91Ó°ÊÓ

The angular velocity is$\mathrm{Ó\neg }$ .

The magnitude of the dipole is $p=2\mathrm{Ï€}R$.

The charges are$+q$ and $-q$.

The position of the charges$Â\pm q=Â\pm R(\cos \mathrm{Ó\neg }t\widehat{x}+\sin \mathrm{Ó\neg }t\widehat{y})$ .

The expression for the potential due to osculation dipole along $z-$axis is given as follows.

role="math" localid="1658821706455" $\mathit{V}\mathbf{=}\frac{\mathbf{-}{\mathbf{p}}_{\mathbf{0}}\mathbf{Ó\neg }}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}\mathbf{c}}\frac{\mathbf{r}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{θ}}{{\mathbf{r}}^{\mathbf{2}}}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left[}\mathbf{Ó\neg }\left(t-\frac{r}{c}\right)\mathbf{\right]}$ …… (1)

The expression for the power radiated is given as follows.

$\mathit{P}\mathbf{=}∫⟨SâŸ\copyright â‹\dots da$ …… (2)

The dipole moment of an electric dipole is given by,

$p=2q{r}_{Â\pm }$

The rotating electric dipole can be taken as super position of dipole moments of two oscillating.

The dipole moment can be expressed as,

$p=2qR(\cos \mathrm{Ó\neg }t\widehat{x}+\sin \mathrm{Ó\neg }t\widehat{y})$

The total charge is divided into two halves separated by the distance; therefore, the radiation reaction force is given by,

$\begin{array}{l}{\mathrm{F}}_{\mathrm{rad}}^{\mathrm{end}}=\frac{{\mathrm{μ}}_0{\left(\frac{\mathrm{q}}{2}\right)}^2}{6\mathrm{Ï€³¦}}\stackrel{·}{\mathrm{a}}\\ {\mathrm{F}}_{\mathrm{rad}}^{\mathrm{end}}=\frac{{\mathrm{μ}}_0{\mathrm{q}}^2}{24\mathrm{Ï€³¦}}\stackrel{·}{\mathrm{a}}\end{array}$

The interaction force is given by,

localid="1658821862972" $F_{rad}^{\mathrm{int}}=\frac{{\mathrm{μ}}_0{q}^2}{12\mathrm{π}c}\stackrel{·}{a}$

The total radiation force is the sum of the radiation force at both the end.

$F_{rad}=F_{rad}^{end}+F_{rad}^{\mathrm{int}}$

Substitute $\frac{{\mathrm{μ}}_0{q}^2}{24\mathrm{π}c}\stackrel{·}{a}$ for localid="1658821934982" $F_{rad}^{end}$ and $\frac{{\mathrm{μ}}_0{q}^2}{12\mathrm{π}c}\stackrel{·}{a}$ for $F_{rad}^{\mathrm{int}}$into above equation.

localid="1658825657309" $\begin{array}{l}{F}_{rad}=\frac{{\mathrm{μ}}_0{q}^2}{24\mathrm{π}c}\stackrel{·}{a}+\frac{{\mathrm{μ}}_0{q}^2}{12\mathrm{π}c}\stackrel{·}{a}\\ F_{rad}=\frac{{\mathrm{μ}}_0{q}^2}{12\mathrm{π}c}\stackrel{·}{a}×\frac{3}{2}\\ F_{rad}=\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{π}c}\stackrel{·}{a}\end{array}$…… (3)

If the total force is $F\left(q\right)=F^{\mathrm{int}}\left(q\right)+2F\left(\frac{q}{2}\right)$. The force$F\left(\frac{q}{2}\right)$is independent of distance then the interaction force$F^{\mathrm{int}}\left(q\right)$part.

From the equation 11.1, the interaction force is given by,

localid="1658822132351" $F^{\mathrm{int}}\left(q\right)=\frac{{\mathrm{μ}}_0{q}^2}{12\mathrm{π}c}\stackrel{·}{a}$ …… (4)

Now compare the equation (3), the interaction term in the self-torque is$F_{rad}^{\mathrm{int}}=\frac{{\mathrm{μ}}_0{q}^2}{12\mathrm{π}c}\stackrel{·}{a}$which is analogous to equation (4).

