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Chapter 22: Problem 54

A constant current of \(I=15 \mathrm{A}\) exists in a solenoid whose inductance is \(L=3.1 \mathrm{H}\). The current is then reduced to zero in a certain amount of time. (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 \(0 \mathrm{A}\) in a time of \(75 \mathrm{ms}\) ? Note that the rate at which energy is removed is the power.

Short Answer

Expert verified
(a) 620 V, (b) 348.75 J, (c) 4650 W.

Step by step solution

01

Understanding the Problem

We are given a solenoid with its inductance \( L = 3.1 \, \text{H} \) and an initial current \( I = 15 \, \text{A} \). The current is reduced to 0 A in 75 ms. We need to find the induced emf, the electrical energy stored, and the rate at which energy is removed.
02

Calculating the Induced EMF (Faraday's Law)

The emf (\( \varepsilon \)) induced in a solenoid is given by Faraday's Law: \( \varepsilon = -L \frac{dI}{dt} \). First, we calculate the rate of change of current \( \frac{dI}{dt} \). Since the current changes from 15 A to 0 A in 0.075 s, we have \( \frac{dI}{dt} = \frac{0 - 15}{0.075} = -200 \, \text{A/s} \). Substituting into the formula, \( \varepsilon = -3.1 \times (-200) = 620 \, \text{V} \).
03

Calculating Electrical Energy Stored

The energy (\( W \)) stored in an inductor is given by \( W = \frac{1}{2} L I^2 \). Substituting the given values, \( W = \frac{1}{2} \times 3.1 \times 15^2 = 348.75 \, \text{J} \).
04

Calculating the Power (Rate of Energy Removal)

The power \( P \), or the rate at which energy is removed, is given by \( P = \frac{\Delta W}{\Delta t} \), where \( \Delta W \) is the energy change. Since all energy is removed when the current goes to zero, \( \Delta W = 348.75 \, \text{J} \) and \( \Delta t = 0.075 \, \text{s} \). Thus, \( P = \frac{348.75}{0.075} = 4650 \, \text{W} \).

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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 is a fundamental principle that describes how a change in magnetic field within a closed loop induces an electromotive force (emf) in the loop. It's one of the basic principles of electromagnetism, stating that the induced emf in any closed circuit is equal to the negative rate of change of the magnetic flux through the circuit. In mathematical terms, it is given by:\[\varepsilon = - \frac{d\Phi}{dt}\]where \(\varepsilon\) is the induced emf and \(\frac{d\Phi}{dt}\) represents the rate of change of magnetic flux. In the context of a solenoid with changing current, this law helps to calculate the induced emf based on how quickly the current changes within the solenoid.
emf induced in solenoid
The electromotive force (emf) induced in a solenoid is a direct application of Faraday's Law. A solenoid is a coil of wire that becomes magnetized when electrical current passes through it. When the current changes, a changing magnetic field occurs, inducing an emf according to the formula:\[\varepsilon = -L \frac{dI}{dt}\]where:
  • \(\varepsilon\) is the induced emf,
  • \(L\) represents the inductance of the solenoid,
  • \(\frac{dI}{dt}\) is the rate of change of current.
In the given exercise, when the current decreases from 15 A to 0 A in 75 ms, we calculate the rate of change of current as \(-200\, \text{A/s}\). Substituting this and the inductance value (3.1 H) into the equation gives an induced emf of 620 V.
electrical energy stored
The electrical energy stored in an inductor, such as a solenoid, is a measure of the energy contained within the magnetic field created by the current flowing through it. This is given by the formula:\[W = \frac{1}{2} L I^2\]where:
  • \(W\) is the stored energy,
  • \(L\) is the inductance,
  • \(I\) is the current.
In this exercise, with the initial current of 15 A and inductance of 3.1 H, the stored energy calculates to 348.75 Joules. This energy is crucial for applications in systems that require a temporary energy store, such as in inductors used in power supplies or radio transmitters.
rate of energy removal
The rate of energy removal in a solenoid, when the current is brought to zero, is a crucial aspect in understanding power handling. This rate, also known as power (\(P\)), is calculated using the energy and the time over which the energy is released. The formula is:\[P = \frac{\Delta W}{\Delta t}\]where:
  • \(\Delta W\) is the change in energy, i.e., the total energy removed.
  • \(\Delta t\) is the time in which the energy is removed.
In the problem at hand, the energy of 348.75 J is removed in 0.075 s, resulting in a power of 4650 Watts. This high power output emphasizes how rapidly energy can be withdrawn from a solenoid, highlighting the need for careful design in high energy applications.

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Most popular questions from this chapter

A constant magnetic field passes through a single rectangular loop whose dimensions are \(0.35 \mathrm{m} \times 0.55 \mathrm{m} .\) The magnetic field has a magnitude of \(2.1 \mathrm{T}\) and is inclined at an angle of \(65^{\circ}\) with respect to the normal to the plane of the loop. (a) If the magnetic field decreases to zero in a time of \(0.45 \mathrm{s},\) what is the magnitude of the average emf induced in the loop? (b) If the magnetic field remains constant at its initial value of \(2.1 \mathrm{T},\) what is the magnitude of the rate \(\Delta A / \Delta t\) at which the area should change so that the average emf has the same magnitude as in part (a)?

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. The batteries 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 average power delivered by the wall socket and the average power sent to the batteries.

A conducting coil of 1850 turns is connected to a galvanometer, and the total resistance of the circuit is \(45.0 \Omega\). The area of each turn is \(4.70 \times\) \(10^{-4} \mathrm{m}^{2} .\) This coil is moved from a region where the magnetic field is zero into a region where it is nonzero, the normal to the coil being kept parallel to the magnetic field. The amount of charge that is induced to flow around the circuit is measured to be \(8.87 \times 10^{-3} \mathrm{C} .\) Find the magnitude of the magnetic field.

A magnetic field is passing through a loop of wire whose area is \(0.018 \mathrm{m}^{2}\). The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of \(0.20 \mathrm{T} / \mathrm{s}\) (a) Determine the magnitude of the emf induced in the loop. (b) Suppose that the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in \(\mathrm{m}^{2} / \mathrm{s}\) ) should the area be changed at the instant when \(B=1.8 \mathrm{T}\) if the induced emf is to be zero? Explain whether the area is to be enlarged or shrunk.

At its normal operating speed, an electric fan motor draws only \(15.0 \%\) of the current it draws when it just begins to turn the fan blade. The fan is plugged into a \(120.0-\mathrm{V}\) socket. What back emf does the motor generate at its normal operating speed?
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