91影视

Radius of the ring is a.

The line charge density is${位}_0/\sin 胃/2$ .

The angular velocity is $蝇$.

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

$V=\frac{1}{4蟺{蔚}_0}鈭�\frac{位}{r}dl$ 鈥︹�� (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.

$\mathit{I}\mathbf{=}\mathit{位}\mathit{v}$ 鈥︹�� (2)

Here is the linear velocity.

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

$\mathit{A}\left(t\right)\mathbf{=}\frac{{\mathbf{渭}}_{\mathbf{o}}}{\mathbf{4}\mathbf{蟺}}\mathbf{鈭�}\frac{\mathbf{I}}{\mathbf{r}}\mathit{d}\mathit{l}$ 鈥︹�� (3)

Consider the diagram of the ring as,

Calculate the scaler potential.

Substitute for and for into equation (1).

$$\begin{gathered} V=\frac{1}{4蟺{蔚}_0}鈭�\frac{{位}_0\left|\sin \left(\frac{胃}{2}\right)\right|}{a}dl \\ V=\frac{1}{4蟺{蔚}_0}鈭�\frac{{位}_0\sin \left(\frac{胃}{2}\right)}{a}dl \end{gathered}$$

Substitute$胃$for into above equation.

localid="1657882718219" $$\begin{gathered} V=\frac{1}{4蟺{蔚}_0}{鈭�}_0^{2蟺}\frac{{位}_0\sin \left(\frac{胃}{2}\right)}{a}ad蠒 \\ V=\frac{{位}_0}{4蟺{蔚}_0}{鈭�}_0^{2蟺}\sin \left(\frac{胃}{2}\right)d蠒 \\ V=\frac{{位}_0}{4蟺{蔚}_0}{\left[-2\cos \left(\frac{胃}{2}\right)\right]}_0^{2蟺} \\ V=-\frac{{位}_0}{2蟺{蔚}_0}\left[\cos \left(\frac{2蟺}{2}\right)-\cos \left(\frac{0}{2}\right)\right] \end{gathered}$$

Solve further as,

$$\begin{gathered} V=-\frac{{位}_0}{2蟺{蔚}_0}\left[\cos \left(蟺\right)-\cos \left(0\right)\right] \\ V=-\frac{{位}_0}{2蟺{蔚}_0}\left[-1-1\right] \\ V=\frac{{位}_0}{蟺{蔚}_0} \end{gathered}$$

Hence the scalar potential at the centre of the ring is $\frac{{位}_0}{蟺{蔚}_0}$.

Calculate the current density through the line,

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

Substitute for into above equation.

$I={位}_o\left|\sin \left(\frac{胃}{2}\right)\right|a蝇\widehat{蠒}$

Calculate the vector potential as,

Substitute for and for into equation (3).

$I={位}_o\left|\sin \left(\frac{蠒-蝇t_r}{2}\right)\right|a蝇\widehat{蠒}$

Here is constant.

Substitute ${位}_o\left|\sin \left(\frac{蠒-蝇t_r}{2}\right)\right|a蝇\widehat{蠒}$for into above equation.

$A\left(t\right)=\frac{{渭}_0}{4蟺}{鈭�}_0^{2蟺}\frac{a蝇{位}_0\left|\sin \frac{蠒-蝇t_r}{2}\right|\widehat{蠒}}{a}ad蠒$

Here$$\begin{gathered} 胃=蠒-蝇t_{r鈭�} \\ d胃=d蠒 \end{gathered}$$

Substitute for and for into above equation.

$$\begin{gathered} A\left(t\right)=\frac{{位}_0{渭}_{0}a蝇}{4蟺}{鈭�}_0^{2蟺}\left|\sin \frac{胃}{2}\right|\left(-\sin 蠒\hat{x}+\cos 蠒\hat{y}\right)d胃 \\ A\left(t\right)=\frac{{位}_0{渭}_{0}a蝇}{4蟺}{鈭�}_0^{2蟺}\sin \left(\frac{胃}{2}\right)\left(-\sin 蠒\hat{x}+\cos 蠒\hat{y}\right)d胃 \\ A\left(t\right)=\frac{{位}_0{渭}_{0}a蝇}{4蟺}\left[{鈭�}_0^{2蟺}-\sin \left(\frac{胃}{2}\right)\sin 蠒\hat{x}d胃+{鈭�}_0^{2蟺}\sin \left(\frac{胃}{2}\right)\cos 蠒\hat{y}d胃\right] \\ A\left(t\right)=\frac{{位}_0{渭}_{0}a蝇}{4蟺}\left[-\frac{1}{2}{鈭�}_0^{2蟺}\left\{-\cos \left(\frac{3胃}{2}+蝇t_r\right)+\cos \left(蝇t_r+\frac{胃}{2}\right)\right\}\hat{x}d胃 \\ +\frac{1}{2}{鈭�}_0^{2蟺}\left\{-\sin \left(\frac{胃}{2}+蝇t_r\right)+\sin \left(\frac{3胃}{2}+蝇t_r\right)\right\}\hat{y}d胃 \\ \right] \end{gathered}$$

Solve further as,

$$\begin{gathered} \frac{{位}_0{渭}_{0}a蝇}{4蟺}\left\{-\frac{1}{2}{\left[\frac{\sin \left(\frac{胃}{2}+蝇t_r\right)}{\frac{1}{2}}-\frac{\sin \left(\frac{3胃}{2}+蝇t_r\right)}{\frac{3}{2}}\right]}_0^{2蟺}\hat{x} \\ +\frac{1}{2}{\left[\frac{\cos \left(\frac{胃}{2}+蝇t_r\right)}{\frac{1}{2}}-\frac{\sin \left(\frac{3胃}{2}+蝇t_r\right)}{\frac{3}{2}}\right]}_0^{2蟺}\hat{y} \\ \right\} \\ A\left(t\right)=\frac{{位}_0{渭}_{0}a蝇}{4蟺}\left\{-{\left[\sin \left(\frac{胃}{2}+蝇t_r\right)-\frac{1}{3}\sin \left(\frac{3胃}{2}+蝇t_r\right)\right]}_0^{2蟺}\hat{x} \\ +{\left[\cos \left(\frac{胃}{2}+蝇t_r\right)-\frac{1}{3}\sin \left(\frac{3胃}{2}+蝇t_r\right)\right]}_0^{2蟺}\hat{y} \\ \right\} \\ A\left(t\right)=\frac{{位}_0{渭}_{0}a蝇}{4蟺}\left[\frac{4}{3}\sin \left(蝇t_r\right)\hat{x}\right]+\left[-\frac{4}{3}\cos \left(蝇t_r\right)\hat{y}\right] \\ A\left(t\right)=\frac{{位}_0{渭}_{0}a蝇}{3蟺}\left[\sin \left(蝇t_r\right)\hat{x}-\cos \left(蝇t_r\right)\hat{y}\right] \end{gathered}$$

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