Question

Given the vector field \(\mathbf{E}=4 z y^{2} \cos 2 x \mathbf{a}_{x}+2 z y \sin 2 x \mathbf{a}_{y}+y^{2} \sin 2 x \mathbf{a}_{z}\) for the region \(|x|,|y|\), and \(|z|\) less than 2, find \((a)\) the surfaces on which \(E_{y}=0 ;(b)\) the region in which \(E_{y}=E_{z} ;(c)\) the region in which \(\mathbf{E}=0 .\)

Short answer

Question: Determine the conditions for the following cases in the given vector field \(\mathbf{E}=4 z y^{2} \cos 2 x \mathbf{a}_{x}+2 z y \sin 2 x \mathbf{a}_{y}+y^{2} \sin 2 x \mathbf{a}_{z}\): (a) The surfaces on which \(E_y = 0\). (b) The region in which \(E_y = E_z\). (c) The region in which \(\mathbf{E} = 0\). Answer: (a) The surfaces where \(E_y = 0\) are given by: 1. \(y = 0\). 2. \(z = 0\). 3. \(x = \frac{n\pi}{2}\), where \(n\) is any integer. (b) The region where \(E_y = E_z\) is described by the equation \(y = 2z\). (c) The region where \(\mathbf{E} = 0\) is given by the following intersections: 1. Intersection of the surfaces \(y=0\) and \(x = \frac{(2n+1)\pi}{4}\). 2. Intersection of the surfaces \(z=0\) and \(x = \frac{(2n+1)\pi}{4}\). 3. Intersection of the surfaces \(y=0\) and \(z=0\).

Step-by-step solution

Part a: Surfaces where \(E_y = 0\)

To find the surfaces where \(E_y = 0\), we examine the \(y\) component of the given vector field, given by: \(E_y = 2 z y \sin 2x\). To make this component zero, we need to find the values of \(x\), \(y\), and \(z\) that satisfy: \(2zy\sin(2x) = 0\). There are three possible conditions for this to occur: 1. \(y = 0\). 2. \(z = 0\). 3. \(\sin(2x) = 0\). In the case of (3), we know that the sine function is zero at integer multiples of \(\pi\), so \(\sin(2x) = 0\) when \(2x = n\pi\) for any integer \(n\). This means that \(x = \frac{n\pi}{2}\). Therefore, the surfaces where \(E_y = 0\) are given by: 1. \(y = 0\). 2. \(z = 0\). 3. \(x = \frac{n\pi}{2}\), where \(n\) is any integer.

Part b: Region where \(E_y = E_z\)

To find the region in which the \(y\) component of the vector field is equal to the \(z\) component, we set \(E_y = E_z\): \(2 z y \sin 2 x = y^2 \sin 2 x\). Divide through by \(y\sin(2x)\) to cancel out equal terms: \(2z = y\). So, the region where \(E_y = E_z\) is described by the equation \(y = 2z\).

Part c: Region where \(\mathbf{E} = 0\)

To find the region where the entire vector field is zero, we need to find when each component of the vector field is zero: $\begin{cases} 4 z y^2 \cos 2 x = 0 \\ 2 z y \sin 2 x = 0 \\ y^2 \sin 2 x = 0 \end{cases}$. The first equation is zero when \(y=0\), \(z=0\), or \(\cos(2x)=0\). The last two equations have already been analyzed in part (a). The cosine function is zero at odd multiples of \(\frac{\pi}{2}\), so \(\cos(2x) = 0\) when \(2x = (2n+1)\frac{\pi}{2}\) for any integer \(n\). This means that \(x = \frac{(2n+1)\pi}{4}\). Therefore, the region where \(\mathbf{E} = 0\) is given by the following intersections: 1. Intersection of the surfaces \(y=0\) and \(x = \frac{(2n+1)\pi}{4}\). 2. Intersection of the surfaces \(z=0\) and \(x = \frac{(2n+1)\pi}{4}\). 3. Intersection of the surfaces \(y=0\) and \(z=0\).

Key concepts

Electromagnetic Field

An electromagnetic field (EMF) is a physical field produced by electrically charged objects, affecting the behavior of charged objects in the vicinity of its origin. In vector calculus, this field is represented as a vector field consisting of both an electric field and a magnetic field. In our vector field analysis, we encounter the electric field component, represented by the vector field \(\mathbf{E}\), which varies with respect to position and consists of x, y, and z components.

Electromagnetic fields are fundamental to the workings of many electrical devices and are at the heart of the concepts such as induction and radiation. In this context, understanding how to manipulate vector fields like \(\mathbf{E}\) can shed light on the behavior of EMFs within defined regions and under various conditions. The exercise provided explores the properties of this field within a specific region, offering insight into the practical aspects of EMF analysis.

Vector Calculus

Vector calculus is a branch of mathematics concerned with differentiation and integration of vector fields, often dealing with functions that take vectors as inputs and outputs. It is essential for solving problems in fields such as physics and engineering, especially when dealing with vector fields like electromagnetic fields.

In this example, we use vector calculus techniques to determine the behavior of the electric field \(\mathbf{E}\) within a certain region. Through this analysis, we dissect the vector field into its components along the x, y, and z axes to understand where the field displays certain properties or is equal to zero. Students should be familiar with operations like setting vector components equal to zero, equating components, and identifying significant points or surfaces where these conditions are met, tasks common in vector calculus.

Surface Equations

Surface equations represent mathematical relationships that define surfaces in three-dimensional space. In the context of vector fields, these equations help in locating the specific geometric places where the field has particular values or properties. As seen in our example, setting the y-component of \(\mathbf{E}\) to zero reveals surfaces described by simple equations involving x, y, and z.

Understanding Surface Equations

When dealing with surface equations, students should pay attention to how changes in one variable may or may not affect the others. For instance, \(y=0\) is a plane that extends indefinitely along the x and z axes. Recognizing these surfaces allows for a spatial understanding of where vector field components become zero, leading to better comprehension of physical phenomena such as electrical null points or boundaries.

Sine and Cosine Functions

Sine and cosine functions are foundational in mathematics, particularly in trigonometry and the analysis of periodic phenomena. These functions relate the angles of a right-angled triangle to the ratios of two of its sides and are also interpreted as projections of uniform circular motion. In the vector field exercise, the sine and cosine functions help identify points and surfaces where the field’s magnitude changes direction or becomes zero.

Students should know that because these functions are periodic, they become zero at specific intervals. For \(\sin(2x)\), it's zero at multiples of \(\pi\), and for \(\cos(2x)\), it's zero at odd multiples of \(\frac{\pi}{2}\). This periodic nature is pivotal in determining the structure of the electromagnetic field over space, playing a significant role in the evaluation of vector fields as shown in the solution to the given exercise.