91Ó°ÊÓ

Dipole moment is a property of dipole which develops when a certain distance separates two charged particles. The main reason behind the rise of this property is the electronegativity and difference between chemically bonded atoms or elements.

(a)

From equation 11.4, we can write,

$\mathrm{p}\left(\mathrm{t}\right)=\mathrm{q}\left(\mathrm{t}\right)\mathrm{d}\widehat{\mathrm{z}}$

And we know that,

$\mathrm{q}\left(\mathrm{t}\right)={\mathrm{q}}_0\text{cos}\left(\mathrm{Ó\neg ³Ù}\right)={\mathrm{k}}_{\mathrm{d}}{\mathrm{t}}^2...\text{(i)}$

Here,

$k=\frac{1}{2}{\stackrel{¨}{p}}_0$

From equation 11.5, we get:

$V(r,t)=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left\{\frac{q_0\text{​}\cos \left[\mathrm{Ó\neg }\left(t−\frac{r_{+}}{c}\right)\right]\text{​}}{r_{+}}−\frac{q_0\text{​}\cos \left[\mathrm{Ó\neg }\left(t−\frac{r_{−}}{c}\right)\right]\text{​}}{r_{−}}\right\}\mathrm{...}\text{(ii)}$

Now, by using equation (i), we can modify equation (ii) as:

$\mathrm{V}(\mathrm{r},\mathrm{t})=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\left\{\frac{\mathrm{k}{\left(\mathrm{t}−\frac{{\mathrm{r}}_{+}}{\mathrm{c}}\right)}^2\text{​}}{{\mathrm{r}}_{+}}−\frac{\mathrm{k}{\left(\mathrm{t}−\frac{{\mathrm{r}}_{−}}{\mathrm{c}}\right)}^2\text{​}}{{\mathrm{r}}_{−}}\right\}$

Taking common the term k, we get:

$\begin{array}{c}\mathrm{V}(\mathrm{r},\mathrm{t})=\frac{\mathrm{k}}{4{\mathrm{Ï€Î\mu }}_0}\left\{\frac{{\left(\mathrm{t}−\frac{{\mathrm{r}}_{+}}{\mathrm{c}}\right)}^2\text{​}}{{\mathrm{r}}_{+}}−\frac{{\left(\mathrm{t}−\frac{{\mathrm{r}}_{−}}{\mathrm{c}}\right)}^2\text{​}}{{\mathrm{r}}_{−}}\right\}\\ =\frac{\mathrm{k}}{4{\mathrm{Ï€Î\mu }}_0}\left\{\frac{\mathrm{t}{}^2+{\left(\frac{{\mathrm{r}}_{+}}{\mathrm{c}}\right)}^2−2\mathrm{t}\frac{{\mathrm{r}}_{+}}{\mathrm{c}}\text{​}}{{\mathrm{r}}_{+}}−\frac{\mathrm{t}{}^2+{\left(\frac{{\mathrm{r}}_{−}}{\mathrm{c}}\right)}^2−2\mathrm{t}\frac{{\mathrm{r}}_{−}}{\mathrm{c}}}{{\mathrm{r}}_{−}}\right\}\\ =\frac{\mathrm{k}}{4{\mathrm{Ï€Î\mu }}_0}\left[\mathrm{t}{}^2\left(\frac{1}{{\mathrm{r}}_{+}}−\frac{1}{{\mathrm{r}}_{−}}\right)+\frac{1}{{\mathrm{c}}^2}({\mathrm{r}}_{+}−{\mathrm{r}}_{−})\right]...\text{(iii)}\end{array}$

By using equations 11.8 and 11.9, we can write:

${\mathrm{r}}_{Â\pm }â‰\dots \mathrm{r}\left(1∓\frac{\mathrm{d}}{2\mathrm{r}}\mathrm{³¦´Ç²õθ}\right)$

And

$\frac{1}{{\mathrm{r}}_{Â\pm }}â‰\dots \mathrm{r}\left(1Â\pm \frac{\mathrm{d}}{2\mathrm{r}}\mathrm{³¦´Ç²õθ}\right)$

Now, putting the values in equation (iii), we get:

role="math" localid="1658838599556" $\begin{array}{c}\mathrm{V}(\mathrm{r},\mathrm{t})=\frac{\mathrm{k}}{4{\mathrm{Ï€Î\mu }}_0}\left[\frac{\mathrm{t}{}^2}{\mathrm{r}}\left(\frac{\mathrm{d}}{\mathrm{r}}\mathrm{³¦´Ç²õθ}\right)+\frac{\mathrm{r}}{{\mathrm{c}}^2}\left(\frac{\mathrm{d}}{\mathrm{r}}\mathrm{³¦´Ç²õθ}\right)\right]\\ =\frac{\mathrm{k}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{c}}^2}\mathrm{d³¦´Ç²õθ}\left[{\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)}^2−1\right]\end{array}$

