/*! 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 7 At the instant when the current ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 30: Problem 7

At the instant when the current in an inductor is increasing at a rate of 0.0640 \(\mathrm{A} / \mathrm{s}\) , the magnitude of the self-induced emf is 0.0160 \(\mathrm{V}\) . (a) What is the inductance of the inductor? (b) If the inductor is a solenoid with 400 turns, what is the average magnetic flux through each turn when the current is 0.720 \(\mathrm{A} ?\)

Short Answer

Expert verified
(a) Inductance is 0.25 H. (b) Average magnetic flux is 0.00045 Wb.

Step by step solution

01

Identify Known Values

The rate of change of current \( \frac{di}{dt} = 0.0640 \, \text{A/s} \) and the self-induced emf \( \varepsilon = 0.0160 \, \text{V} \) are given. We need to find the inductance \( L \). We also have a solenoid with 400 turns, and the current is \( i = 0.720 \, \text{A} \).
02

Use The Formula for Inductance

The magnitude of the self-induced emf is given by the formula \( \varepsilon = L \frac{di}{dt} \). We need to solve for \( L \).
03

Solve for Inductance \( L \)

Rearrange the formula \( \varepsilon = L \frac{di}{dt} \) to \( L = \frac{\varepsilon}{\frac{di}{dt}} \). Substitute the known values: \( L = \frac{0.0160}{0.0640} \).
04

Calculate Inductance \( L \)

Perform the division: \( L = \frac{0.0160}{0.0640} = 0.25 \, \text{H} \). Thus, the inductance of the inductor is \( 0.25 \, \text{H} \).
05

Find Formula for Magnetic Flux

The magnetic flux \( \Phi \) through one turn is related to inductance by \( \Phi = \frac{L \cdot i}{N} \), where \( N \) is the number of turns and \( i \) is the current.
06

Calculate Magnetic Flux

Substitute the known values into the formula: \( \Phi = \frac{0.25 \cdot 0.720}{400} \).
07

Perform Calculation for \( \Phi \)

Calculate \( \frac{0.25 \cdot 0.720}{400} = 0.00045 \, \text{Wb} \). Thus, the average magnetic flux through each turn is \( 0.00045 \, \text{Wb} \).

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.

Self-induced emf
When we talk about self-induced emf, we're diving into the fascinating world of electromagnetism. This concept revolves around the idea that an electric current flowing through an inductor can induce an electromotive force (emf) within itself. This phenomenon occurs due to the changing magnetic field as the current varies.
  • When the current flowing through an inductor changes, it creates a changing magnetic field.
  • This change in the magnetic field induces a voltage, known as the self-induced emf.
  • The direction of this induced emf always opposes the change in current, following Lenz's Law.
The relationship between self-induced emf (\( \varepsilon \)) and inductance (\( L \)) is given by the formula:
\( \varepsilon = L \frac{di}{dt} \)
where \( \frac{di}{dt} \) is the rate of change of current. This formula implies that higher inductance or faster changes in current result in a greater self-induced emf.
Solenoid
A solenoid is one of the most common inductor forms. It's essentially a coil of wire, and when an electric current passes through, it produces a magnetic field.
  • They are used in creating magnetic fields for a wide range of applications, from electronic circuits to electromagnets.
  • The magnetic field inside a solenoid is often uniform and parallel to the axis of the coil when the solenoid is long compared to its diameter.
  • The field strength can be increased by adding more turns of wire or increasing the current.
In our exercise, a solenoid with 400 turns is used. The inductance of a solenoid depends on factors such as the number of turns, the cross-sectional area, and the length of the coil. By rearranging the windings tightly, a stronger magnetic field is achieved, enhancing its ability to store energy in the magnetic field.
Magnetic Flux
Magnetic flux refers to the number of magnetic field lines passing through a given area. It's a critical concept when discussing how inductors work.
  • The flux, often denoted by \( \Phi \), is a measure of the strength and distribution of a magnetic field over a certain area.
  • It's measured in Webers (Wb).
  • In a solenoid, the average magnetic flux through each loop can be calculated by using the formula: \( \Phi = \frac{L \cdot i}{N} \)
Here, \( L \) is the inductance, \( i \) is the current, and \( N \) is the number of turns. By applying this formula, one can determine the average magnetic flux through each loop, showing how the energy is stored in the magnetic field within the solenoid. Understanding magnetic flux helps us grasp the dynamics of energy transformation and storage within inductive systems.

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

A toroidal solenoid has 500 turns, cross-sectional area \(6.25 \mathrm{cm}^{2},\) and mean radius \(4.00 \mathrm{cm} .\) (a) Calcualte the coil's self-inductance. (b) If the current decreases uniformly from 5.00 \(\mathrm{A}\) to 2.00 \(\mathrm{A}\) in 3.00 \(\mathrm{ms}\) , calculate the self- induced emf in the coil. (c) The current is directed from terminal \(a\) of the coil to terminal \(b\) . Is the direction of the induced emf from \(a\) to \(b\) or from \(b\) to \(a ?\)

An Electromagnetic CarAlarm. Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3500 \(\mathrm{Hz}\) . To do this, the car-alarm circuitry must produce an altermating electric current of the same frequency. That's why your design includes an inductor and a capacitor in series. The maximum voltage across the capacitor is to be 12.0 \(\mathrm{V}\) (the same voltage as the car battery). To produce a sufficiently loud sound, the capacitor must store 0.0160 \(\mathrm{J}\) of energy. What values of capacitance and inductance should you choose for your car-alarm circuit?

An air-filled toroidal solenoid has a mean radius of 15.0 \(\mathrm{cm}\) and a cross-sectional area of \(5.00 \mathrm{cm}^{2} .\) When the current is 12.0 \(\mathrm{A}\) , the energy stored is 0.390 \(\mathrm{J}\) . How many turns does the winding have?

A \(0.250-\mathrm{H}\) inductor carries a time-varying current given by the expression \(i=(124 \mathrm{mA}) \cos [(240 \pi / \mathrm{s}) t] .\) (a) Find an expression for the induced emf as a function of time. Graph the current and induced emf as functions of time for \(t=0\) to \(t=\frac{1}{60} \mathrm{s}\) . (b) What is the maximum emf? What is the current when the induced emf is a maximum? (c) What is the maximum current? What is the induced emf when the current is a maximum?

An inductor used in a de power supply has an inductance of 12.0 \(\mathrm{H}\) and a resistance of \(180 \Omega .\) It carries a current of 0.300 \(\mathrm{A}\) . (a) What is the energy stored in the magnetic field? (b) At what rate is thermal energy developed in the inductor?(c) Does your answer to part (b) mean that the magnetic-field energy is decreasing with time? Explain.
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Waves Physics

Read Explanation

Modern Physics

Read Explanation

Electromagnetism

Read Explanation

Work Energy and Power

Read Explanation

Measurements

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.