91Ó°ÊÓ

Total scalar potential of two oppositely oriented electric dipoles is given from the section 11.1.2.

$V_{Â\pm }=∓\frac{p_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}c}\left(\frac{\cos {\mathrm{θ}}_{Â\pm }}{r_{Â\pm }}\right)\sin \left[\mathrm{Ó\neg }(t−\frac{r_{Â\pm }}{c})\right]$ …………. (1)

Total vector potential of two oppositely oriented electric dipoles is given from the section 11.1.2.

$A_{Â\pm }=∓\frac{{\mathrm{μ}}_0{p}_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{r}_{Â\pm }}\sin \left[\mathrm{Ó\neg }(t−\frac{r_{Â\pm }}{c})\right]\widehat{z}$

The electric field is given from the section 11.1.2.

$E=−∇V−\frac{∂A}{∂t}$ ……….…… (3)

The law of cosines is given as follows:

$r_{Â\pm }=\sqrt{r^2+{\left(\frac{d}{2}\right)}^2+2r\left(\frac{d}{2}\right)cos\mathrm{θ}}$……….…… (4)

Here, is the distance.

(a)

Apply the law of cosines in the given figure.

$\begin{array}{c}{r}_{Â\pm }=\sqrt{r^2+{\left(\frac{d}{2}\right)}^2+2r\left(\frac{d}{2}\right)\cos \mathrm{θ}}\\ â‰\dots r\sqrt{1∓\left(\frac{d}{r}\right)\cos \mathrm{θ}}\\ â‰\dots r{(1∓\left(\frac{d}{r}\right)\cos \mathrm{θ})}^{\frac{1}{2}}\\ â‰\dots r(1∓\frac{d}{2r}\cos \mathrm{θ})\end{array}$

It follows that

$\begin{array}{c}\frac{1}{r_{Â\pm }}=\frac{1}{r(1∓\frac{d}{2r}\cos \mathrm{θ})}\\ \frac{1}{r_{Â\pm }}=\frac{1}{r}{(1∓\frac{d}{2r}\cos \mathrm{θ})}^{−1}\\ \frac{1}{r_{Â\pm }}=\frac{1}{r}(1Â\pm \frac{d}{2r}\cos \mathrm{θ})\end{array}$ …………. (5)

Now, the angle is

$\cos {\mathrm{θ}}_{Â\pm }=\frac{r\cos ∓\left(\frac{d}{2}\right)}{r_{Â\pm }}$

Now substitute the value of $\frac{1}{r_{Â\pm }}=\frac{1}{r}(1Â\pm \frac{d}{2r}\cos \mathrm{θ})$.

role="math" localid="1658756836865" $\begin{array}{c}\cos {\mathrm{θ}}_{Â\pm }=\frac{1}{r_{Â\pm }}(r\cos \mathrm{θ}∓\left(\frac{d}{2}\right))\\ =\frac{1}{r}(1Â\pm \frac{d}{2r}\cos \mathrm{θ})(r\cos \mathrm{θ}∓\frac{d}{2r})\\ =(\cos \mathrm{θ}∓\frac{d}{2r})(1Â\pm \frac{d}{2r}\cos \mathrm{θ})\\ =\cos \mathrm{θ}Â\pm \frac{d}{2r}{\cos }^2\mathrm{θ}∓\frac{d}{2r}\end{array}$

Solve further as

$\begin{array}{c}\cos {\mathrm{θ}}_{Â\pm }=\cos \mathrm{θ}∓\frac{d}{2r}(1−{\cos }^2\mathrm{θ})\\ =\cos \mathrm{θ}∓\frac{d}{2r}{\sin }^2\mathrm{θ}\end{array}$

Let us take

$\sin \left[\mathrm{Ó\neg }(t−\frac{r_{Â\pm }}{c})\right]$

Substitute the value of $r_{Â\pm }=r(1∓\frac{d}{2r}\cos \mathrm{θ})$.

