/*! 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 13 The armature of a small generato... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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 field 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 of the coil is approximately 10 rad/s.

Step by step solution

01

Understand the Problem

We need to find the angular speed of the coil in a generator. We're given the number of turns (120), side length of the coil (1.60 cm), the magnetic field strength (0.0750 T), and the maximum electromotive force (emf) produced (24.0 mV). We are required to find the angular speed which results in this maximum emf.
02

Convert Units if Necessary

Convert the side length of the coil from centimeters to meters to get uniform units for calculations. Since 1 cm = 0.01 m, the side length is 1.60 cm \( = 0.0160 \) m. Convert the emf from millivolts to volts: 24.0 mV = 0.024 V.
03

Calculate the Area of One Turn of the Coil

The coil is square, so its area \( A \) is \( 0.0160 \text{ m} \times 0.0160 \text{ m} = 0.000256 \text{ m}^2 \).
04

Use the Formula for Maximum EMF

The formula for the maximum emf \( \epsilon_{max} \) induced in a rotating coil is \( \epsilon_{max} = NAB \omega \). Where \( N = 120 \) is the number of turns, \( A = 0.000256 \text{ m}^2 \) is the area, \( B = 0.0750 \text{ T} \) is the magnetic field strength, and \( \omega \) is the angular speed. We need to solve for \( \omega \).
05

Rearrange and Solve the Equation

Rearrange the formula to solve for \( \omega \): \( \omega = \frac{\epsilon_{max}}{NAB} \). Plug in the values: \( \omega = \frac{0.024 \text{ V}}{120 \times 0.000256 \text{ m}^2 \times 0.0750 \text{ T}} \).
06

Calculate the Angular Speed

Carry out the calculations. \( \omega = \frac{0.024}{120 \times 0.000256 \times 0.0750} \approx 10 \text{ rad/s} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Angular Speed
Angular speed is a measure of how fast an object rotates or spins. It tells us how many radians an object rotates through per second. In the context of electromagnetic induction, understanding angular speed is crucial as it directly impacts the electromotive force (emf) generated by a rotating coil. The angular speed is typically denoted by the symbol \( \omega \), and measured in radians per second (rad/s).

In a generator, a coil rotates within a magnetic field. The angular speed of the coil determines how often the magnetic lines of force cross the coil, which directly influences the magnitude of the induced emf. A higher angular speed means more frequent changes in the magnetic field experienced by the coil, leading to a larger emf.

Understanding angular speed can help in designing rotating equipment used in generating electricity. Operators can adjust the angular speed to control the output voltage, optimizing performance based on specific requirements of the system.
Magnetic Field
A magnetic field is an invisible force field surrounding magnets and current-carrying wires, and it exerts force on other nearby magnets and currents. It is represented by the symbol \( B \), and its strength is measured in teslas (T).

In the context of our exercise, the magnetic field is a critical component as it interacts with the rotating coil in the generator. The stronger the magnetic field, the greater the force exerted on the moving charges within the coil. This is essential for inducing an emf, as changes in the magnetic field with respect to the coil generate electricity.

The magnetic field's direction also matters, as it determines the direction of the induced current in the coil. By utilizing this field effectively, devices like electric generators can convert mechanical energy into electrical energy efficiently. Understanding how the magnetic field operates helps designers and engineers optimize the design of electromagnetic devices, ensuring they work effectively and are safe to use.
Electromotive Force (emf)
Electromotive force (emf) is the energy per unit charge produced by a source of electric power. It is the potential difference that drives electric current around a circuit, even though it is not an actual force. Denoted by \( \epsilon \), it is usually measured in volts (V).

In electromagnetic induction, emf is generated when a conductor, such as a coil, experiences a changing magnetic field. In our context, a rotating coil in a generator moves through a magnetic field, which induces an emf. The maximum emf is produced when the coil is rotating at its optimal angular speed and positioned for maximum change in magnetic flux.

