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At the instant when the current in an inductor is increasing at a rate of 0.0640 A/s, the magnitude of the self-induced emf is 0.0160 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 A?

Short Answer

Expert verified
(a) Inductance is 0.25 H. (b) Magnetic flux is 0.00045 Wb per turn.

Step by step solution

01

Understand the Problem and Identify Formulas

To solve for the inductance \( L \), we have the formula for the self-induced emf, \( \epsilon = -L \frac{dI}{dt} \), where \( \frac{dI}{dt} \) is the rate of change of current. We need to calculate \( L \) since \( \epsilon = 0.0160 \) V and \( \frac{dI}{dt} = 0.0640 \) A/s.
02

Solve for the Inductance

Rearrange the formula for self-induced emf to solve for \( L \): \( L = \frac{-\epsilon}{\frac{dI}{dt}} \). Substitute the given values: \( L = \frac{-0.0160}{0.0640} = -0.25 \). The inductance \( L \) is 0.25 H (since the value of the inductance is generally considered as positive).
03

Set Up for Flux Calculation in a Solenoid

To find the average magnetic flux \( \Phi \) through each turn when the current is 0.720 A, use the relation \( \Phi = \frac{L \cdot I}{N} \), where \( L \) is the inductance, \( I \) is the current, and \( N \) is the number of turns in the solenoid.
04

Calculate the Magnetic Flux

Substitute the known values into the formula: \( \Phi = \frac{0.25 \times 0.720}{400} \). Simplify to find \( \Phi = 0.00045 \text{ Wb (Webers)} \) as the average magnetic flux through each turn when the current is 0.720 A.

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Key Concepts

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

Self-Induced EMF
Self-induced EMF is a phenomenon in which an electromotive force (EMF) is generated within a circuit due to the change in current flowing through it. This occurs because the changing current produces a varying magnetic field, which induces a voltage back into the circuit itself. This concept is described by Faraday's Law of Electromagnetic Induction.
  • This induced EMF opposes the change in current because of Lenz's law.
  • The formula used to calculate self-induced EMF is \( \epsilon = -L \frac{dI}{dt} \), where \( \epsilon \) is the EMF, \( L \) is the inductance, and \( \frac{dI}{dt} \) is the rate of change of current.
A key point to remember is that the negative sign in the formula indicates opposition to changes in the current flow. Understanding how self-induced EMF operates helps in designing circuits, especially those containing inductors.
Inductance
Inductance is a measure of how effective an inductor is at inducing an EMF due to a change in current. It represents the ability of an inductor to store energy in its magnetic field.
  • Inductance \( L \) is measured in henries (H).
  • The larger the inductance, the greater the ability to oppose changes in current.
The inductance is affected by factors such as:
  • The number of turns in the coil or solenoid.
  • The cross-sectional area of the coil.
  • The material of the core around which the coil is wound.
In practical applications, inductors with high inductance are used to control the flow of AC or store energy in some forms of power supplies.
Magnetic Flux
Magnetic flux \( \Phi \) quantifies the total magnetic field passing through a given area, often measured in webers (Wb). It's a fundamental concept in electromagnetics to describe the strength and orientation of the magnetic field through a surface.
  • Magnetic flux is calculated using the formula \( \Phi = \frac{L \cdot I}{N} \), where \( L \) is the inductance, \( I \) is the current, and \( N \) is the number of turns.
For a solenoid, the magnetic flux through each loop is relevant in understanding how effectively it can store energy. Changes in magnetic flux are crucial for inducing EMF in nearby circuits, emphasizing its role in transformers and induction processes.
Solenoid
A solenoid is a type of inductor arranged in a helical coil. It generates a magnetic field when an electric current flows through it, and is commonly used in various applications such as electromagnets, inductors, and motors.
  • The strength of the magnetic field created by a solenoid is largely dependent on the number of turns and the current passing through it.
  • A solenoid with a larger number of turns or a greater current will have a stronger magnetic field.
Solenoids are essential in electromagnetic systems where it is crucial to control the magnetic field, such as in switches and relays. The magnetic field inside a solenoid is generally uniform and concentrated along the axis, making them ideal for controlled induction applications.

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Most popular questions from this chapter

An inductor used in a dc power supply has an inductance of 12.0 H and a resistance of 180 \(\Omega\). It carries a current of 0.500 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.

When the current in a toroidal solenoid is changing at a rate of 0.0260 A/s, the magnitude of the induced emf is 12.6 mV. When the current equals 1.40 A, the average flux through each turn of the solenoid is 0.00285 Wb. How many turns does the solenoid have?

A 2.50-mH toroidal solenoid has an average radius of 6.00 cm and a cross- sectional area of 2.00 cm\(^2\). (a) How many coils does it have? (Make the same assumption as in Example 30.3.) (b) At what rate must the current through it change so that a potential difference of 2.00 V is developed across its ends?

Two coils have mutual inductance \(M = 3.25 \times 10^{-4}\) H. The current \(i_1\) in the first coil increases at a uniform rate of 830 A/s. (a) What is the magnitude of the induced emf in the second coil? Is it constant? (b) Suppose that the current described is in the second coil rather than the first. What is the magnitude of the induced emf in the first coil?

A long, straight solenoid has 800 turns. When the current in the solenoid is 2.90 A, the average flux through each turn of the solenoid is \(3.25 \times 10^{-3}\) Wb. What must be the magnitude of the rate of change of the current in order for the self-induced emf to equal 6.20 mV?

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