91Ó°ÊÓ

Radius of the ring is a.

The line charge density is${\mathrm{λ}}_0/\sin \mathrm{θ}/2$ .

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

The expression to calculate the scaler potential at the centre is given as follows.

$V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}∫\frac{\mathrm{λ}}{r}dl$ …… (1)

Here is the linear charge density and is the small element of the ring.

The expression to calculate the current density is given as follows.

$\mathit{I}\mathbf{=}\mathit{λ}\mathit{v}$ …… (2)

Here is the linear velocity.

The expression to calculate the vector potential is given as follows.

$\mathit{A}\left(t\right)\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{o}}}{\mathbf{4}\mathbf{π}}\mathbf{∫}\frac{\mathbf{I}}{\mathbf{r}}\mathit{d}\mathit{l}$ …… (3)

Consider the diagram of the ring as,

Calculate the scaler potential.

Substitute for and for into equation (1).

$$\begin{gathered} V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}∫\frac{{\mathrm{λ}}_0\left|\sin \left(\frac{\mathrm{θ}}{2}\right)\right|}{a}dl \\ V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}∫\frac{{\mathrm{λ}}_0\sin \left(\frac{\mathrm{θ}}{2}\right)}{a}dl \end{gathered}$$

Substitute$\mathrm{θ}$for into above equation.

localid="1657882718219" $$\begin{gathered} V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{∫}_0^{2\mathrm{Ï€}}\frac{{\mathrm{λ}}_0\sin \left(\frac{\mathrm{θ}}{2}\right)}{a}ad\mathrm{Ï•} \\ V=\frac{{\mathrm{λ}}_0}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{∫}_0^{2\mathrm{Ï€}}\sin \left(\frac{\mathrm{θ}}{2}\right)d\mathrm{Ï•} \\ V=\frac{{\mathrm{λ}}_0}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left[-2\cos \left(\frac{\mathrm{θ}}{2}\right)\right]}_0^{2\mathrm{Ï€}} \\ V=-\frac{{\mathrm{λ}}_0}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left[\cos \left(\frac{2\mathrm{Ï€}}{2}\right)-\cos \left(\frac{0}{2}\right)\right] \end{gathered}$$

Solve further as,

$$\begin{gathered} V=-\frac{{\mathrm{λ}}_0}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left[\cos \left(\mathrm{Ï€}\right)-\cos \left(0\right)\right] \\ V=-\frac{{\mathrm{λ}}_0}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left[-1-1\right] \\ V=\frac{{\mathrm{λ}}_0}{\mathrm{Ï€}{\mathrm{Î\mu }}_0} \end{gathered}$$

Hence the scalar potential at the centre of the ring is $\frac{{\mathrm{λ}}_0}{\mathrm{Ï€}{\mathrm{Î\mu }}_0}$.

Calculate the current density through the line,

Substitute ${\mathrm{λ}}_o\left|\sin \left(\frac{\mathrm{θ}}{2}\right)\right|$for into equation (2).

Substitute for into above equation.

$I={\mathrm{λ}}_o\left|\sin \left(\frac{\mathrm{θ}}{2}\right)\right|a\mathrm{Ó\neg }\widehat{\mathrm{Ï•}}$

Calculate the vector potential as,

Substitute for and for into equation (3).

$I={\mathrm{λ}}_o\left|\sin \left(\frac{\mathrm{Ï•}-\mathrm{Ó\neg }{t}_r}{2}\right)\right|a\mathrm{Ó\neg }\widehat{\mathrm{Ï•}}$

Here is constant.

Substitute ${\mathrm{λ}}_o\left|\sin \left(\frac{\mathrm{Ï•}-\mathrm{Ó\neg }{t}_r}{2}\right)\right|a\mathrm{Ó\neg }\widehat{\mathrm{Ï•}}$for into above equation.

$A\left(t\right)=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}{∫}_0^{2\mathrm{Ï€}}\frac{a\mathrm{Ó\neg }{\mathrm{λ}}_0\left|\sin \frac{\mathrm{Ï•}-\mathrm{Ó\neg }{t}_r}{2}\right|\widehat{\mathrm{Ï•}}}{a}ad\mathrm{Ï•}$

Here$$\begin{gathered} \mathrm{θ}=\mathrm{Ï•}-\mathrm{Ó\neg }{t}_{r∂} \\ d\mathrm{θ}=d\mathrm{Ï•} \end{gathered}$$

Substitute for and for into above equation.

