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A constant current of \(I=15 \mathrm{A}\) exists in a solenoid whose in- ductance 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 A in a time of 75 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) Induced emf = 620 V; (b) Stored energy = 348.75 J; (c) Power = 4650 W.

Step by step solution

01

Identify given values and required formulas

We are given the following values: current \( I = 15 \text{ A} \), inductance \( L = 3.1 \text{ H} \), and time \( \Delta t = 75 \text{ ms} = 0.075 \text{ s} \). We need to find: (a) induced emf \( \mathcal{E} \); (b) stored electrical energy \( U \); and (c) rate of energy removal (power \( P \)). The relevant formulas are:- Induced emf: \( \mathcal{E} = -L \frac{\Delta I}{\Delta t} \)- Energy stored: \( U = \frac{1}{2}LI^2 \)- Power: \( P = \frac{U}{\Delta t} \).
02

Calculate the induced emf

Using the formula for induced emf:\[\mathcal{E} = -L \frac{\Delta I}{\Delta t}\]Substitute the values:\[\mathcal{E} = -3.1 \text{ H} \times \frac{15 \text{ A} - 0 \text{ A}}{0.075 \text{ s}} = -3.1 \times \frac{15}{0.075}\]This gives:\[\mathcal{E} = -620 \text{ V}\]The negative sign indicates the direction of the induced emf, opposing the change.
03

Calculate the electrical energy stored in the solenoid

Use the formula for stored energy:\[U = \frac{1}{2}L I^2\]Substitute the values:\[U = \frac{1}{2} \times 3.1 \text{ H} \times (15 \text{ A})^2\]Calculating gives:\[U = \frac{1}{2} \times 3.1 \times 225 = 348.75 \text{ J}\]
04

Determine the rate of energy removal (power)

Use the formula for power:\[P = \frac{U}{\Delta t}\]Substitute the values we've calculated and been given:\[P = \frac{348.75 \text{ J}}{0.075 \text{ s}}\]This results in:\[P = 4650 \text{ W}\]

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Key Concepts

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

Induced Electromotive Force
In physics, the concept of induced electromotive force (emf) is crucial for understanding interactions in circuits with changing current. The inductance of a solenoid, denoted by the symbol \( L \), plays a significant role in generating this emf. When the current flowing through a solenoid changes, the inductance causes a voltage to be induced along the solenoid's coil. This principle is captured by Faraday's law of electromagnetic induction.

The formula used to calculate the induced emf is \( \mathcal{E} = -L \frac{\Delta I}{\Delta t} \). Here, \( \Delta I \) represents the change in current, and \( \Delta t \) is the time over which the current changes. The negative sign signifies Lenz's Law, which tells us that the induced emf will act in a direction to oppose the change in current. This opposition helps maintain equilibrium in the circuit.

To illustrate, consider a solenoid with an inductance of 3.1 H, where the current drops from 15 A to 0 A in 75 ms. Substituting these values into the formula gives an induced emf of \( -620 \text{ V} \). The negative sign indicates that this induced voltage works against the decrease in current.
Stored Electrical Energy
The energy stored in a solenoid due to its inductance is another fascinating aspect of electromagnetic fields. This stored energy can be calculated using the formula \( U = \frac{1}{2} L I^2 \). It depends on both the inductance \( L \) and the square of the current \( I \) flowing through the solenoid.

When current flows through an inductor like a solenoid, it creates a magnetic field. This field is essentially where the electrical energy gets stored. Once the current stops, the magnetic field collapses, releasing the stored energy back into the circuit. This property is essential in many practical applications, such as electrical circuits in power supplies and transformers.

For instance, using a solenoid with an inductance of 3.1 H and a current of 15 A, the stored energy is computed as \( 348.75 \text{ J} \). This value highlights the potential energy available to do work or to convert into other forms of energy under the right circumstances.
Power in Circuits
Power in electric circuits refers to the rate at which energy is used or transferred. It is particularly important when considering how quickly energy is removed from an inductor, like a solenoid, as it ceases to carry current. This is because the stored energy in the solenoid must be rapidly dissipated when the circuit changes.

We calculate power, \( P \), from electrical energy using the formula \( P = \frac{U}{\Delta t} \), where \( U \) is the stored energy, and \( \Delta t \) is the time over which the energy is dissipated. Knowing the rate at which energy is released makes it easier to manage the energy flow in complex power systems.

In our scenario, the stored energy of 348.75 J is released over a brief period of 75 ms, resulting in a high power release of \( 4650 \text{ W} \). This demonstrates how quickly significant amounts of power can be delivered from a solenoid, emphasizing the need for efficient circuit design and safety measures in electrical engineering. Harnessing such power effectively ensures minimal energy loss and protection for the rest of the circuit.

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Most popular questions from this chapter

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 vacuum cleaner is plugged into a \(120.0-\mathrm{V}\) socket and uses \(3.0 \mathrm{A}\) of current in normal operation when the back emf generated by the electric motor is \(72.0 \mathrm{V}\). Find the coil resistance of the motor.

A flat circular coil with 105 turns, a radius of \(4.00 \times 10^{-2} \mathrm{m}\), and a resistance of \(0.480 \Omega\) is exposed to an external magnetic field that is directed perpendicular to the plane of the coil. The magnitude of the external magnetic field is changing at a rate of \(\Delta B / \Delta t=0.783 \mathrm{T} / \mathrm{s}\), thereby inducing a current in the coil. Find the magnitude of the magnetic field at the center of the coil that is produced by the induced current.

When its coil rotates at a frequency of \(280 \mathrm{Hz},\) a certain generator has a peak emf of 75 V. (a) What is the peak emf of the generator when its coil rotates at a frequency of 45 Hz? (b) Determine the frequency of the coil's rotation when the peak emf of the generator is \(180 \mathrm{V}\).

A generating station is producing \(1.2 \times 10^{6} \mathrm{W}\) of power that is to be sent to a small town located \(7.0 \mathrm{km}\) away. Each of the two wires that comprise the transmission line has a resistance per kilometer of \(5.0 \times 10^{-2} \Omega / \mathrm{km}\) (a) Find the power used to heat the wires if the power is transmitted at \(1200 \mathrm{V}\). (b) A 100: 1 step-up transformer is used to raise the voltage before the power is transmitted. How much power is now used to heat the wires?

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