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Coil 1 has \(L_{1}=25 \mathrm{mH}\) and \(N_{1}=100\) turns. Coil 2 has \(L_{2}=40\) \(\mathrm{mH}\) and \(N_{2}=200\) turns. The coils are fixed in place; their mutual inductance \(M\) is \(3.0 \mathrm{mH}\). A \(6.0 \mathrm{~mA}\) current in coil 1 is changing at the rate of \(4.0 \mathrm{~A} / \mathrm{s}\). (a) What magnetic flux \(\Phi_{12}\) links coil 1, and \((\mathrm{b})\) what self-induced emf appears in that coil? (c) What magnetic flux \(\Phi_{21}\) links coil 2 , and \((\mathrm{d})\) what mutually induced emf appears in that coil?

Short Answer

Expert verified
(a) \(0.00015\,\mathrm{Wb}\); (b) \(0.1\,\mathrm{V}\); (c) \(0.000018\,\mathrm{Wb}\); (d) \(0.012\,\mathrm{V}\).

Step by step solution

01

Understanding Magnetic Flux in Coil 1

The magnetic flux \( \Phi_{12} \) linked to coil 1 can be calculated using the formula \( \Phi_{12} = L_{1} \times I_1 \), where \( I_1 \) is the current in coil 1. Substitute \( L_1 = 25 \; \mathrm{mH} = 0.025 \; \mathrm{H} \) and \( I_1 = 6.0 \; \mathrm{mA} = 0.006 \; \mathrm{A} \). Hence, \( \Phi_{12} = 0.025 \times 0.006 \; \mathrm{Wb} = 0.00015 \; \mathrm{Wb}. \)
02

Calculating Self-Induced EMF in Coil 1

The self-induced electromotive force (emf) \( \varepsilon_1 \) in coil 1 is given by \( \varepsilon_1 = -L_1 \frac{dI_1}{dt} \), where \( \frac{dI_1}{dt} = 4.0 \; \mathrm{A/s} \). Substituting, \( \varepsilon_1 = -0.025 \times 4.0 = -0.1 \; \mathrm{V} \). This means the self-induced emf is \( 0.1 \; \mathrm{V} \).
03

Understanding Magnetic Flux in Coil 2

The magnetic flux \( \Phi_{21} \) linked to coil 2 is calculated using the mutual inductance \( M \). It is given by \( \Phi_{21} = M \times I_1 \). Substitute \( M = 3.0 \; \mathrm{mH} = 0.003 \; \mathrm{H} \) and \( I_1 = 0.006 \; \mathrm{A} \). Hence, \( \Phi_{21} = 0.003 \times 0.006 = 0.000018 \; \mathrm{Wb}. \)
04

Calculating Mutually Induced EMF in Coil 2

The mutual induced emf \( \varepsilon_2 \) in coil 2 is calculated as \( \varepsilon_2 = -M \frac{dI_1}{dt} \). Substitute \( M = 0.003 \; \mathrm{H} \) and \( \frac{dI_1}{dt} = 4.0 \; \mathrm{A/s} \). Therefore, \( \varepsilon_2 = -0.003 \times 4.0 = -0.012 \; \mathrm{V} \). The mutually induced emf is \( 0.012 \; \mathrm{V} \).

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Key Concepts

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

Mutual Inductance
Mutual inductance is a key concept in electromagnetic induction that describes the ability of one coil to induce an electromotive force (EMF) in another nearby coil as the current in the first coil changes. This principle is central in transformers and many types of electrical machinery.

The mutual inductance, denoted by the symbol \( M \), depends on the geometrical arrangement of the coils and the permeability of the medium that the magnetic field passes through. In the given problem, the mutual inductance between the two coils is given as \( M = 3.0 \, \text{mH} \) (or \( 0.003 \, \text{H} \)).

Mutual inductance is affected by:
  • Number of turns in the coils
  • Distance between the coils
  • The core material (air or iron)
It is calculated so that when the current in the first coil changes, it creates a magnetic field that links to the second coil, inducing an EMF in it. If the current in coil 1 changes at a rate of 4.0 A/s, the mutually induced EMF in coil 2 can be found using the formula \( \varepsilon_2 = -M \frac{dI_1}{dt} \).

Clever application of these principles allows devices like transformers to efficiently transfer energy between electrical circuits.
Magnetic Flux
Magnetic flux is a measure of the magnetic field passing through a given area, which in this context is the area of the coil. It is symbolized by \( \Phi \) and has units of Weber (Wb). Understanding magnetic flux helps us visualize and calculate how the magnetic field lines interact with the coil.

