Question

An ideal electric dipole is situated at the origin; its dipole moment points in the z direction and is quadratic in time:

$\mathbf{p}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\ddot{\mathbf{p}}}_{\mathbf{0}}{\mathbf{t}}^{\mathbf{2}}\widehat{\mathbf{z}}\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{<}\mathbf{t}\mathbf{<}\mathbf{\infty }\mathbf{)}$

where${\ddot{\mathbf{p}}}_{\mathbf{0}}$is a constant.

1. Use the method of Section 11.1.2 to determine the (exact) electric and magnetic fields for all r > 0 (there's also a delta-function term at the origin, but we're not concerned with that).
2. Calculate the power, $\mathit{P}\mathbf{(}\mathit{r}\mathbf{,}\mathit{t}\mathbf{)}$, passing through a sphere of radius r.
3. Find the total power radiated (Eq. 11.2), and check that your answer is consistent with Eq. 11.60.21

Short answer

1. The (exact) electric and magnetic fields for all r > 0 is $V=\frac{{\mu }_0{\ddot{p}}_0}{8\pi }\cos \left[{\left(\frac{ct}{r}\right)}^2-1\right]$,
2. the power, P(r, t), passing through a sphere of radius r. is $\frac{{\mathrm{\mu }}_0{{\ddot{\mathrm{p}}}_0}^2}{12{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}\mathrm{t}\left[{\mathrm{t}}^2+{\left(\frac{\mathrm{r}}{\mathrm{c}}\right)}^2\right]$
3. The radiated power is ${\mathrm{P}}_{\mathrm{rad}}=\frac{{\mathrm{\mu }}_0{{\ddot{\mathrm{p}}}^2}_0}{6\mathrm{\pi c}\text{}}$

Step-by-step solution

Understanding of dipole moment

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.

Determination of the Electric field and Magnetic field

(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,

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

Here,

$k=\frac{1}{2}{\ddot{p}}_0$

From equation 11.5, we get:

$V(r,t)=\frac{1}{4\pi {\epsilon }_0}\left\{\frac{q_0\text{}\cos \left[\omega \left(t-\frac{r_{+}}{c}\right)\right]\text{}}{r_{+}}-\frac{q_0\text{}\cos \left[\omega \left(t-\frac{r_{-}}{c}\right)\right]\text{}}{r_{-}}\right\}\mathrm{...}\text{(ii)}$

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

$\mathrm{V}(\mathrm{r},\mathrm{t})=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left\{\frac{\mathrm{k}{\left(\mathrm{t}-\frac{{\mathrm{r}}_{+}}{\mathrm{c}}\right)}^2\text{}}{{\mathrm{r}}_{+}}-\frac{\mathrm{k}{\left(\mathrm{t}-\frac{{\mathrm{r}}_{-}}{\mathrm{c}}\right)}^2\text{}}{{\mathrm{r}}_{-}}\right\}$

Taking common the term k, we get:

$\begin{array}{c}\mathrm{V}(\mathrm{r},\mathrm{t})=\frac{\mathrm{k}}{4{\mathrm{\pi \epsilon }}_0}\left\{\frac{{\left(\mathrm{t}-\frac{{\mathrm{r}}_{+}}{\mathrm{c}}\right)}^2\text{}}{{\mathrm{r}}_{+}}-\frac{{\left(\mathrm{t}-\frac{{\mathrm{r}}_{-}}{\mathrm{c}}\right)}^2\text{}}{{\mathrm{r}}_{-}}\right\}\\ =\frac{\mathrm{k}}{4{\mathrm{\pi \epsilon }}_0}\left\{\frac{\mathrm{t}{}^2+{\left(\frac{{\mathrm{r}}_{+}}{\mathrm{c}}\right)}^2-2\mathrm{t}\frac{{\mathrm{r}}_{+}}{\mathrm{c}}\text{}}{{\mathrm{r}}_{+}}-\frac{\mathrm{t}{}^2+{\left(\frac{{\mathrm{r}}_{-}}{\mathrm{c}}\right)}^2-2\mathrm{t}\frac{{\mathrm{r}}_{-}}{\mathrm{c}}}{{\mathrm{r}}_{-}}\right\}\\ =\frac{\mathrm{k}}{4{\mathrm{\pi \epsilon }}_0}\left[\mathrm{t}{}^2\left(\frac{1}{{\mathrm{r}}_{+}}-\frac{1}{{\mathrm{r}}_{-}}\right)+\frac{1}{{\mathrm{c}}^2}({\mathrm{r}}_{+}-{\mathrm{r}}_{-})\right]...\text{(iii)}\end{array}$

By using equations 11.8 and 11.9, we can write:

