/*! 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 73 A rod of pure silicon (resistivi... [FREE SOLUTION] | 91Ó°ÊÓ

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A rod of pure silicon (resistivity \(\rho=2300 \Omega \cdot \mathrm{m} )\) is carry-ing a current. The electric field varies sinusoidally with time according to \(E=E_{0} \sin \omega t,\) where \(E_{0}=0.450 \mathrm{V} / \mathrm{m}, \omega=2 \pi f,\) and the frequency \(f=120 \mathrm{Hz}\) (a) Find the magnitude of the maximum conduction current density in the wire. (b) Assuming \(\epsilon=\epsilon_{0}\) , find the maximum displacement current density in the wire, and compare with the result of part (a). (c) At what frequency \(f\) would the maximum conduction and displacement densitics become equal if \(\epsilon=\epsilon_{0}\) (which is not actually the case)? (d) At the frequency determined in part (c), what is the relative phase of the conduction and displacement currents?

Short Answer

Expert verified
Maximum conduction density: \( 1.957 \times 10^{-4} \, \mathrm{A/m}^2 \). Displacement density: \( 3.2 \times 10^{-11} \, \mathrm{A/m}^2 \). Equal density frequency: \( 7.91 \times 10^6 \, \mathrm{Hz} \); Phase difference: 90°.

Step by step solution

01

Understanding the given data

We are given these values: resistivity \( \rho = 2300 \, \Omega \cdot \mathrm{m} \), maximum electric field \( E_0 = 0.450 \, \mathrm{V}/\mathrm{m} \), frequency \( f = 120 \, \mathrm{Hz} \), and \( \omega = 2\pi f \). We need to find conduction current density and displacement current density under these conditions.
02

Calculate maximum conduction current density

We use Ohm's law for a uniform field, \( \mathbf{J} = \sigma \mathbf{E} \), where \( \sigma = \frac{1}{\rho} \) is the conductivity. We find the conductivity as \( \sigma = \frac{1}{2300} \, \Omega^{-1} \cdot \mathrm{m}^{-1} \). Then, the maximum conduction current density is \( J_{\text{max}} = \sigma E_0 = \left( \frac{1}{2300} \right) \times 0.45 = 1.957 \times 10^{-4} \, \mathrm{A/m}^2 \).
03

Calculate maximum displacement current density

The displacement current density is given by \( J_{d} = \epsilon \frac{dE}{dt} \). With \( E = 0.450 \sin(\omega t) \), we find \( \frac{dE}{dt} = 0.450 \omega \cos(\omega t) \). The maximum value of \( \cos(\omega t) \) is 1, so \( J_{d, \text{max}} = \epsilon_0 \times 0.450 \times \omega = \epsilon_0 \times 0.450 \times 2\pi \times 120 = 3.2 \times 10^{-11} \, \mathrm{A/m}^2 \).
04

Calculate frequency for equal conduction and displacement densities

For conduction current density and displacement current density to be equal, \( \sigma E_0 = \epsilon_0 \omega E_0 \). This simplifies to \( \sigma = \epsilon_0 \omega \) \( \Rightarrow \) \( \omega = \frac{\sigma}{\epsilon_0} \). Solving for \( \omega \) gives \( \omega = \frac{1}{2300 \times 8.854 \times 10^{-12}} = 4.97 \times 10^7 \, \mathrm{rad/s} \). So \( f = \frac{\omega}{2\pi} \approx 7.91 \times 10^6 \, \mathrm{Hz} \).
05

Phase between conduction and displacement currents at this frequency

Conduction current \( J = J_0 \sin(\omega t) \) and displacement current \( J_d = J_{d,0} \cos(\omega t) \). Thus, the conduction current lags the displacement current by 90 degrees (or \( \frac{\pi}{2} \) radians).

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Key Concepts

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

Ohm's Law
Ohm's Law is a foundational principle in electromagnetism and electrical engineering. It relates the voltage, current, and resistance in a conductive material. The basic form of Ohm's Law is expressed as:
  • \( V = IR \)
where
  • \( V \) is the voltage across the conductor in volts,
  • \( I \) is the current through the conductor in amperes, and
  • \( R \) is the resistance of the conductor in ohms.
This equation demonstrates how the current flow through a conductor relates to the applied voltage and its inherent resistance. A high resistance material will restrict the current flow, much like a narrow pipe restricts water flow.
In electromagnetism, especially for uniform fields within materials, we often use a variant of Ohm's Law:
  • \( \mathbf{J} = \sigma \mathbf{E} \)
where:
  • \( \mathbf{J} \) represents the current density, describing how much electric current is flowing per unit area of the material,
  • \( \sigma \) is the conductivity of the material (inverse of resistivity, \( \rho \)), and
  • \( \mathbf{E} \) is the electric field applied across the material.
Here, the current density \( \mathbf{J} \) is directly proportional to the electric field \( \mathbf{E} \) and is contingent upon the material's conductivity. This relationship is crucial for understanding how materials conduct electricity and is especially pertinent when working with different materials like silicon in electronics.
Conduction Current Density
Conduction current density is a key concept in electromagnetism, describing how electric current flows through a conductor per unit area. It's particularly important in analyzing how efficiently materials like silicon conduct electrical currents. The conduction current density \( \mathbf{J} \) is expressed through the equation:
  • \( \mathbf{J} = \sigma \mathbf{E} \)
where:
  • \( \sigma \) represents the material's conductivity,
  • \( \mathbf{E} \) is the electric field.
Conductivity \( \sigma \) is a measure of how easily a material allows current to pass through it. For any given electric field, a high conductivity implies a high current density.
In the context of the original exercise, the conductivity of silicon is calculated using its resistivity: \( \sigma = \frac{1}{\rho} \). Once we know \( \sigma \), we use that value with the electric field \( E_0 \) to find the maximum conduction current density. This is pivotal to understanding how currents behave in electrical components made from different materials and affects how we design circuits.
Displacement Current Density
Displacement current density is a fundamental concept introduced to extend the applicability of Ampere's law in electrodynamics. Unlike conduction current, displacement current does not involve actual charge carriers moving through the material. Instead, it is related to the time-varying electric field and is significant in scenarios where fields change over time, such as in capacitors and electromagnetic waves.
The displacement current density \( J_d \) is defined by the equation:
  • \( J_d = \epsilon_0 \frac{dE}{dt} \)
where:
  • \( \epsilon_0 \) is the permittivity of free space,
  • \( \frac{dE}{dt} \) is the time derivative of the electric field.
In essence, displacement current fills in the gaps in Maxwell's equations, particularly in non-conductive regions where there's a changing electric field but no conduction current.
In the exercise, the electric field \( E \) is given as a sinusoidal function of time. By taking the derivative with respect to time, we calculate the rate of change of the electric field, which in turn helps us find the displacement current density. Understanding this concept is essential for analyzing circuits with alternating currents and electromagnetic radiation phenomena, ensuring comprehensive coverage of electrodynamics principles.