The magnetic flux through a coil depends on:
  • The strength of the magnetic field
  • The area the field penetrates
  • The angle at which the field lines pass through the surface of the coil
  • The number of turns of the coil
In our problem, the magnetic flux linking coil 1, \( \Phi_{12} \), can be calculated using \( \Phi_{12} = L_1 \times I_1 \).

For coil 1, we use \( L_1 = 25 \, \text{mH} \) and \( I_1 = 6.0 \, \text{mA} \), getting \( \Phi_{12} = 0.00015 \, \text{Wb} \). Similarly, the flux linking coil 2 due to coil 1 is rendered as \( \Phi_{21} = M \times I_1 \) yielding \( 0.000018 \, \text{Wb} \).

These calculations illustrate how the mutual configurations of the coils affect the flux linking them, playing a crucial role in electromagnetic induction.
Self-Inductance
Self-inductance is the property of a coil to induce an EMF in itself due to a change in the current flowing through it. It is represented by the letter \( L \) and is measured in Henry (H). This concept ensures that any change in current within a coil is restricted by the coil itself generating an opposing EMF.

Self-inductance arises because a changing current in the coil creates a changing magnetic flux that links with its own turns. This induces a voltage in opposition to the change in current, following Lenz's law.

The self-induced EMF can be calculated using the formula:
  • \( \varepsilon_1 = -L_1 \frac{dI_1}{dt} \)
In the provided problem, with parameters \( L_1 = 25 \, \text{mH} \) and \( \frac{dI_1}{dt} = 4.0 \, \text{A/s} \), the calculation yields an EMF of \( 0.1 \, \text{V} \).

This process prevents abrupt changes in current and helps maintain stable circuit operation, a crucial aspect of electrical design and analysis.
Electromotive Force (EMF)
Electromotive Force, commonly abbreviated as EMF, is not a force in the traditional sense, but rather a potential difference produced by a source of energy, such as a battery or generator, that can drive electric charge around a circuit. In electromagnetic induction, EMF is generated when there is a change in magnetic flux through a coil.

The magnitude of the induced EMF depends on:
  • The rate of change of the flux
  • The number of turns in the coil
This relationship is described by Faraday's law of electromagnetic induction, which states that the induced EMF in a coil is equal to the negative rate of change of magnetic flux through the coil:

\[ \text{EMF} = -\frac{d\Phi}{dt} \]

In our scenario, for coil 1, the change in current \( \frac{dI_1}{dt} = 4.0 \, \text{A/s} \) results in a self-induced EMF of \( 0.1 \, \text{V} \). Meanwhile, the changing current in Coil 1 induces an EMF \( \varepsilon_2 = 0.012 \, \text{V} \) in Coil 2.

This phenomenon allows for the conversion and regulation of power in electrical circuits, fundamental for the operation of everything from simple household tools to complex industrial machinery.

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Most popular questions from this chapter

The inductance of a closely packed coil of 400 turns is \(8.0\) \(\mathrm{mH}\). Calculate the magnetic flux through the coil when the current is \(5.0 \mathrm{~mA}\).

Ttwo parallel loops of wire having a common axis. The smaller loop (radius \(r\) ) is above the larger loop (radius \(R\) ) by a distance \(x \geqslant R\). Consequently, the magnetic field due to the counterclockwise current \(i\) in the larger loop is nearly uniform throughout the smaller loop. Suppose that \(x\) is increasing at the constant rate \(d x / d t=v .\) (a) Find an expression for the magnetic flux through the area of the smaller loop as a function of \(x\).

A certain elastic conducting material is stretched into a circular loop of \(12.0 \mathrm{~cm}\) radius. It is placed with its plane perpendicular to a uniform \(0.800 \mathrm{~T}\) magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of \(75.0 \mathrm{~cm} / \mathrm{s}\). What emf is induced in the loop at that instant?

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A square wire loop \(20 \mathrm{~cm}\) on a side, with resistance \(20 \mathrm{~m} \Omega\), has its plane normal to a uniform magnetic field of magnitude \(B=2.0 \mathrm{~T}\). If you pull two opposite sides of the loop away from each other, the other two sides automatically draw toward each other, reducing the area enclosed by the loop. If the area is reduced to zero in time \(\Delta t=0.20 \mathrm{~s}\), what are (a) the average emf and (b) the average current induced in the loop during \(\Delta t\) ?

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