Hence the interaction term in the self-torque is $F_{rad}^{\mathrm{int}}=\frac{{\mathrm{μ}}_0{q}^2}{12\mathrm{π}c}\stackrel{·}{a}$.

The magnitude of the position vector is given by,

$\begin{array}{c}\mathrm{r}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2}\\ \mathrm{°ù³¦´Ç²õθ}=\mathrm{z}\end{array}$

Calculate the expression for the potential,

Substitute $\sqrt{x^2+y^2+z^2}$for $r$and $z$for $r\cos \mathrm{θ}$into equation (1).

$V=\frac{−p_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}c}\frac{x}{x^2+y^2+z^2}\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]$

The potential due to rotating electrical dipole is given by,

$\begin{array}{l}V=\frac{−p_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}c}\frac{x}{x^2+y^2+z^2}\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]−\frac{−p_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}c}\frac{y}{x^2+y^2+z^2}\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\\ V=\frac{−p_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}c}\left\{\frac{\sin \mathrm{θ}\cos \mathrm{Ï•}}{r}\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]−\frac{\sin \mathrm{θ}\sin \mathrm{Ï•}}{r}\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\right\}\end{array}$

The vector potential due to osculation dipole is given by,

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

The vector potential due to rotating dipole is given by,

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

The intensity is given by.

$S=\frac{1}{{\mathrm{μ}}_0}(E×B)$

Substitute $\frac{1}{c}(\widehat{r}×E)$ for $B$into above equation.

$\begin{array}{c}S=\frac{1}{{\mathrm{μ}}_0}\left(E×\frac{1}{c}(\widehat{r}×E)\right)\\ S=\frac{1}{{\mathrm{μ}}_{0}c}(E^2â‹\dots \widehat{r}−(Eâ‹\dots \widehat{r})E)\\ S=\frac{1}{{\mathrm{μ}}_{0}c}(E^2â‹\dots \widehat{r}−\left(0\right)E)\\ S=\frac{1}{{\mathrm{μ}}_{0}c}(E^2â‹\dots \widehat{r})\end{array}$ ……. (5)

The electric strength is given by,

$\begin{array}{l}E=−∇V−\frac{∂A}{∂t}\\ E=\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}r}\left(\frac{\sin \mathrm{θ}}{r}\right)\cos \left(\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right)\widehat{\mathrm{θ}}\end{array}$

The electric strength can be expressed as,

$\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\left(\widehat{x}−\frac{r}{c}\widehat{r}\right)+\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\left(\widehat{y}−\frac{y}{r}\widehat{r}\right)$

Take a square of both the sides of the above equation,

$\begin{array}{l}{E}^2={\left(\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}r}\right)}^2{\left\{\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\left(\widehat{x}−\frac{r}{c}\widehat{r}\right)+\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\left(\widehat{y}−\frac{y}{r}\widehat{r}\right)\right\}}^2\\ E^2={\left(\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}r}\right)}^2\left\{\begin{array}{l}{\left(\widehat{x}−\frac{r}{c}\widehat{r}\right)}^2{\cos }^2\left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]+\left(\widehat{y}−\frac{y}{r}\widehat{r}\right){\sin }^2\left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\\ −2\left(\widehat{x}−\frac{r}{c}\widehat{r}\right)â‹\dots \left(\widehat{y}−\frac{y}{r}\widehat{r}\right)\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\end{array}\right\}\end{array}$ …… (6)

Calculate the value of ${\left(\widehat{x}−\frac{r}{c}\widehat{r}\right)}^2$.

$\begin{array}{c}{\left(\widehat{x}−\frac{r}{c}\widehat{r}\right)}^2=1+\frac{x^2}{r^2}−2\frac{x^2}{r^2}\\ {\left(\widehat{x}−\frac{r}{c}\widehat{r}\right)}^2=1−\frac{x^2}{r^2}\end{array}$

Similarly,

${\left(\widehat{y}−\frac{y}{r}\widehat{r}\right)}^2=1−\frac{y^2}{r^2}$

Calculate the value of $\left(\widehat{x}−\frac{r}{c}\widehat{r}\right)â‹\dots \left(\widehat{y}−\frac{y}{r}\widehat{r}\right)$.

