/*! 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 27 A solenoid has a cross-sectional... [FREE SOLUTION] | 91Ó°ÊÓ

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A solenoid has a cross-sectional area of \(6.0 \times 10^{-4} \mathrm{~m}^{2},\) consists of 400 turns per meter, and carries a current of 0.40 A. A 10 -turn coil is wrapped tightly around the circumference of the solenoid. The ends of the coil are connected to a \(1.5-\Omega\) resistor. Suddenly, a switch is opened, and the current in the solenoid dies to zero in a time of \(0.050 \mathrm{~s} .\) Find the average current induced in the coil.

Short Answer

Expert verified
The average current induced in the coil is \( 1.6 \times 10^{-5} \ \text{A} \).

Step by step solution

01

Find initial magnetic field inside solenoid

The magnetic field inside a solenoid is given by the formula \( B = \mu_0 nI \), where \( \mu_0 = 4\pi \times 10^{-7} \ \text{T}\, \text{m/A} \) is the permeability of free space, \( n = 400 \ \text{turns/m} \) is the number of turns per unit length, and \( I = 0.40 \ \text{A} \) is the current in the solenoid. Thus, \( B = 4\pi \times 10^{-7} \times 400 \times 0.40 \). Calculating this, \( B = 2.0 \times 10^{-4} \ \text{T} \).
02

Calculate initial magnetic flux through coil

The magnetic flux \( \Phi \) through the coil is \( \Phi = B \times A \times N \), where \( A = 6.0 \times 10^{-4} \ \text{m}^2 \) is the cross-sectional area of the solenoid, and \( N = 10 \) is the number of turns in the coil. Substituting the values, \( \Phi = 2.0 \times 10^{-4} \times 6.0 \times 10^{-4} \times 10 \). Calculating this, \( \Phi = 1.2 \times 10^{-6} \ \text{Wb} \).
03

Determine change in magnetic flux

Since the current in the solenoid drops to zero, the final magnetic flux \( \Phi_f \) is zero. The change in magnetic flux \( \Delta \Phi \) is equal to the initial magnetic flux, \( \Delta \Phi = 1.2 \times 10^{-6} \ \text{Wb} \).
04

Calculate induced emf in the coil

According to Faraday’s law of electromagnetic induction, the induced emf \( \varepsilon \) is given by \( \varepsilon = -\frac{\Delta \Phi}{\Delta t} \), where \( \Delta t = 0.050 \ \text{s} \) is the time over which the change occurs. Substituting the values, \( \varepsilon = -\frac{1.2 \times 10^{-6} \ \text{Wb}}{0.050 \ \text{s}} \). Calculating this gives \( \varepsilon = -2.4 \times 10^{-5} \ \text{V} \).
05

Calculate average induced current through resistor

The average current \( I_{ ext{induced}} \) through the resistor is given by Ohm's law: \( I = \frac{\varepsilon}{R} \), where \( R = 1.5 \ \Omega \) is the resistance. Substituting the values, \( I_{ ext{induced}} = \frac{2.4 \times 10^{-5} \ \text{V}}{1.5 \ \Omega} \). Calculating this gives \( I_{ ext{induced}} = 1.6 \times 10^{-5} \ \text{A} \).

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Key Concepts

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

Faraday's Law
Faraday's Law of Electromagnetic Induction is a fundamental principle in physics that explains how electric current can be generated by changing magnetic fields. According to this law, the electromotive force (emf) induced in a circuit is equal to the rate of change of magnetic flux through the circuit. This means if you have a coil of wire and you change the magnetic field around it, either by moving a magnet or altering the current in a nearby solenoid, an emf is induced in the coil.

This concept is mathematically expressed as:
  • \( \varepsilon = -\frac{d\Phi}{dt} \)
Where \( \varepsilon \) is the induced emf, \( \Phi \) is the magnetic flux, and \( \frac{d\Phi}{dt} \) denotes the rate of change of magnetic flux over time. The negative sign indicates the direction of the induced emf as per Lenz's Law, which states that the induced current will oppose the change in flux causing it. Understanding this law helps to analyze how electric generators work and how transformers efficiently transfer electricity.
Magnetic Flux
Magnetic flux, symbolized by \( \Phi \), represents the quantity of magnetic field passing through a given area. Think of it like counting the number of magnetic field lines that pass through an area perpendicular to the direction of the magnetic field.

