Question

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 \(\mathbf{H}\).

Short answer

b) What is the expression for the total current crossing the surface within a cylindrical boundary of radius \(\rho_0\) at \(z=0\)? c) What is the expression for the magnetic field \(\mathbf{H}\)?

Step-by-step solution

Find the current density \(\mathbf{J}\)

According to Ohm's law, the current density \(\mathbf{J}\) is related to the conductivity \(\sigma\) and the electric field \(\mathbf{E}\) as: \(\mathbf{J} = \sigma \mathbf{E}\) The given conductivity is \(\sigma = 1.5 e^{-150 \rho} \mathrm{kS} / \mathrm{m}\) and the electric field is \(\mathbf{E} = 30 \mathbf{a}_{z} \mathrm{V} / \mathrm{m}\). Substitute these values into the equation: \(\mathbf{J} = (1.5 e^{-150 \rho} \mathrm{kS} / \mathrm{m}) \cdot (30 \mathbf{a}_{z} \mathrm{V} / \mathrm{m})\) \(\mathbf{J} = 45 e^{-150 \rho} \mathbf{a}_{z} \mathrm{A} / \mathrm{m}^2\)

Find the total current crossing the surface \(\rho < \rho_{0}\), \(z=0\)

To find the total current crossing the surface, we need to integrate the current density \(\mathbf{J}\) over the given surface. In cylindrical coordinates, the surface integral is given by: \(I = \int\int_S \mathbf{J} \cdot d\mathbf{S} = \int_{0}^{2\pi}\int_{0}^{\rho_{0}} (45 e^{-150 \rho} \mathbf{a}_{z}) \cdot (\rho d\rho d\phi \mathbf{a}_{z})\) \(I = 45 \int_{0}^{2\pi} d\phi \int_{0}^{\rho_{0}} e^{-150 \rho} \rho d\rho\) Evaluate the integrals separately: \(\int_{0}^{2\pi} d\phi = 2\pi\) \(\int_{0}^{\rho_{0}} e^{-150 \rho} \rho d\rho = \frac{1}{150^2} \left[-e^{-150 \rho} (150 \rho + 1)\right]_{0}^{\rho_{0}} = \frac{1}{150^2}(1 - e^{-150 \rho_0}(150\rho_0 + 1))\) Combine the results from both integrals: \(I = 45(2\pi) \frac{1}{150^2}(1 - e^{-150 \rho_0}(150\rho_0 + 1))\)

Use Ampère's law to find the magnetic field \(\mathbf{H}\)

We'll use Ampère's circuital law in cylindrical coordinates to find the magnetic field \(\mathbf{H}\): \(\oint_{C} \mathbf{H} \cdot d\mathbf{l} = I_{enc}\) Here, \(I_{enc}\) is the total current enclosed by a circle of radius \(\rho\). From step 2, we have an expression for the total current enclosed within a circle of radius \(\rho_0\): \(I_{enc} = 45(2\pi) \frac{1}{150^2}(1 - e^{-150 \rho}(150\rho + 1))\) In cylindrical coordinates, the Ampère's law equation is: \(H_{\phi}(\rho) 2 \pi \rho = 45(2\pi) \frac{1}{150^2}(1 - e^{-150 \rho}(150\rho + 1))\) Solve for \(H_{\phi}(\rho)\): \(H_{\phi}(\rho) = \frac{45}{150^2} \frac{1 - e^{-150 \rho}(150\rho + 1)}{\rho}\) So, the magnetic field in cylindrical coordinates is: \(\mathbf{H} = H_{\phi}(\rho) \mathbf{a}_{\phi} = \frac{45}{150^2} \frac{1 - e^{-150 \rho}(150\rho + 1)}{\rho} \mathbf{a}_{\phi}\)

Key concepts

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional coordinate system that extends polar coordinates into three dimensions. It's especially useful in scenarios involving symmetry around a central axis, such as in problems dealing with cylinders or circular patterns. The three descriptors in this system are: radial distance \( \rho \), which measures how far from the origin the point is horizontally; azimuthal angle \( \phi \), which indicates the angle in the xy-plane from the x-axis; and axial distance (z), which is the same as in Cartesian coordinates.

In our exercise, the use of cylindrical coordinates simplifies the process of integration when finding the total current over a circular surface. The symmetry inherent in a problem like this means cylindrical coordinates are often the most natural choice for computational ease.

Ohm's Law

Ohm's law is a fundamental principle in electromagnetism that describes the relationship between electric current, voltage, and resistance in a conductor. It states that the current \( \mathbf{I} \) flowing through a conductor is directly proportional to the voltage \( \mathbf{V} \) and inversely proportional to the resistance \( \mathbf{R} \), given by the formula \( \mathbf{I} = \frac{\mathbf{V}}{\mathbf{R}} \).

In the context of fields and current density, Ohm’s law takes the form \( \mathbf{J}= \sigma \mathbf{E} \) where \( \mathbf{J} \) is the current density, \( \sigma \) is the conductivity, and \( \mathbf{E} \) is the electric field. This allows us to calculate the current density given the electric field and conductivity, as seen in the step-by-step solution where \( \mathbf{J} = 45 e^{-150 \rho} \mathbf{a}_{z} \mathrm{A} / \mathrm{m}^2 \) is found by applying Ohm's law.

Current Density

Current density is a measure of the electric current \( \mathbf{I} \) per unit area \( \mathbf{A} \) flowing through a material at a given point. It is a vector quantity, represented as \( \mathbf{J} \), and is defined as the amount of charge that passes through a specific area in a specific direction per unit time. The unit for current density is amperes per square meter (\( \mathrm{A}/\mathrm{m}^2 \)).

In the exercise, we calculated the current density using Ohm's law adapted for fields, which involved the given electric field and conductivity. Understanding current density is crucial as it leads to calculations of the total current flowing through a given surface area, like the circular surface in the exercise, which was found by an integration process over the area of interest.

Ampère's Circuital Law

Ampère's circuital law is a fundamental law in magnetostatics that relates magnetic fields to the electric currents that produce them. In integral form, the law states that the line integral of the magnetic field \( \mathbf{H} \) around a closed loop \( C \) is equal to the free current \( I_{enc} \) passing through the area enclosed by the loop: \( \oint_{C} \mathbf{H} \cdot d\mathbf{l} = I_{enc} \).

In our exercise, Ampère’s law revealed the relationship between the magnetic field and the current density we previously derived. As the current density changed with \( \rho \) due to the cylindrical symmetry, we used the law in the context of cylindrical coordinates to find an expression for \( \mathbf{H} \) involving the azimuthal component \( H_{\phi}(\rho) \) which accounts for the circular path of integration.