Now, putting the value of k, we get:

role="math" localid="1658838650158" $\begin{array}{c}\mathrm{V}(\mathrm{r},\mathrm{t})=\frac{\frac{1}{2}{\stackrel{¨}{\mathrm{p}}}_0}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{c}}^2\text{​}}\mathrm{³¦´Ç²õθ}\left[{\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)}^2−1\right]\\ =\frac{{\mathrm{μ}}_0{\stackrel{¨}{\mathrm{p}}}_0}{8\mathrm{Ï€}\text{​}}\mathrm{³¦´Ç²õθ}\left[{\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)}^2−1\right]\end{array}$

From equation 11.15, we can write:

$\mathrm{I}\left(\mathrm{t}\right)=\left(\frac{\mathrm{dq}}{\mathrm{dt}}\right)\widehat{\mathrm{z}}=2\mathrm{kt}\widehat{\mathrm{z}}$

Again from equations 11.16 and 11.17, we get:

$\mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\underset{−\frac{\mathrm{d}}{2}}{\overset{+\frac{\mathrm{d}}{2}}{∫}}\frac{−{\mathrm{q}}_0\mathrm{Ó\neg sin}\left[\mathrm{Ó\neg }\left(\mathrm{t}−\frac{\mathrm{r}}{\mathrm{c}}\right)\right]\widehat{\mathrm{z}}}{\mathrm{r}}\mathrm{dz}\mathrm{...}\left(11.16\right)$

And

$\mathrm{A}(\mathrm{r},\mathrm{θ},\mathrm{t})=−\frac{{\mathrm{μ}}_0{\mathrm{p}}_0\mathrm{Ó\neg }}{4\mathrm{Ï€°ù}}\sin \left[\mathrm{Ó\neg }\right(\mathrm{t}−\frac{\mathrm{r}}{\mathrm{c}}\left)\right]\widehat{\mathrm{z}}$

Therefore, from equation (11.16), we can further calculate as:

$\begin{array}{c}\mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\widehat{\mathrm{z}}\underset{−\frac{\mathrm{d}}{2}}{\overset{+\frac{\mathrm{d}}{2}}{∫}}\frac{2\mathrm{k}\left(\mathrm{t}−\frac{\mathrm{r}}{\mathrm{c}}\right)}{\mathrm{r}}\mathrm{dz}\\ =\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\widehat{\mathrm{z}}\text{ }\frac{2\mathrm{k}\left(\mathrm{t}−\frac{\mathrm{r}}{\mathrm{c}}\right)}{\mathrm{r}}\\ =\frac{{\mathrm{μ}}_0{\stackrel{¨}{\mathrm{p}}}_0}{4\mathrm{Ï€³¦}}\left[\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)−1\right]\widehat{\mathrm{z}}\end{array}$

We know that the formulae of Electric field intensity can be written as

$\mathrm{E}=−∇\mathrm{V}−\frac{∂\mathrm{A}}{∂\mathrm{t}}$

Therefore,

$\begin{array}{c}E=−\frac{{\mathrm{μ}}_0{\stackrel{¨}{p}}_0}{8\mathrm{π}\text{​}}\left\{\cos \mathrm{θ}\left[\left[−2\frac{{\left(ct\right)}^2}{r^3}\right]\right]\widehat{r}−\frac{1}{r}\sin \mathrm{θ}\left[{\left(\frac{ct}{r}\right)}^2−1\right]\widehat{\mathrm{θ}}\right\}−\frac{{\mathrm{μ}}_0{\stackrel{¨}{p}}_0}{4\mathrm{π}c\text{​}}\left(\frac{c}{r}\right)\widehat{z}\\ =\frac{{\mathrm{μ}}_0{\stackrel{¨}{p}}_0}{4\mathrm{π}r\text{​}}\left\{\left[{\left(\frac{ct}{r}\right)}^2−1\right]\cos \mathrm{θ}\widehat{r}+\frac{1}{2}\left[{\left(\frac{ct}{r}\right)}^2+1\right]\sin \mathrm{θ}\widehat{\mathrm{θ}}\right\}\end{array}$

We know that the formulae for calculating magnetic flux density (B) are:

$\begin{array}{c}\mathrm{B}=−∇×\mathrm{A}\\ =−∇×\left(\frac{{\mathrm{μ}}_0{\stackrel{¨}{\mathrm{p}}}_0}{4\mathrm{Ï€³¦}}\left[\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)−1\right]\widehat{\mathrm{z}}\right)\\ =\frac{{\mathrm{μ}}_0{\stackrel{¨}{\mathrm{p}}}_0}{4\mathrm{Ï€³¦}}∇×\left\{\left[\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)−1\right](\mathrm{³¦´Ç²õθ}\widehat{\mathrm{r}}−\mathrm{²õ¾\pm ²Ôθ}\widehat{\mathrm{θ}})\right\}\\ =\frac{{\mathrm{μ}}_0{\stackrel{¨}{\mathrm{p}}}_0\mathrm{t}}{4{\mathrm{Ï€°ù}}^2}\mathrm{²õ¾\pm ²Ôθ}\widehat{\mathrm{Ï•}}\end{array}$

