Question: Suppose you take a plastic ring of radius and glue charge on it, so that the line charge density is . Then you spin the loop about its axis at an angular velocity . Find the (exact) scalar and vector potentials at the center of the ring. [Answer:]

Radius of the ring is a.

The line charge density is${\lambda }_0/\sin \theta /2$ .

The angular velocity is $\omega$.

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

$V=\frac{1}{4\pi {\epsilon }_0}\int \frac{\lambda }{r}dl$ …… (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.

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

Here is the linear velocity.

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

$\mathit{A}\left(t\right)\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{o}}}{\mathbf{4}\mathbf{\pi }}\mathbf{\int }\frac{\mathbf{I}}{\mathbf{r}}\mathit{d}\mathit{l}$ …… (3)

Consider the diagram of the ring as,

Calculate the scaler potential.

Substitute for and for into equation (1).

$$\begin{gathered} V=\frac{1}{4\pi {\epsilon }_0}\int \frac{{\lambda }_0\left|\sin \left(\frac{\theta }{2}\right)\right|}{a}dl \\ V=\frac{1}{4\pi {\epsilon }_0}\int \frac{{\lambda }_0\sin \left(\frac{\theta }{2}\right)}{a}dl \end{gathered}$$

Substitute$\theta$for into above equation.

localid="1657882718219" $$\begin{gathered} V=\frac{1}{4\pi {\epsilon }_0}{\int }_0^{2\pi }\frac{{\lambda }_0\sin \left(\frac{\theta }{2}\right)}{a}ad\varphi \\ V=\frac{{\lambda }_0}{4\pi {\epsilon }_0}{\int }_0^{2\pi }\sin \left(\frac{\theta }{2}\right)d\varphi \\ V=\frac{{\lambda }_0}{4\pi {\epsilon }_0}{\left[-2\cos \left(\frac{\theta }{2}\right)\right]}_0^{2\pi } \\ V=-\frac{{\lambda }_0}{2\pi {\epsilon }_0}\left[\cos \left(\frac{2\pi }{2}\right)-\cos \left(\frac{0}{2}\right)\right] \end{gathered}$$

Solve further as,

$$\begin{gathered} V=-\frac{{\lambda }_0}{2\pi {\epsilon }_0}\left[\cos \left(\pi \right)-\cos \left(0\right)\right] \\ V=-\frac{{\lambda }_0}{2\pi {\epsilon }_0}\left[-1-1\right] \\ V=\frac{{\lambda }_0}{\pi {\epsilon }_0} \end{gathered}$$

Hence the scalar potential at the centre of the ring is $\frac{{\lambda }_0}{\pi {\epsilon }_0}$.

Calculate the current density through the line,

Substitute ${\lambda }_o\left|\sin \left(\frac{\theta }{2}\right)\right|$for into equation (2).

Substitute for into above equation.

$I={\lambda }_o\left|\sin \left(\frac{\theta }{2}\right)\right|a\omega \hat{\varphi }$

Calculate the vector potential as,

Substitute for and for into equation (3).

$I={\lambda }_o\left|\sin \left(\frac{\varphi -\omega t_r}{2}\right)\right|a\omega \hat{\varphi }$

Here is constant.

Substitute ${\lambda }_o\left|\sin \left(\frac{\varphi -\omega t_r}{2}\right)\right|a\omega \hat{\varphi }$for into above equation.

$A\left(t\right)=\frac{{\mu }_0}{4\pi }{\int }_0^{2\pi }\frac{a\omega {\lambda }_0\left|\sin \frac{\varphi -\omega t_r}{2}\right|\hat{\varphi }}{a}ad\varphi$

Here$$\begin{gathered} \theta =\varphi -\omega t_{r\partial } \\ d\theta =d\varphi \end{gathered}$$

Substitute for and for into above equation.

$$\begin{gathered} A\left(t\right)=\frac{{\lambda }_0{\mu }_{0}a\omega }{4\pi }{\int }_0^{2\pi }\left|\sin \frac{\theta }{2}\right|\left(-\sin \varphi \hat{x}+\cos \varphi \hat{y}\right)d\theta \\ A\left(t\right)=\frac{{\lambda }_0{\mu }_{0}a\omega }{4\pi }{\int }_0^{2\pi }\sin \left(\frac{\theta }{2}\right)\left(-\sin \varphi \hat{x}+\cos \varphi \hat{y}\right)d\theta \\ A\left(t\right)=\frac{{\lambda }_0{\mu }_{0}a\omega }{4\pi }\left[{\int }_0^{2\pi }-\sin \left(\frac{\theta }{2}\right)\sin \varphi \hat{x}d\theta +{\int }_0^{2\pi }\sin \left(\frac{\theta }{2}\right)\cos \varphi \hat{y}d\theta \right] \\ A\left(t\right)=\frac{{\lambda }_0{\mu }_{0}a\omega }{4\pi }\left[-\frac{1}{2}{\int }_0^{2\pi }\left\{-\cos \left(\frac{3\theta }{2}+\omega t_r\right)+\cos \left(\omega t_r+\frac{\theta }{2}\right)\right\}\hat{x}d\theta \\ +\frac{1}{2}{\int }_0^{2\pi }\left\{-\sin \left(\frac{\theta }{2}+\omega t_r\right)+\sin \left(\frac{3\theta }{2}+\omega t_r\right)\right\}\hat{y}d\theta \\ \right] \end{gathered}$$

Solve further as,

$$\begin{gathered} \frac{{\lambda }_0{\mu }_{0}a\omega }{4\pi }\left\{-\frac{1}{2}{\left[\frac{\sin \left(\frac{\theta }{2}+\omega t_r\right)}{\frac{1}{2}}-\frac{\sin \left(\frac{3\theta }{2}+\omega t_r\right)}{\frac{3}{2}}\right]}_0^{2\pi }\hat{x} \\ +\frac{1}{2}{\left[\frac{\cos \left(\frac{\theta }{2}+\omega t_r\right)}{\frac{1}{2}}-\frac{\sin \left(\frac{3\theta }{2}+\omega t_r\right)}{\frac{3}{2}}\right]}_0^{2\pi }\hat{y} \\ \right\} \\ A\left(t\right)=\frac{{\lambda }_0{\mu }_{0}a\omega }{4\pi }\left\{-{\left[\sin \left(\frac{\theta }{2}+\omega t_r\right)-\frac{1}{3}\sin \left(\frac{3\theta }{2}+\omega t_r\right)\right]}_0^{2\pi }\hat{x} \\ +{\left[\cos \left(\frac{\theta }{2}+\omega t_r\right)-\frac{1}{3}\sin \left(\frac{3\theta }{2}+\omega t_r\right)\right]}_0^{2\pi }\hat{y} \\ \right\} \\ A\left(t\right)=\frac{{\lambda }_0{\mu }_{0}a\omega }{4\pi }\left[\frac{4}{3}\sin \left(\omega t_r\right)\hat{x}\right]+\left[-\frac{4}{3}\cos \left(\omega t_r\right)\hat{y}\right] \\ A\left(t\right)=\frac{{\lambda }_0{\mu }_{0}a\omega }{3\pi }\left[\sin \left(\omega t_r\right)\hat{x}-\cos \left(\omega t_r\right)\hat{y}\right] \end{gathered}$$

Hence the expression for the vector potential is $A\left(t\right)=\frac{{\lambda }_0{\mu }_{0}a\omega }{3\pi }\left[\sin \left(\omega t_r\right)\hat{x}-\cos \left(\omega t_r\right)\hat{y}\right]$.