91Ó°ÊÓ

The point charge moves along the axis.

The expression for electric field due to moving charge is given as follows.

$\mathit{E}\left(r,t\right)\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{\mathbf{r}}{{\left(r×u\right)}^{\mathbf{3}}}\left[\left(c^2-v^2\right)u+r×\left(u×a\right)\right]$ …… (1)

Here, is the charge, is the speed, and is the velocity of light.

The expression for the magnetic field strength is given as follows.

$\mathit{B}\mathbf{=}\frac{\mathbf{1}}{\mathbf{c}}\left(\hat{r}×E\right)$ …… (2)

The electric field due to moving point charge is given by,

$E\left(r,t\right)=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{r}{{\left(r·u\right)}^3}\left[\left(c^2-v^2\right)u+r×\left(u×a\right)\right]$ …… (3)

Here particle in constrained to moves along the axis.

Therefore, the points on the right side of the charge,

localid="1657878176173" $$\begin{gathered} v=v\hat{x}\backslash \\ a=a\hat{x} \\ \hat{r}=\hat{x} \\ r·u=r\left(c-v\right) \end{gathered}$$

Since,

$$\begin{gathered} u=\left(c-v\right)\hat{x} \\ r=r\hat{x} \\ u×a=\left(c-v\right)\hat{x}×a\hat{x} \\ u×a=0 \end{gathered}$$

Recall equation (3),

$E\left(r,t\right)=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{r}{{\left(r·u\right)}^3}\left[\left(c^2-v^2\right)u+r×\left(u×a\right)\right]$

Substitute $\left(c-v\right)\hat{x}$for u,0, $r\left(c-v\right)$for r.u for into above equation.

$$\begin{gathered} E\left(r,t\right)=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{r}{{\left(r\left(c-v\right)\right)}^3}\left[\left(c^2-v^2\right)\left(c-v\right)\hat{x}+r×\left(0\right)\right] \\ E\left(r,t\right)=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{r}{r^3{\left(c-v\right)}^3}\left[\left(c+v\right)\left(c-v\right)\left(c-v\right)\hat{x}\right] \\ E\left(r,t\right)=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^2{\left(c-v\right)}^3}\left[\left(c+v\right){\left(c-v\right)}^2\hat{x}\right] \\ E\left(r,t\right)=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^2}\frac{\left(c+v\right)}{\left(c-v\right)}\hat{x}\backslash \end{gathered}$$

Calculate the expression for magnetic field,

Substitute $\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^2}\frac{\left(c+v\right)}{\left(c-v\right)}\hat{x}$for $E\left(r,t\right)$and for into equation (3).

$$\begin{gathered} B=\frac{1}{c}\left(\hat{x}×\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^2}\frac{\left(c+v\right)}{\left(c-v\right)}\hat{x}\right) \\ B=0 \end{gathered}$$

The points on the right side of the charge,

$$\begin{gathered} \hat{r}=-\hat{x} \\ u=-\left(c-v\right)\hat{x}\backslash \\ r·u=\left(c+v\right)r\backslash \\ u×a=0 \end{gathered}$$

Recall equation (3),

$E\left(r,t\right)=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{r}{{\left(r·u\right)}^3}\left[\left(c^2-v^2\right)u+r×\left(u×a\right)\right]$

Substitute $-\left(c-v\right)\hat{x}$for ,u,0 for uxa and $r\left(c+v\right)$ for r.u into above equation.

$$\begin{gathered} E\left(r,t\right)=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{r}{{\left(r\left(c+v\right)\right)}^3}\left[-\left(c^2-v^2\right)\left(c+v\right)\hat{x}+r×\left(0\right)\right] \\ E\left(r,t\right)=-\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^2{\left(c+v\right)}^3}\left[\left(c+v\right)\left(c-v\right)\left(c+v\right)\hat{x}\right] \\ E\left(r,t\right)=-\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^2{\left(c+v\right)}^3}\left[{\left(c+v\right)}^2\left(c-v\right)\hat{x}\right] \\ E\left(r,t\right)=-\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^2}\frac{\left(c-v\right)}{\left(c+v\right)}\hat{x} \end{gathered}$$

Calculate the expression for magnetic field,

Substitute$-\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^2}\frac{\left(c-v\right)}{\left(c+v\right)}\hat{x}$ for$E\left(r,t\right)$and for into equation (3).

$$\begin{gathered} B=\frac{1}{c}\left(-\hat{x}×\left(-\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^2}\frac{\left(c-v\right)}{\left(c+v\right)}\right)\hat{x}\right) \\ B=0 \end{gathered}$$

Hence the expression for electric and magnetic field on the right side is showed and the electric and magnetic field on left side are$-\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^2}\frac{\left(c-v\right)}{\left(c+v\right)}\hat{x}$ and 0 respectively.