Question

Given the current density $\mathbf{J}=-{10}^4[\sin (2x)e^{-2y}{\mathbf{a}}_x+\cos (2x)e^{-2y}{\mathbf{a}}_y]\mathrm{kA}/{\mathrm{m}}^2$ (a) Find the total current crossing the plane $y=1$ in the ${\mathbf{a}}_y$ direction in the region \(0

Short answer

Answer: The total current leaving the region calculated using the divergence theorem is approximately -17518 kA.

Step-by-step solution

Identify the ay component of the current density

We are given the current density J as $$\mathbf{J}=-{10}^4[\sin (2x)e^{-2y}{\mathbf{a}}_x+\cos (2x)e^{-2y}{\mathbf{a}}_y]$$The ay component is the one multiplying ${\mathbf{a}}_y$:$$J_y=-{10}^4\cos (2x)e^{-2y}$$ #Step 2: Integrate the ay component over the given region in x and z#

Integrate the current density in the ay direction

Since the region is 0<x<1 and 0<z<2, and the problem specifies the plane y=1 in the ${\mathbf{a}}_y$ direction, we need to evaluate the following integral:$$I={\int }_0^1{\int }_0^2{J}_y\mathrm{d}z\mathrm{d}x={\int }_0^1{\int }_0^2-{10}^4\cos (2x)e^{-2(1)}\mathrm{d}z\mathrm{d}x$$ #Step 3: Compute the integral and find the total current#

Compute the integral

Evaluate the integral:$$I={\int }_0^1{\int }_0^2-{10}^4\cos (2x)e^{-2}\mathrm{d}z\mathrm{d}x=-{10}^4{e}^{-2}[{\int }_0^1\cos (2x)\mathrm{d}x]\left[{\int }_0^2\mathrm{d}z\right]$$Using the result, we get:$$I=-{10}^4{e}^{-2}{[\frac{1}{2}\sin (2x)]}_0^1[z{]}_0^2=-{10}^4{e}^{-2}[\frac{1}{2}\sin (2)](2)$$Hence, the total current crossing the plane y=1 in the ay direction is$$I\approx -3015\mathrm{kA}$$ (b) Total current leaving the region using surface integral #Step 1: Find the dot product between the current density and dS#

Compute the dot product of J and dS

The total current leaving the region is given by the surface integral of the dot product between the current density and the differential surface element dS. For the given cube, the dot product can be computed over the six faces, which are parallel to the coordinate planes:$$\begin{array}{rl}{\iint }_S\mathbf{J}\cdot d\mathbf{S}& ={\iint }_{S_1}\mathbf{J}\cdot d{\mathbf{S}}_1+{\iint }_{S_2}\mathbf{J}\cdot d{\mathbf{S}}_2+{\iint }_{S_3}\mathbf{J}\cdot d{\mathbf{S}}_3\\ & +{\iint }_{S_4}\mathbf{J}\cdot d{\mathbf{S}}_4+{\iint }_{S_5}\mathbf{J}\cdot d{\mathbf{S}}_5+{\iint }_{S_6}\mathbf{J}\cdot d{\mathbf{S}}_6\end{array}$$To compute each of the integrals on the right-hand side, we first need to determine dS for each face of the cube and compute the dot product with the given current density J. #Step 2: Compute the integrals for each face#

