91影视

The linear increasing current is the wire,$I\left(t\right)=kt$

Sudden burst of current,$I\left(t\right)=q_0未\left(t\right)$

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

$\mathbf{A}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{{\mathbf{渭}}_{\mathbf{0}}}{\mathbf{4}\mathbf{蟺}}\widehat{\mathbf{z}}{鈭�}_0^{\sqrt{{\left(\mathrm{ct}\right)}^2-r^2}}\frac{I\left(t_r\right)}{r}\mathbf{dz}$ 鈥︹�� (1)

The expression to calculate the electric field is given as follows.

$\mathbf{E}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{-}\frac{\mathbf{鈭�}\mathbf{A}}{\mathbf{鈭�}\mathbf{t}}$ 鈥︹�� (2)

The expression to calculate the magnetic field is given as follows.

$\mathbf{B}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{-}\frac{\mathbf{鈭�}{\mathbf{A}}_{\mathbf{z}}}{\mathbf{鈭�}\mathbf{r}}\widehat{\mathbf{蠒}}$ 鈥︹�� (3)

(a)

Calculate the vector potential for $\mathrm{t}>\frac{\mathrm{r}}{\mathrm{c}}$,

Substitute $k\left(t鈭�\frac{\text{r}}{c}\right)$for $I\left(t_r\right)$into equation (1).

$A(r,t)=\frac{{渭}_0}{4蟺}\widehat{z}{鈭�}_0^{\sqrt{{\left(ct\right)}^2鈭�r^2}}\frac{\text{k}\left(t鈭�\frac{r}{c}\right)}{r}dz$

Substitute $\sqrt{r^2+z^2}$for $r$into above equation.

$\begin{array}{l}A(r,t)=\frac{{渭}_0}{4蟺}\widehat{z}{鈭�}_0^{\sqrt{{\left(ct\right)}^2鈭�r^2}}\frac{\text{k}\left(t鈭�\frac{\sqrt{r^2+z^2}}{c}\right)}{\sqrt{r^2+z^2}}dz\\ A(r,t)=\frac{{渭}_0}{4蟺}\widehat{z}\left[t{鈭�}_0^{\sqrt{{\left(ct\right)}^2鈭�r^2}}\frac{dz}{\sqrt{r^2+z^2}}鈭�\frac{1}{c}{鈭�}_0^{\sqrt{{\left(ct\right)}^2鈭�r^2}}dz\right]\\ A(r,t)=\frac{{渭}_0}{4蟺}\widehat{z}\left[t\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2鈭�r^2}}{r}\right)鈭�\frac{1}{c}\sqrt{{\left(ct\right)}^2鈭�r^2}\right]\end{array}$

Calculate the electric field in terms of vector potential.

Substitute $\frac{{渭}_0}{4蟺}\widehat{z}\left[t\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2鈭�r^2}}{r}\right)鈭�\frac{1}{c}\sqrt{{\left(ct\right)}^2鈭�r^2}\right]$for $A(r,t)$into equation (2).

$\begin{array}{l}E\left(\text{r,t}\right)=鈭�\frac{鈭�}{鈭�t}\left\{\frac{{渭}_0}{4蟺}\widehat{z}\left[t\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2鈭�r^2}}{r}\right)鈭�\frac{1}{c}\sqrt{{\left(ct\right)}^2鈭�r^2}\right]\right\}\\ E\left(\text{r,t}\right)=\frac{{渭}_0}{2蟺}\widehat{z}\left[\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2鈭�r^2}}{r}\right)+t\left(\frac{r}{ct+\sqrt{{\left(ct\right)}^2鈭�r^2}}\right)\frac{1}{r}\left(c+\frac{1}{2}\frac{2c^{2}t}{\sqrt{{\left(ct\right)}^2鈭�r^2}}\right)鈭�\frac{1}{2c}\frac{2c^{2}t}{\sqrt{{\left(ct\right)}^2鈭�r^2}}\right]\\ E\left(\text{r,t}\right)=\frac{{渭}_0}{2蟺}\widehat{z}\left[\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2鈭�r^2}}{r}\right)+\left(\frac{ct}{\sqrt{{\left(ct\right)}^2鈭�r^2}}\right)鈭�\frac{ct}{\sqrt{{\left(ct\right)}^2鈭�r^2}}\right]\\ E\left(\text{r,t}\right)=\frac{{渭}_0}{2蟺}\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2鈭�r^2}}{r}\right)\widehat{z}\end{array}$

