An electric dipole rotates at constant angular velocity $\mathbf{\omega }$in the$\mathbf{xy}$ plane. (The charges,$\mathbf{\pm }\mathbf{q}$ , are at ${\mathbf{r}}_{\mathbf{\pm }}\mathbf{=}\mathbf{\pm }\mathbf{R}\mathbf{(}\mathbf{cos\omega t}\widehat{\mathbf{x}}\mathbf{+}\mathbf{sin\omega t}\widehat{\mathbf{y}}\mathbf{)}$; the magnitude of the dipole moment is $\mathbf{p}\mathbf{=}\mathbf{2}\mathbf{qR}$.)

(a) Find the interaction term in the self-torque (analogous to Eq. 11.99). Assume the motion is nonrelativistic ( $\mathbf{\omega R}\mathbf{<}\mathbf{<}\mathbf{c}$).

(b) Use the method of Prob. 11.20(a) to obtain the total radiation reaction torque on this system. [answer: $\mathbf{-}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{p}}^{\mathbf{2}}{\mathbf{\omega }}^{\mathbf{3}}}{\mathbf{6}\mathbf{\pi c}}\widehat{\mathbf{z}}$]

(c) Check that this result is consistent with the power radiated (Eq. 11.60).

The angular velocity is$\omega$ .

The magnitude of the dipole is $p=2\pi R$.

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

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

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

role="math" localid="1658821706455" $\mathit{V}\mathbf{=}\frac{\mathbf{-}{\mathbf{p}}_{\mathbf{0}}\mathbf{\omega }}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{c}}\frac{\mathbf{r}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }}{{\mathbf{r}}^{\mathbf{2}}}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left[}\mathbf{\omega }\left(t-\frac{r}{c}\right)\mathbf{\right]}$ …… (1)

The expression for the power radiated is given as follows.

$\mathit{P}\mathbf{=}\int ⟨S⟩\cdot da$ …… (2)

The dipole moment of an electric dipole is given by,

$p=2q{r}_{\pm }$

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

The dipole moment can be expressed as,

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

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

$\begin{array}{l}{\mathrm{F}}_{\mathrm{rad}}^{\mathrm{end}}=\frac{{\mathrm{\mu }}_0{\left(\frac{\mathrm{q}}{2}\right)}^2}{6\mathrm{\pi c}}\stackrel{·}{\mathrm{a}}\\ {\mathrm{F}}_{\mathrm{rad}}^{\mathrm{end}}=\frac{{\mathrm{\mu }}_0{\mathrm{q}}^2}{24\mathrm{\pi c}}\stackrel{·}{\mathrm{a}}\end{array}$

The interaction force is given by,

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

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

localid="1658825657309" $\begin{array}{l}{F}_{rad}=\frac{{\mu }_0{q}^2}{24\pi c}\stackrel{·}{a}+\frac{{\mu }_0{q}^2}{12\pi c}\stackrel{·}{a}\\ F_{rad}=\frac{{\mu }_0{q}^2}{12\pi c}\stackrel{·}{a}\times \frac{3}{2}\\ F_{rad}=\frac{{\mu }_0{q}^2}{6\pi c}\stackrel{·}{a}\end{array}$…… (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" $F^{\mathrm{int}}\left(q\right)=\frac{{\mu }_0{q}^2}{12\pi c}\stackrel{·}{a}$ …… (4)

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

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

The magnitude of the position vector is given by,

$\begin{array}{c}\mathrm{r}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2}\\ \mathrm{rcos\theta }=\mathrm{z}\end{array}$

Calculate the expression for the potential,

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

$V=\frac{-p_0\omega }{4\pi {\epsilon }_{0}c}\frac{x}{x^2+y^2+z^2}\sin \left[\omega \left(t-\frac{r}{c}\right)\right]$

