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A piece of copper wire is formed into a single circular loop of radius \(12 \mathrm{~cm}\). A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to \(0.60 \mathrm{~T}\) in a time of \(0.45 \mathrm{~s}\). The wire has a resistance per unit length of \(3.3 \times 10^{-2} \Omega / \mathrm{m}\). What is the average electrical energy dissipated in the resistance of the wire?

Short Answer

Expert verified
The average electrical energy dissipated is approximately 0.0657 J.

Step by step solution

01

Calculate the Circumference of the Loop

First, we need to find the total length of the wire, which is the circumference of the circular loop. The formula for the circumference of a circle is given by \( C = 2\pi r \), where \( r \) is the radius. With a radius \( r = 12 \text{ cm} = 0.12 \text{ m} \), the circumference is \( C = 2 \pi \times 0.12 \text{ m} = 0.24\pi \text{ m} \approx 0.7536 \text{ m} \).
02

Calculate the Total Resistance of the Wire

The resistance of the wire is calculated by multiplying the resistance per unit length by the length of the wire. Therefore, \( R = \text{resistance per unit length} \times \text{length of wire} = 3.3 \times 10^{-2} \Omega/\text{m} \times 0.7536 \text{ m} \approx 0.02486 \Omega \).
03

Calculate the Change in Magnetic Flux

The magnetic flux \( \Phi \) through the loop is given by \( \Phi = B \times A \), where \( B \) is the magnetic field and \( A \) is the area of the loop. The area \( A \) is \( \pi r^2 \), giving \( A = \pi \times (0.12)^2 \text{ m}^2 \approx 0.04524 \text{ m}^2 \). The change in magnetic field is from 0 to \( 0.60 \text{ T} \), so the change in flux is \( \Delta \Phi = 0.60 \text{ T} \times 0.04524 \text{ m}^2 = 0.027144 \text{ Wb} \).
04

Calculate the Induced EMF

According to Faraday's Law, the induced electromotive force (EMF) is equal to the rate of change of magnetic flux through the loop, expressed as \( \varepsilon = -\frac{\Delta \Phi}{\Delta t} \). Substituting the known values, \( \varepsilon = -\frac{0.027144 \text{ Wb}}{0.45 \text{ s}} \approx -0.06032 \text{ V} \).
05

Calculate the Average Power Dissipated

The power dissipated in the loop, \( P \), is given by \( P = \frac{\varepsilon^2}{R} \). Substituting the known EMF and resistance, we get \( P = \frac{(0.06032)^2}{0.02486} \approx 0.146 \text{ W} \).
06

Calculate the Total Energy Dissipated

The total energy dissipated is the average power multiplied by the time duration: \( E = P \times \Delta t = 0.146 \text{ W} \times 0.45 \text{ s} \approx 0.0657 \text{ J} \).