$\begin{array}{c}\sin \left[\mathrm{Ó\neg }(t−\frac{r(1∓\frac{d}{2r}\cos \mathrm{θ})}{c})\right]=\sin \left[\mathrm{Ó\neg }(t−\frac{r}{c}(1∓\frac{d}{2r}\cos \mathrm{θ}))\right]\\ =\sin \left\{\mathrm{Ó\neg }[t−\frac{r}{c}Â\pm \frac{d}{2c}\cos \mathrm{θ}]\right\}\\ =\sin \left\{\mathrm{Ó\neg }[(t−\frac{r}{c})Â\pm \frac{d}{2c}\cos \mathrm{θ}]\right\}\end{array}$ ….… (6)

Put $t_0=t−\frac{r}{c}$in the equation (6).

$\begin{array}{c}\sin \left\{\mathrm{Ó\neg }[(t−\frac{r}{c})Â\pm \frac{d}{2c}\cos \mathrm{θ}]\right\}=\sin \left\{\mathrm{Ó\neg }[t_0Â\pm \frac{d}{2c}\cos \mathrm{θ}]\right\}\\ =\sin [\mathrm{Ó\neg }{t}_0Â\pm \frac{\mathrm{Ó\neg }d}{2c}\cos \mathrm{θ}]\\ =\sin \left(\mathrm{Ó\neg }{t}_0\right)\cos \left(\frac{\mathrm{Ó\neg }d}{2c}\cos \mathrm{θ}\right)Â\pm \cos \left(\mathrm{Ó\neg }{t}_0\right)\sin \left(\frac{\mathrm{Ó\neg }d}{2c}\cos \mathrm{θ}\right)\\ =\sin \left(\mathrm{Ó\neg }{t}_0\right)Â\pm \frac{\mathrm{Ó\neg }d}{2c}\cos \mathrm{θ}\cos \left(\mathrm{Ó\neg }{t}_0\right)\end{array}$… (7)

Substituting all these values in the equation (1)

$\begin{array}{c}{V}_{Â\pm }=∓\frac{p_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}c}\left(\frac{\cos {\mathrm{θ}}_{Â\pm }}{r_{Â\pm }}\right)\sin \left[\mathrm{Ó\neg }(t−\frac{r_{Â\pm }}{c})\right]\\ V_{Â\pm }=∓\frac{p_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}c}\left[(\cos \mathrm{θ}∓\frac{d}{2r}{\sin }^2\mathrm{θ})\frac{1}{r}(1Â\pm \frac{d}{2r}\cos \mathrm{θ})\right]\left[\sin \left({\mathrm{Ó\neg }}_{0}t\right)Â\pm \frac{\mathrm{Ó\neg }d}{2c}\cos \mathrm{θ}\cos \left(\mathrm{Ó\neg }{t}_0\right)\right]\\ V_{Â\pm }=∓\frac{p_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}cr}\left[(\cos \mathrm{θ}∓\frac{d}{2r}{\sin }^2\mathrm{θ})(1Â\pm \frac{d}{2r}\cos \mathrm{θ})\right]\left[\sin \left({\mathrm{Ó\neg }}_{0}t\right)Â\pm \frac{\mathrm{Ó\neg }d}{2c}\cos \mathrm{θ}\cos \left(\mathrm{Ó\neg }{t}_0\right)\right]\\ V_{Â\pm }=∓\frac{p_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}cr}\left[(\cos \mathrm{θ}Â\pm \frac{d}{2r}{\cos }^2\mathrm{θ}∓\frac{d}{2r}{\sin }^2\mathrm{θ})\right]\left[\sin \left({\mathrm{Ó\neg }}_{0}t\right)Â\pm \frac{\mathrm{Ó\neg }d}{2c}\cos \mathrm{θ}\cos \left(\mathrm{Ó\neg }{t}_0\right)\right]\end{array}$