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Most popular questions from this chapter

A rectangle measuring 30.0 \(\mathrm{cm}\) by 40.0 \(\mathrm{cm}\) is located inside a region of a spatially uniform magnetic field of 1.25 \(\mathrm{T}\) , with the field perpendicular to the plane of the coil (Fig. 29.29 ). The coil is pulled out at a steady rate of 2.00 \(\mathrm{cm} / \mathrm{s}\) traveling perpendicular to the field lines. The region of the field ends abruptly as shown. Find the emf induced in this coil when it is (a) all inside the field: (b) partly inside the field; (c) all outside the field.

Motional emfs in Transportation. Airplanes and trains move through the earth's magnetic field at rather high speeds, so it is reasonable to wonder whether this field can have a substantial effect on them. We shall use a typical value of 0.50 \(\mathrm{G}\) for the earth's field (a) The French TGV train and the Japanese "bullet train" reach speeds of up to 180 \(\mathrm{mph}\) moving on tracks about 1.5 \(\mathrm{m}\) apart. At top speed moving perpendicular to the earth's magnetic field, what potential difference is induced across the tracks as the wheels roll? Does this seem large enough to produce noticeable effects? (b) The Boeing \(747-400\) aircraft has a wingspan of 64.4 \(\mathrm{m}\) and a cruising speed of 565 \(\mathrm{mph}\) . If there is no wind blowing (so that this is also their speed relative to the ground), what is the maximum potential difference that could be induced between the opposite tips of the wings? Does this seem large enough to cause problems with the plane?

In a region of space, a magnetic ficld points in the \(+x\) -direction (toward the right). Its magnitude varies with position according to the formula \(B_{x}=B_{0}+b x,\) where \(B_{0}\) and \(b\) are positive constants, for \(x \geq 0\) . A flat coil of area \(A\) moves with uniform speed \(v\) from right to left with the plane of its area always perpendicular to this field. (a) What is the emf induced in this coil while it is to the right of the origin? (b) As viewed from the origin, what is the direction (clockwise or counterclockwise) of the current induced in the coil? (c) If instead the coil moved from left to right, what would be the answers to parts (a) and (b)?

It is impossible to have a uniform electric field that abruptly drops to zero in a region of space in which the magnetic field is constant and in which there are no electric charges. To prove this statement, use the method of contradiction: Assume that such a case is possible and then show that your assumption contradicts a law of nature. (a) In the bottom half of a piece of paper, draw evenly spaced horizontal lines representing a uniform electric field to your right. Use dashed lines to draw a rectangle abcda with horizontal side ab in the electric-field region and horizontal side \(c d\) in the top half of your paper where \(E=0 .\) (b) Show that integration around your rectangle contradicts Faraday's law, Eq. \((29.21) .\)

A circular wire loop of radius \(a\) and resistance \(R\) initially has a magnetic flux through it due to an external magnetic field. The extermal field then decreases to zero. A current is induced in the loop while the external field is changing; however, this current does not stop at the instant that the external field stops changing. The reason is that the current itself generates a magnetic field, which gives rise to a flux through the loop. If the current changes, the flux through the loop changes as well, and an induced emf appears in the loop to oppose the change. (a) The magnetic field at the center of the loop of radius a produced by a current i in the loop is given by \(B=\mu_{0} i / 2 a\) . If we use the crode approximation that the field has this same value at all points within the loop, what is the flux of this field through the loop? (b) By using Faraday's law, Eq. \((29.3),\) and the relationship \(\mathcal{E}=i R,\) show that after the external field has stopped changing, the current in the loop obeys the differential equation $$ \frac{d i}{d t}=-\left(\frac{2 R}{\pi \mu_{0} a}\right) i $$ (c) If the current has the value \(i_{0}\) at \(t=0\) , the instant that the external field stops changing. snive the equation in part \((b)\) to find \(i\) as a function of time for \(t>0 .\) (Hint: In Section 26.4 we encountered a similar differential equation, Eq. \((26.15),\) for the quantity \(q .\) This equation for \(i\) may be solved in the same way. (d) If the loop has radius \(a=50 \mathrm{cm}\) and resistance \(R=0.10 \Omega,\) how long after the external field stops changing will the current be equal to 0.010 \(\mathrm{o}\) (that is, \(\frac{1}{100}\) of its initial value)? (e) In solving the examples in this chapter, we ignored the effects described in this problem. Explain why this is a good approximation.

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