Question

Given \(\mathrm{J}=-10^{-4}\left(\mathrm{ya}_{x}+x \mathbf{a}_{y}\right) \mathrm{A} / \mathrm{m}^{2}\), find the current crossing the \(y=0\) plane in the \(-\mathbf{a}_{y}\) direction between \(z=0\) and 1 , and \(x=0\) and 2 .

Short answer

Answer: \(-2 \times 10^{-4} \, A\)

Step-by-step solution

Set up the integral

Since we are interested in the current crossing the \(y=0\) plane in the -\(\mathbf{a_y}\) direction, we only need to consider the component of the current density in the -\(\mathbf{a_y}\) direction. So, our integral will be with respect to the -\(\mathbf{a_y}\) component of current density, i.e., \(-10^{-4}x\mathbf{a_y}\). The area we are integrating over is a rectangle in the \(xz\) plane, with x varying from 0 to 2 and z varying from 0 to 1. Therefore, the integral will be: \(\displaystyle \int_{0}^{2} \int_{0}^{1} -10^{-4}x \, \mathrm{d}z\, \mathrm{d}x\)

Integrate with respect to z

First, let's integrate the inner integral with respect to z: \(\displaystyle \int_{0}^{1} -10^{-4}x \, \mathrm{d}z = -10^{-4}x \int_{0}^{1} \, \mathrm{d}z\) When integrating with respect to z, the equation becomes: \(-10^{-4}x \left[z\right]_0^1 = -10^{-4}x\) Now, we have the following integral: \(\displaystyle \int_{0}^{2} -10^{-4}x \, \mathrm{d}x\)

Integrate with respect to x

Now, let's integrate the remaining integral with respect to x: \(\displaystyle \int_{0}^{2} -10^{-4}x \, \mathrm{d}x\) When integrating with respect to x, we get: \(-10^{-4} \left[\frac{1}{2}x^2\right]_0^2\)

Evaluate the integral

Evaluating the integral, we get: \(-10^{-4} \left[\frac{1}{2}(2^2)\right] = -10^{-4} \cdot 2 = -2 \times 10^{-4} \, A\) So, the current crossing the \(y=0\) plane in the \(-\mathbf{a}_{y}\) direction between \(z=0\) and 1, and \(x=0\) and 2 is \(-2 \times 10^{-4} \, A\).

Key concepts

Current Crossing Calculations

Understanding current crossing calculations is a practical application of electromagnetics that helps us predict the flow of electric current through specified boundaries. In the provided exercise, we are focusing on the current crossing the y=0 plane.

In this context, we need to consider the area of the plane through which the current passes, and the direction of current density vector. The given current density vector, \(\mathrm{J}=-10^{-4}(\mathrm{ya}_{x}+x \mathbf{a}_{y})\mathrm{A} / \mathrm{m}^{2}\), must be analyzed to extract the relevant component that contributes to the flow across the plane. Since we are looking at the plane at y=0, ya_{x} becomes zero and we're solely interested in the x \mathbf{a}_{y} component - particularly, its value in the negative \mathbf{a}_{y} direction.

Setting up an integral for the specific boundaries (from x=0 to x=2, and z=0 to z=1) enables us to calculate the total current. By carefully defining these bounds, we ensure accuracy and appropriateness of the results required for the calculation of current crossing in electromagnetics.

Electromagnetic Theory

The exercise taps into the fundamental aspects of electromagnetic theory, which provides a framework for understanding how electric and magnetic fields interact with matter. Current density, a key concept in this theory, is defined as the amount of electric charge passing through a specific area of a conductor in a set direction per unit time.

Electromagnetic theory explains that current density has both magnitude and direction, typically represented by vector fields. The concept of current density is crucial because it helps in picturing how currents distribute in different geometries and under diverse conditions. The negative sign in the given current density vector \(\mathrm{J}\) implicates that the current flows in the opposite direction of the indicated axes vector. When we consider the electromagnetism principles involved, applying the right-hand rule or understanding the orientation of vectors becomes essential in visualizing and solving problems related to current flow in a specific direction.

Integration in Electromagnetics

Integration is a powerful mathematical tool utilized in electromagnetics to calculate various physical quantities over a space or surface. In our exercise, integration allows us to find the total current crossing a certain area by summing up the contributions of current density at each infinitesimal element of the area.

The integral is set up over a plane considering the specific limits for x and z axes, forming a bounded region in space. The step-by-step solution demonstrates the importance of approaching the problem methodically, starting with integrating with respect to z since the integrand is independent of z, followed by integrating the result with respect to x. Finally, evaluating the definite integral yields the current. This process highlights how integration is utilized in electromagnetics to solve for currents, fields, and other quantities of interest across volumes, surfaces, or paths within electromagnetic fields.