Calculate the magnetic field in terms of vector potential,

Substitute $\frac{{渭}_0}{4蟺}\widehat{z}\left[t\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2鈭�r^2}}{r}\right)鈭�\frac{1}{c}\sqrt{{\left(ct\right)}^2鈭�r^2}\right]$for $A(r,t)$into equation (3).

$\begin{array}{c}B\left(\text{r,t}\right)=鈭�\frac{鈭�}{鈭�r}\left\{\frac{{渭}_0}{4蟺}\widehat{z}\left[t\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2鈭�r^2}}{r}\right)鈭�\frac{1}{c}\sqrt{{\left(ct\right)}^2鈭�r^2}\right]\right\}\\ B\left(\text{r,t}\right)=\frac{{渭}_0}{2蟺}\left[t\left(\frac{r}{ct+\sqrt{{\left(ct\right)}^2鈭�r^2}}\right)\frac{r\frac{1}{2}\frac{(鈭�2r)}{\sqrt{{\left(ct\right)}^2鈭�r^2}}鈭�ct鈭�\sqrt{{\left(ct\right)}^2鈭�r^2}}{r^2}鈭�\frac{1}{2c}\frac{(鈭�2r)}{\sqrt{{\left(ct\right)}^2鈭�r^2}}\right]\widehat{蠒}\\ B\left(\text{r,t}\right)=\frac{{渭}_0}{2蟺}\left[\frac{鈭�c{t}^2}{r\sqrt{{\left(ct\right)}^2鈭�r^2}}+\frac{r}{c\sqrt{{\left(ct\right)}^2鈭�r^2}}\right]\widehat{蠒}\\ B\left(\text{r,t}\right)=\frac{{渭}_{0}k}{2蟺rc}\sqrt{{\left(ct\right)}^2鈭�r^2}\widehat{蠒}\end{array}$

Hence the electric field$E\left(\text{r,t}\right)$ is$\frac{{渭}_0}{2蟺}\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2鈭�r^2}}{r}\right)\widehat{z}$ for $r>\frac{r}{c}$ and magnetic field$B\left(\text{r,t}\right)$ is $\frac{{渭}_{0}k}{2蟺rc}\sqrt{{\left(ct\right)}^2鈭�r^2}\widehat{蠒}$.

(b)

Calculate the vector potential.

Substitute$q_o未\left(t鈭�\frac{\text{r}}{c}\right)$for$I\left(t_r\right)$into equation (1).

$A(r,t)=\frac{{渭}_0}{4蟺}\widehat{z}{鈭�}_0^{\mathrm{鈭�}}\frac{q_o未\left(t鈭�\frac{\text{r}}{c}\right)}{r}dz$

$A(r,t)=\frac{{渭}_0{q}_0}{4蟺}\widehat{z}{鈭�}_0^{\mathrm{鈭�}}\frac{未\left(t鈭�\frac{\text{r}}{c}\right)}{r}dz$ 鈥︹��. (4)

Let assume,

$\begin{array}{c}\text{r=}\sqrt{r^2+z^2}\\ z=\sqrt{{\text{r}}^2鈭�r^2}\\ dz=\frac{1}{2}\frac{2\text{rdr}}{\sqrt{{\text{r}}^2鈭�r^2}}\end{array}$

At $z=0,\text{鈥塺鈥�=}r$and $z=\mathrm{鈭�},\text{鈥塺鈥�=鈥�}r$

From equation (1),

$A(r,t)=\frac{{渭}_0{q}_0}{4蟺}\widehat{z}{鈭�}_r^{\mathrm{鈭�}}未\left(t鈭�\frac{\text{r}}{c}\right)\frac{\text{rdr}}{\sqrt{{\text{r}}^2鈭�r^2}}dz$