Further solved as

$\begin{array}{l}{V}_{Â\pm }â‰\dots ∓\frac{p_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}cr}\left(\cos \mathrm{θ}\sin \left({\mathrm{Ó\neg }}_{0}t\right)Â\pm \frac{\mathrm{Ó\neg }d}{2c}{\cos }^2\mathrm{θ}\cos \left(\mathrm{Ó\neg }{t}_0\right)Â\pm \frac{d}{2r}{\cos }^2\mathrm{θ}\sin \left({\mathrm{Ó\neg }}_{0}t\right)∓\frac{d}{2r}{\sin }^2\mathrm{θ}\sin \left({\mathrm{Ó\neg }}_{0}t\right)\right)\\ V_{Â\pm }=∓\frac{p_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}cr}\left(\cos \mathrm{θ}\sin \left({\mathrm{Ó\neg }}_{0}t\right)Â\pm \frac{\mathrm{Ó\neg }d}{2c}{\cos }^2\mathrm{θ}\cos \left(\mathrm{Ó\neg }{t}_0\right)Â\pm \frac{d}{2r}({\cos }^2\mathrm{θ}−{\sin }^2\mathrm{θ})\sin \left({\mathrm{Ó\neg }}_{0}t\right)\right)\end{array}$

Therefore, the expression for the total voltage is derived as

$\begin{array}{l}{V}_{tot}=V_{+}+V_{−}\\ V_{tot}=−\frac{p_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}cr}[\frac{\mathrm{Ó\neg }d}{c}{\cos }^2\mathrm{θ}\cos \left({\mathrm{Ó\neg }}_{0}t\right)+\frac{d}{r}({\cos }^2\mathrm{θ}−{\sin }^2\mathrm{θ})\sin \left(\mathrm{Ó\neg }{t}_0\right)]\\ V_{tot}=−\frac{p_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}cr}×\frac{\mathrm{Ó\neg }d}{c}[{\cos }^2\mathrm{θ}\cos \left({\mathrm{Ó\neg }}_{0}t\right)+\frac{c}{\mathrm{Ó\neg }d}\frac{d}{r}({\cos }^2\mathrm{θ}−{\sin }^2\mathrm{θ})\sin \left(\mathrm{Ó\neg }{t}_0\right)]\\ V_{tot}=−\frac{p_0{\mathrm{Ó\neg }}^{2}d}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{c}^{2}r}[{\cos }^2\mathrm{θ}\cos \left({\mathrm{Ó\neg }}_{0}t\right)+\frac{c}{\mathrm{Ó\neg }r}({\cos }^2\mathrm{θ}−{\sin }^2\mathrm{θ})\sin \left(\mathrm{Ó\neg }{t}_0\right)]\end{array}$

In the radiation zone $r>>\frac{\mathrm{Ó\neg }}{c}$then the second term in the above square bracket can be neglected then the above equation becomes

$V_{tot}=−\frac{p_0{\mathrm{Ó\neg }}^{2}d}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{c}^{2}r}{\cos }^2\mathrm{θ}\cos \left(\mathrm{Ó\neg }{t}_0\right)$ ………. (8)

Now, put $t_0=t−\frac{r}{c}$in the equation (8).

$V_{tot}=−\frac{p_0{\mathrm{Ó\neg }}^{2}d}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{c}^{2}r}{\cos }^2\mathrm{θ}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]$

Hence, the scalar potential is $V_{tot}=−\frac{p_0{\mathrm{Ó\neg }}^{2}d}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{c}^{2}r}{\cos }^2\mathrm{θ}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]$.

Now calculate the vector potential from equation (2).

$A_{Â\pm }=∓\frac{{\mathrm{μ}}_0{p}_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{r}_{Â\pm }}\sin \left[\mathrm{Ó\neg }(t−\frac{r_{Â\pm }}{c})\right]\widehat{z}$

Substitute equation (5) and (7) in equation (2).

