91Ó°ÊÓ

Concept Questions A constant current \(I\) exists in a solenoid whose inductance is \(L .\) The current is then reduced to zero in a certain amount of time. (a) If the wire from which the solenoid is made has no resistance, is there a voltage across the solenoid during the time when the current is constant? (b) If the wire from which the solenoid is made has no resistance, is there an emf across the solenoid during the time that the current is being reduced to zero? (c) Does the solenoid store electrical energy when the current is constant? If so, express this energy in terms of the current and the inductance. (d) When the current is reduced from its constant value to zero, what is the rate at which energy is removed from the solenoid? Express your answer in terms of the initial current, the inductance, and the time during which the current goes to zero. Problem A solenoid has an inductance of \(L=3.1 \mathrm{H}\) and carries a current of \(I=15 \mathrm{~A}\). (a) If the current goes from 15 to \(0 \mathrm{~A}\) in a time of \(75 \mathrm{~ms}\), what is the emf induced in the solenoid? (b) How much electrical energy is stored in the solenoid? (c) At what rate must the electrical energy be removed from the solenoid when the current is reduced to zero in \(75 \mathrm{~ms} ?\)

Step by step solution

01

Understanding the Problem

We are given the inductance \(L=3.1 \, \text{H}\) and initial current \(I=15 \, \text{A}\). The current decreases to zero in \(75 \, \text{ms}\). We need to find the induced emf during this change, the energy stored initially, and the rate at which energy is removed.

02

Calculating the Induced EMF

Use Faraday's law of induction which states \( \text{emf} = -L \frac{dI}{dt} \). Since the current changes from 15 to 0 A over \(75 \, \text{ms}\), the rate of change is \( \frac{dI}{dt} = \frac{0 - 15}{0.075} = -200 \, \text{A/s} \). Substitute into the formula: \( \text{emf} = -3.1 \times (-200) = 620 \, \text{V} \). The emf is 620 V.

03

Determining the Stored Energy

The energy stored in an inductor is given by \( U = \frac{1}{2} LI^2 \). Substitute the given values: \( U = \frac{1}{2} \times 3.1 \times 15^2 = 348.75 \, \text{J} \). Thus, the solenoid stores 348.75 J of energy when the current is constant.

04

Calculating the Rate of Energy Removal

The power or rate at which energy is removed as the current decreases can be calculated using \( P = \frac{U}{t} \) where \(t=0.075 \, \text{s}\). Therefore, \( P = \frac{348.75}{0.075} = 4650 \, \text{W} \). The rate of energy removal from the solenoid is 4650 W.

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Key Concepts

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

Solenoid Inductance

Inductance is a fundamental property of a solenoid, which is essentially a coil of wire. It depends on the number of turns of the coil, the area of the coil, and the length of the solenoid. Inductance, represented as \(L\), is measured in henrys (H).
A solenoid with a higher number of turns or one that is more closely wound will have greater inductance. This is because the magnetic field generated by the solenoid is directly linked with the number of turns.
A solenoid functions effectively as an inductor, which means it can store energy in the form of a magnetic field when an electric current passes through it. This is crucial in circuits where controlling the rate of current change is necessary.

Energy Storage in Inductors

Inductors can store energy when a current flows through them, akin to how a capacitor stores energy in an electric field. The energy stored in an inductor is due to the magnetic field created by the current.
The energy \(U\) in an inductor is calculated using the formula:\[ U = \frac{1}{2} LI^2 \]where \(L\) is the inductance and \(I\) is the current flowing through the inductor.
When current flows steadily, the inductor maintains a constant level of energy. However, when the current is reduced to zero, this energy is dissipated — potentially generating heat or causing other forms of power dissipation in the circuit.

Electromotive Force (EMF)

EMF, or electromotive force, is a crucial concept in understanding how voltage is induced in an electrical circuit. It essentially refers to the voltage generated by a source such as a battery or, in this case, a solenoid when experiencing a change in current.
According to Faraday's Law of Electromagnetic Induction, a change in magnetic flux through a coil induces an EMF. For a solenoid with inductance \(L\), when the current changes at a rate \(\frac{dI}{dt}\), the induced EMF is given by:\[ \text{emf} = -L \frac{dI}{dt} \]Here, the negative sign indicates the direction of the induced EMF opposes the change in current, which is consistent with Lenz's law.

Faraday's Law

Faraday’s Law of Electromagnetic Induction is pivotal in explaining how and why EMF is induced in electrical circuits. It states that the induced EMF in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.
Mathematically, it can be expressed as:\[ \text{emf} = - \frac{d\Phi}{dt} \]where \(\Phi\) is the magnetic flux.
In simple terms, if the magnetic environment of a coil changes, either because the magnetic field itself changes or the coil moves within a magnetic field, EMF is induced.

• This principle is fundamental in the operation of transformers, motors, and generators.
• It helps explain the behavior of inductors in transient state circuits.

Understanding this law helps in analyzing circuits where inductors play a critical role, particularly when currents are changing.