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Chapter 7: Problem 30

(An inversion of Problem 7.20.) A solid, nonmagnetic conductor of circular cross section has a radius of \(2 \mathrm{~mm}\). The conductor is inhomogeneous, with \(\sigma=10^{6}\left(1+10^{6} \mathrm{\rho}^{2}\right) \mathrm{S} / \mathrm{m}\). If the conductor is \(1 \mathrm{~m}\) in length and has a voltage of \(1 \mathrm{mV}\) between its ends, find: \((a) \mathbf{H}\) inside; \((b)\) the total magnetic flux inside the conductor.

Short Answer

Expert verified
Answer: The key steps in calculating the magnetic field and total magnetic flux inside a circular conductor with an inhomogeneous resistivity distribution are: 1. Convert the given resistivity to conductivity using the equation: \(\sigma=10^6 \left(1+10^6 \rho^2\right) S/m\). 2. Find the current density vector using Ohm's Law: \(\textbf{J} = \sigma \textbf{E}\), and calculate the electric field inside the conductor using the given voltage. 3. Find the magnetic field \(\textbf{H}\) inside the conductor by applying Ampere's Circuital Law: \(\oint \textbf{H} \cdot d\textbf{l} = I_{enc}\). Calculate the total current enclosed by a circular path inside the conductor with radius \(r\). 4. Calculate the total magnetic flux inside the conductor using the equation: \(\Phi_B = \int (\textbf{B} \cdot d\textbf{A})\). Obtain the magnetic flux density \(\textbf{B}\) using the relation: \(\textbf{B} = \mu_0 \textbf{H}\), and integrate it over the cross-section of the conductor.

Step by step solution

01

Convert the given resistivity to conductivity

We are given the resistivity \(\rho\) of the conductor, which is related to conductivity \(\sigma\) by the equation \(\sigma=10^6 \left(1+10^6 \rho^2\right) S/m\).
02

Find the current density vector

The current density vector \(J\) can be found using Ohm's Law. Ohm's Law states that \(\textbf{J} = \sigma \textbf{E}\). To find the electric field inside the conductor, we can use the equation \(V = \int \textbf{E} \cdot d\textbf{l} \) where V is the voltage and \(d\textbf{l}\) is a differential length element. Since the voltage is given, we can find the electric field inside the conductor. Then, we can find the current density vector using Ohm’s law.
03

Find the magnetic field H inside the conductor

To find the magnetic field \(\textbf{H}\) inside the conductor, we can use Ampere's Circuital Law, which states that: \(\oint \textbf{H} \cdot d\textbf{l} = I_{enc}\) where \(I_{enc}\) is the current enclosed by the path, and \(d\textbf{l}\) is a differential length element. By knowing the current density vector and the dimensions of the conductor, we can calculate the total current enclosed by a circular path inside the conductor with radius \(r\). Then, we can use Ampere's Circuital Law to find the \(\textbf{H}\) inside the conductor.
04

Calculate total magnetic flux inside the conductor

To calculate the total magnetic flux inside the conductor, we can use the equation: \(\Phi_B = \int (\textbf{B} \cdot d\textbf{A})\) where \(\Phi_B\) is the total magnetic flux and \(\textbf{B}\) is the magnetic flux density, which is related to the magnetic field \(\textbf{H}\) as: \(\textbf{B} = \mu_0 \textbf{H}\) and \(d\textbf{A}\) is a differential area element. By integrating this equation over the cross-section of the conductor, we can find the total magnetic flux inside the conductor.

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Key Concepts

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

Ampere's Circuital Law
Understanding Ampere's Circuital Law is crucial when dealing with magnetic fields in materials like conductors. Essentially, this law relates the magnetic field in a loop to the electric current passing through it. The law is mathematically expressed as \[\begin{equation} \oint \textbf{H} \cdot d\textbf{l} = I_{\text{enc}} \end{equation}\], where \(\oint\) represents a line integral over a closed loop, \(\textbf{H}\) is the magnetic field, \(d\textbf{l}\) is the differential element of the loop, and \(I_{\text{enc}}\) refers to the current enclosed by the loop.

In the context of our exercise, where we have a conductor with current flowing through it, Ampere's Law enables us to find the magnetic field inside the conductor. By selecting an appropriate loop within the conductor, we can calculate the total current enclosed and subsequently find the magnetic field, \(\textbf{H}\), at any point inside the material.
Ohm's Law
Ohm's Law is foundational in the study of circuits and conductive materials. This principle states that the current through a conductor is directly proportional to the voltage across it and inversely proportional to the resistance: \(V = IR\), where \(V\) is voltage, \(I\) current, and \(R\) resistance.