$\begin{array}{l}{A}_{Â\pm }=∓\frac{{\mathrm{μ}}_0{p}_0\mathrm{Ó\neg }}{4\mathrm{Ï€}}×\left\{\frac{1}{r}(1Â\pm \frac{d}{2r}\cos \mathrm{θ})[\sin \left(\mathrm{Ó\neg }{t}_0\right)Â\pm \frac{\mathrm{Ó\neg }d}{2c}\cos \mathrm{θ}\cos \left(\mathrm{Ó\neg }{t}_0\right)]\right\}\\ A_{Â\pm }=∓\frac{{\mathrm{μ}}_0{p}_0\mathrm{Ó\neg }}{4\mathrm{Ï€}r}[\sin \left(\mathrm{Ó\neg }{t}_0\right)Â\pm \frac{\mathrm{Ó\neg }d}{2c}\cos \mathrm{θ}\cos \left(\mathrm{Ó\neg }{t}_0\right)Â\pm \frac{d}{2r}\cos \mathrm{θ}\sin \left(\mathrm{Ó\neg }{t}_0\right)]\widehat{z}\end{array}$

Hence, the total scalar potential is

$\begin{array}{l}{A}_{tot}=A^{+}+A^{−}\\ A_{tot}=∓\frac{{\mathrm{μ}}_0{p}_0\mathrm{Ó\neg }}{4\mathrm{Ï€}r}[\frac{\mathrm{Ó\neg }d}{c}\cos \mathrm{θ}\cos \left(\mathrm{Ó\neg }{t}_0\right)Â\pm \frac{d}{r}\cos \mathrm{θ}\sin \left(\mathrm{Ó\neg }{t}_0\right)]\widehat{z}\\ A_{tot}=∓\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}cr}[\cos \mathrm{θ}\cos \left(\mathrm{Ó\neg }{t}_0\right)Â\pm \frac{c}{r\mathrm{Ó\neg }}\cos \mathrm{θ}\sin \left(\mathrm{Ó\neg }{t}_0\right)]\widehat{z}\end{array}$

In the radiation zone $r>>\frac{\mathrm{Ó\neg }}{c}$then the second term in the above square bracket can be neglected then the above equation becomes

$\begin{array}{l}{A}_{tot}=−\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^{2}d}{4\mathrm{Ï€}cr}\cos \mathrm{θ}\cos \left(\mathrm{Ó\neg }{t}_0\right)\widehat{z}\\ A_{tot}=−\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^{2}d}{4\mathrm{Ï€}cr}\cos \mathrm{θ}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]\widehat{z}\end{array}$

Hence, the total vector potential is $A_{tot}=−\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^{2}d}{4\mathrm{Ï€}cr}\cos \mathrm{θ}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]\widehat{z}$.

(b)

Consider the expression for the total voltage drop as

$V_{tot}=−\frac{p_0{\mathrm{Ó\neg }}^{2}d}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{c}^{2}r}{\cos }^2\mathrm{θ}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]$

As we know

$c=\frac{1}{\sqrt{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0}}$

where ${\mathrm{Î\mu }}_0$is the vacuum permittivity and ${\mathrm{μ}}_0$ is the vacuum permeability.

$\begin{array}{l}{V}_{tot}=−\frac{p_0{\mathrm{Ó\neg }}^{2}d}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0\left(\frac{1}{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0}\right)r}{\cos }^2\mathrm{θ}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]\\ V_{tot}=−\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^{2}d}{4\mathrm{Ï€}r}{\cos }^2\mathrm{θ}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]\end{array}$

Let us consider $\mathrm{Î\pm }=−\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^{2}d}{4\mathrm{Ï€}}$; then the above equation become

$\begin{array}{l}{V}_{tot}=\frac{\mathrm{Î\pm }}{r}{\cos }^2\mathrm{θ}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]\\ V_{tot}=\frac{\mathrm{Î\pm }{\cos }^2\mathrm{θ}}{r}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]\end{array}$

Now vector potential in the co-ordinate form

$∇V=\frac{∂V_{tot}}{∂r}\widehat{r}+\frac{1}{r}\frac{∂V_{tot}}{∂\mathrm{θ}}\widehat{\mathrm{θ}}$ ………. (9)

Put $V_{tot}=\frac{\mathrm{Î\pm }{\cos }^2\mathrm{θ}}{r}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]$ in the equation (9).