The potential due to rotating electrical dipole is given by,

$\begin{array}{l}V=\frac{-p_0\omega }{4\pi {\epsilon }_{0}c}\frac{x}{x^2+y^2+z^2}\sin \left[\omega \left(t-\frac{r}{c}\right)\right]-\frac{-p_0\omega }{4\pi {\epsilon }_{0}c}\frac{y}{x^2+y^2+z^2}\cos \left[\omega \left(t-\frac{r}{c}\right)\right]\\ V=\frac{-p_0\omega }{4\pi {\epsilon }_{0}c}\left\{\frac{\sin \theta \cos \varphi }{r}\sin \left[\omega \left(t-\frac{r}{c}\right)\right]-\frac{\sin \theta \sin \varphi }{r}\cos \left[\omega \left(t-\frac{r}{c}\right)\right]\right\}\end{array}$

The vector potential due to osculation dipole is given by,

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

The vector potential due to rotating dipole is given by,

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

The intensity is given by.

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

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

$\begin{array}{c}S=\frac{1}{{\mu }_0}\left(E\times \frac{1}{c}(\widehat{r}\times E)\right)\\ S=\frac{1}{{\mu }_{0}c}(E^2\cdot \widehat{r}-(E\cdot \widehat{r})E)\\ S=\frac{1}{{\mu }_{0}c}(E^2\cdot \widehat{r}-\left(0\right)E)\\ S=\frac{1}{{\mu }_{0}c}(E^2\cdot \widehat{r})\end{array}$ ……. (5)

The electric strength is given by,

$\begin{array}{l}E=-\nabla V-\frac{\partial A}{\partial t}\\ E=\frac{{\mu }_0{p}_0{\omega }^2}{4\pi r}\left(\frac{\sin \theta }{r}\right)\cos \left(\omega \left(t-\frac{r}{c}\right)\right)\widehat{\theta }\end{array}$

The electric strength can be expressed as,

$\cos \left[\omega \left(t-\frac{r}{c}\right)\right]\left(\widehat{x}-\frac{r}{c}\widehat{r}\right)+\sin \left[\omega \left(t-\frac{r}{c}\right)\right]\left(\widehat{y}-\frac{y}{r}\widehat{r}\right)$

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

$\begin{array}{l}{E}^2={\left(\frac{{\mu }_0{p}_0{\omega }^2}{4\pi r}\right)}^2{\left\{\cos \left[\omega \left(t-\frac{r}{c}\right)\right]\left(\widehat{x}-\frac{r}{c}\widehat{r}\right)+\sin \left[\omega \left(t-\frac{r}{c}\right)\right]\left(\widehat{y}-\frac{y}{r}\widehat{r}\right)\right\}}^2\\ E^2={\left(\frac{{\mu }_0{p}_0{\omega }^2}{4\pi r}\right)}^2\left\{\begin{array}{l}{\left(\widehat{x}-\frac{r}{c}\widehat{r}\right)}^2{\cos }^2\left[\omega \left(t-\frac{r}{c}\right)\right]+\left(\widehat{y}-\frac{y}{r}\widehat{r}\right){\sin }^2\left[\omega \left(t-\frac{r}{c}\right)\right]\\ -2\left(\widehat{x}-\frac{r}{c}\widehat{r}\right)\cdot \left(\widehat{y}-\frac{y}{r}\widehat{r}\right)\sin \left[\omega \left(t-\frac{r}{c}\right)\right]\cos \left[\omega \left(t-\frac{r}{c}\right)\right]\end{array}\right\}\end{array}$ …… (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 $\left(\widehat{x}-\frac{r}{c}\widehat{r}\right)\cdot \left(\widehat{y}-\frac{y}{r}\widehat{r}\right)$.

