Question

A particle of chargeq moves in a circle of radius a at constant angular velocity $\mathit{\omega }$. (Assume that the circle lies in thexy plane, centered at the origin, and at time$\mathit{t}\mathbf{=}\mathbf{0}$ the charge is at role="math" localid="1653885001176" $\left(\mathit{a}\mathbf{,}\mathbf{0}\right)$, on the positive x axis.) Find the Liénard-Wiechert potentials for points on the z-axis.

Short answer

The Lienard-Wiechert potentials for points on the z-axis are $V\left(z,t\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{\sqrt{z^2+a^2}}$and $A\left(z,t\right)=\frac{q\omega a}{4{\mathrm{\pi \epsilon }}_0{\mathrm{c}}^2\sqrt{{\mathrm{z}}^2+{\mathrm{a}}^2}}\left[-\sin \left(\omega t_r\right)\hat{x}+\cos \left(\omega t_r\right)\hat{y}\right]$.

Step-by-step solution

Expression for the position and linear velocity of a particle:

Write the expression for the position of a particle.

$\mathit{r}\left(t\right)\mathbf{=}\mathit{a}\left[cos\left(\omega t\right)\hat{x}+sin\left(\omega t\right)\hat{y}\right]$

Here, t is the retarded time and$\omega t$ is the position of q at time t.

Write the expression for the linear velocity of a particle.

$$\begin{gathered} V\left(t\right)=a\left[-\omega \sin \left(\omega t\right)\hat{x}+\omega \cos \left(\omega t\right)\hat{y}\right] \\ V\left(t\right)=\omega a\left[-\sin \left(\omega t\right)\hat{x}+\cos \left(\omega t\right)\hat{y}\right] \end{gathered}$$

Determine the Lienard-Wiechert potentials for a moving particle:

Write the expression for the retarded position to the field point r.

$$\begin{gathered} r=z\hat{z}-a\left[\cos \left(\omega t_r\right)\hat{x}+\sin \left(\omega t_r\right)\hat{y}\right] \\ {r}^2=\left(z\hat{z}-a\left[\cos \left(\omega t_r\right)\hat{x}+\sin \left(\omega t_r\right)\hat{y}\right]\right)\left(z\hat{z}-a\left[\cos \left(\omega t_r\right)\hat{x}+\sin \left(\omega t_r\right)\hat{y}\right]\right) \\ {r}^2=z^2+a^2 \\ r=\sqrt{z^2+a^2} \end{gathered}$$

Write the relation between retarded position and linear velocity.

$\hat{r}·V=\frac{1}{r}\left(r·V\right)$

Substitute the value of rand Vin the above expression.

$$\begin{gathered} \hat{r}·V=\frac{1}{r}\left[-a\left(-\sin \omega t_r\cos \omega t_r\hat{x}+\sin \omega t_r\cos \omega t_r\right)\right]·\omega a\left[-\sin \omega t_r\hat{x}+\cos \omega t_r\hat{y}\right] \\ \hat{r}·V=\frac{1}{r}\left[-\omega a^2\left(-\sin \omega t_r\cos \omega t_r+\sin \omega t_r\cos \omega t_r\right)\right] \\ \hat{r}·V=0 \end{gathered}$$

Write the expression for the Lienard-Wiechert potentials for a moving particle.

$\mathit{V}\left(r,t\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}\mathbf{c}}{\mathbf{r}\mathbf{c}\mathbf{-}\hat{\mathbf{r}}\mathbf{·}\mathbf{v}}$ …… (1)

Here, v is the velocity of the charge, r is the vector from the retard position to the field point r, c is the speed of light and q is the charge.

For the z-axis, re-write the above expression.

$$\begin{gathered} V\left(z,t\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{r} \\ V\left(z,t\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{\sqrt{z^2+a^2}} \end{gathered}$$

Determine the Lienard-Wiechert potentials for a vector potential:

Write the expression for the vector potential.

$A\left(r,t\right)=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{qcv}{\left(rc-r·v\right)}$ …… (2)

From equations (1) and (2),

$A\left(r,t\right)=\frac{v}{c^2}V\left(r,t\right)$

For the z-axis, re-write the above expression.

$A\left(z,t\right)=\frac{v}{c^2}V\left(z,t\right)$

Substitute$v=\omega a\left(-\sin \omega t_r\right)\hat{x}+\cos \left(\omega t_r\right)\hat{y}$and$r=z\hat{z}-a\left[\cos \left(\omega t_r\right)\hat{x}+\sin \left(\omega t_r\right)\hat{y}\right]$in the above expression.

$$\begin{gathered} A\left(z,t\right)=\frac{\omega a\left(-\sin \omega t_r\right)\hat{x}+\cos \left(\omega t_r\right)\hat{y}}{c^2}\left[\frac{q}{4{\mathrm{\pi \epsilon }}_0\sqrt{{\mathrm{z}}^2+{\mathrm{a}}^2}}\right] \\ A\left(z,t\right)=\frac{q\omega a}{4{\mathrm{\pi \epsilon }}_0{\mathrm{c}}^2\sqrt{{\mathrm{z}}^2+{\mathrm{a}}^2}}\left[-\sin \left(\omega t_r\right)\hat{x}+\cos \left(\omega t_r\right)\hat{y}\right] \end{gathered}$$

Therefore, the Lienard-Wiechert potentials for points on the z-axis are equated as$V\left(z,t\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{\sqrt{z^2+a2}}$ and $A\left(z,t\right)=\frac{q\omega a}{4{\mathrm{\pi \epsilon }}_0{\mathrm{c}}^2\sqrt{{\mathrm{z}}^2+{\mathrm{a}}^2}}\left[-\sin \left(\omega t_r\right)\hat{x}+\cos \left(\omega t_r\right)\hat{y}\right]$.