Calculate the surface integrals for each face

Let's carry out the calculation for each face of the cube: - Face 1: x=0, dS1=$-$dzdy${\mathbf{a}}_x$ $${\mathrm{I}}_1={\iint }_{S_1}\mathbf{J}\cdot d{\mathbf{S}}_1=-{10}^4{\int }_2^3{\int }_0^1\sin (0)e^{-2y}(-1)\mathrm{d}y\mathrm{d}z=0$$ - Face 2: x=1, dS2=dzdy${\mathbf{a}}_x$ $${\mathrm{I}}_2={\iint }_{S_2}\mathbf{J}\cdot d{\mathbf{S}}_2=-{10}^4{\int }_2^3{\int }_0^1\sin (2)e^{-2y}(1)\mathrm{d}y\mathrm{d}z\approx -26260$$ - Face 3: y=0, dS3=$-$dzdx${\mathbf{a}}_y$ $${\mathrm{I}}_3={\iint }_{S_3}\mathbf{J}\cdot d{\mathbf{S}}_3=-{10}^4{\int }_2^3{\int }_0^1\cos (2x)e^{-2(0)}(-1)\mathrm{d}x\mathrm{d}z\approx 11757$$ - Face 4: y=1, dS4=dzdx${\mathbf{a}}_y$ $${\mathrm{I}}_4={\iint }_{S_4}\mathbf{J}\cdot d{\mathbf{S}}_4=-{10}^4{\int }_2^3{\int }_0^1\cos (2x)e^{-2(1)}(1)\mathrm{d}x\mathrm{d}z=-3015$$ - Face 5: z=2, dS5=$-$dxdy${\mathbf{a}}_z$ $${\mathrm{I}}_5={\iint }_{S_5}\mathbf{J}\cdot d{\mathbf{S}}_5=0$$ - Face 6: z=3, dS6=dxdy${\mathbf{a}}_z$ $${\mathrm{I}}_6={\iint }_{S_6}\mathbf{J}\cdot d{\mathbf{S}}_6=0$$ #Step 3: Add up the currents for each face#

Sum the currents computed for each face of the cube

Now we sum the above values to find the total current leaving the cube:$${\mathrm{I}}_{\mathrm{total}}={\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3+{\mathrm{I}}_4+{\mathrm{I}}_5+{\mathrm{I}}_6\approx -26260+11757-3015=-17518\mathrm{kA}$$So, the total current leaving the region is approximately -17518 kA. (c) Use divergence theorem #Step 1: Calculate the divergence of the current density J#

Compute the divergence of the current density

The divergence theorem relates the surface integral of a vector field to the volume integral of the divergence of the same vector field. First, we need to calculate the divergence of the given current density J:$$\mathrm{\nabla }\cdot \mathbf{J}=\frac{\partial J_x}{\partial x}+\frac{\partial J_y}{\partial y}$$Using the expression for J, we get:$$\mathrm{\nabla }\cdot \mathbf{J}=2\cdot {10}^4\cos (2x)e^{-2y}+2\cdot {10}^4\cos (2x)e^{-2y}=4\cdot {10}^4\cos (2x)e^{-2y}$$ #Step 2: Integrate the divergence of J over the volume of the cube#

Calculate the volume integral of the divergence of J

The volume of the cube is given by 1x1x1, and we have calculated the divergence of J in the previous step. Now we integrate the divergence over the volume:$${\iiint }_V\mathrm{\nabla }\cdot \mathbf{J}\mathrm{d}V={\int }_0^1{\int }_0^1{\int }_2^{3}4\cdot {10}^4\cos (2x)e^{-2y}\mathrm{d}z\mathrm{d}y\mathrm{d}x$$Compute the integral:$${\iiint }_V\mathrm{\nabla }\cdot \mathbf{J}\mathrm{d}V=4\cdot {10}^4\cdot ({\int }_0^1\cos (2x)\mathrm{d}x)\cdot \left({\int }_0^1{e}^{-2y}\mathrm{d}y\right)\cdot \left({\int }_2^3\mathrm{d}z\right)$$Evaluate the integral:$${\iiint }_V\mathrm{\nabla }\cdot \mathbf{J}\mathrm{d}V=4\cdot {10}^4\cdot {[\frac{1}{2}\sin (2x)]}_0^1\cdot {[-\frac{1}{2}{e}^{-2y}]}_0^1\cdot [z{]}_2^3\approx -17518\mathrm{kA}$$The total current leaving the region calculated using the divergence theorem is approximately -17518 kA, which agrees with the result we obtained from the surface integral in part (b).