${\mathrm{r}}_{\pm }\cong \mathrm{r}\left(1\mp \frac{\mathrm{d}}{2\mathrm{r}}\mathrm{cos\theta }\right)$

And

$\frac{1}{{\mathrm{r}}_{\pm }}\cong \mathrm{r}\left(1\pm \frac{\mathrm{d}}{2\mathrm{r}}\mathrm{cos\theta }\right)$

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

role="math" localid="1658838599556" $\begin{array}{c}\mathrm{V}(\mathrm{r},\mathrm{t})=\frac{\mathrm{k}}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{\mathrm{t}{}^2}{\mathrm{r}}\left(\frac{\mathrm{d}}{\mathrm{r}}\mathrm{cos\theta }\right)+\frac{\mathrm{r}}{{\mathrm{c}}^2}\left(\frac{\mathrm{d}}{\mathrm{r}}\mathrm{cos\theta }\right)\right]\\ =\frac{\mathrm{k}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{c}}^2}\mathrm{dcos\theta }\left[{\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)}^2-1\right]\end{array}$

Now, putting the value of k, we get:

role="math" localid="1658838650158" $\begin{array}{c}\mathrm{V}(\mathrm{r},\mathrm{t})=\frac{\frac{1}{2}{\ddot{\mathrm{p}}}_0}{4{\mathrm{\pi \epsilon }}_0{\mathrm{c}}^2\text{}}\mathrm{cos\theta }\left[{\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)}^2-1\right]\\ =\frac{{\mathrm{\mu }}_0{\ddot{\mathrm{p}}}_0}{8\mathrm{\pi }\text{}}\mathrm{cos\theta }\left[{\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)}^2-1\right]\end{array}$

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:

$\mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\underset{-\frac{\mathrm{d}}{2}}{\overset{+\frac{\mathrm{d}}{2}}{\int }}\frac{-{\mathrm{q}}_0\mathrm{\omega sin}\left[\mathrm{\omega }\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)\right]\widehat{\mathrm{z}}}{\mathrm{r}}\mathrm{dz}\mathrm{...}\left(11.16\right)$

And

$\mathrm{A}(\mathrm{r},\mathrm{\theta },\mathrm{t})=-\frac{{\mathrm{\mu }}_0{\mathrm{p}}_0\mathrm{\omega }}{4\mathrm{\pi r}}\sin \left[\mathrm{\omega }\right(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\left)\right]\widehat{\mathrm{z}}$

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

$\begin{array}{c}\mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\widehat{\mathrm{z}}\underset{-\frac{\mathrm{d}}{2}}{\overset{+\frac{\mathrm{d}}{2}}{\int }}\frac{2\mathrm{k}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)}{\mathrm{r}}\mathrm{dz}\\ =\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\widehat{\mathrm{z}}\frac{2\mathrm{k}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)}{\mathrm{r}}\\ =\frac{{\mathrm{\mu }}_0{\ddot{\mathrm{p}}}_0}{4\mathrm{\pi c}}\left[\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)-1\right]\widehat{\mathrm{z}}\end{array}$

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

$\mathrm{E}=-\nabla \mathrm{V}-\frac{\partial \mathrm{A}}{\partial \mathrm{t}}$

Therefore,

$\begin{array}{c}E=-\frac{{\mu }_0{\ddot{p}}_0}{8\pi \text{}}\left\{\cos \theta \left[\left[-2\frac{{\left(ct\right)}^2}{r^3}\right]\right]\widehat{r}-\frac{1}{r}\sin \theta \left[{\left(\frac{ct}{r}\right)}^2-1\right]\widehat{\theta }\right\}-\frac{{\mu }_0{\ddot{p}}_0}{4\pi c\text{}}\left(\frac{c}{r}\right)\widehat{z}\\ =\frac{{\mu }_0{\ddot{p}}_0}{4\pi r\text{}}\left\{\left[{\left(\frac{ct}{r}\right)}^2-1\right]\cos \theta \widehat{r}+\frac{1}{2}\left[{\left(\frac{ct}{r}\right)}^2+1\right]\sin \theta \widehat{\theta }\right\}\end{array}$

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

$\begin{array}{c}\mathrm{B}=-\nabla \times \mathrm{A}\\ =-\nabla \times \left(\frac{{\mathrm{\mu }}_0{\ddot{\mathrm{p}}}_0}{4\mathrm{\pi c}}\left[\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)-1\right]\widehat{\mathrm{z}}\right)\\ =\frac{{\mathrm{\mu }}_0{\ddot{\mathrm{p}}}_0}{4\mathrm{\pi c}}\nabla \times \left\{\left[\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)-1\right](\mathrm{cos\theta }\widehat{\mathrm{r}}-\mathrm{sin\theta }\widehat{\mathrm{\theta }})\right\}\\ =\frac{{\mathrm{\mu }}_0{\ddot{\mathrm{p}}}_0\mathrm{t}}{4{\mathrm{\pi r}}^2}\mathrm{sin\theta }\widehat{\mathrm{\varphi }}\end{array}$

Determination of power passing through the sphere.