$\begin{array}{c}\left(\widehat{x}-\frac{r}{c}\widehat{r}\right)\cdot \left(\widehat{y}-\frac{y}{r}\widehat{r}\right)=\widehat{x}\cdot \widehat{y}+\frac{xy}{2}-\frac{x}{r}\widehat{r}\cdot \widehat{y}-\frac{y}{r}\widehat{x}\cdot \widehat{r}\\ \left(\widehat{x}-\frac{r}{c}\widehat{r}\right)\cdot \left(\widehat{y}-\frac{y}{r}\widehat{r}\right)=0+\frac{xy}{r^2}-\frac{xy}{r^2}-\frac{xy}{r^2}\\ \left(\widehat{x}-\frac{r}{c}\widehat{r}\right)\cdot \left(\widehat{y}-\frac{y}{r}\widehat{r}\right)=\frac{-xy}{r^2}\end{array}$

Substitute$\frac{-xy}{r^2}$ for$\left(\widehat{x}-\frac{r}{c}\right)\widehat{r}\cdot \left(\widehat{y}-\frac{y}{r}\widehat{r}\right)$, $1-\frac{y^2}{r^2}$ 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" $\begin{array}{l}{E}^2={\left(\frac{{\mu }_0{p}_0{\omega }^2}{4\pi r}\right)}^2\left\{\begin{array}{l}\left(1-\frac{x^2}{r^2}\right){\cos }^2\left[\omega \left(t-\frac{r}{c}\right)\right]+\left(1-\frac{y^2}{r^2}\right){\sin }^2\left[\omega \left(t-\frac{r}{c}\right)\right]\\ -2\left(\frac{-xy}{r^2}\right)\sin \left[\omega \left(t-\frac{r}{c}\right)\right]\cos \left[\omega \left(t-\frac{r}{c}\right)\right]\end{array}\right\}\\ E^2={\left(\frac{{\mu }_0{p}_0{\omega }^2}{4\pi r}\right)}^2\left\{\begin{array}{l}\left(1-\frac{x^2}{r^2}\right){\cos }^2\left[\omega \left(t-\frac{r}{c}\right)\right]+\left(1-\frac{y^2}{r^2}\right){\sin }^2\left[\omega \left(t-\frac{r}{c}\right)\right]\\ +2\left(\frac{xy}{r^2}\right)\sin \left[\omega \left(t-\frac{r}{c}\right)\right]\cos \left[\omega \left(t-\frac{r}{c}\right)\right]\end{array}\right\}\\ E^2={\left(\frac{{\mu }_0{p}_0{\omega }^2}{4\pi r}\right)}^2\left[1-\frac{1}{r^2}{\left(x\cos \omega \left(t-\frac{r}{c}\right)+y\sin \omega \left(t-\frac{r}{c}\right)\right)}^2\right]\end{array}$

Substitute $r\sin \theta \cos \varphi$for $x$, and $r\sin \theta \sin \varphi$for $y$into above equation.

$\begin{array}{l}{E}^2={\left(\frac{{\mu }_0{p}_0{\omega }^2}{4\pi r}\right)}^2\left[1-\frac{1}{r^2}{\left(r\sin \theta \cos \varphi \cos \omega \left(t-\frac{r}{c}\right)+r\sin \theta \sin \varphi \sin \omega \left(t-\frac{r}{c}\right)\right)}^2\right]\\ E^2={\left(\frac{{\mu }_0{p}_0{\omega }^2}{4\pi r}\right)}^2\left[1-{\sin }^2\theta {\left(\cos \varphi \cos \omega \left(t-\frac{r}{c}\right)+\sin \varphi \sin \omega \left(t-\frac{r}{c}\right)\right)}^2\right]\\ E^2={\left(\frac{{\mu }_0{p}_0{\omega }^2}{4\pi r}\right)}^2\left[1-{\sin }^2\theta {\left(\left(\cos \omega \left(t-\frac{r}{c}\right)\right)-\varphi \right)}^2]\right]\widehat{r}\end{array}$

Substitute ${\left(\frac{{\mu }_0{p}_0{\omega }^2}{4\pi r}\right)}^2\left[1-{\sin }^2\theta {\left(\cos \omega \left(t-\frac{r}{c}\right)-\varphi \right)}^2\right]\widehat{r}$ for$E^2$into equation (5).