However, in the realm of electromagnetics, we often deal with continuous media and hence use a localized, field-based version of Ohm's Law: \(\textbf{J} = \sigma \textbf{E}\). Here \(\textbf{J}\) is the current density, \(\sigma\) is the conductivity, and \(\textbf{E}\) is the electric field inside the material. This formulation is utilized in the solution of our exercise to determine the current density from the known electric field and conductivity of the conductor.
Magnetic Flux
Magnetic Flux, denoted as \(\Phi_B\), is a measure of the quantity of magnetism, considering the strength and the extent of a magnetic field. It is given by the product of the magnetic field and the area it penetrates: \(\Phi_B = \int (\textbf{B} \cdot d\textbf{A})\), where \(\textbf{B}\) is the magnetic flux density, and \(d\textbf{A}\) represents a differential area perpendicular to the flux.

In our exercise, we apply this concept to determine the total magnetic flux within the conductor by integrating the magnetic field across the cross-sectional area of the conductor. Understanding how to calculate magnetic flux is essential for designing and analyzing systems like electric generators and transformers.
Conductivity
Conductivity, symbolized as \(\sigma\), is a scalar quantity that describes a material's ability to conduct electric current. It is the reciprocal of resistivity, \(\rho\), and its unit is siemens per meter (S/m). High conductivity implies that a material allows for easier flow of electric charge.

The provided exercise mentions an inhomogeneous conductor whose conductivity varies with the square of the radial position, \(\sigma = 10^6(1 + 10^6 \rho^2) \ S/m\). Knowing the conductivity is vital to determining not just the current distribution within the conductor, but also the resultant electric and magnetic fields using Ohm's Law and Ampere's Circuital Law, respectively.
Current Density
Current Density, denoted as \(\textbf{J}\), is a vector quantity that represents the amount of electric current flowing per unit cross-sectional area in a material, pointing in the direction of flow. It's crucial for understanding how current is distributed in conductive media.

In the context of our exercise, we calculate the current density by applying a form of Ohm's Law for continuous media, \(\textbf{J} = \sigma \textbf{E}\), where \(\sigma\) is the aforementioned conductivity, and \(\textbf{E}\) the electric field. This step is a precursor to finding the magnetic field inside the conductor via Ampere's Circuital Law.
Electric Field
The Electric Field, represented as \(\textbf{E}\), is a vector field that exerts force on charges, and its strength is measured in volts per meter (V/m). It describes the electrical force per unit charge at any point in space.

In our exercise, we ascertain the electric field strength throughout the conductor by integrating the voltage over the length of the conductor. A precise understanding of the electric field helps us calculate the current density and consequently predict how electricity behaves within the conductor, leading towards understanding the overall electromagnetic behavior.

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Most popular questions from this chapter

Assume that there is a region with cylindrical symmetry in which the conductivity is given by \(\sigma=1.5 e^{-150 \rho} \mathrm{kS} / \mathrm{m}\). An electric field of \(30 \mathbf{a}_{z} \mathrm{~V} / \mathrm{m}\) is present. ( \(a\) ) Find \(\mathbf{J}\). \((b)\) Find the total current crossing the surface \(\rho<\rho_{0}\), \(z=0\), all \(\phi\). ( \(c\) ) Make use of Ampère's circuital law to find \(\mathbf{H}\).

A square filamentary differential current loop, \(d L\) on a side, is centered at the origin in the \(z=0\) plane in free space. The current \(I\) flows generally in the \(\mathbf{a}_{\phi}\) direction. ( \(a\) ) Assuming that \(r>>d L\), and following a method similar to that in Section \(4.7\), show that $$d \mathbf{A}=\frac{\mu_{0} I(d L)^{2} \sin \theta}{4 \pi r^{2}} \mathbf{a}_{\phi}$$ (b) Show that $$d \mathbf{H}=\frac{I(d L)^{2}}{4 \pi r^{3}}\left(2 \cos \theta \mathbf{a}_{r}+\sin \theta \mathbf{a}_{\theta}\right)$$ The square loop is one form of a magnetic dipole.

The free space region defined by \(1

( \(a\) ) Find \(\mathbf{H}\) in rectangular components at \(P(2,3,4)\) if there is a current filament on the \(z\) axis carrying \(8 \mathrm{~mA}\) in the \(\mathbf{a}_{z}\) direction. ( \(b\) ) Repeat if the filament is located at \(x=-1, y=2\). ( \(c\) ) Find \(\mathbf{H}\) if both filaments are present.

Show that the line integral of the vector potential A about any closed path is equal to the magnetic flux enclosed by the path, or \(\oint \mathbf{A} \cdot d \mathbf{L}=\int \mathbf{B} \cdot d \mathbf{S}\).
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