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A constant current of \(I=15\) 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. \((\mathrm{a})\) 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) The induced emf is -620 V. (b) The stored energy is 348.75 J. (c) The power is 4,650 W.

Step by step solution

01

Calculate Induced EMF

The induced electromotive force (emf) in a solenoid due to the change in current can be calculated using Faraday's Law of Induction given by \[ \text{EMF} = -L \frac{\Delta I}{\Delta t} \]Where \( L = 3.1 \text{ H} \) is the inductance, \( \Delta I = 15 \text{ A} \) is the change in current, and \( \Delta t = 75 \text{ ms} = 0.075 \text{ s} \) is the time duration. Substituting in these values, we get:\[\text{EMF} = -3.1 \times \frac{15}{0.075} = -620 \, \text{V}\] The negative sign indicates the emf is in a direction opposing the change in current.
02

Calculate Electrical Energy Stored

The energy stored in an inductor can be calculated using the formula \[ U = \frac{1}{2} L I^2 \]Substituting the values \( L = 3.1 \text{ H} \) and \( I = 15 \text{ A} \), we have:\[U = \frac{1}{2} \times 3.1 \times 15^2 = 348.75 \, \text{J}\]Thus, the electrical energy stored in the solenoid is 348.75 J.
03

Compute Power for Energy Removal

The rate of energy removal, or power, when reducing the current to zero is given by the energy change divided by the time:\[ P = \frac{U}{\Delta t} \]We have already calculated \( U = 348.75 \text{ J} \) and \( \Delta t = 0.075 \text{ s} \):\[P = \frac{348.75}{0.075} = 4,650 \, \text{W}\]Therefore, the rate at which energy must be removed is 4,650 W.

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Key Concepts

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

Inductance
Inductance is a fundamental property of electrical circuits that relates to how they oppose changes in current. It is an essential principle when working with elements like solenoids and inductors, relying on the idea of magnetic fields generated by electric currents. This property is quantitatively expressed as the inductance, denoted by the symbol \(L\). The unit of inductance is the Henry (H), and it signifies the amount of electromotive force (emf) generated per unit change of current over time.

Imagine a solenoid as a tightly coiled wire. When current flows through it, a magnetic field is created. If the current changes, the magnetic field changes, which in turn induces an emf. This induced emf opposes the change in current, in line with Lenz's Law. Faraday's Law of Induction mathematically expresses this phenomenon as \( \text{EMF} = -L \frac{\Delta I}{\Delta t} \). Here:
  • \(\Delta I\) is the change in current.
  • \(\Delta t\) is the time duration of the change.
  • The negative sign reflects the opposing nature of the induced emf.

In practical terms, inductance is crucial for designing circuits that require stability amidst fluctuating currents, like filters and transformers.
Energy Stored in Inductors
Inductors store energy in the form of a magnetic field. This is why they are often used in circuits for energy storage or to smooth out electrical signals. The energy stored in an inductor is derived from its ability to sustain a current flow, even if it's temporary when the current changes.

The amount of energy stored, \(U\), in an inductor is expressed by the equation \( U = \frac{1}{2} L I^2 \):
  • \(L\) is the inductance in Henrys (H).
  • \(I\) is the current in Amperes (A).

This formula reveals that the stored energy is proportional to the square of the current, meaning even a small change in current can lead to a significant change in the energy stored.

For instance, in the problem given, the solenoid stores 348.75 Joules when the current is 15 A, illustrating the high efficiency of energy storage in inductors. This stored energy can be released back into the circuit, providing vital functionality in electrical systems, such as those found in power supplies and RF applications.
Electric Power Calculation
Electric power relates to the rate at which energy is transferred or converted in a circuit. It's essential both for understanding how circuits operate and for ensuring that they function safely and efficiently.

When it comes to inductors, calculating the power can involve determining how quickly stored energy is removed. This is particularly relevant when an inductor's current decreases, and the stored magnetic energy dissipates, meaning it gets transformed, often into another form of energy, like heat.

The formula for power, \(P\), derived from the rate of energy change over time, is given by \( P = \frac{U}{\Delta t} \):
  • \(U\) is the energy change, measured in Joules (J).
  • \(\Delta t\) is the time duration over which the energy change occurs, in seconds (s).

In the example exercise, when the solenoid's current diminishes to zero in 75 milliseconds, the power calculated is 4,650 Watts. This significant figure underscores how power can spike with rapid energy changes, highlighting the importance of accounting for such dynamics in electrical engineering and design to avoid overheating or damage.

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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?

A motor is designed to operate on 117 \(\mathrm{V}\) and draws a current of 12.2 \(\mathrm{A}\) when it first starts up. At its normal operating speed, the motor draws a current of 2.30 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third of the normal speed.

ssm Two 0.68 -m-long conducting rods are rotating at the same speed in opposite directions, and both are perpendicular to a \(4.7-\mathrm{T}\) magnetic field. As the drawing shows, the ends of these rods come to within 1.0 \(\mathrm{mm}\) of each other as they rotate. Moreover, the fixed ends about which the rods are rotating are connected by a wire, so these ends are at the same electric potential. If a potential difference of \(4.5 \times 10^{3} \mathrm{V}\) is required to cause a \(1.0-\mathrm{mm}\) spark in air, what is the angular speed (in rad/s) of the rods when a spark jumps across the gap?

A loop of wire has the shape shown in the drawing. The top part of the wire is bent into a semicircle of radius \(r=0.20 \mathrm{m}\) . The normal to the plane of the loop is parallel to a constant magnetic field \(\left(\phi=0^{\circ}\right)\) of magnitude 0.75 \(\mathrm{T}\) . What is the change \(\Delta \Phi\) in the magnetic flux that passes through the loop when, starting with the position shown in the drawing, the semicircle is rotated through half a revolution?

A 120.0 -V motor draws a current of 7.00 \(\mathrm{A}\) when running at normal speed. The resistance of the armature wire is 0.720\(\Omega .\) (a) Determine the back emf generated by the motor. (b) What is the current at the instant when the motor is just turned on and has not begun to rotate? (c) What series resistance must be added to limit the starting current to 15.0 \(\mathrm{A}\) ?

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