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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 6.327 J.

Step by step solution

01

Calculate the Circumference of the Loop

The copper wire is formed into a circular loop with a radius of \(12 \text{ cm}\). First, calculate the circumference of the loop, which is also the length of the wire: \[ C = 2\pi r \]where \(r = 12 \text{ cm} = 0.12 \text{ m}\). Thus, \[ C = 2\pi (0.12) = 0.24\pi \text{ m} \approx 0.754 \text{ m} \].
02

Calculate the Resistance of the Wire

The resistance of the wire can be calculated using the resistance per unit length and the total circumference of the loop:\[ R = \text{Resistance per unit length} \times \text{Circumference} \]Given that the resistance per unit length is \(3.3 \times 10^{-2} \Omega/\text{m}\),\[ R = 3.3 \times 10^{-2} \times 0.754 \approx 0.0249 \Omega \].
03

Determine the Induced EMF

Next, calculate the electromotive force (EMF) induced in the loop due to the changing magnetic field using Faraday's law of induction:\[ \text{EMF} = -\frac{\Delta \Phi_B}{\Delta t} \]where \( \Phi_B = B \times A \) is the magnetic flux, \(B\) is the magnetic field, and \(A=\pi r^2\) is the area of the loop.The change in magnetic flux, \(\Delta \Phi_B\), as \(B\) changes from 0 to 0.6 T, is:\[ \Delta \Phi_B = (0.60) \times \pi (0.12)^2 - 0 \]\[ = 0.60 \times \pi \times 0.0144 \approx 0.0272\pi \text{ Tm}^2 \].Therefore, the induced EMF is:\[ \text{EMF} = \frac{0.0272\pi}{0.45} \approx 0.190\pi \text{ V} \].
04

Calculate the Average Power Dissipated

Using Ohm's Law, relate the EMF to the current \(I\):\[ \text{EMF} = I \times R \]Therefore, \( I = \frac{\text{EMF}}{R} \).The power \(P\) dissipated can be calculated using the formula:\[ P = I^2R \]Plugging the values:\[ I = \frac{0.190\pi}{0.0249} \approx 23.73 \text{ A} \]\[ P = (23.73)^2 \times 0.0249 \approx 14.06 \text{ W} \].
05

Calculate the Average Electrical Energy Dissipated

The average electrical energy \(E\) dissipated over time \(t\) is given by:\[ E = P \times t \]With \(P = 14.06 \text{ W}\), the energy dissipated over \(0.45 \text{ s}\) is:\[ E = 14.06 \times 0.45 \approx 6.327 \text{ J} \].

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Key Concepts

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

Faraday's Law of Induction
Faraday's law of induction is a fundamental principle in electromagnetism that explains how a change in magnetic field within a closed loop induces an electromotive force (EMF) in the loop. This phenomenon is crucial in understanding devices like transformers and generators. In simpler terms, when the magnetic environment of a coil of wire changes, an electric current is induced in the wire. According to Faraday's law, the induced EMF is given by:\[\text{EMF} = - \frac{\Delta \Phi_B}{\Delta t}\]where \( \Phi_B = B \times A \) is the magnetic flux, \( A \) is the area of the loop, and \( B \) is the magnetic field. The negative sign indicates the direction of the induced EMF is such that it opposes the change in flux (Lenz's law).
To put it into practice, consider a magnetic field that increases from 0 T to 0.6 T over 0.45 s in a circular loop of radius 0.12 m. Calculate the change in flux, \( \Delta \Phi_B \), then divide by the time, \( \Delta t \), to find the EMF. Understanding this induction principle is crucial for solving problems involving magnetic fields and induced currents.
Resistance Calculation
Resistance plays a crucial role in determining how much current flows in a circuit. To find the resistance of a particular material, it's important to know both the resistance per unit length and the total length of the material. The formula for resistance is:\[ R = \text{Resistance per unit length} \times \text{Length of wire} \]In the example, we have a copper wire bent into a loop with a circumference of approximately 0.754 m.
Given a resistance per unit length of \(3.3 \times 10^{-2} \, \Omega/\text{m}\), the total resistance of the wire can be calculated as:\[ R = 3.3 \times 10^{-2} \times 0.754 \approx 0.0249 \, \Omega \]
By understanding how to calculate resistance, you can determine how much opposition there will be to the flow of current in a circuit. This value is essential when analyzing circuits and predicting how they will behave when activated.
Average Electrical Energy Dissipation
Average electrical energy dissipation in a circuit refers to the energy lost over time due to resistance. This is an essential concept when considering how much energy is utilized or wasted within an electrical system. The power dissipated as heat in the wire, due to its resistance, can be calculated using the formula:\[ P = I^2R \]where \( I \) is the current and \( R \) is the resistance. The average energy dissipated over a time period \( t \) is determined by:\[ E = P \times t \]
In the example problem, the power dissipated is approximately 14.06 W and the time is 0.45 s. Therefore, the energy lost, or dissipated, is:\[ E = 14.06 \times 0.45 \approx 6.327 \, J \]
By understanding average electrical energy dissipation, you can assess how efficiently a device operates. It also allows you to calculate the cost in terms of energy loss, which is critical for designing energy-efficient systems. This concept is particularly important in everyday applications, from household electronics to large-scale power systems.

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Most popular questions from this chapter

Mutual induction can be used as the basis for a metal detector. A typical setup uses two large coils that are parallel to each other and have a common axis. Because of mutual induction, the ac generator connected to the primary coil causes an emf of \(0.46 \mathrm{V}\) to be induced in the secondary coil. When someone without metal objects walks through the coils, the mutual inductance and, thus, the induced emf do not change much. But when a person carrying a handgun walks through, the mutual inductance increases. The change in emf can be used to trigger an alarm. If the mutual inductance increases by a factor of three, find the new value of the induced emf.

Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.47 T. When viewed edge-on, the first surface is tilted at an angle of \(12^{\circ}\) from the horizontal, and a net magnetic flux of \(8.4 \times 10^{-3} \mathrm{Wb}\) passes through it. The same net magnetic flux passes through the second surface. (a) Determine the area of the first surface. (b) Find the smallest possible value for the area of the second surface.

The plane of a flat, circular loop of wire is horizontal. An external magnetic field is directed perpendicular to the plane of the loop. The magnitude of the external magnetic field is increasing with time. Because of this increasing magnetic field, an induced current is flowing clockwise in the loop, as viewed from above. What is the direction of the external magnetic field? Justify your conclusion.

Two coils of wire are placed close together. Initially, a current of 2.5 A exists in one of the coils, but there is no current in the other. The current is then switched off in a time of \(3.7 \times 10^{-2}\) s. During this time, the average emf induced in the other coil is \(1.7 \mathrm{V}\). What is the mutual inductance of the two-coil system?

A long solenoid of length \(8.0 \times 10^{-2} \mathrm{m}\) and cross-sectional area \(5.0 \times 10^{-5} \mathrm{m}^{2}\) contains 6500 turns per meter of length. Determine the emf induced in the solenoid when the current in the solenoid changes from 0 to 1.5 A during the time interval from 0 to 0.20 s.

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