Question

Let \(\mathrm{J}=400 \sin \theta /\left(r^{2}+4\right)\) a \(_{r} \mathrm{~A} / \mathrm{m}^{2} \cdot(a)\) Find the total current flowing through that portion of the spherical surface \(r=0.8\), bounded by \(0.1 \pi<\theta<0.3 \pi, 0<\phi<2 \pi .(b)\) Find the average value of \(\mathbf{J}\) over the defined area.

Short answer

Question: Find the total current flowing through a portion of a spherical surface and the average value of the current density J over the defined area, given the current density function \(J = \frac{400 \sin{\theta}}{(r^2 + 4)}\) A/m², where r is fixed at 0.8 meters, and the bounds for θ and φ are \(0.1\pi\) to \(0.3\pi\) and 0 to \(2\pi\), respectively. Answer: The total current flowing through the portion of the spherical surface is approximately 55.172 A, and the average value of the current density J over the defined area is approximately 17.56 A/m².

Step-by-step solution

Convert J to spherical coordinates

Given the current density function \(J = \frac{400 \sin{\theta}}{(r^2 + 4)}\) A/m², we see that it is already in spherical coordinates, and we have that \(r = 0.8\).

Find the total current flowing (Part A)

In spherical coordinates, we have the current density vector \(\mathbf{J} = J\mathbf{e}_r = \frac{400 \sin{\theta}}{(0.8^2 + 4)}\mathbf{e}_r\) A/m². The total current I flowing through the portion of surface is given by the integral of the current density over the surface. We have \(\displaystyle I= \iiint_{\text{sphere}} \mathbf{J} \cdot d\mathbf{S}\), where dS is the area element in spherical coordinates given by \(r^2 \sin{\theta} dr d\theta d\phi\). We will now integrate the given function \(\displaystyle I = \int_0^{2\pi}\int_{0.1\pi}^{0.3\pi}\int_0^{0.8} \frac{400(0.8)^2\sin{\theta}}{(0.8^2 + 4)}dr d\theta d\phi\). Since r is fixed at 0.8, we only need to perform the integration with respect to \(\theta\) and \(\phi\). So, \(I = (0.8)^2\int_0^{2\pi}\int_{0.1\pi}^{0.3\pi}\frac{400\sin{\theta}}{(0.8^2 + 4)} d\theta d\phi\). Now we can separate the integrals: \(I = \frac{400(0.8)^2}{(0.8^2 + 4)}\left(\int_{0.1\pi}^{0.3\pi}\sin{\theta} d\theta\right) \left(\int_0^{2\pi} d\phi\right)\). Calculating the integrals, we find: \(\displaystyle I = \frac{400 \cdot 0.64}{4.64}\left[\left(-\cos{(0.3\pi)}+\cos{(0.1\pi)}\right)\left(2\pi\right)\right] \approx 55.172\) A. So, the total current flowing through the portion of the spherical surface is approximately \(55.172\) A.

Calculate the average value of J (Part B)

To find the average value of J over the defined area, we will divide the total current obtained in part (a) by the total area of the surface. The area of the surface element in spherical coordinates is given by: \(dA= r^2 \sin{\theta} d\theta d\phi\). For the given bounds, the total area A can be calculated as: \(A = \int_0^{2\pi} \int_{0.1\pi}^{0.3\pi} (0.8)^2\sin{\theta} d\theta d\phi\). Separating the integrals, \(A = (0.8)^2\left(\int_{0.1\pi}^{0.3\pi} \sin{\theta}d\theta\right)\left(\int_0^{2\pi} d\phi\right)\). Calculating the integrals, we find: \(A = (0.8)^2\left[-\cos{(0.3\pi)}+\cos{(0.1\pi)}\right]\left(2\pi\right) \approx 3.1416\) m². Now, to find the average value of J, we take the total current from Part A, and divide it by the total area: \(\displaystyle \bar{J} = \frac{I}{A} =\frac{55.172}{3.1416} \approx 17.56\) A/m². Thus, the average value of the current density J over the defined area is approximately \(17.56\) A/m².

Key concepts

Spherical Coordinates

Spherical coordinates are a way to define a point in three-dimensional (3D) space using three values: the radius ( ), polar angle ( heta), and azimuthal angle (\phi). This system converts the traditional Cartesian coordinates (x, y, z) into a more convenient form for problems with symmetry about a point, like those involving spheres.
In our exercise, the given current density function, \( J = \frac{400 \sin{\theta}}{(r^2 + 4)} \) A/m², is already expressed in spherical coordinates. Here, \( r \) denotes the radial distance from the origin (e.g., \( r = 0.8 \) is given), while \( \theta \) and \( \phi \) are angular measurements.

• \( \theta \): the polar angle, measured from the positive z-axis.
• \( \phi \): the azimuthal angle, swept from the positive x-axis in the xy-plane.

The utility of spherical coordinates becomes clear when dealing with problems involving symmetry or radial fields, as they simplify integration and computation.

Current Density

Current density is crucial in electromagnetics as it describes how electric current flows in a given area. Analytically, it is defined as the electric current per unit area, typically expressed in amperes per square meter (A/m²).
In our problem, the expression of current density is \( J = \frac{400 \sin{\theta}}{(r^2 + 4)} \) A/m². This indicates that the current density not only varies with the direction (angle \( \theta \)) but also inversely with the square of the radius plus a constant.

• The use of \( \sin{\theta} \) indicates that the density is highest at the equatorial plane (maximum at \( \theta = \pi/2 \)) and decreases towards the poles (where \( \theta = 0 \) or \( \pi \)).
• By evaluating the density at \( r = 0.8 \), we focus on a specific radial distance, simplifying further integrations.

Ultimately, current density is an invaluable concept in defining how charges flow over a surface.

Surface Integral

Surface integrals allow us to calculate the total quantity of a field that passes through a specific surface. For current flow in electromagnetics, it gives the total current passing through a surface by integrating the current density over that area.
In spherical coordinates, the surface element \( dS \) is defined as \( r^2 \sin{\theta}\, d\theta \, d\phi \). This accounts for the radial distance and angular components on the spherical surface.
Integration Steps:

• Using the bounds from our problem \, \( 0.1\pi < \theta < 0.3\pi \) and \( 0 < \phi < 2\pi \), it provides the angular limits for integration.
• Because \( r \) is constant at 0.8, it simplifies our calculation, making this a two-dimensional integration over \( \theta \) and \( \phi \) only.

Through this integration, we derived the total current through the defined spherical surface portion, yielding a result of approximately 55.172 A.

Average Value Calculation

Average value calculations help us determine a typical substitute for variable quantities over a range. In our exercise, this involves finding the average current density over a specified surface area.
To achieve this, we need both the total integrated current \( I \), obtained using the surface integral, and the total area \( A \) of the designated surface.

• Total Area Calculation: The area of interest is given by \( A = (0.8)^2\int_{0.1\pi}^{0.3\pi} \sin{\theta} d\theta\ \int_0^{2\pi} d\phi \). Performing the integration provides \( A \approx 3.1416 \) m².
• Average Current Density: The average is found by dividing the total current flow by the area, \( \bar{J} = \frac{I}{A} \). This results in an approximate value of 17.56 A/m².

This average gives insight into how densely the current flows through the specified surface area, smoothing out the irregularities of the real distribution.