Chapter 9, Problem 40P

To determine

Prove that the given magnetic vector potential satisfies the expression ${\nabla }^{2}A-\mu \epsilon \frac{{\partial }^{2}A}{\partial t^2}=-\mu J$ in free space and find the corresponding electric scalar potential.

Answer to Problem 40P

The given magnetic vector potential satisfies the expression ${\nabla }^{2}A-\mu \epsilon \frac{{\partial }^{2}A}{\partial t^2}=-\mu J$ in free space and the electric scalar potential is $\underset{\_}{\frac{\cos \theta }{j4\pi \omega {\epsilon }_{\text{o}}r}\left(j\beta +\frac{1}{r}\right)e^{j\left(\omega t-\beta r\right)}}$.

Explanation of Solution

Calculation:

Write the expression has to be satisfied by the given magnetic vector potential.

${\nabla }^{2}A-\mu \epsilon \frac{{\partial }^{2}A}{\partial t^2}=-\mu J$

Rewrite the expression for free space region.

$\begin{array}{l}{\nabla }^{2}A-{\mu }_{\text{o}}{\epsilon }_{\text{o}}\frac{{\partial }^{2}A}{\partial t^2}=-{\mu }_{\text{o}}\left(0\right)\left\{\because J=0\right\}\\ {\nabla }^{2}A={\mu }_{\text{o}}{\epsilon }_{\text{o}}\frac{{\partial }^{2}A}{\partial t^2}\end{array}$

Substitute $\frac{{\mu }_{\text{o}}}{4\pi r}\left[\cos \left(\theta \right)a_r-\sin \left(\theta \right)a_{\theta }\right]e^{j\omega \left(t-\frac{r}{c}\right)}$ for $A$ on the right side of the expression.

$\begin{array}{c}{\nabla }^{2}A={\mu }_{\text{o}}{\epsilon }_{\text{o}}\frac{{\partial }^2}{\partial t^2}\left\{\frac{{\mu }_{\text{o}}}{4\pi r}\left[\cos \left(\theta \right)a_r-\sin \left(\theta \right)a_{\theta }\right]e^{j\omega \left(t-\frac{r}{c}\right)}\right\}\\ ={\mu }_{\text{o}}{\epsilon }_{\text{o}}\frac{\partial }{\partial t}\left(\frac{\partial }{\partial t}\left\{\frac{{\mu }_{\text{o}}}{4\pi r}\left[\cos \left(\theta \right)a_r-\sin \left(\theta \right)a_{\theta }\right]e^{j\omega \left(t-\frac{r}{c}\right)}\right\}\right)\\ ={\mu }_{\text{o}}{\epsilon }_{\text{o}}\frac{\partial }{\partial t}\left(j\omega \left\{\frac{{\mu }_{\text{o}}}{4\pi r}\left[\cos \left(\theta \right)a_r-\sin \left(\theta \right)a_{\theta }\right]e^{j\omega \left(t-\frac{r}{c}\right)}\right\}\right)\\ ={\mu }_{\text{o}}{\epsilon }_{\text{o}}\left(j\omega \right)\left(j\omega \right)\left\{\frac{{\mu }_{\text{o}}}{4\pi r}\left[\cos \left(\theta \right)a_r-\sin \left(\theta \right)a_{\theta }\right]e^{j\omega \left(t-\frac{r}{c}\right)}\right\}\end{array}$

Simplify the expression.

$\begin{array}{c}{\nabla }^{2}A=-{\mu }_{\text{o}}{\epsilon }_{\text{o}}{\omega }^{2}A\left\{\begin{array}{l}\because \frac{{\mu }_{\text{o}}}{4\pi r}\left[\cos \left(\theta \right)a_r-\sin \left(\theta \right)a_{\theta }\right]e^{j\omega \left(t-\frac{r}{c}\right)}=A\\ \because j^2=-1\end{array}\right\}\\ =-{\beta }^{2}A\left\{\because {\beta }^2={\mu }_{\text{o}}{\epsilon }_{\text{o}}{\omega }^2\right\}\end{array}$

Therefore, the expression ${\nabla }^{2}A=-{\beta }^{2}A$ has to be satisfied by the given magnetic vector potential.

${\nabla }^{2}A=-{\beta }^{2}A$        (1)

Rewrite the given magnetic vector potential.