Substitute $c未\left(\text{r-}ct\right)$for $未\left(t鈭�\frac{\text{r}}{c}\right)$into above equation.

role="math" localid="1657799163827" $\begin{array}{l}A(r,t)=\frac{{渭}_0{q}_0}{4蟺}\widehat{z}{鈭�}_r^{\mathrm{鈭�}}\frac{1}{\text{r}}未\left(t鈭�\frac{\text{r}}{c}\right)\frac{\text{rdr}}{\sqrt{{\text{r}}^2鈭�r^2}}d\text{r}\\ A(r,t)=\frac{{渭}_0{q}_0}{2蟺}\widehat{z}{鈭�}_r^{\mathrm{鈭�}}\frac{未\left(\text{r-}ct\right)}{\sqrt{{\text{r}}^2鈭�r^2}}d\text{r}\\ A(r,t)=\frac{{渭}_0{q}_{0}c}{2蟺}\frac{1}{\sqrt{{\left(ct\right)}^2鈭�r^2}}\widehat{z}\end{array}$

Calculate the electric field in terms of vector potential.

Substitute $\frac{{渭}_0{q}_{0}c}{2蟺}\frac{1}{\sqrt{{\left(ct\right)}^2鈭�r^2}}\widehat{z}$ for $A(r,t)$into equation (2).

$\begin{array}{l}E\left(\text{r,t}\right)=鈭�\frac{鈭�}{鈭�t}\left[\frac{{渭}_0{q}_{0}c}{2蟺}\frac{1}{\sqrt{{\left(ct\right)}^2鈭�r^2}}\widehat{z}\right]\\ E\left(\text{r,t}\right)=\frac{{渭}_0{q}_{0}c}{2蟺}\left(鈭�\frac{1}{2}\right)\left(\frac{2c^{2}t}{{\left[{\left(ct\right)}^2鈭�r^2\right]}^{\frac{3}{2}}}\widehat{z}\right)\\ E\left(\text{r,t}\right)=\frac{{渭}_0{q}_0{c}^{3}t}{2蟺{\left[{\left(ct\right)}^2鈭�r^2\right]}^{\frac{3}{2}}}\widehat{z}\end{array}$

Calculate the magnetic field in terms of vector potential,

Substitute$\frac{{渭}_0{q}_{0}c}{2蟺}\frac{1}{\sqrt{{\left(ct\right)}^2鈭�r^2}}\widehat{z}$ for$A(r,t)$ into equation (3).

$\begin{array}{l}B\left(\text{r,t}\right)=鈭�\frac{鈭�}{鈭�r}\left[\frac{{渭}_0{q}_{0}c}{2蟺}\frac{1}{\sqrt{{\left(ct\right)}^2鈭�r^2}}\widehat{z}\right]\widehat{蠒}\\ B\left(\text{r,t}\right)=鈭�\frac{{渭}_0{q}_{0}c}{2蟺}\left(鈭�\frac{1}{2}\right)\left(\frac{鈭�2r}{{{\left(ct\right)}^2鈭�r^2)}^{\frac{3}{2}}}\right)\widehat{蠒}\\ B\left(\text{r,t}\right)=鈭�\frac{{渭}_0{q}_{0}c}{2蟺{\left({\left(ct\right)}^2鈭�r^2\right)}^{\frac{3}{2}}}\widehat{蠒}\end{array}$

Hence the electric field$E\left(\text{r,t}\right)$ is $\frac{{渭}_0{q}_0{c}^{3}t}{2蟺{\left[{\left(ct\right)}^2鈭�r^2\right]}^{\frac{3}{2}}}\widehat{z}$and magnetic field$B\left(\text{r,t}\right)$ is $鈭�\frac{{渭}_0{q}_{0}c}{2蟺{({\left(ct\right)}^2鈭�r^2)}^{\frac{3}{2}}}\widehat{蠒}$.