Chapter 9, Problem 8RQ

To determine

Identify the fields that are not Maxwellian in free space.

Answer to Problem 8RQ

The fields that are not Maxwellian in free space are $\underset{\_}{\left(\text{b}\right)E=100\cos \left(\omega t\right)a_x}$ and $\underset{\_}{\left(\text{d}\right)B=0.4\sin \left({10}^{4}t\right)a_z}$.

Explanation of Solution

Calculation:

Write the generalized forms of Maxwell’s four Equations.

$\nabla \cdot D={\rho }_v$        (1)

$\nabla \cdot B=0$        (2)

$\nabla \times E=-\frac{\partial B}{\partial t}$        (3)

$\nabla \times H=J+\frac{\partial D}{\partial t}$        (4)

Rewrite Equation (1) for source-free region.

$\nabla \cdot D=0\left\{\because {\rho }_v=0\right\}$        (5)

Rewrite Equations (3) and (4) for source-free region.

$\nabla \times E\ne 0$        (6)

$\nabla \times H\ne 0$        (7)

$\left(\text{a}\right)H=\cos x\cos \left({10}^{6}t\right)a_y$:

Rewrite Equation (2) for source-free region.

${\mu }_{\text{o}}\nabla \cdot H=0\left\{\because B={\mu }_{\text{o}}H\right\}$

$\nabla \cdot H=0$        (8)

Find $\nabla \cdot H$.

$\begin{array}{c}\nabla \cdot H=\nabla \cdot \left[\cos x\cos \left({10}^{6}t\right)a_y\right]\\ =\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left[\cos x\cos \left({10}^{6}t\right)a_y\right]\\ =\frac{\partial }{\partial y}\left[\cos x\cos \left({10}^{6}t\right)\right]\\ =0\end{array}$

Find $\left(\nabla \times H\right)$.

$\begin{array}{c}\nabla \times H=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 0& \cos x\cos \left({10}^{6}t\right)& 0\end{array}\right|\\ =\left\{0-\frac{\partial }{\partial z}\left[\cos x\cos \left({10}^{6}t\right)\right]\right\}{a}_x-0a_y+\frac{\partial }{\partial x}\left[\cos x\cos \left({10}^{6}t\right)\right]a_z\\ \ne 0\end{array}$

As Equations (7) and (8) are satisfied, the field $H=\cos x\cos \left({10}^{6}t\right)a_y$ is a Maxwellian field.

$\left(\text{b}\right)E=100\cos \left(\omega t\right)a_x$:

Rewrite Equation (5).

${\epsilon }_{\text{o}}\nabla \cdot E=0\left\{\because D={\epsilon }_{\text{o}}E\right\}$

$\nabla \cdot E=0$        (9)

Find $\nabla \cdot E$.

$\begin{array}{c}\nabla \cdot E=\nabla \cdot \left[100\cos \left(\omega t\right)a_x\right]\\ =\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left[100\cos \left(\omega t\right)a_x\right]\\ =\frac{\partial }{\partial x}\left[100\cos \left(\omega t\right)\right]\\ =0\end{array}$

Find $\left(\nabla \times E\right)$.

$\begin{array}{c}\nabla \times E=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 100\cos \left(\omega t\right)& 0& 0\end{array}\right|\\ =0a_x-\left\{0-\frac{\partial }{\partial z}\left[100\cos \left(\omega t\right)\right]\right\}{a}_y+\left\{0-\frac{\partial }{\partial y}\left[100\cos \left(\omega t\right)\right]\right\}{a}_z\\ =0\end{array}$

Equation (9) is satisfied but, Equation (6) is not satisfied. Therefore, the field $E=100\cos \left(\omega t\right)a_x$ is not a Maxwellian field.

$\left(\text{c}\right)D=e^{-10y}\sin \left({10}^{5}t-10y\right)a_z$:

Find $\nabla \cdot D$.