$\begin{array}{c}\left(\widehat{x}−\frac{r}{c}\widehat{r}\right)â‹\dots \left(\widehat{y}−\frac{y}{r}\widehat{r}\right)=\widehat{x}â‹\dots \widehat{y}+\frac{xy}{2}−\frac{x}{r}\widehat{r}â‹\dots \widehat{y}−\frac{y}{r}\widehat{x}â‹\dots \widehat{r}\\ \left(\widehat{x}−\frac{r}{c}\widehat{r}\right)â‹\dots \left(\widehat{y}−\frac{y}{r}\widehat{r}\right)=0+\frac{xy}{r^2}−\frac{xy}{r^2}−\frac{xy}{r^2}\\ \left(\widehat{x}−\frac{r}{c}\widehat{r}\right)â‹\dots \left(\widehat{y}−\frac{y}{r}\widehat{r}\right)=\frac{−xy}{r^2}\end{array}$

Substitute$\frac{−xy}{r^2}$ for$\left(\widehat{x}−\frac{r}{c}\right)\widehat{r}â‹\dots \left(\widehat{y}−\frac{y}{r}\widehat{r}\right)$, $1−\frac{y^2}{r^2}$ for${\left(\widehat{y}−\frac{y}{r}\widehat{r}\right)}^2$and$1−\frac{x^2}{r^2}$for${\left(\widehat{x}−\frac{r}{c}\widehat{r}\right)}^2$into equation (6).

localid="1658824474855" $\begin{array}{l}{E}^2={\left(\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}r}\right)}^2\left\{\begin{array}{l}\left(1−\frac{x^2}{r^2}\right){\cos }^2\left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]+\left(1−\frac{y^2}{r^2}\right){\sin }^2\left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\\ −2\left(\frac{−xy}{r^2}\right)\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\end{array}\right\}\\ E^2={\left(\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}r}\right)}^2\left\{\begin{array}{l}\left(1−\frac{x^2}{r^2}\right){\cos }^2\left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]+\left(1−\frac{y^2}{r^2}\right){\sin }^2\left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\\ +2\left(\frac{xy}{r^2}\right)\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\end{array}\right\}\\ E^2={\left(\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}r}\right)}^2\left[1−\frac{1}{r^2}{\left(x\cos \mathrm{Ó\neg }\left(t−\frac{r}{c}\right)+y\sin \mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right)}^2\right]\end{array}$

Substitute $r\sin \mathrm{θ}\cos \mathrm{ϕ}$for $x$, and $r\sin \mathrm{θ}\sin \mathrm{ϕ}$for $y$into above equation.

$\begin{array}{l}{E}^2={\left(\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}r}\right)}^2\left[1−\frac{1}{r^2}{\left(r\sin \mathrm{θ}\cos \mathrm{Ï•}\cos \mathrm{Ó\neg }\left(t−\frac{r}{c}\right)+r\sin \mathrm{θ}\sin \mathrm{Ï•}\sin \mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right)}^2\right]\\ E^2={\left(\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}r}\right)}^2\left[1−{\sin }^2\mathrm{θ}{\left(\cos \mathrm{Ï•}\cos \mathrm{Ó\neg }\left(t−\frac{r}{c}\right)+\sin \mathrm{Ï•}\sin \mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right)}^2\right]\\ E^2={\left(\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}r}\right)}^2\left[1−{\sin }^2\mathrm{θ}{\left(\left(\cos \mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right)−\mathrm{Ï•}\right)}^2]\right]\widehat{r}\end{array}$

Substitute ${\left(\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}r}\right)}^2\left[1−{\sin }^2\mathrm{θ}{\left(\cos \mathrm{Ó\neg }\left(t−\frac{r}{c}\right)−\mathrm{Ï•}\right)}^2\right]\widehat{r}$ for$E^2$into equation (5).