(b)

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

$\mathrm{S}=\frac{1}{{\mathrm{\mu }}_0}(\mathrm{E}\times \mathrm{B})$

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

$\begin{array}{c}S=\frac{1}{{\mu }_0}(E\times B)\\ =\frac{1}{{\mu }_0}\left[\left(\frac{{\mu }_0{\ddot{p}}_0}{4\pi r\text{}}\left[{\left(\frac{ct}{r}\right)}^2-1\right]\cos \theta \widehat{r}+\frac{1}{2}\left[{\left(\frac{ct}{r}\right)}^2+1\right]\sin \theta \widehat{\theta }\right)\times \left(\frac{{\mu }_0{\ddot{p}}_{0}t}{4\pi r^2}\sin \theta \widehat{\varphi }\right)\right]\\ =\frac{{\mu }_0{{\ddot{p}}^2}_{0}t}{32{\pi }^2\text{}}\left[{\left(\frac{ct}{r}\right)}^2-1\right]\frac{{\sin }^2\theta }{r^3}\left(r^2\sin \theta d\theta d\varphi \right)\end{array}$

Therefore, power radiated can be calculated as

$\begin{array}{c}\mathrm{P}(\mathrm{r},{\mathrm{t}}_0)=\underset{0}{\overset{\mathrm{\pi }}{\int }}\mathrm{sd\theta }\\ =\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{{\mathrm{\mu }}_0{{\ddot{\mathrm{p}}}^2}_0\mathrm{t}}{32{\mathrm{\pi }}^2\text{}}\left[{\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)}^2-1\right]\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{r}}^3}\left({\mathrm{r}}^2\mathrm{sin\theta d\theta d\varphi }\right)\mathrm{d\theta }\\ =\frac{{\mathrm{\mu }}_0{{\ddot{\mathrm{p}}}^2}_0\mathrm{t}}{32{\mathrm{\pi }}^2\text{}}\left[{\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)}^2-1\right]2\mathrm{\pi }\underset{0}{\overset{\mathrm{\pi }}{\int }}{\sin }^3\mathrm{\theta d\theta }\\ =\frac{{\mathrm{\mu }}_0{{\ddot{\mathrm{p}}}^2}_0\mathrm{t}}{12\mathrm{\pi r}\text{}}\left[{\left(\frac{\mathrm{ct}}{\mathrm{r}}\right)}^2-1\right]...\text{(iv)}\end{array}$

To check the consistency of the answer.

(c)

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

From equation (iv), we can write:

$\begin{array}{c}\mathrm{P}(\mathrm{r},{\mathrm{t}}_0\text{}+\mathrm{r}/\mathrm{c})=\frac{{\mathrm{\mu }}_0{{\ddot{\mathrm{p}}}^2}_0}{12\mathrm{\pi r}\text{}}\left({\mathrm{t}}_0\text{}+\frac{\mathrm{r}}{\mathrm{c}}\right)\left[{\left(\frac{\mathrm{c}}{\mathrm{r}}\right)}^2({{\mathrm{t}}_0}^2\text{}+2{\mathrm{t}}_0\text{}\frac{\mathrm{r}}{\mathrm{c}}+{\left(\frac{\mathrm{r}}{\mathrm{c}}\right)}^2+1\right]\\ =\frac{{\mathrm{\mu }}_0{{\ddot{\mathrm{p}}}^2}_0}{12\mathrm{\pi c}\text{}}\left(1+\frac{{\mathrm{ct}}_0}{\mathrm{r}}\right)\left[2+2\frac{{\mathrm{ct}}_0}{\mathrm{r}}+{\left(\frac{{\mathrm{ct}}_0}{\mathrm{r}}\right)}^2\right]\end{array}$

Therefore,

The radiated power:

$\begin{array}{c}{\mathrm{P}}_{\mathrm{rad}}=\underset{\mathrm{r}\to \mathrm{\infty }}{\lim }\mathrm{P}\left(\mathrm{r},{\mathrm{t}}_0\text{}+\frac{\mathrm{r}}{\mathrm{c}}\right)\\ =\frac{{\mathrm{\mu }}_0{{\ddot{\mathrm{p}}}^2}_0}{6\mathrm{\pi c}\text{}}\end{array}$