$\begin{array}{c}\nabla \cdot D=\nabla \cdot \left[e^{-10y}\sin \left({10}^{5}t-10y\right)a_z\right]\\ =\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left[e^{-10y}\sin \left({10}^{5}t-10y\right)a_z\right]\\ =\frac{\partial }{\partial z}\left[e^{-10y}\sin \left({10}^{5}t-10y\right)\right]\\ =0\end{array}$

Rewrite Equation (6).

$\begin{array}{l}\nabla \times E\ne 0\\ \frac{1}{{\epsilon }_{\text{o}}}\nabla \times D\ne 0\left\{\because E=\frac{D}{{\epsilon }_{\text{o}}}\right\}\end{array}$

$\nabla \times D\ne 0$        (10)

Find $\left(\nabla \times D\right)$.

$\begin{array}{c}\nabla \times D=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 0& 0& e^{-10y}\sin \left({10}^{5}t-10y\right)\end{array}\right|\\ =\frac{\partial }{\partial y}\left[e^{-10y}\sin \left({10}^{5}t-10y\right)\right]a_x-\frac{\partial }{\partial x}\left[e^{-10y}\sin \left({10}^{5}t-10y\right)\right]a_y+0a_z\\ =\frac{\partial }{\partial y}\left[e^{-10y}\sin \left({10}^{5}t-10y\right)\right]a_x-0a_y+0a_z\\ \ne 0\end{array}$

As Equations (5) and (10) are satisfied, the field $D=e^{-10y}\sin \left({10}^{5}t-10y\right)a_z$ is a Maxwellian field.

$\left(\text{d}\right)B=0.4\sin \left({10}^{4}t\right)a_z$:

Find $\nabla \cdot B$.

$\begin{array}{c}\nabla \cdot B=\nabla \cdot \left[0.4\sin \left({10}^{4}t\right)a_z\right]\\ =\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left[0.4\sin \left({10}^{4}t\right)a_z\right]\\ =\frac{\partial }{\partial z}\left[0.4\sin \left({10}^{4}t\right)\right]\\ =0\end{array}$

Rewrite Equation (7).

$\frac{1}{{\mu }_{\text{o}}}\nabla \times B\ne 0\left\{\because H=\frac{B}{{\mu }_{\text{o}}}\right\}$

$\nabla \times B\ne 0$        (11)

Find $\left(\nabla \times B\right)$.

$\begin{array}{c}\nabla \times B=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 0& 0& 0.4\sin \left({10}^{4}t\right)\end{array}\right|\\ =\frac{\partial }{\partial y}\left[0.4\sin \left({10}^{4}t\right)\right]a_x-\frac{\partial }{\partial z}\left[0.4\sin \left({10}^{4}t\right)\right]a_y+0a_z\\ =0\end{array}$

Equation (2) is satisfied but, Equation (11) is not satisfied. Therefore, the field $B=0.4\sin \left({10}^{4}t\right)a_z$ is not a Maxwellian field.

$\left(\text{e}\right)H=10\cos \left({10}^{5}t-\frac{z}{10}\right)a_x$:

Find $\nabla \cdot H$.

$\begin{array}{c}\nabla \cdot H=\nabla \cdot \left[10\cos \left({10}^{5}t-\frac{z}{10}\right)a_x\right]\\ =\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left[10\cos \left({10}^{5}t-\frac{z}{10}\right)a_x\right]\\ =\frac{\partial }{\partial x}\left[10\cos \left({10}^{5}t-\frac{z}{10}\right)\right]\\ =0\end{array}$

Find $\left(\nabla \times H\right)$.