$\begin{array}{c}∇V=\frac{∂\left[\frac{\mathrm{Î\pm }{\cos }^2\mathrm{θ}}{r}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]\right]}{∂r}\widehat{r}+\frac{1}{r}\frac{∂\left[\frac{\mathrm{Î\pm }{\cos }^2\mathrm{θ}}{r}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]\right]}{∂\mathrm{θ}}\widehat{\mathrm{θ}}\\ ∇V=\mathrm{Î\pm }{\cos }^2\mathrm{θ}\frac{∂}{∂r}\left[\frac{1}{r}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]\right]\widehat{r}+\frac{1}{r}\frac{\mathrm{Î\pm }}{r}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]\frac{∂}{∂\mathrm{θ}}\left[{\cos }^2\mathrm{θ}\right]\widehat{\mathrm{θ}}\\ ∇V=\left\{\begin{array}{l}\mathrm{Î\pm }{\cos }^2\mathrm{θ}\{−\frac{1}{r^2}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]+\frac{1}{r}[−\sin \left(\mathrm{Ó\neg }(t−\frac{r}{c})\right)(−\frac{\mathrm{Ó\neg }}{c})]\}\\ +\frac{\mathrm{Î\pm }}{r^2}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right](−2\cos \mathrm{θ}\sin \mathrm{θ})\widehat{\mathrm{θ}}\end{array}\right\}\\ ∇V=\left\{\begin{array}{l}\mathrm{Î\pm }{\cos }^2\mathrm{θ}\{−\frac{1}{r^2}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]+\frac{\mathrm{Ó\neg }}{rc}\left[\sin \left(\mathrm{Ó\neg }(t−\frac{r}{c})\right)\right]\}\\ +\frac{\mathrm{Î\pm }}{r^2}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right](−2\cos \mathrm{θ}\sin \mathrm{θ})\widehat{\mathrm{θ}}\end{array}\right\}\end{array}$

In the radiation zone

$\begin{array}{l}\mathrm{Δ}V=\mathrm{Î\pm }{\cos }^2\mathrm{θ}\frac{\mathrm{Ó\neg }}{rc}\left[\sin \left(\mathrm{Ó\neg }(t−\frac{r}{c})\right)\right]\widehat{r}\\ \mathrm{Δ}V=\mathrm{Î\pm }\frac{\mathrm{Ó\neg }}{c}\frac{{\cos }^2\mathrm{θ}}{r}\left[\sin \left(\mathrm{Ó\neg }(t−\frac{r}{c})\right)\right]\widehat{r}\end{array}$ ……… (10)

Since

$A=\frac{\mathrm{Î\pm }}{c}\frac{\cos \mathrm{θ}}{r}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right](\cos \mathrm{θ}\widehat{r}−\sin \mathrm{θ}\widehat{\mathrm{θ}})$

Differentiate the equation with respect to time .

$\begin{array}{l}\frac{∂A}{∂t}=\frac{∂}{∂t}\left\{\frac{\mathrm{Î\pm }}{c}\frac{\cos \mathrm{θ}}{r}\cos \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right](\cos \mathrm{θ}\widehat{r}−\sin \mathrm{θ}\widehat{\mathrm{θ}})\right\}\\ \frac{∂A}{∂t}=\frac{\mathrm{Î\pm }}{c}\frac{\cos \mathrm{θ}}{r}\{−\sin \left[\mathrm{Ó\neg }(t−\frac{r}{c})\left(\mathrm{Ó\neg }\right)\right](\cos \mathrm{θ}\widehat{r}−\sin \mathrm{θ}\widehat{\mathrm{θ}})\}\\ \frac{∂A}{∂t}=−\frac{\mathrm{Î\pm }\mathrm{Ó\neg }}{c}\frac{\cos \mathrm{θ}}{r}\sin \left[\mathrm{Ó\neg }(t−\frac{r}{c})\left(\mathrm{Ó\neg }\right)\right](\cos \mathrm{θ}\widehat{r}−\sin \mathrm{θ}\widehat{\mathrm{θ}})\end{array}$

From equation (3)

$E=−∇V−\frac{∂A}{∂t}$ ……… (11)

Put the value of $\mathrm{Δ}V$ and $\frac{∂A}{∂t}$in equation (11).

