Chapter 5, Problem 5.1P

To determine

(a)

The total current crossing the plane $y=1$ in the ${\text{a}}_y$ direction in the region $0<x<1,0<z<2$.

Answer to Problem 5.1P

The total current is, $I=-1.23\text{ MA}$.

Explanation of Solution

Given Information:

The current density is,

   $\text{J}=-{10}^4\left[\sin \left(2x\right)e^{-2y}{\text{a}}_x+\cos \left(2x\right)e^{-2y}{\text{a}}_y\right]{\text{ kA/m}}^2$.

Calculation:

The current density is given as,

   $\begin{array}{l}\text{J}=-{10}^4\left[\sin \left(2x\right)e^{-2y}{\text{a}}_x+\cos \left(2x\right)e^{-2y}{\text{a}}_y\right]\\ \Rightarrow \text{J}\cdot {\text{a}}_y=-{10}^4\cos \left(2x\right)e^{-2y}\end{array}$

The total current,

   $\begin{array}{l}I={\displaystyle \underset{S}{\iint }{\text{J}\cdot \text{n}|}_{S}da}={\displaystyle \underset{0}{\overset{2}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{\text{J}\cdot {\text{a}}_y|}_{y=1}dx}dz}\\ ={\displaystyle \underset{0}{\overset{2}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{-{10}^4\cos \left(2x\right)e^{-2y}|}_{y=1}dx}dz}\\ ={\displaystyle \underset{0}{\overset{2}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}-{10}^4\cos \left(2x\right)e^{-2}dx}dz}={\displaystyle \underset{0}{\overset{2}{\int }}{\left[-\frac{1}{2}{10}^4{e}^{-2}\sin \left(2x\right)\right]}_0^{1}dz}\\ ={\displaystyle \underset{0}{\overset{2}{\int }}-5000e^{-2}\sin 2dz}=-5000e^{-2}\sin 2{\left[z\right]}_0^2\\ =-10000e^{-2}\sin 2=-1230.6\text{ kA}\\ =-1.23\text{ MA}\end{array}$

Conclusion:

The total current is, $I=-1.23\text{ MA}$.

To determine

(b)

The total current leaving the region $0<x,y<1,2<z<3$ by integrating $\text{J}\cdot d\text{S}$ over the surface of cube.

Answer to Problem 5.1P

The total current leaving over the surface is zero.

Explanation of Solution

Given Information:

The current density is,

   $\text{J}=-{10}^4\left[\sin \left(2x\right)e^{-2y}{\text{a}}_x+\cos \left(2x\right)e^{-2y}{\text{a}}_y\right]{\text{ kA/m}}^2$.

Calculation:

The current density is given as,

   $\begin{array}{l}\text{J}=-{10}^4\left[\sin \left(2x\right)e^{-2y}{\text{a}}_x+\cos \left(2x\right)e^{-2y}{\text{a}}_y\right]\\ \Rightarrow \text{J}\cdot {\text{a}}_x=-{10}^4\sin \left(2x\right)e^{-2y}\\ \text{J}\cdot {\text{a}}_y=-{10}^4\cos \left(2x\right)e^{-2y}\\ \text{J}\cdot {\text{a}}_z=0\end{array}$

The total current leaving over the surface of cube,

   $I={\displaystyle \underset{2}{\overset{3}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{\text{J}\cdot \left(-{\text{a}}_x\right)|}_{x=0}dy}dz}+{\displaystyle \underset{2}{\overset{3}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{\text{J}\cdot {\text{a}}_x|}_{x=1}dy}dz}+{\displaystyle \underset{2}{\overset{3}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{\text{J}\cdot \left(-{\text{a}}_y\right)|}_{y=0}dx}dz}+{\displaystyle \underset{2}{\overset{3}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{\text{J}\cdot {\text{a}}_y|}_{y=1}dx}dz}$

   $+{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{\text{J}\cdot \left(-{\text{a}}_z\right)|}_{z=2}dx}dz}+{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{\text{J}\cdot {\text{a}}_z|}_{z=3}dx}dz}$

   $=0+{\displaystyle \underset{2}{\overset{3}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}-{10}^4\sin \left(2\right)e^{-2y}dy}dz}+{\displaystyle \underset{2}{\overset{3}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{10}^4\cos \left(2x\right)dx}dz}+{\displaystyle \underset{2}{\overset{3}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}-{10}^4\cos \left(2x\right)e^{-2}dx}dz}+0+0$

   $=-{10}^4\sin 2{\displaystyle \underset{2}{\overset{3}{\int }}{\left[-\frac{e^{-2y}}{2}\right]}_0^{1}dz}+{10}^4\left(1-e^{-2}\right){\displaystyle \underset{2}{\overset{3}{\int }}{\left[\frac{\sin \left(2x\right)}{2}\right]}_0^{1}dz}$

   $=\frac{{10}^4}{2}\left(e^{-2}-1\right)\sin 2+\frac{{10}^4}{2}\left(1-e^{-2}\right)\sin 2$

   $=0$

Conclusion:

The total current leaving over the surface is zero.

To determine

(c)

The total current leaving the region $0<x,y<1,2<z<3$ over the surface of cube using divergence theorem.

Answer to Problem 5.1P

The total current leaving over the surface is zero, as expected.

Explanation of Solution

Given Information:

The current density is,

   $\text{J}=-{10}^4\left[\sin \left(2x\right)e^{-2y}{\text{a}}_x+\cos \left(2x\right)e^{-2y}{\text{a}}_y\right]{\text{ kA/m}}^2$.

Calculation:

The current density is given as,

   $\begin{array}{l}\text{J}=-{10}^4\left[\sin \left(2x\right)e^{-2y}{\text{a}}_x+\cos \left(2x\right)e^{-2y}{\text{a}}_y\right]\\ \Rightarrow J_x=-{10}^4\sin \left(2x\right)e^{-2y}\\ J_y=-{10}^4\cos \left(2x\right)e^{-2y}\end{array}$

By using divergence theorem, the total current,

   $\begin{array}{l}I=\nabla \cdot \text{J}=\frac{\partial J_x}{\partial x}+\frac{\partial J_y}{\partial y}\\ =\frac{\partial \left(-{10}^4\sin \left(2x\right)e^{-2y}\right)}{\partial x}+\frac{\partial \left(-{10}^4\cos \left(2x\right)e^{-2y}\right)}{\partial y}\\ =-2\times {10}^4\cos \left(2x\right)e^{-2y}+-2\times -{10}^4\cos \left(2x\right)e^{-2y}\\ =0\end{array}$

Conclusion:

The total current leaving over the surface is zero, as expected.