91Ó°ÊÓ

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

Sudden burst of current,$I\left(t\right)=q_0\mathrm{δ}\left(t\right)$

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

$\mathbf{A}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}}{\mathbf{4}\mathbf{π}}\widehat{\mathbf{z}}{∫}_0^{\sqrt{{\left(\mathrm{ct}\right)}^2-r^2}}\frac{I\left(t_r\right)}{r}\mathbf{dz}$ …… (1)

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

$\mathbf{E}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{-}\frac{\mathbf{∂}\mathbf{A}}{\mathbf{∂}\mathbf{t}}$ …… (2)

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

$\mathbf{B}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{-}\frac{\mathbf{∂}{\mathbf{A}}_{\mathbf{z}}}{\mathbf{∂}\mathbf{r}}\widehat{\mathbf{ϕ}}$ …… (3)

(a)

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

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

$A(r,t)=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\widehat{z}{∫}_0^{\sqrt{{\left(ct\right)}^2−r^2}}\frac{\text{k}\left(t−\frac{r}{c}\right)}{r}dz$

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

$\begin{array}{l}A(r,t)=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\widehat{z}{∫}_0^{\sqrt{{\left(ct\right)}^2−r^2}}\frac{\text{k}\left(t−\frac{\sqrt{r^2+z^2}}{c}\right)}{\sqrt{r^2+z^2}}dz\\ A(r,t)=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\widehat{z}\left[t{∫}_0^{\sqrt{{\left(ct\right)}^2−r^2}}\frac{dz}{\sqrt{r^2+z^2}}−\frac{1}{c}{∫}_0^{\sqrt{{\left(ct\right)}^2−r^2}}dz\right]\\ A(r,t)=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\widehat{z}\left[t\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2−r^2}}{r}\right)−\frac{1}{c}\sqrt{{\left(ct\right)}^2−r^2}\right]\end{array}$

Calculate the electric field in terms of vector potential.

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

$\begin{array}{l}E\left(\text{r,t}\right)=−\frac{∂}{∂t}\left\{\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\widehat{z}\left[t\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2−r^2}}{r}\right)−\frac{1}{c}\sqrt{{\left(ct\right)}^2−r^2}\right]\right\}\\ E\left(\text{r,t}\right)=\frac{{\mathrm{μ}}_0}{2\mathrm{π}}\widehat{z}\left[\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2−r^2}}{r}\right)+t\left(\frac{r}{ct+\sqrt{{\left(ct\right)}^2−r^2}}\right)\frac{1}{r}\left(c+\frac{1}{2}\frac{2c^{2}t}{\sqrt{{\left(ct\right)}^2−r^2}}\right)−\frac{1}{2c}\frac{2c^{2}t}{\sqrt{{\left(ct\right)}^2−r^2}}\right]\\ E\left(\text{r,t}\right)=\frac{{\mathrm{μ}}_0}{2\mathrm{π}}\widehat{z}\left[\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2−r^2}}{r}\right)+\left(\frac{ct}{\sqrt{{\left(ct\right)}^2−r^2}}\right)−\frac{ct}{\sqrt{{\left(ct\right)}^2−r^2}}\right]\\ E\left(\text{r,t}\right)=\frac{{\mathrm{μ}}_0}{2\mathrm{π}}\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2−r^2}}{r}\right)\widehat{z}\end{array}$

Calculate the magnetic field in terms of vector potential,

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

$\begin{array}{c}B\left(\text{r,t}\right)=−\frac{∂}{∂r}\left\{\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\widehat{z}\left[t\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2−r^2}}{r}\right)−\frac{1}{c}\sqrt{{\left(ct\right)}^2−r^2}\right]\right\}\\ B\left(\text{r,t}\right)=\frac{{\mathrm{μ}}_0}{2\mathrm{π}}\left[t\left(\frac{r}{ct+\sqrt{{\left(ct\right)}^2−r^2}}\right)\frac{r\frac{1}{2}\frac{(−2r)}{\sqrt{{\left(ct\right)}^2−r^2}}−ct−\sqrt{{\left(ct\right)}^2−r^2}}{r^2}−\frac{1}{2c}\frac{(−2r)}{\sqrt{{\left(ct\right)}^2−r^2}}\right]\widehat{\mathrm{ϕ}}\\ B\left(\text{r,t}\right)=\frac{{\mathrm{μ}}_0}{2\mathrm{π}}\left[\frac{−c{t}^2}{r\sqrt{{\left(ct\right)}^2−r^2}}+\frac{r}{c\sqrt{{\left(ct\right)}^2−r^2}}\right]\widehat{\mathrm{ϕ}}\\ B\left(\text{r,t}\right)=\frac{{\mathrm{μ}}_{0}k}{2\mathrm{π}rc}\sqrt{{\left(ct\right)}^2−r^2}\widehat{\mathrm{ϕ}}\end{array}$

Hence the electric field$E\left(\text{r,t}\right)$ is$\frac{{\mathrm{μ}}_0}{2\mathrm{π}}\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2−r^2}}{r}\right)\widehat{z}$ for $r>\frac{r}{c}$ and magnetic field$B\left(\text{r,t}\right)$ is $\frac{{\mathrm{μ}}_{0}k}{2\mathrm{π}rc}\sqrt{{\left(ct\right)}^2−r^2}\widehat{\mathrm{ϕ}}$.