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Key Concepts

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

Magnetic Flux
Magnetic flux is a measure of the quantity of magnetism, considering the strength and extent of a magnetic field. It is often denoted by the symbol \( \Phi \) and is calculated by the product of the magnetic field \( B \) and the perpendicular area \( A \) through which it passes.
In mathematical terms, magnetic flux is given by the formula:
  • \( \Phi = B \times A \)
For a circular loop with a radius, the area \( A \) is calculated using the formula for the area of a circle, \( A = \pi r^2 \). In our example, this results in \( 0.04524 \text{ m}^2 \), where \( r = 0.12 \text{ m}\).
As the magnetic field changes from 0 to 0.60 T, the change in magnetic flux \( \Delta \Phi \) becomes \( 0.027144 \text{ Wb} \) (Weber is the unit for magnetic flux).
This change in magnetic flux is key to understanding the induced electrical effects in conductive loops.
Faraday's Law
Faraday's Law of electromagnetic induction plays a crucial role in determining how electrical energy is generated via a changing magnetic field.
This law states that the induced electromotive force (EMF) in any closed circuit is equal to the negative rate of change of the magnetic flux through the circuit. The expression for Faraday's Law is:
  • \( \varepsilon = -\frac{\Delta \Phi}{\Delta t} \)
The negative sign indicates the direction of the induced EMF and current resulting from Lenz's Law, which opposes the change in magnetic flux.
In our scenario, the change in magnetic flux \( (0.027144 \text{ Wb}) \) over the period \( (0.45 \text{ s}) \) results in an induced EMF of approximately \( -0.06032 \text{ V} \).
This induced EMF is what drives current through the wire, allowing us to explore further electrical phenomena.
EMF (Electromotive Force)
The electromotive force, or EMF, is not actually a force but a potential difference that drives current in a circuit. It is a critical concept when considering circuits where magnetic fields change, as it represents the energy per unit charge that is made available by the power source.
From Faraday's Law, we know that the EMF is directly linked to the change in magnetic flux. The formula, \( \varepsilon = -\frac{\Delta \Phi}{\Delta t} \), helps calculate this induced EMF.
For our example, this turns out to be \( -0.06032 \text{ V} \). This represents the average voltage generated across the loop as the magnetic field varies.
Understanding EMF is vital because it tells us how much energy is available to drive the current around the loop, overcoming the circuit's total resistance and leading to energy dissipation.
Resistance
Resistance is a measure of the opposition to the flow of current in an electric circuit. It is denoted by the symbol \( R \) and measured in Ohms \( \Omega \). In our problem, the wire's resistance per unit length affects how much overall resistance the circuit has.
To find the total resistance of the loop, you multiply the resistance per unit length by the circumference of the loop:
  • Total resistance formula: \( R = \text{resistance per unit length} \times \text{length of wire} \)
Given a resistance per unit length of \( 3.3 \times 10^{-2} \Omega/\text{m} \) and a circumference of \( 0.7536 \text{ m} \), the total resistance of the loop is approximately \( 0.02486 \Omega \).
This resistance impacts how efficiently electrical power is converted into other forms such as thermal energy when current flows through the wire.
Power Dissipation
Power dissipation refers to the process of electric energy being converted into heat in a resistive element, resulting in energy being 'dissipated' as waste heat. This occurs in any circuit with resistance, and the power dissipated is calculated using the formula:
  • \( P = \frac{\varepsilon^2}{R} \)
Here, \( P \) is the power (in watts), \( \varepsilon \) is the induced EMF, and \( R \) is the resistance. For our circuit, substituting \( \varepsilon = 0.06032 \text{ V} \) and \( R = 0.02486 \Omega \), the power dissipated is about \( 0.146 \text{ W} \).
The total energy dissipated over time can then be found by multiplying the power by the time duration \( \Delta t \), giving us a total energy dissipation of \( 0.0657 \text{ J} \).
Understanding power dissipation is important in all electrical and electronic applications as it helps gauge efficiency and reliability over time.

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Most popular questions from this chapter

The resistances of the primary and secondary coils of a transformer are 56 and \(14 \Omega\), respectively. Both coils are made from lengths of the same copper wire. The circular turns of each coil have the same diameter. Find the turns ratio \(N_{\mathrm{s}} / N_{\mathrm{p}}\).

A long, current-carrying solenoid with an air core has 1750 turns per meter of length and a radius of \(0.0180 \mathrm{~m}\). A coil of 125 turns is wrapped tightly around the outside of the solenoid. What is the mutual inductance of this system?

Multiple-Concept Example 13 reviews the concepts used in this problem. A long solenoid (cross-sectional area \(=1.0 \times 10^{-6} \mathrm{~m}^{2}\), number of turns per unit length \(=2400\) turns \(/ \mathrm{m}\) ) is bent into a circular shape so it looks like a doughnut. This wire-wound doughnut is called a toroid. Assume that the diameter of the solenoid is small compared to the radius of the toroid, which is \(0.050 \mathrm{~m}\). Find the emf induced in the toroid when the current decreases to \(1.1\) A from \(2.5 \mathrm{~A}\) in a time of \(0.15 \mathrm{~s}\).

A house has a floor area of \(112 \mathrm{~m}^{2}\) and an outside wall that has an area of \(28 \mathrm{~m}^{2}\). The earth's magnetic field here has a horizontal component of \(2.6 \times 10^{-5} \mathrm{~T}\) that points due north and a vertical component of \(4.2 \times 10^{-5} \mathrm{~T}\) that points straight down, toward the earth. Determine the magnetic flux through the wall if the wall faces (a) north and (b) east. (c) Calculate the magnetic flux that passes through the floor.

Two \(0.68\) -m-long conducting rods are rotating at the same speed in opposite directions, and both are perpendicular to a 4.7-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 \(1.0\) -mm spark in air, what is the angular speed (in \(\mathrm{rad} / \mathrm{s}\) ) of the rods when a spark jumps across the gap?

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