$\begin{array}{l}E=−\mathrm{Î\pm }\frac{\mathrm{Ó\neg }}{c}\frac{{\cos }^2\mathrm{θ}}{r}\left[\sin \left(\mathrm{Ó\neg }(t−\frac{r}{c})\right)\right]\widehat{r}+\frac{\mathrm{Î\pm }\mathrm{Ó\neg }}{c}\frac{\cos \mathrm{θ}}{r}\sin \left[\mathrm{Ó\neg }(t−\frac{r}{c})\left(\mathrm{Ó\neg }\right)\right](\cos \mathrm{θ}\widehat{r}−\sin \mathrm{θ}\widehat{\mathrm{θ}})\\ E=\frac{\mathrm{Î\pm }\mathrm{Ó\neg }}{c}\frac{\cos \mathrm{θ}}{r}\left[\sin \mathrm{Ó\neg }(t−\frac{r}{c})\right](\cos \mathrm{θ}\widehat{r}−\sin \mathrm{θ}\widehat{\mathrm{θ}})−\frac{\mathrm{Î\pm }\mathrm{Ó\neg }}{c}\frac{{\cos }^2\mathrm{θ}}{r}\sin \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]\widehat{r}\\ E=\frac{\mathrm{Î\pm }\mathrm{Ó\neg }}{c}\frac{\cos \mathrm{θ}}{r}\left[\sin \mathrm{Ó\neg }(t−\frac{r}{c})\right][\cos \mathrm{θ}\widehat{r}−\sin \mathrm{θ}\widehat{\mathrm{θ}}−\cos \mathrm{θ}\widehat{r}]\\ E=\frac{\mathrm{Î\pm }\mathrm{Ó\neg }}{c}\frac{\cos \mathrm{θ}}{r}\left[\sin \mathrm{Ó\neg }(t−\frac{r}{c})\right][−\sin \mathrm{θ}\widehat{\mathrm{θ}}]\end{array}$

Further solve as

$\begin{array}{l}E=\frac{\mathrm{Î\pm }\mathrm{Ó\neg }}{c}\frac{\cos \mathrm{θ}\sin \mathrm{θ}}{r}\left[\sin \mathrm{Ó\neg }(t−\frac{r}{c})\right]\widehat{\mathrm{θ}}\\ E=−\frac{\mathrm{Î\pm }\mathrm{Ó\neg }}{cr}\cos \mathrm{θ}\sin \mathrm{θ}\left[\sin \mathrm{Ó\neg }(t−\frac{r}{c})\right]\widehat{\mathrm{θ}}\end{array}$ ……… (12)

Thus, the electric field is .$E=−\frac{\mathrm{Î\pm }\mathrm{Ó\neg }}{cr}\cos \mathrm{θ}\sin \mathrm{θ}\left[\sin \mathrm{Ó\neg }(t−\frac{r}{c})\right]\widehat{\mathrm{θ}}$

The magnetic field is given by

$\begin{array}{l}B=∇×A\\ B=\frac{1}{r}[\frac{∂}{∂r}\left(r{A}_0\right)−\frac{∂A_r}{∂\mathrm{θ}}]\widehat{\mathrm{Ï•}}\\ B=\frac{\mathrm{Î\pm }}{cr}\{\frac{∂}{∂r}\left[\cos \mathrm{θ}\cos \left(\mathrm{Ó\neg }(t−\frac{r}{c})\right)(−\sin \mathrm{θ})\right]−\frac{∂}{∂\mathrm{θ}}\frac{{\cos }^2\mathrm{θ}}{r}\cos \left(\mathrm{Ó\neg }(t−\frac{r}{c})\right)\}\widehat{\mathrm{Ï•}}\\ B=\frac{\mathrm{Î\pm }}{cr}\left\{(−\sin \mathrm{θ}\cos \mathrm{θ})[−\sin \left(\mathrm{Ó\neg }(t−\frac{r}{c})\right)]\left(\frac{−\mathrm{Ó\neg }}{c}\right)−\frac{∂}{∂\mathrm{θ}}\frac{{\cos }^2\mathrm{θ}}{r}\cos \left(\mathrm{Ó\neg }(t−\frac{r}{c})\right)\right\}\widehat{\mathrm{Ï•}}\end{array}$