$\begin{array}{l}S=\frac{1}{{\mathrm{μ}}_{0}c}\left({\left(\frac{{\mathrm{μ}}_0{p}_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}r}\right)}^2\left[1−{\sin }^2\mathrm{θ}{\left(\cos \mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right)−\mathrm{Ï•})}^2\right]â‹\dots \widehat{r}\right)\\ S=\frac{{\mathrm{μ}}_0}{c}{\left(\frac{p_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}{r}^2}\right)}^2\left[1−\frac{1}{2}{\sin }^2\mathrm{θ}\right]\widehat{r}\end{array}$

Calculate the power radiated dipole,

Substitute localid="1658825122009" $\frac{{\mathrm{μ}}_0}{c}{\left(\frac{p_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}{r}^2}\right)}^2\left[1−\frac{1}{2}{\sin }^2\mathrm{θ}\right]\widehat{r}$for $S$into equation (3).

localid="1658825212175" $\begin{array}{l}P=∫\frac{{\mathrm{μ}}_0}{c}{\left(\frac{p_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}{r}^2}\right)}^2\left(1−\frac{1}{2}{\sin }^2\mathrm{θ}\right)â‹\dots da\\ P=\frac{{\mathrm{μ}}_0}{c}{\left(\frac{p_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}{r}^2}\right)}^2{∫}_{2\mathrm{Ï€}}\left(1−\frac{1}{2}{\sin }^2\mathrm{θ}\right)r^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}\\ P=\frac{{\mathrm{μ}}_0}{c}{\left(\frac{p_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}{r}^2}\right)}^2\underset{0}{\overset{\mathrm{Ï€}}{∫}}\underset{0}{\overset{2\mathrm{Ï€}}{∫}}\left(1−\frac{1}{2}{\sin }^2\mathrm{θ}\right)\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}\\ P=\frac{{\mathrm{μ}}_0}{c}{\left(\frac{p_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}{r}^2}\right)}^{2}2\mathrm{Ï€}\underset{0}{\overset{\mathrm{Ï€}}{∫}}\left(1−\frac{1}{2}{\sin }^2\mathrm{θ}\right)\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}\end{array}$

Solve further as,

$\begin{array}{l}P=\frac{{\mathrm{μ}}_0{p}_0^2{\mathrm{Ó\neg }}^4}{8\mathrm{Ï€}c}\left[\underset{0}{\overset{\mathrm{Ï€}}{∫}}\sin \mathrm{θ}d\mathrm{θ}−\frac{1}{2}\underset{0}{\overset{\mathrm{Ï€}}{∫}}{\sin }^3\mathrm{θ}d\mathrm{θ}\right]\\ P=\frac{{\mathrm{μ}}_0{p}_0^2{\mathrm{Ó\neg }}^4}{8\mathrm{Ï€}c}\left(2−\frac{1}{2}â‹\dots \frac{4}{3}\right)\\ P=\frac{{\mathrm{μ}}_0{p}_0^2{\mathrm{Ó\neg }}^4}{6\mathrm{Ï€}c}\end{array}$

The relationship between the power, torque and velocity is given by,

$\mathrm{Ï„}=\frac{R}{v}P$

Substitute $R\mathrm{Ó\neg }$for $v$into above equation.

$\begin{array}{l}\mathrm{Ï„}=\frac{R}{R\mathrm{Ó\neg }}P\\ \mathrm{Ï„}=\frac{1}{\mathrm{Ó\neg }}P\end{array}$

Substitute $\frac{{\mathrm{μ}}_0{p}_0^2{\mathrm{Ó\neg }}^4}{6\mathrm{Ï€}c}$for $P$into above equation.

$\begin{array}{c}\mathrm{Ï„}=\frac{1}{\mathrm{Ó\neg }}\left(\frac{{\mathrm{μ}}_0{p}_0^2{\mathrm{Ó\neg }}^4}{6\mathrm{Ï€}c}\right)\\ \mathrm{Ï„}=\frac{{\mathrm{μ}}_0{p}_0^2{\mathrm{Ó\neg }}^3}{6\mathrm{Ï€}c}\end{array}$

Hence the total radiation reaction torque on this system $−\frac{{\mathrm{μ}}_0{p}_0^2{\mathrm{Ó\neg }}^3}{6\mathrm{Ï€}c}$

(c)

The total power radiated from the equation 11.60 is given by,

$P_{rad}\left(t_0\right)≈\frac{{\mathrm{μ}}_0}{6\mathrm{π}c}{\left[\stackrel{·\text{ }·}{p}\left(t_0\right)\right]}^2$ …… (7)

The calculated total power radiated is given by,

$P=\frac{{\mathrm{μ}}_0{p}_0^2{\mathrm{Ó\neg }}^4}{6\mathrm{Ï€}c}$ …… (8)

By comparing the equations (7) and (8), the power radiated by the electrical dipole is consistent with power radiated by the accelerated charge.