/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Shows a circular region of radiu... [FREE SOLUTION] | 91Ó°ÊÓ

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Shows a circular region of radius \(R=3.00 \mathrm{~cm}\) in which an electric flux is directed out of the plane of the page. The flux encircled by a concentric circle of radius \(r\) is given by \(\Phi_{E, \text { enc }}=(0.600 \mathrm{~V} \cdot \mathrm{m} / \mathrm{s})\) \((r / R) t,\) where \(r \leq R\) and \(t\) is in seconds. What is the magnitude of the induced magnetic field at radial distances (a) \(2.00 \mathrm{~cm}\) and (b) \(5.00 \mathrm{~cm} ?\)

Short Answer

Expert verified
At 2.00 cm, the magnitude of induced magnetic field is approximately 3.18 µT; at 5.00 cm, it is 0.

Step by step solution

01

Understanding the Problem

We are given an expression for electric flux \( \Phi_{E, \text{enc}} \) as a function of time \( t \) and radius \( r \), specifically \( \Phi_{E, \text{enc}} = (0.600 \, \mathrm{V} \cdot \mathrm{m/s}) \frac{r}{R} t \) for \( r \leq R \). The goal is to find the induced magnetic field at different radial distances \( r \).
02

Use Faraday's Law of Induction

Faraday's Law states that the electromotive force \( \varepsilon \) around a closed loop is equal to the negative rate of change of electric flux through that loop, i.e., \( \varepsilon = -\frac{d\Phi_{E, \text{enc}}}{dt} \). We use this to calculate \( \varepsilon \).
03

Evaluate the Rate of Change of Flux

Substitute the expression for electric flux into Faraday's Law: \[ \varepsilon = -\frac{d}{dt}\left((0.600 \, \mathrm{V} \cdot \mathrm{m/s}) \frac{r}{R} t\right) = -(0.600 \, \mathrm{V} \cdot \mathrm{m/s}) \frac{r}{R}, \]since the derivative of \( t \) with respect to \( t \) is 1.
04

Calculate the Induced Magnetic Field for r = 2.00 cm

First, consider \( r = 2.00 \, \mathrm{cm} \). Note \( r \leq R \). The induced magnetic field \( B \) around a circle of radius \( r \) is given by \( B(2\pi r) = \varepsilon \). Solve for \( B \):\[ B = \frac{-(0.600 \, \mathrm{V} \cdot \mathrm{m/s}) \frac{r}{R}}{2\pi r} = -\frac{0.600}{2\pi R} \].Substitute \( R = 3.00 \, \text{cm} \) (convert to meters): \[ B = -\frac{0.600}{2\pi \times 0.03} = -\frac{0.600}{0.1885} \approx -3.18 \, \mu\mathrm{T} \].
05

Calculate the Induced Magnetic Field for r = 5.00 cm

Since \( r = 5.00 \, \mathrm{cm} > R \), the flux through the area \( r = 5.00 \, \mathrm{cm} \) is the same as for \( r = 3.00 \, \mathrm{cm} \). Thus, \( \varepsilon \) and \( B \) are both zero because there's no change in total flux (outside the circle of radius \( R \)).Therefore, \( B = 0 \, \mathrm{T} \).

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Key Concepts

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

Electric Flux
Electric flux is a measure of the electric field passing through a given area. Imagine it as a flow of electric field lines through a surface. This concept is especially useful when dealing with variations in electric fields, like in this exercise.
Formula Insight: The given formula for the electric flux in a circular region is \( \Phi_{E, \text{enc}} = (0.600 \, \mathrm{V} \cdot \mathrm{m/s}) \frac{r}{R} t \). Here, \( r \) is the radius of the encircling loop, \( R \) is the maximum region radius, and \( t \) represents time. This informs us that the flux is directly proportional to both \( r \) and \( t \), indicating flux grows as these values increase.
Understanding electric flux is crucial because its changes often induce other effects, like magnetic fields. It represents how electric energy is transmitted through a space, which helps engineers and scientists analyze and predict the behavior dynamically changing electric fields.
Induced Magnetic Field
An induced magnetic field is generated as a result of changes in electric flux. According to Faraday's Law of Induction, changes in the magnetic environment of a circuit produce an electromotive force, leading to an induced magnetic field. This is a fascinating phenomenon where a simple change in one type of field creates another.
Application in the Exercise: In the problem, as the electric flux through a circular region changes, the induced magnetic field is calculated using the relation \( \varepsilon = -\frac{d\Phi_{E, \text{enc}}}{dt} \). Once we calculate the electromotive force (EMF), we can directly find the magnetic field \( B \). For instance, at \( r = 2.00 \, \text{cm} \), the induced field is around \(-3.18 \, \mu\mathrm{T} \). This illustrates Lenz's Law, which points to how the induced field acts to oppose the change in flux, maintaining balance in the system.
Understanding these interactions gives a solid foundation for exploring electromagnetic applications, from electric generators to transformers.
Rate of Change of Flux
The rate of change of flux is a critical component in applying Faraday's Law. It quantifies how quickly the electric flux through a particular area is changing over time, which directly affects the induced EMF and, consequently, the magnetic field.
Steps to Determine the Rate:
  • Use the given formula for flux: \( \Phi_{E, \text{enc}} = (0.600 \, \mathrm{V} \cdot \mathrm{m/s}) \frac{r}{R} t \).
  • Differentiate with respect to \( t \): The result is \(-0.600 \, \mathrm{V} \cdot \mathrm{m/s} \frac{r}{R} \), which is constant for a fixed \( r \).
This negative sign indicates the direction of induced EMF apprehends the change, attempting to maintain equilibrium. When assessing at \( r = 5.00 \, \text{cm} \), for instance, the change in total flux does not produce a new magnetic field outside, showing that only the flux change within \( R \) can cause induction.
Grasping how flux changes guide induction phenomena is fundamental for various scientific and engineering disciplines including atmospheric science and telecommunication equipment design.

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Most popular questions from this chapter

The saturation magnetization \(M_{\max }\) of the ferromagnetic metal nickel is \(4.70 \times 10^{5} \mathrm{~A} / \mathrm{m} .\) Calculate the magnetic dipole moment of a single nickel atom. (The density of nickel is \(8.90 \mathrm{~g} / \mathrm{cm}^{3}\), and its molar mass is \(58.71 \mathrm{~g} / \mathrm{mol} .)\)

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