Chapter 9, Problem 35P

(a)

To determine

Find whether the field $A=40\sin \left(\omega t+10x\right)a_z$ is a genuine EM field or not.

Answer to Problem 35P

The field $A=40\sin \left(\omega t+10x\right)a_z$ is a genuine EM field.

Explanation of Solution

Calculation:

Find $\nabla \cdot A$.

$\begin{array}{c}\nabla \cdot A=\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left[40\sin \left(\omega t+10x\right)a_z\right]\\ =\frac{\partial }{\partial z}\left[40\sin \left(\omega t+10x\right)\right]\\ =0\end{array}$

Find $\nabla \times A$.

$\begin{array}{c}\nabla \times A=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 0& 0& 40\sin \left(\omega t+10x\right)\end{array}\right|\\ =\left(\begin{array}{l}\left\{\frac{\partial }{\partial y}\left[40\sin \left(\omega t+10x\right)\right]\right\}{a}_x\\ -\left\{\frac{\partial }{\partial x}\left[40\sin \left(\omega t+10x\right)\right]\right\}{a}_y+0a_z\end{array}\right)\\ =0a_x-400\cos \left(\omega t+10x\right)a_y+0a_z\\ =-400\cos \left(\omega t+10x\right)a_y\end{array}$

As $\nabla \cdot A=0$ and $\nabla \times A\ne 0$, the given field is a genuine electromagnetic field.

Conclusion:

Thus, the field $A=40\sin \left(\omega t+10x\right)a_z$ is a genuine EM field.

(b)

To determine

Find whether the field $B=\frac{10}{\rho }\cos \left(\omega t-2\rho \right)a_{\varphi }$ is a genuine EM field or not.

Answer to Problem 35P

The field $B=\frac{10}{\rho }\cos \left(\omega t-2\rho \right)a_{\varphi }$ is a genuine EM field.

Explanation of Solution

Calculation:

Find $\nabla \cdot B$.

$\begin{array}{c}\nabla \cdot B=\left(\frac{\partial }{\partial \rho }{a}_{\rho }+\frac{\partial }{\partial \theta }{a}_{\theta }+\frac{\partial }{\partial \varphi }{a}_{\varphi }\right)\cdot \left[\frac{10}{\rho }\cos \left(\omega t-2\rho \right)a_{\varphi }\right]\\ =\frac{\partial }{\partial \varphi }\left[\frac{10}{\rho }\cos \left(\omega t-2\rho \right)\right]\\ =0\end{array}$

Find $\nabla \times B$.

$\begin{array}{c}\nabla \times B=\left|\begin{array}{ccc}{a}_{\rho }& a_{\theta }& a_{\varphi }\\ \frac{\partial }{\partial \rho }& \frac{\partial }{\partial \theta }& \frac{\partial }{\partial \varphi }\\ 0& 0& \frac{10}{\rho }\cos \left(\omega t-2\rho \right)\end{array}\right|\\ =\left(\begin{array}{l}\left\{\frac{\partial }{\partial \theta }\left[\frac{10}{\rho }\cos \left(\omega t-2\rho \right)\right]\right\}{a}_{\rho }\\ -\left\{\frac{\partial }{\partial \rho }\left[\frac{10}{\rho }\cos \left(\omega t-2\rho \right)\right]\right\}{a}_{\theta }+0a_{\varphi }\end{array}\right)\\ =0a_{\rho }-\left\{\frac{\partial }{\partial \rho }\left[\frac{10}{\rho }\cos \left(\omega t-2\rho \right)\right]\right\}{a}_{\theta }+0a_{\varphi }\\ \ne 0\end{array}$

As $\nabla \cdot B=0$ and $\nabla \times B\ne 0$, the given field is a genuine electromagnetic field.

Conclusion:

Thus, the field $B=\frac{10}{\rho }\cos \left(\omega t-2\rho \right)a_{\varphi }$ is a genuine EM field.

(c)

To determine

Find whether the field $C=\left(3{\rho }^2\cot \varphi a_{\rho }+\frac{\cos \varphi }{\rho }{a}_{\varphi }\right)\sin \left(\omega t\right)$ is not a genuine EM field or not.

Answer to Problem 35P

The field $C=\left(3{\rho }^2\cot \varphi a_{\rho }+\frac{\cos \varphi }{\rho }{a}_{\varphi }\right)\sin \left(\omega t\right)$ is not a genuine EM field.

Explanation of Solution

Calculation:

Find $\nabla \cdot C$.