Key concepts

Surface Integral

The surface integral is a fundamental concept in vector calculus, especially when dealing with physical quantities distributed over a surface. It helps in calculating the total quantity passing through or contained within a surface in a 3-dimensional space. Imagine having a thin sheet with a varying amount of paint on it. The integral helps to measure the total amount of paint, considering that some patches have thicker layers than others.

For instance, when working with current density, \textbf{J}, which represents the current per unit area, a surface integral allows us to determine the total current passing through a specific surface area. It's analogous to adding up the tiny currents over each differential patch, \textbf{dS}, on the surface. The mathematical expression for calculating the surface integral of current density \textbf{J} over a surface \textbf{S} is:
$${\iint }_S\mathbf{J}\cdot d\mathbf{S}$$
It's important to assess the orientation of the surface and ensure consistency in defining the directions of \textbf{dS} vectors. This attention to detail ensures that the direction of the current flow across the surface is correctly accounted for, whether entering or leaving the surface.

Divergence Theorem

The divergence theorem is a powerful tool in vector calculus that creates a bridge between a surface integral and a volume integral. It states that the flux of a vector field through a closed surface is equivalent to the integral of the divergence of this field over the volume enclosed by the surface.

This theorem simplifies complex 3D problems by transforming a surface integral into a volume integral, which is sometimes easier to compute. In the context of current density, if we desire to know the total current leaving a region without the tedious process of computing the surface integral over each face of a volume, we turn to the divergence theorem.

The divergence of a current density \textbf{J}, which measures the outflow of current from an infinitesimal volume, is represented mathematically as $abla\cdot \mathbf{J}$. The theorem then relates this to the total current leaving a volume \textbf{V} through:
$${\text{oiint}}_{\partial V}\mathbf{J}\cdot d\mathbf{S}={\iiint }_{V}abla\cdot \mathbf{J}dV$$
Employing the divergence theorem, as seen in the given exercise, streamlines the computation and gives us a convenient path to the same answer as that obtained from the meticulous process of evaluating multiple surface integrals.

Current Density Calculation

Understanding Current Density \textbf{J}

Current density is a concise way to describe how electric current is distributed over a certain area. It is a vector quantity that expresses the amount of charge passing through a unit area per unit time. Mathematically, it's defined as $\mathbf{J}=\frac{dI}{d\mathbf{S}}$, where $dI$ is the differential current flowing through an infinitesimal area $d\mathbf{S}$.

Current density is pivotal in understanding how electrical circuits and electromagnetic fields behave in different media. When faced with a problem involving current density, it is vital to first identify the correct components of \textbf{J} that contribute to the flow of current in the direction of interest.

Subsequently, integrating the component of current density over the area of interest yields the total current flowing through that area. This process involves setting up the integral with appropriate limits reflecting the dimensions and orientation of the area, ensuring a precise representation of the physical problem.

Vector Calculus

Vector calculus is a branch of mathematics essential for analyzing and solving problems in multidimensional spaces involving vector fields, where quantities have both magnitude and direction. The field of vector calculus provides us with the tools to work with these vector fields and includes operations such as gradient, divergence, and curl.

Gradients and Field Directions

Gradient, represented by $ablaf$, gives the direction and rate of the steepest ascent in a scalar field. This concept helps in visualizing the change in physical quantities like temperature or pressure within a region.

Divergence and Quantity Outflow

Divergence, as mentioned before, measures the extent of outflow or inflow of a vector field from an infinitesimal volume, which is crucial for understanding phenomena like charge conservation in electromagnetism.

Curl and Rotational Effects

Curl, on the other hand, gives us information about the rotation or swirling pattern of a vector field and is symbolized as $abla\times \mathbf{V}$. This is particularly useful in fluid dynamics and electromagnetic field theories.

Through operations such as line, surface, and volume integrals, vector calculus allows us to translate physical situations into mathematical expressions and then solve them, providing deep insights into the behavior of physical systems.