$\begin{array}{l}S=\frac{1}{{\mu }_{0}c}\left({\left(\frac{{\mu }_0{p}_0{\omega }^2}{4\pi r}\right)}^2\left[1-{\sin }^2\theta {\left(\cos \omega \left(t-\frac{r}{c}\right)\right)-\varphi )}^2\right]\cdot \widehat{r}\right)\\ S=\frac{{\mu }_0}{c}{\left(\frac{p_0{\omega }^2}{4\pi r^2}\right)}^2\left[1-\frac{1}{2}{\sin }^2\theta \right]\widehat{r}\end{array}$

Calculate the power radiated dipole,

Substitute localid="1658825122009" $\frac{{\mu }_0}{c}{\left(\frac{p_0{\omega }^2}{4\pi r^2}\right)}^2\left[1-\frac{1}{2}{\sin }^2\theta \right]\widehat{r}$for $S$into equation (3).

localid="1658825212175" $\begin{array}{l}P=\int \frac{{\mu }_0}{c}{\left(\frac{p_0{\omega }^2}{4\pi r^2}\right)}^2\left(1-\frac{1}{2}{\sin }^2\theta \right)\cdot da\\ P=\frac{{\mu }_0}{c}{\left(\frac{p_0{\omega }^2}{4\pi r^2}\right)}^2{\int }_{2\pi }\left(1-\frac{1}{2}{\sin }^2\theta \right)r^2\sin \theta d\theta d\varphi \\ P=\frac{{\mu }_0}{c}{\left(\frac{p_0{\omega }^2}{4\pi r^2}\right)}^2\underset{0}{\overset{\pi }{\int }}\underset{0}{\overset{2\pi }{\int }}\left(1-\frac{1}{2}{\sin }^2\theta \right)\sin \theta d\theta d\varphi \\ P=\frac{{\mu }_0}{c}{\left(\frac{p_0{\omega }^2}{4\pi r^2}\right)}^{2}2\pi \underset{0}{\overset{\pi }{\int }}\left(1-\frac{1}{2}{\sin }^2\theta \right)\sin \theta d\theta d\varphi \end{array}$

Solve further as,

$\begin{array}{l}P=\frac{{\mu }_0{p}_0^2{\omega }^4}{8\pi c}\left[\underset{0}{\overset{\pi }{\int }}\sin \theta d\theta -\frac{1}{2}\underset{0}{\overset{\pi }{\int }}{\sin }^3\theta d\theta \right]\\ P=\frac{{\mu }_0{p}_0^2{\omega }^4}{8\pi c}\left(2-\frac{1}{2}\cdot \frac{4}{3}\right)\\ P=\frac{{\mu }_0{p}_0^2{\omega }^4}{6\pi c}\end{array}$

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

$\tau =\frac{R}{v}P$

Substitute $R\omega$for $v$into above equation.

$\begin{array}{l}\tau =\frac{R}{R\omega }P\\ \tau =\frac{1}{\omega }P\end{array}$

Substitute $\frac{{\mu }_0{p}_0^2{\omega }^4}{6\pi c}$for $P$into above equation.

$\begin{array}{c}\tau =\frac{1}{\omega }\left(\frac{{\mu }_0{p}_0^2{\omega }^4}{6\pi c}\right)\\ \tau =\frac{{\mu }_0{p}_0^2{\omega }^3}{6\pi c}\end{array}$

Hence the total radiation reaction torque on this system $-\frac{{\mu }_0{p}_0^2{\omega }^3}{6\pi c}$

(c)

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

$P_{rad}\left(t_0\right)\approx \frac{{\mu }_0}{6\pi c}{\left[\stackrel{··}{p}\left(t_0\right)\right]}^2$ …… (7)

The calculated total power radiated is given by,

$P=\frac{{\mu }_0{p}_0^2{\omega }^4}{6\pi c}$ …… (8)

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