The formula for calculating magnetic flux is:
  • \( \Phi = B \times A \times \cos(\theta) \)
Where \( B \) is the magnetic field strength, \( A \) is the area through which the field lines pass, and \( \theta \) is the angle between the magnetic field and the perpendicular to the surface. When the surface is aligned perpendicular to the field lines, \( \cos(\theta) \) equals 1, making the flux simply \( B \times A \).

Understanding magnetic flux is crucial when evaluating the effectiveness of electromagnetic induction, as it determines how many field lines are influencing the coil, affecting the induced emf.
Solenoid
A solenoid is a long coil of wire, which can be used to generate a uniform magnetic field in its interior when an electric current passes through it. The magnetic field inside the solenoid is relatively strong and uniform because of the tight twists of the wire that effectively create a series of loops of current.

The magnetic field \( B \) produced inside a solenoid is given by:
  • \( B = \mu_0 nI \)
Where \( \mu_0 \) is the permeability of free space, \( n \) is the number of turns per unit length of the solenoid, and \( I \) is the current passing through it.

Solenoids are widely used in applications like electromagnetic switches and inductors because they provide a controllable and concentrated magnetic field. They ensure all of the field lines are well-aligned, maximizing the area of magnetic flux through any coil wrapped around them, such as in the original exercise.
Induced Current
When a changing magnetic field is present, such as one formed by a solenoid whose current is suddenly switched off, an induced current is generated in nearby conductors. This induced current can be calculated using Ohm’s Law in conjunction with Faraday’s Law.

The average current \( I_{\text{induced}} \) induced in a nearby coil or conductor with a known resistance \( R \) is given by:
  • \( I_{\text{induced}} = \frac{\varepsilon}{R} \)
Where \( \varepsilon \) is the induced emf. This formula shows how the induced current can be determined by the resistance of the connected circuit or components.

In practical terms, the direction of the induced current will be such that it creates a magnetic field opposing the original change, illustrating Lenz's Law in action. This concept is pivotal in designing circuits and devices like generators and transformers, as it allows electrical engineers to predict the effects of magnetic fields on nearby coils and conductors.

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Most popular questions from this chapter

During a 72 -ms interval, a change in the current in a primary coil occurs. This change leads to the appearance of a \(6.0\) -mA current in a nearby secondary coil. The secondary coil is part of a circuit in which the resistance is \(12 \Omega\). The mutual inductance between the two coils is \(3.2 \mathrm{mH}\). What is the change in the primary current?

Concept Questions The rechargeable batteries for a laptop computer need a much smaller voltage than what a wall socket provides. Therefore, a transformer is plugged into the wall socket and produces the necessary voltage for charging the batteries. (a) Is the transformer a step-up or a step-down transformer? (b) Is the current that goes through the batteries greater than, equal to, or smaller than the current coming from the wall socket? (c) If the transformer has a negligible resistance, is the electric power delivered to the batteries greater than, equal to, or less than the power coming from the wall socket? In all cases, provide a reason for your answer. Problem The batteries of a laptop computer are rated at \(9.0 \mathrm{~V},\) and a current of \(225 \mathrm{~mA}\) is used to charge them. The wall socket provides a voltage of \(120 \mathrm{~V}\). (a) Determine the turns ratio of the transformer, (b) What is the current coming from the wall socket? (c) Find the power delivered by the wall socket and the power sent to the batteries. Be sure your answers are consistent with your answers to the Concept Questions.

A circular loop of wire rests on a table. A long, straight wire lies on this loop, directly over its center, as the drawing illustrates. The current \(I\) in the straight wire is decreasing. In what direction is the induced current, if any, in the loop? Give your reasoning.

Interactive Solution \(\underline{22.55}\) at offers one approach to problems such as this one. The secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter. The turns ratio of the transformer is 50: 1 . The primary coil is plugged into a standard \(120-\mathrm{V}\) outlet. The current in the secondary coil is \(1.7 \times 10^{-3} \mathrm{~A} .\) Find the power consumed by the air filter.

One generator uses a magnetic field of \(0.10 \mathrm{~T}\) and has a coil area per turn of \(0.045 \mathrm{~m}^{2} . \mathrm{A}\) second generator has a coil area per turn of \(0.015 \mathrm{~m}^{2}\). The generator coils have the same number of turns and rotate at the same angular speed. What magnetic field should be used in the second generator so that its peak emf is the same as that of the first generator?

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