$\begin{array}{c}\nabla \times H=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 10\cos \left({10}^{5}t-\frac{z}{10}\right)& 0& 0\end{array}\right|\\ =0a_x-\left\{0-\frac{\partial }{\partial z}\left[10\cos \left({10}^{5}t-\frac{z}{10}\right)\right]\right\}{a}_y+\left\{0-\frac{\partial }{\partial y}\left[10\cos \left({10}^{5}t-\frac{z}{10}\right)\right]\right\}{a}_z\\ =0a_x+\frac{\partial }{\partial z}\left[10\cos \left({10}^{5}t-\frac{z}{10}\right)\right]a_y+0a_z\\ \ne 0\end{array}$

As Equations (7) and (8) are satisfied, the field $H=10\cos \left({10}^{5}t-\frac{z}{10}\right)a_x$ is a Maxwellian field.

$\left(\text{f}\right)E=\frac{\sin \theta }{r}\cos \left(\omega t-r\omega \sqrt{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}\right)a_{\theta }$:

Find $\nabla \cdot E$.

$\begin{array}{c}\nabla \cdot E=\nabla \cdot \left[\frac{\sin \theta }{r}\cos \left(\omega t-r\omega \sqrt{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}\right)a_{\theta }\right]\\ =\left(\frac{\partial }{\partial r}{a}_r+\frac{\partial }{\partial \theta }{a}_{\theta }+\frac{\partial }{\partial \varphi }{a}_{\varphi }\right)\cdot \left[\frac{\sin \theta }{r}\cos \left(\omega t-r\omega \sqrt{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}\right)a_{\theta }\right]\\ =\frac{\partial }{\partial \theta }\left[\frac{\sin \theta }{r}\cos \left(\omega t-r\omega \sqrt{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}\right)\right]\\ \ne 0\end{array}$

As Equation (9) is not satisfied, the field $E=\frac{\sin \theta }{r}\cos \left(\omega t-r\omega \sqrt{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}\right)a_{\theta }$ is not a Maxwellian field.

$\left(\text{g}\right)B=\left(1-{\rho }^2\right)\sin \left(\omega t\right)a_z$:

Find $\nabla \cdot B$.

$\begin{array}{c}\nabla \cdot B=\nabla \cdot \left[\left(1-{\rho }^2\right)\sin \left(\omega t\right)a_z\right]\\ =\left(\frac{\partial }{\partial \rho }{a}_{\rho }+\frac{\partial }{\partial \varphi }{a}_{\varphi }+\frac{\partial }{\partial z}{a}_z\right)\cdot \left[\left(1-{\rho }^2\right)\sin \left(\omega t\right)a_z\right]\\ =\frac{\partial }{\partial z}\left[\left(1-{\rho }^2\right)\sin \left(\omega t\right)\right]\\ =0\end{array}$

Find $\left(\nabla \times B\right)$.

$\begin{array}{c}\nabla \times B=\left|\begin{array}{ccc}{a}_{\rho }& a_{\varphi }& a_z\\ \frac{\partial }{\partial \rho }& \frac{1}{\rho }\frac{\partial }{\partial \varphi }& \frac{\partial }{\partial z}\\ 0& 0& \left(1-{\rho }^2\right)\sin \left(\omega t\right)\end{array}\right|\\ =\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left[\left(1-{\rho }^2\right)\sin \left(\omega t\right)\right]a_{\rho }-\frac{\partial }{\partial \rho }\left[\left(1-{\rho }^2\right)\sin \left(\omega t\right)\right]a_{\varphi }+0a_z\\ =0a_{\rho }-\frac{\partial }{\partial \rho }\left[\left(1-{\rho }^2\right)\sin \left(\omega t\right)\right]a_{\varphi }+0a_z\\ \ne 0\end{array}$

As Equations (2) and (11) are satisfied, the field $B=\left(1-{\rho }^2\right)\sin \left(\omega t\right)a_z$ is a Maxwellian field.

Conclusion:

Thus, the fields that are not Maxwellian in free space are $\underset{\_}{\left(\text{b}\right)E=100\cos \left(\omega t\right)a_x}$ and $\underset{\_}{\left(\text{d}\right)B=0.4\sin \left({10}^{4}t\right)a_z}$.