical energy into electrical energy.
Magnetic Field Strength
Magnetic field strength, denoted by \( B \), quantifies how strong a magnetic field is. In the International System of Units (SI), it is measured in teslas (T).
This physical quantity is essential in electromagnetic applications. In our exercise, we have a magnetic field strength of 0.0750 T. A stronger magnetic field can induce a higher maximum emf in a generator.
When a coil of wire rotates within a magnetic field, it "cuts" through the magnetic lines of force. This movement causes a change in magnetic flux through the coil. Due to electromagnetic induction, this change induces an emf.
The relationship between magnetic field strength and the generator's output is captured by the formula \( \varepsilon_{max} = NAB\omega \). This shows how a strong magnetic field, along with other factors like area and angular speed, impacts emf production. In understanding magnetic field strength, one appreciates its pivotal role in determining the performance of devices like motors and generators.

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Most popular questions from this chapter

Make a Generator? You are shipwrecked on a deserted tropical island. You have some electrical devices that you could uperate using a generalor but you have nu maguels. The eardis magnetic field at your location is horizontal and has magnitude 8.0 \(\times 10^{-5} \mathrm{T}\) , and you decide to try to use this field for a generator by rotating a large circular coil of wire at a high rate. You need to produce a peak emf of 9.0 \(\mathrm{V}\) and estimate that you can rotate the coil at 30 \(\mathrm{rpm}\) by turning a crank handle. You also decide that to have an acceptable coil resistance, the maximum mumber of turns the coil can have is 2000 . (a) What area must the coil have? (b) If the coil is circular, what is the maximum translational speed of a point on the coil as it rotates? Do you this device is feasible? Explain.

A closely wound search coil (Exercise 29.3) has an area of \(3.20 \mathrm{cm}^{2}, 120\) turns, and a resistance of \(60.0 \Omega .\) It is connected to a charge-measuring instrument whose resistance is 45.0\(\Omega\) . When the coil is rotated quickly from a position parallel to a uniform magnetic field to a position perpendicular to the field, the instrument indicates a charge of \(3.56 \times 10^{-5} \mathrm{C}\) . What is the magnitude of the field?

A metal ring 4.50 \(\mathrm{cm}\) in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic ficld. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.250 \(\mathrm{T} / \mathrm{s}\) (a) What is the magnitude of the electric field induced in the ring? (b) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?

A diclectric of permitivity \(3.5 \times 10^{-11} \mathrm{F} / \mathrm{m}\) completely fills the volume between two capacitor plates. For \(t > 0\) the electric flux through the dielectric is \(\left(8.0 \times 10^{3} \mathrm{V} \cdot \mathrm{m} / \mathrm{s}^{3}\right) t^{3}\) . The dielectric is ideal and nonmagnetic; the conduction current in the dielectric is zero. At what time does the displacement current in the dielectric equal 21\(\mu \mathrm{A} ?\)

In a physics laboratory experiment, a coil with 200 tums enclosing an area of 12 \(\mathrm{cm}^{2}\) is rotated in 0.040 s from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is \(6.0 \times 0^{-5} \mathrm{T}\) . (a) What is the total magnetic flux through the coil before it is rotated? After it is rotated? b) What is the average emf induced in the coil?
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