Further solved as

$B=\frac{\mathrm{Î\pm }}{cr}\left\{(−\sin \mathrm{θ}\cos \mathrm{θ})\frac{\mathrm{Ó\neg }}{c}\sin \left(\mathrm{Ó\neg }(t−\frac{r}{c})\right)−\frac{∂}{∂\mathrm{θ}}\frac{{\cos }^2\mathrm{θ}}{r}\cos \left(\mathrm{Ó\neg }(t−\frac{r}{c})\right)\right\}\widehat{\mathrm{Ï•}}$

Consider the radiation zone.

$\begin{array}{l}B=\frac{\mathrm{Î\pm }}{cr}(−\sin \mathrm{θ}\cos \mathrm{θ})\frac{\mathrm{Ó\neg }}{c}\sin \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]\widehat{\mathrm{Ï•}}\\ B=−\frac{\mathrm{Î\pm }\mathrm{Ó\neg }}{c^{2}r}\sin \mathrm{θ}\cos \mathrm{θ}\sin \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]\widehat{\mathrm{Ï•}}\end{array}$

Thus, the magnetic field is .$B=−\frac{\mathrm{Î\pm }\mathrm{Ó\neg }}{c^{2}r}\sin \mathrm{θ}\cos \mathrm{θ}\sin \left[\mathrm{Ó\neg }(t−\frac{r}{c})\right]\widehat{\mathrm{Ï•}}$

(c)

Consider the formula for the magnetic field as

$B=\frac{1}{c}(\widehat{r}×E)$

Here, $Eâ‹\dots \widehat{r}=0$.

Then, the pointing vector is given as

$\begin{array}{l}S=\frac{1}{{\mathrm{μ}}_0}(E×B)\\ S=\frac{1}{{\mathrm{μ}}_0}[E×\frac{1}{c}(\widehat{r}×E)]\\ S=\frac{1}{{\mathrm{μ}}_{0}c}[E×(\widehat{r}×E)]\\ S=\frac{1}{{\mathrm{μ}}_{0}c}(E^2\widehat{r}−(E.\widehat{r})E)\end{array}$

Substitute $E.\widehat{r}=0$ in the above expression.

$\begin{array}{l}S=\frac{1}{{\mathrm{μ}}_{0}c}[E^2\widehat{r}−\left(0\right)E]\\ S=\frac{E^2}{{\mathrm{μ}}_{0}c}\widehat{r}\end{array}$

Substitute equation (12) in the expression.

$S=\frac{{\{−\frac{\mathrm{Î\pm }\mathrm{Ó\neg }}{cr}\cos \mathrm{θ}\sin \mathrm{θ}\left[\sin \mathrm{Ó\neg }(t−\frac{r}{c})\right]\}}^2}{{\mathrm{μ}}_{0}c}\widehat{r}$

Thus, the pointing vector is

$S=\frac{{\{−\frac{\mathrm{Î\pm }\mathrm{Ó\neg }}{cr}\cos \mathrm{θ}\sin \mathrm{θ}\left[\sin \mathrm{Ó\neg }(t−\frac{r}{c})\right]\}}^2}{{\mathrm{μ}}_{0}c}\widehat{r}$

