91影视

Write the expression for the retarded vector potential.

$\mathit{A}\mathbf{=}\frac{{\mathbf{渭}}_{\mathbf{0}}}{\mathbf{4}\mathbf{蟺}}鈭�\frac{\mathbf{I}\left(t_r\right)}{\mathbf{r}}\mathit{d}\mathit{I}$ 鈥︹�� (1)

Here, I is the current flowing through the loop, dI is the length of the current element, r is distance, and $t_r$ is the retarded time.

Write the expression for the retarded time.

${\mathit{t}}_{\mathbf{r}}\mathbf{=}\mathit{t}\mathbf{-}\frac{\mathbf{r}}{\mathbf{c}}$ 鈥︹�� (2)

Here, t is the present time, c is the speed of light, and r is the distance travelled.

Write the given current equation in terms of the present time.

$I\left(t_r\right)=k{t}_r$ 鈥︹�� (3)

Substitute the value of equation (3) in equation (2).

$I\left(t_r\right)=k\left(t-\frac{r}{c}\right)$

Substitute $I\left(t_r\right)=k\left(t-\frac{r}{c}\right)$ in equation (1).

$$\begin{gathered} A=\frac{{渭}_0}{4\mathrm{蟺}}鈭�\frac{k\left(t-{\displaystyle \frac{r}{c}}\right)}{r}dI \\ A=\frac{{渭}_{0}k}{4\mathrm{蟺}}鈭�\frac{\left(t-{\displaystyle \frac{r}{c}}\right)}{r}dI \\ A=\frac{{渭}_{0}k}{4\mathrm{蟺}}\left[t鈭�\frac{dI}{r}-\frac{1}{c}鈭�dI\right] \end{gathered}$$

As for the complete loop $鈭�dI=0$, express the retarded potential.

$A=\frac{{渭}_{0}kt}{4\mathrm{蟺}}\left[\frac{1}{a}{鈭�}_{1}dI+\frac{1}{b}{鈭�}_{2}dI+2\hat{x}{鈭�}_a^b\frac{dx}{x}\right]$ .......( 4 )

Here, ${鈭�}_{1}dI=2a\hat{x}$ is for the inner circle and ${鈭�}_{2}dI=-2b\hat{x}$is for the outer circle.

Substitute ${鈭�}_{1}dI=2a\hat{x}$and ${鈭�}_{2}dI=-2b\hat{x}$in equation (4).

$$\begin{gathered} A=\frac{{渭}_{0}kt}{4\mathrm{蟺}}\left[\frac{1}{a}\left(2a\right)+\frac{1}{b}\left(-2b\right)+2In\left(\frac{b}{a}\right)\right]\hat{x} \\ A=\frac{{渭}_{0}kt}{4\mathrm{蟺}}\left[2In\left(\frac{b}{a}\right)\right]\hat{x} \\ A=\frac{{渭}_{0}kt}{4\mathrm{蟺}}In\left(\frac{b}{a}\right)\hat{x} \end{gathered}$$

Write the expression for the electric field at the center of the loop.

$E=-\frac{鈭�A}{鈭�t}$

Substitute $A=\frac{{渭}_{0}kt}{4\mathrm{蟺}}In\left(\frac{b}{a}\right)\hat{x}$in the above expression.

$$\begin{gathered} E=-\frac{鈭�}{鈭�t}\left[\frac{{渭}_{0}kt}{4\mathrm{蟺}}In\left(\frac{b}{a}\right)\hat{x}\right] \\ E=-\frac{{渭}_{0}k}{2\mathrm{蟺}}In\left(\frac{b}{a}\right)\hat{x} \end{gathered}$$

From the above expression, it can be observed that there will be a development of an electric field due to the alteration in the magnetic field. Hence, the magnetic field cannot be analyzed because it is known that the vector potential is found at the center only.

Therefore, the retarded vector potential and the electric field at the center is $A=\frac{{渭}_{0}kt}{4\mathrm{蟺}}In\left(\frac{b}{a}\right)\hat{x}$and $E=-\frac{{渭}_{0}k}{2\mathrm{蟺}}In\left(\frac{b}{a}\right)\hat{x}$ , respectively.