$$\begin{gathered} A\left(t\right)=\frac{{\mathrm{λ}}_0{\mathrm{μ}}_{0}a\mathrm{Ó\neg }}{4\mathrm{Ï€}}{∫}_0^{2\mathrm{Ï€}}\left|\sin \frac{\mathrm{θ}}{2}\right|\left(-\sin \mathrm{Ï•}\hat{x}+\cos \mathrm{Ï•}\hat{y}\right)d\mathrm{θ} \\ A\left(t\right)=\frac{{\mathrm{λ}}_0{\mathrm{μ}}_{0}a\mathrm{Ó\neg }}{4\mathrm{Ï€}}{∫}_0^{2\mathrm{Ï€}}\sin \left(\frac{\mathrm{θ}}{2}\right)\left(-\sin \mathrm{Ï•}\hat{x}+\cos \mathrm{Ï•}\hat{y}\right)d\mathrm{θ} \\ A\left(t\right)=\frac{{\mathrm{λ}}_0{\mathrm{μ}}_{0}a\mathrm{Ó\neg }}{4\mathrm{Ï€}}\left[{∫}_0^{2\mathrm{Ï€}}-\sin \left(\frac{\mathrm{θ}}{2}\right)\sin \mathrm{Ï•}\hat{x}d\mathrm{θ}+{∫}_0^{2\mathrm{Ï€}}\sin \left(\frac{\mathrm{θ}}{2}\right)\cos \mathrm{Ï•}\hat{y}d\mathrm{θ}\right] \\ A\left(t\right)=\frac{{\mathrm{λ}}_0{\mathrm{μ}}_{0}a\mathrm{Ó\neg }}{4\mathrm{Ï€}}\left[-\frac{1}{2}{∫}_0^{2\mathrm{Ï€}}\left\{-\cos \left(\frac{3\mathrm{θ}}{2}+\mathrm{Ó\neg }{t}_r\right)+\cos \left(\mathrm{Ó\neg }{t}_r+\frac{\mathrm{θ}}{2}\right)\right\}\hat{x}d\mathrm{θ} \\ +\frac{1}{2}{∫}_0^{2\mathrm{Ï€}}\left\{-\sin \left(\frac{\mathrm{θ}}{2}+\mathrm{Ó\neg }{t}_r\right)+\sin \left(\frac{3\mathrm{θ}}{2}+\mathrm{Ó\neg }{t}_r\right)\right\}\hat{y}d\mathrm{θ} \\ \right] \end{gathered}$$

Solve further as,

$$\begin{gathered} \frac{{\mathrm{λ}}_0{\mathrm{μ}}_{0}a\mathrm{Ó\neg }}{4\mathrm{Ï€}}\left\{-\frac{1}{2}{\left[\frac{\sin \left(\frac{\mathrm{θ}}{2}+\mathrm{Ó\neg }{t}_r\right)}{\frac{1}{2}}-\frac{\sin \left(\frac{3\mathrm{θ}}{2}+\mathrm{Ó\neg }{t}_r\right)}{\frac{3}{2}}\right]}_0^{2\mathrm{Ï€}}\hat{x} \\ +\frac{1}{2}{\left[\frac{\cos \left(\frac{\mathrm{θ}}{2}+\mathrm{Ó\neg }{t}_r\right)}{\frac{1}{2}}-\frac{\sin \left(\frac{3\mathrm{θ}}{2}+\mathrm{Ó\neg }{t}_r\right)}{\frac{3}{2}}\right]}_0^{2\mathrm{Ï€}}\hat{y} \\ \right\} \\ A\left(t\right)=\frac{{\mathrm{λ}}_0{\mathrm{μ}}_{0}a\mathrm{Ó\neg }}{4\mathrm{Ï€}}\left\{-{\left[\sin \left(\frac{\mathrm{θ}}{2}+\mathrm{Ó\neg }{t}_r\right)-\frac{1}{3}\sin \left(\frac{3\mathrm{θ}}{2}+\mathrm{Ó\neg }{t}_r\right)\right]}_0^{2\mathrm{Ï€}}\hat{x} \\ +{\left[\cos \left(\frac{\mathrm{θ}}{2}+\mathrm{Ó\neg }{t}_r\right)-\frac{1}{3}\sin \left(\frac{3\mathrm{θ}}{2}+\mathrm{Ó\neg }{t}_r\right)\right]}_0^{2\mathrm{Ï€}}\hat{y} \\ \right\} \\ A\left(t\right)=\frac{{\mathrm{λ}}_0{\mathrm{μ}}_{0}a\mathrm{Ó\neg }}{4\mathrm{Ï€}}\left[\frac{4}{3}\sin \left(\mathrm{Ó\neg }{t}_r\right)\hat{x}\right]+\left[-\frac{4}{3}\cos \left(\mathrm{Ó\neg }{t}_r\right)\hat{y}\right] \\ A\left(t\right)=\frac{{\mathrm{λ}}_0{\mathrm{μ}}_{0}a\mathrm{Ó\neg }}{3\mathrm{Ï€}}\left[\sin \left(\mathrm{Ó\neg }{t}_r\right)\hat{x}-\cos \left(\mathrm{Ó\neg }{t}_r\right)\hat{y}\right] \end{gathered}$$

Hence the expression for the vector potential is $A\left(t\right)=\frac{{\mathrm{λ}}_0{\mathrm{μ}}_{0}a\mathrm{Ó\neg }}{3\mathrm{Ï€}}\left[\sin \left(\mathrm{Ó\neg }{t}_r\right)\hat{x}-\cos \left(\mathrm{Ó\neg }{t}_r\right)\hat{y}\right]$.