The power radiated is given as

$\begin{array}{c}P=∫⟨SâŸ\copyright .da\\ =\frac{1}{{\mathrm{μ}}_{0}c}{\left(\frac{\mathrm{Î\pm }\mathrm{Ó\neg }}{c}\right)}^2∫{\sin }^2\mathrm{θ}{\cos }^2\mathrm{θ}\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}\\ =\frac{1}{2{\mathrm{μ}}_{0}c}{\left(\frac{\mathrm{Î\pm }\mathrm{Ó\neg }}{c}\right)}^2\left(2\mathrm{Ï€}\right)\underset{0}{\overset{\mathrm{Ï€}}{∫}}(1−{\cos }^2\mathrm{θ}){\cos }^2\mathrm{θ}\sin \mathrm{θ}d\mathrm{θ}\end{array}$

Let$u=\cos \mathrm{θ}$; then $du=−\sin \mathrm{θ}d\mathrm{θ}$.

Upper limit is $\mathrm{θ}=\mathrm{π}$; then

$\begin{array}{l}u=\cos \mathrm{π}\\ u=−1\end{array}$

Lower limit is $\mathrm{θ}=0$; then

$\begin{array}{l}u=\cos 0\\ u=1\end{array}$

Therefore

$\begin{array}{c}\underset{1}{\overset{−1}{∫}}(1−u^2)u^2(−du)=\underset{−1}{\overset{1}{∫}}(1−u^2)u^2\left(du\right)\\ =\underset{−1}{\overset{1}{∫}}(u^2−u^4)\left(du\right)\\ =\underset{−1}{\overset{1}{∫}}{u}^2\left(du\right)−\underset{−1}{\overset{1}{∫}}{u}^4\left(du\right)\end{array}$

Now integrate the expression

$\begin{array}{c}\underset{−1}{\overset{1}{∫}}{u}^2\left(du\right)−\underset{−1}{\overset{1}{∫}}{u}^4\left(du\right)={{\left[\frac{u^3}{3}\right]}^{+1}}_{−1}−{{\left[\frac{u^5}{5}\right]}_{−1}}^{+1}\\ =[\frac{1}{3}−(−\frac{1}{3})]−[\frac{1}{5}−(−\frac{1}{5})]\\ =\frac{2}{3}−\frac{2}{5}\\ =\frac{4}{15}\end{array}$

Now, power radiated is

$\begin{array}{l}P=\frac{1}{2{\mathrm{μ}}_{0}c}\frac{{\mathrm{Ó\neg }}^2}{c}\left(2\mathrm{Ï€}\right)\left(\frac{4}{15}\right)\\ P=\frac{1}{2{\mathrm{μ}}_{0}c}\frac{{\mathrm{Ó\neg }}^2}{c}\left(2\mathrm{Ï€}\right)\left(\frac{4}{15}\right){\mathrm{Î\pm }}^2\end{array}$

Since $\mathrm{Î\pm }=−\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^{2}d}{4\mathrm{Ï€}}$, then the above equation becomes

$\begin{array}{c}P=\frac{1}{2{\mathrm{μ}}_{0}c}\frac{{\mathrm{Ó\neg }}^2}{c}\left(2\mathrm{Ï€}\right)\left(\frac{4}{15}\right){[−\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^{2}d}{4\mathrm{Ï€}}]}^2\\ =\frac{1}{2{\mathrm{μ}}_{0}c}\frac{{\mathrm{Ó\neg }}^2}{c}\left(2\mathrm{Ï€}\right)\left(\frac{4}{15}\right)\frac{{{\mathrm{μ}}_0}^2{p_0}^2{\mathrm{Ó\neg }}^4{d}^2}{16{\mathrm{Ï€}}^2}\\ =\frac{{\mathrm{μ}}_0}{60\mathrm{Ï€}{c}^3}{\left(p_{0}d\right)}^2{\mathrm{Ó\neg }}^6\end{array}$

Thus, the power radiated is $P=\frac{{\mathrm{μ}}_0}{60\mathrm{Ï€}{c}^3}{\left(p_{0}d\right)}^2{\mathrm{Ó\neg }}^6$.