(b)

For the determination of power, we have to use a pointing vector which is:

$\mathrm{S}=\frac{1}{{\mathrm{μ}}_0}(\mathrm{E}×\mathrm{B})$

Putting the respective expressions of E and B, we get:

$\begin{array}{c}S=\frac{1}{{\mathrm{μ}}_0}(E×B)\\ =\frac{1}{{\mathrm{μ}}_0}\left[\left(\frac{{\mathrm{μ}}_0{\stackrel{¨}{p}}_0}{4\mathrm{π}r\text{​}}\left[{\left(\frac{ct}{r}\right)}^2−1\right]\cos \mathrm{θ}\widehat{r}+\frac{1}{2}\left[{\left(\frac{ct}{r}\right)}^2+1\right]\sin \mathrm{θ}\widehat{\mathrm{θ}}\right)×\left(\frac{{\mathrm{μ}}_0{\stackrel{¨}{p}}_{0}t}{4\mathrm{π}{r}^2}\sin \mathrm{θ}\widehat{\mathrm{ϕ}}\right)\right]\\ =\frac{{\mathrm{μ}}_0{{\stackrel{¨}{p}}^2}_{0}t}{32{\mathrm{π}}^2\text{​}}\left[{\left(\frac{ct}{r}\right)}^2−1\right]\frac{{\sin }^2\mathrm{θ}}{r^3}\left(r^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{ϕ}\right)\end{array}$

Therefore, power radiated can be calculated as

$\begin{array}{c}\mathrm{P}(\mathrm{r},{\mathrm{t}}_0)=\underset{0}{\overset{\mathrm{Ï€}}{∫}}\mathrm{²õ»åθ}\\ =\underset{0}{\overset{\mathrm{Ï€}}{∫}}\frac{{\mathrm{μ}}_0{{\stackrel{¨}{\mathrm{p}}}^2}_0\mathrm{t}}{32{\mathrm{Ï€}}^2\text{​}}\left[{\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)}^2−1\right]\frac{{\sin }^2\mathrm{θ}}{{\mathrm{r}}^3}\left({\mathrm{r}}^2\mathrm{²õ¾\pm ²Ôθ»åθdÏ•}\right)\mathrm{»åθ}\\ =\frac{{\mathrm{μ}}_0{{\stackrel{¨}{\mathrm{p}}}^2}_0\mathrm{t}}{32{\mathrm{Ï€}}^2\text{​}}\left[{\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)}^2−1\right]2\mathrm{Ï€}\underset{0}{\overset{\mathrm{Ï€}}{∫}}{\sin }^3\mathrm{θ»åθ}\\ =\frac{{\mathrm{μ}}_0{{\stackrel{¨}{\mathrm{p}}}^2}_0\mathrm{t}}{12\mathrm{Ï€°ù}\text{​}}\left[{\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)}^2−1\right]...\text{(iv)}\end{array}$

(c)

To check the consistency with the equation Eq. 11.60.21, we can do the following:

From equation (iv), we can write:

$\begin{array}{c}\mathrm{P}(\mathrm{r},{\mathrm{t}}_0\text{​}+\mathrm{r}/\mathrm{c})=\frac{{\mathrm{μ}}_0{{\stackrel{¨}{\mathrm{p}}}^2}_0}{12\mathrm{Ï€°ù}\text{​}}\left({\mathrm{t}}_0\text{​}+\frac{\mathrm{r}}{\mathrm{c}}\right)\left[{\left(\frac{\mathrm{c}}{\mathrm{r}}\right)}^2({{\mathrm{t}}_0}^2\text{​}+2{\mathrm{t}}_0\text{​}\frac{\mathrm{r}}{\mathrm{c}}+{\left(\frac{\mathrm{r}}{\mathrm{c}}\right)}^2+1\right]\\ =\frac{{\mathrm{μ}}_0{{\stackrel{¨}{\mathrm{p}}}^2}_0}{12\mathrm{Ï€³¦}\text{​}}\left(1+\frac{{\mathrm{ct}}_0}{\mathrm{r}}\right)\left[2+2\frac{{\mathrm{ct}}_0}{\mathrm{r}}+{\left(\frac{{\mathrm{ct}}_0}{\mathrm{r}}\right)}^2\right]\end{array}$

Therefore,

The radiated power:

$\begin{array}{c}{\mathrm{P}}_{\mathrm{rad}}=\underset{\mathrm{r}→\mathrm{∞}}{\lim }\mathrm{P}\left(\mathrm{r},{\mathrm{t}}_0\text{​}+\frac{\mathrm{r}}{\mathrm{c}}\right)\\ =\frac{{\mathrm{μ}}_0{{\stackrel{¨}{\mathrm{p}}}^2}_0}{6\mathrm{Ï€³¦}\text{​}}\end{array}$