(b)

Calculate the vector potential.

Substitute$q_o\mathrm{δ}\left(t−\frac{\text{r}}{c}\right)$for$I\left(t_r\right)$into equation (1).

$A(r,t)=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\widehat{z}{∫}_0^{\mathrm{∞}}\frac{q_o\mathrm{δ}\left(t−\frac{\text{r}}{c}\right)}{r}dz$

$A(r,t)=\frac{{\mathrm{μ}}_0{q}_0}{4\mathrm{π}}\widehat{z}{∫}_0^{\mathrm{∞}}\frac{\mathrm{δ}\left(t−\frac{\text{r}}{c}\right)}{r}dz$ ……. (4)

Let assume,

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

At $z=0,\text{ r =}r$and $z=\mathrm{∞},\text{ r = }r$

From equation (1),

$A(r,t)=\frac{{\mathrm{μ}}_0{q}_0}{4\mathrm{π}}\widehat{z}{∫}_r^{\mathrm{∞}}\mathrm{δ}\left(t−\frac{\text{r}}{c}\right)\frac{\text{rdr}}{\sqrt{{\text{r}}^2−r^2}}dz$

Substitute $c\mathrm{δ}\left(\text{r-}ct\right)$for $\mathrm{δ}\left(t−\frac{\text{r}}{c}\right)$into above equation.

role="math" localid="1657799163827" $\begin{array}{l}A(r,t)=\frac{{\mathrm{μ}}_0{q}_0}{4\mathrm{π}}\widehat{z}{∫}_r^{\mathrm{∞}}\frac{1}{\text{r}}\mathrm{δ}\left(t−\frac{\text{r}}{c}\right)\frac{\text{rdr}}{\sqrt{{\text{r}}^2−r^2}}d\text{r}\\ A(r,t)=\frac{{\mathrm{μ}}_0{q}_0}{2\mathrm{π}}\widehat{z}{∫}_r^{\mathrm{∞}}\frac{\mathrm{δ}\left(\text{r-}ct\right)}{\sqrt{{\text{r}}^2−r^2}}d\text{r}\\ A(r,t)=\frac{{\mathrm{μ}}_0{q}_{0}c}{2\mathrm{π}}\frac{1}{\sqrt{{\left(ct\right)}^2−r^2}}\widehat{z}\end{array}$

Calculate the electric field in terms of vector potential.

Substitute $\frac{{\mathrm{μ}}_0{q}_{0}c}{2\mathrm{π}}\frac{1}{\sqrt{{\left(ct\right)}^2−r^2}}\widehat{z}$ for $A(r,t)$into equation (2).

$\begin{array}{l}E\left(\text{r,t}\right)=−\frac{∂}{∂t}\left[\frac{{\mathrm{μ}}_0{q}_{0}c}{2\mathrm{π}}\frac{1}{\sqrt{{\left(ct\right)}^2−r^2}}\widehat{z}\right]\\ E\left(\text{r,t}\right)=\frac{{\mathrm{μ}}_0{q}_{0}c}{2\mathrm{π}}\left(−\frac{1}{2}\right)\left(\frac{2c^{2}t}{{\left[{\left(ct\right)}^2−r^2\right]}^{\frac{3}{2}}}\widehat{z}\right)\\ E\left(\text{r,t}\right)=\frac{{\mathrm{μ}}_0{q}_0{c}^{3}t}{2\mathrm{π}{\left[{\left(ct\right)}^2−r^2\right]}^{\frac{3}{2}}}\widehat{z}\end{array}$

Calculate the magnetic field in terms of vector potential,

Substitute$\frac{{\mathrm{μ}}_0{q}_{0}c}{2\mathrm{π}}\frac{1}{\sqrt{{\left(ct\right)}^2−r^2}}\widehat{z}$ for$A(r,t)$ into equation (3).

$\begin{array}{l}B\left(\text{r,t}\right)=−\frac{∂}{∂r}\left[\frac{{\mathrm{μ}}_0{q}_{0}c}{2\mathrm{π}}\frac{1}{\sqrt{{\left(ct\right)}^2−r^2}}\widehat{z}\right]\widehat{\mathrm{ϕ}}\\ B\left(\text{r,t}\right)=−\frac{{\mathrm{μ}}_0{q}_{0}c}{2\mathrm{π}}\left(−\frac{1}{2}\right)\left(\frac{−2r}{{{\left(ct\right)}^2−r^2)}^{\frac{3}{2}}}\right)\widehat{\mathrm{ϕ}}\\ B\left(\text{r,t}\right)=−\frac{{\mathrm{μ}}_0{q}_{0}c}{2\mathrm{π}{\left({\left(ct\right)}^2−r^2\right)}^{\frac{3}{2}}}\widehat{\mathrm{ϕ}}\end{array}$

Hence the electric field$E\left(\text{r,t}\right)$ is $\frac{{\mathrm{μ}}_0{q}_0{c}^{3}t}{2\mathrm{π}{\left[{\left(ct\right)}^2−r^2\right]}^{\frac{3}{2}}}\widehat{z}$and magnetic field$B\left(\text{r,t}\right)$ is $−\frac{{\mathrm{μ}}_0{q}_{0}c}{2\mathrm{π}{({\left(ct\right)}^2−r^2)}^{\frac{3}{2}}}\widehat{\mathrm{ϕ}}$.