The formula for calculating maximum emf involves the number of turns \( N \), the area of the coil \( A \), the magnetic field strength \( B \), and the angular speed \( \omega \):
  • \( \epsilon_{max} = NAB\omega \)
Understanding and calculating emf allows for precise design in devices that harness electrical energy, ensuring efficient operation and energy conversion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

CALC In a region of space, a magnetic field points in the \(+x\) -direction (toward the right). Its magnitude varies with position according to the formula \(B_{x}=B_{0}+b x,\) where \(B_{0}\) and \(b\) are positive constants, for \(x \geq 0 .\) A flat coil of area \(A\) moves with uniform speed \(v\) from right to left with the plane of its area always perpendicular to this field. (a) What is the emf induced in this coil while it is to the right of the origin? (b) As viewed from the origin, what is the direction (clockwise or counterclockwise) of the current induced in the coil? (c) If instead the coil moved from left to right, what would be the answers to parts (a) and (b)?

The compound \(\mathrm{SiV}_{3}\) is a type-II superconductor. At temperatures near absolute zero the two critical fields are \(B_{\mathrm{cl}}=55.0 \mathrm{mT}\) and \(B_{\mathrm{c} 2}=15.0 \mathrm{T}\) . The normal phase of \(\mathrm{Si} \mathrm{V}_{3}\) has a magnetic susceptibility close to zero. A long, thin \(\mathrm{SiV}_{3}\) cylinder has its axis parallel to an external magnetic field \(\vec{\boldsymbol{B}}_{0}\) in the \(+x\) -direction. At points far from the ends of the cylinder, by symmetry, all the magnetic vectors are parallel to the \(x\) -axis. At a temperature near absolute zero, the external magnetic field is slowly increased from zero. What are the resultant magnetic field \(\vec{\boldsymbol{B}}\) and the magnetization \(\vec{M}\) inside the cylinder at points far from its ends (a) just before the magnetic flux begins to penetrate the material, and (b) just after the material becomes completely normal?

The magnetic field within a long, straight solenoid with a circular cross section and radius \(R\) is increasing at a rate of \(d B / d t\) . (a) What is the rate of change of flux through a circle with radius \(r_{1}\) inside the solenoid, normal to the axis of the solenoid, and with center on the solenoid axis? (b) Find the magnitude of the induced electric field inside the solenoid, at a distance \(r_{1}\) from its axis. Show the direction of this field in a diagram. (c) What is the magnitude of the induced electric field outside the solenoid, at a distance \(r_{2}\).from the axis? (d) Graph the magnitude of the induced electric field as a function of the distance \(r\) from the axis from \(r=0\) to \(r=2 R\) (e) What is the magnitude of the induced emf in a circular turn of radius \(R / 2\) that has its center on the solenoid axis? (f) What is the magnitude of the induced emf if the radius in part (e) is \(R ?\) (g) What is the induced emf if the radius in part (e) is 2R?

CP CALC A capacitor has two parallel plates with area \(A\) separated by a distance \(d .\) The space between plates is filled with a material having dielectric constant \(K\) . The material is not a perfect insulator but has resistivity \(\rho .\) The capacitor is initially charged with charge of magnitude \(Q_{0}\) on each plate that gradually discharges by conduction through the dielectric. (a) Calculate the conduction current density \(j_{C}(t)\) in the dielectric. (b) Show that at any instant the displacement current density in the dielectric is equal in magnitude to the conduction current density but opposite in direction, so the total current density is zero at every instant.

CALC A very long, straight solenoid with a cross-sectional area of 2.00 \(\mathrm{cm}^{2}\) is wound with 90.0 turns of wire per centimeter. Starting at \(t=0,\) the current in the solenoid is increasing according to \(i(t)=\left(0.160 \mathrm{A} / \mathrm{s}^{2}\right) t^{2}\) . A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.20 \(\mathrm{A}\) ?
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Fields in Physics

Read Explanation

Measurements

Read Explanation

Rotational Dynamics

Read Explanation

Torque and Rotational Motion

Read Explanation

Particle Model of Matter

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.