$\begin{array}{c}A=\frac{{\mu }_{\text{o}}}{4\pi r}{e}^{j\omega t}{e}^{-j\left(\frac{\omega }{c}\right)r}\left[\cos \left(\theta \right)a_r-\sin \left(\theta \right)a_{\theta }\right]\\ =\frac{{\mu }_{\text{o}}}{4\pi r}{e}^{j\omega t}{e}^{-j\beta r}\left[\cos \left(\theta \right)a_r-\sin \left(\theta \right)a_{\theta }\right]\left\{\because \frac{\omega }{c}=\beta \right\}\\ =\frac{{\mu }_{\text{o}}}{4\pi r}{e}^{j\omega t}{e}^{-j\beta r}{a}_z\\ =\frac{{\mu }_{\text{o}}}{4\pi }{e}^{j\omega t}\left(\frac{e^{-j\beta r}}{r}\right)a_z\end{array}$

Consider $\frac{e^{-j\beta r}}{r}$ as R and rewrite the expression.

$A=\frac{{\mu }_{\text{o}}}{4\pi }{e}^{j\omega t}R{a}_z$        (2)

Find ${\nabla }^{2}R$.

$\begin{array}{c}{\nabla }^{2}R=\frac{1}{r^2\sin \left(\theta \right)}\left\{\frac{\partial }{\partial r}\left[r^2\sin \left(\theta \right)\frac{\partial R}{\partial r}\right]\right\}\\ =\frac{1}{r^2}\left\{\frac{\partial }{\partial r}\left(r^2\right)\left(\frac{-j\beta }{r}-\frac{1}{r^2}\right)e^{-j\beta r}\right\}\left\{\because R=\frac{e^{-j\beta r}}{r}\right\}\\ =\frac{1}{r^2}\left(-{\beta }^{2}r+j\beta -j\beta \right)e^{-j\beta r}\\ =-{\beta }^2\left(\frac{e^{-j\beta r}}{r}\right)\end{array}$

Substitute R for $\frac{e^{-j\beta r}}{r}$.

${\nabla }^{2}R=-{\beta }^{2}R$

As ${\nabla }^{2}R=-{\beta }^{2}R$, rewrite Equation (2).

${\nabla }^{2}A=-{\beta }^{2}A$

Therefore, Equation (1) is satisfied.

Find the electric scalar potential using Lorentz gauge expression.

$V=-\frac{1}{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}{\displaystyle \int \nabla \cdot Adt}$

Substitute $\frac{{\mu }_{\text{o}}}{4\pi r}\left[\cos \left(\theta \right)a_r-\sin \left(\theta \right)a_{\theta }\right]e^{j\omega t}{e}^{-j\beta r}$ for $A$.

$\begin{array}{c}V=-\frac{1}{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}{\displaystyle \int \nabla \cdot \left[\frac{{\mu }_{\text{o}}}{4\pi r}\left[\cos \left(\theta \right)a_r-\sin \left(\theta \right)a_{\theta }\right]e^{j\omega t}{e}^{-j\beta r}\right]dt}\\ =-\frac{1}{j\omega {\mu }_{\text{o}}{\epsilon }_{\text{o}}}\nabla \cdot \left[\frac{{\mu }_{\text{o}}}{4\pi r}\left[\cos \left(\theta \right)a_r-\sin \left(\theta \right)a_{\theta }\right]e^{j\omega t}{e}^{-j\beta r}\right]\\ =-\frac{1}{j\omega {\mu }_{\text{o}}{\epsilon }_{\text{o}}}\frac{\partial }{\partial r}\left[\frac{{\mu }_{\text{o}}}{4\pi r}{e}^{j\omega t}{e}^{-j\beta r}\right]\\ =-\frac{1}{j4\pi \omega {\epsilon }_{\text{o}}}\left(\frac{-j\beta }{r}-\frac{1}{r^2}\right)e^{-j\beta r}{e}^{j\omega t}\cos \left(\theta \right)\end{array}$

Simplify the expression.

$V=\frac{\cos \left(\theta \right)}{j4\pi \omega {\epsilon }_{\text{o}}r}\left(j\beta +\frac{1}{r}\right)e^{j\left(\omega t-\beta r\right)}$

Conclusion:

Thus, the given magnetic vector potential satisfies the expression ${\nabla }^{2}A-\mu \epsilon \frac{{\partial }^{2}A}{\partial t^2}=-\mu J$ in free space and the electric scalar potential is $\underset{\_}{\frac{\cos \theta }{j4\pi \omega {\epsilon }_{\text{o}}r}\left(j\beta +\frac{1}{r}\right)e^{j\left(\omega t-\beta r\right)}}$.