$\begin{array}{c}\nabla \cdot C=\left(\frac{\partial }{\partial \rho }{a}_{\rho }+\frac{\partial }{\partial \theta }{a}_{\theta }+\frac{\partial }{\partial \varphi }{a}_{\varphi }\right)\cdot \left[\left(3{\rho }^2\cot \varphi a_{\rho }+\frac{\cos \varphi }{\rho }{a}_{\varphi }\right)\sin \left(\omega t\right)\right]\\ =\frac{\partial }{\partial \rho }\left[3{\rho }^2\cot \varphi \sin \left(\omega t\right)\right]+\frac{\partial }{\partial \varphi }\left[\frac{\cos \varphi }{\rho }\sin \left(\omega t\right)\right]\\ \ne 0\end{array}$

Find $\nabla \times C$.

$\begin{array}{c}\nabla \times C=\left|\begin{array}{ccc}{a}_{\rho }& a_{\theta }& a_{\varphi }\\ \frac{\partial }{\partial \rho }& \frac{\partial }{\partial \theta }& \frac{\partial }{\partial \varphi }\\ 3{\rho }^2\cot \varphi \sin \left(\omega t\right)& 0& \frac{\cos \varphi }{\rho }\sin \left(\omega t\right)\end{array}\right|\\ =\left(\begin{array}{l}\left\{\frac{\partial }{\partial \theta }\left[\frac{\cos \varphi }{\rho }\sin \left(\omega t\right)\right]\right\}{a}_{\rho }\\ -\left\{\frac{\partial }{\partial \rho }\left[\frac{\cos \varphi }{\rho }\sin \left(\omega t\right)\right]-\frac{\partial }{\partial \varphi }\left[3{\rho }^2\cot \varphi \sin \left(\omega t\right)\right]\right\}{a}_{\theta }\\ +\frac{\partial }{\partial \theta }3{\rho }^2\cot \varphi \sin \left(\omega t\right)a_{\varphi }\end{array}\right)\\ \ne 0\end{array}$

As $\nabla \cdot C\ne 0$ and $\nabla \times C\ne 0$, the given field is not a genuine electromagnetic field.

Conclusion:

Thus, the field $C=\left(3{\rho }^2\cot \varphi a_{\rho }+\frac{\cos \varphi }{\rho }{a}_{\varphi }\right)\sin \left(\omega t\right)$ is not a genuine EM field.

(d)

To determine

Find whether the field $D=\frac{1}{r}\sin \theta \sin \left(\omega t-5r\right)a_{\theta }$ is not a genuine EM field or not.

Answer to Problem 35P

The field $D=\frac{1}{r}\sin \theta \sin \left(\omega t-5r\right)a_{\theta }$ is not a genuine EM field.

Explanation of Solution

Calculation:

Find $\nabla \cdot D$.

$\begin{array}{c}\nabla \cdot D=\left(\frac{\partial }{\partial r}{a}_r+\frac{\partial }{\partial \theta }{a}_{\theta }+\frac{\partial }{\partial \varphi }{a}_{\varphi }\right)\cdot \left[\frac{1}{r}\sin \theta \sin \left(\omega t-5r\right)a_{\theta }\right]\\ =\frac{\partial }{\partial \theta }\left[\frac{1}{r}\sin \theta \sin \left(\omega t-5r\right)\right]\\ \ne 0\end{array}$

Find $\nabla \times D$.

$\begin{array}{c}\nabla \times D=\left|\begin{array}{ccc}{a}_r& a_{\theta }& a_{\varphi }\\ \frac{\partial }{\partial r}& \frac{\partial }{\partial \theta }& \frac{\partial }{\partial \varphi }\\ 0& \frac{1}{r}\sin \theta \sin \left(\omega t-5r\right)& 0\end{array}\right|\\ =\left(\begin{array}{l}\left\{-\frac{\partial }{\partial \varphi }\left[\frac{1}{r}\sin \theta \sin \left(\omega t-5r\right)\right]\right\}{a}_r\\ -0a_{\theta }+\frac{\partial }{\partial r}\left[\frac{1}{r}\sin \theta \sin \left(\omega t-5r\right)\right]a_{\varphi }\end{array}\right)\\ \ne 0\end{array}$

As $\nabla \cdot D\ne 0$ and $\nabla \times D\ne 0$, the given field is not a genuine electromagnetic field.

Conclusion:

Thus, the field $D=\frac{1}{r}\sin \theta \sin \left(\omega t-5r\right)a_{\theta }$ is not a genuine EM field.