/*! 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 129 When the current changes from \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 129

When the current changes from \(+2 \mathrm{~A}\) to \(-2 \mathrm{~A}\) in \(0.05\) second, an EMF of \(8 \mathrm{~V}\) is induced in a coil. The coefficient of self-induction of the coil is (A) \(0.2 \mathrm{H}\) (B) \(0.4 \mathrm{H}\) (C) \(0.8 \mathrm{H}\) (D) \(0.1 \mathrm{H}\)

Short Answer

Expert verified
The coefficient of self-induction of the coil is \(0.1 H\) (option D). This is determined by using Faraday's law of electromagnetic induction to calculate the induced EMF in the coil and solving for the coefficient of self-induction.

Step by step solution

01

Write down the known values

We have the following values: - Initial current, \(I_1 = +2 A\) - Final current, \(I_2 = -2 A\) - Time interval, \(\Delta t = 0.05 s\) - Induced EMF, \(E = 8 V\)
02

Calculate change in current

We can calculate the change in current as follows: \(\Delta I = I_2 - I_1\) Substitute the values and calculate \(\Delta I\): \(\Delta I = -2 A - (+2 A) = -4 A\)
03

Use Faraday's law of electromagnetic induction

Faraday's law states that the induced EMF (\(E\)) is equal to the negative rate of change of the magnetic flux. In terms of self-induction, we can write it as follows: \(E = -L \frac{\Delta I}{\Delta t}\) Where \(L\) is the coefficient of self-induction, which we want to find.
04

Solve for L

Rearrange the Faraday's law equation to get the value of \(L\): \(L = -\frac{E}{\frac{\Delta I}{\Delta t}}\) Now, substitute the known values and find the value of L: \(L = -\frac{8 V}{\frac{-4 A}{0.05 s}}\) \(L = \frac{8 V}{80 A/s}\) \(L = 0.1 H\) So, the coefficient of self-induction of the coil is 0.1 H, which corresponds to option (D).

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-induction
When we talk about self-induction in an electrical circuit, we're referring to the phenomenon where a changing current in a coil induces an electromotive force (EMF) in the same coil. This happens due to the magnetic field that the current generates.

Here's how it works:
  • The coil reacts to any change in the current flowing through it by producing its own EMF, which opposes the change in current.
  • This opposition is guided by Lenz's Law, which helps conserve the energy in the circuit by opposing the change that caused it.
This opposition is what we call self-inductance, and it is expressed in henrys (H).

The coefficient of self-induction, often simply referred to as inductance, is a measure of how effective the coil is at generating this induced EMF. The larger the inductance, the stronger the induced EMF for a given rate of change of current.
Faraday's law
Faraday's law of electromagnetic induction is the cornerstone of understanding how currents can generate magnetic fields and vice versa. It states that a change in the magnetic environment of a coil will induce an EMF in the coil.

Here's the essence of Faraday's law:
  • An induced EMF is produced in a coil whenever there is a change in the magnetic flux.
  • The equation representing Faraday's law is: \[E = - \frac{d\Phi}{dt}\] where \(E\) is the induced EMF, and \(\Phi\) is the magnetic flux.
  • In cases of self-induction, this principle is applied to changes in current, with the formula \[E = -L \frac{\Delta I}{\Delta t}\] where \(L\) is the self-inductance.
The negative sign in Faraday's law ensures that the direction of the induced EMF opposes the change in current that caused it, in line with Lenz's Law.
induced EMF
Induced electromotive force (EMF) is a fundamental concept in electromagnetism, defined as the voltage produced by a change in magnetic flux in an electrical circuit.

Here's how induced EMF plays a role:
  • Changing magnetic fields within a coil, whether caused by varying current or an external magnetic influence, lead to the generation of an EMF.
  • This EMF is what drives current through the circuit even when no external voltage source is connected. It "induces" a current flow.
In scenarios involving self-induction, such as the exercise you solved, the induced EMF is a response to changes in the circuit's own current.

Understanding induced EMF within the framework of Faraday鈥檚 Law helps highlight how current and magnetic fields interact, as well as how circuits maintain energy balance through induced resistances.

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 flat coil carrying a current has a magnetic moment \(\vec{\mu}\). It is placed in a magnetic field \(\vec{B}\). The torque on the coil is \(\vec{\tau}\), then (A) \(\vec{\tau}=\vec{\mu} \cdot \vec{B}\) (B) \(\vec{\tau}=\vec{B} \times \vec{\mu}\) (C) \(|\vec{\tau}|=\vec{\mu} \cdot \vec{B}\) (D) \(\vec{\tau}\) is perpendicular to both \(\vec{\mu}\) and \(\vec{B}\).

Two particles each of mass \(m\) and charge \(q\) are attached to the two ends of a light rigid rod of length \(2 \ell\). The rod is rotated at a constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is (A) \(\frac{q}{2 m}\) (B) \(\frac{q}{m}\) (C) \(\frac{2 q}{m}\) (D) \(\frac{q}{\pi m}\)

In an AC circuit, the voltage applied is \(E=E_{0} \sin \omega t\). Theresulting currentinthecircuitis \(I=I_{0} \sin \left(\omega t-\frac{\pi}{2}\right)\). The power consumption in the circuit is given by (A) \(P=\sqrt{2} E_{0} I_{0}\) (B) \(P=\frac{E_{0} I_{0}}{\sqrt{2}}\) (C) \(P=0\) (D) \(P=\frac{E_{0} I_{0}}{2}\)

The magnetic field perpendicular to the plane of conducting ring of radius \(r\) changes at the rate \(\frac{d B}{d t}=\alpha\). Then (A) EMF induced in the ring is \(\pi r^{2} \alpha\). (B) EMF induced in the ring is \(2 \pi r \alpha\). (C) The potential difference between diametrically opposite points on the ring is half of induced EMF. (D) All points on the ring are at same potential.

A rectangular coil of 100 turns and size \(0.1 \mathrm{~m} \times 0.05 \mathrm{~m}\) is placed perpendicular to a magnetic field of \(0.1 \mathrm{~T}\). The induced EMF when the field drops to \(0.05 \mathrm{~T}\) is \(0.05 \mathrm{~s}\) is (A) \(0.5 \mathrm{~V}\) (B) \(1.0 \mathrm{~V}\) (C) \(1.5 \mathrm{~V}\) (D) \(2.0 \mathrm{~V}\)
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Particle Model of Matter

Read Explanation

Solid State Physics

Read Explanation

Wave Optics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Fundamentals of Physics

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.