Question

(a) Suppose the wire in Ex. 10.2 carries a linearly increasing current

$\mathbf{I}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{kt}$

for$\mathbf{t}\mathbf{>}\mathbf{0}$ . Find the electric and magnetic fields generated.

(b) Do the same for the case of a sudden burst of current:

$\mathit{I}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathit{q}}_{\mathbf{0}}\mathit{\delta }\mathbf{\left(}\mathbf{t}\mathbf{\right)}$

Short answer

(a) The electric field $E\left(\text{r,t}\right)$is $\frac{{\mu }_0}{2\pi }\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2-r^2}}{r}\right)\widehat{z}$for $r>\frac{r}{c}$ and magnetic field $B\left(\text{r,t}\right)$is $\frac{{\mu }_{0}k}{2\pi rc}\sqrt{{\left(ct\right)}^2-r^2}\widehat{\varphi }$.

(b) The electric field $E\left(\text{r,t}\right)$ is$\frac{{\mu }_0{q}_0{c}^{3}t}{2\pi {[{\left(ct\right)}^2-r^2]}^{\frac{3}{2}}}\widehat{z}$ and magnetic field$B\left(\text{r,t}\right)$ is $-\frac{{\mu }_0{q}_{0}c}{2\pi {({\left(ct\right)}^2-r^2)}^{\frac{3}{2}}}\widehat{\varphi }$.

Step-by-step solution

Write the given data from the question.

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

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

Determine the formulas to calculate the electric and magnetic fields generated.

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

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

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

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

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

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

Determine the electric and magnetic field when wire having the linear increasing current.

(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).

$A(r,t)=\frac{{\mu }_0}{4\pi }\widehat{z}{\int }_0^{\sqrt{{\left(ct\right)}^2-r^2}}\frac{\text{k}\left(t-\frac{r}{c}\right)}{r}dz$

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

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

Calculate the electric field in terms of vector potential.

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

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

Calculate the magnetic field in terms of vector potential,

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

$\begin{array}{c}B\left(\text{r,t}\right)=-\frac{\partial }{\partial r}\left\{\frac{{\mu }_0}{4\pi }\widehat{z}\left[t\ln \left(\frac{ct+\sqrt{{\left(ct\right)}^2-r^2}}{r}\right)-\frac{1}{c}\sqrt{{\left(ct\right)}^2-r^2}\right]\right\}\\ B\left(\text{r,t}\right)=\frac{{\mu }_0}{2\pi }\left[t\left(\frac{r}{ct+\sqrt{{\left(ct\right)}^2-r^2}}\right)\frac{r\frac{1}{2}\frac{(-2r)}{\sqrt{{\left(ct\right)}^2-r^2}}-ct-\sqrt{{\left(ct\right)}^2-r^2}}{r^2}-\frac{1}{2c}\frac{(-2r)}{\sqrt{{\left(ct\right)}^2-r^2}}\right]\widehat{\varphi }\\ B\left(\text{r,t}\right)=\frac{{\mu }_0}{2\pi }\left[\frac{-c{t}^2}{r\sqrt{{\left(ct\right)}^2-r^2}}+\frac{r}{c\sqrt{{\left(ct\right)}^2-r^2}}\right]\widehat{\varphi }\\ B\left(\text{r,t}\right)=\frac{{\mu }_{0}k}{2\pi rc}\sqrt{{\left(ct\right)}^2-r^2}\widehat{\varphi }\end{array}$

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

Determine the electric and magnetic field when wire having the burst of current.

(b)

Calculate the vector potential.

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

$A(r,t)=\frac{{\mu }_0}{4\pi }\widehat{z}{\int }_0^{\infty }\frac{q_o\delta \left(t-\frac{\text{r}}{c}\right)}{r}dz$

$A(r,t)=\frac{{\mu }_0{q}_0}{4\pi }\widehat{z}{\int }_0^{\infty }\frac{\delta \left(t-\frac{\text{r}}{c}\right)}{r}dz$ ……. (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 $z=0,\text{\hspace{0.17em}r\hspace{0.17em}=}r$and $z=\infty ,\text{\hspace{0.17em}r\hspace{0.17em}=\hspace{0.17em}}r$

From equation (1),

$A(r,t)=\frac{{\mu }_0{q}_0}{4\pi }\widehat{z}{\int }_r^{\infty }\delta \left(t-\frac{\text{r}}{c}\right)\frac{\text{rdr}}{\sqrt{{\text{r}}^2-r^2}}dz$

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

role="math" localid="1657799163827" $\begin{array}{l}A(r,t)=\frac{{\mu }_0{q}_0}{4\pi }\widehat{z}{\int }_r^{\infty }\frac{1}{\text{r}}\delta \left(t-\frac{\text{r}}{c}\right)\frac{\text{rdr}}{\sqrt{{\text{r}}^2-r^2}}d\text{r}\\ A(r,t)=\frac{{\mu }_0{q}_0}{2\pi }\widehat{z}{\int }_r^{\infty }\frac{\delta \left(\text{r-}ct\right)}{\sqrt{{\text{r}}^2-r^2}}d\text{r}\\ A(r,t)=\frac{{\mu }_0{q}_{0}c}{2\pi }\frac{1}{\sqrt{{\left(ct\right)}^2-r^2}}\widehat{z}\end{array}$

Calculate the electric field in terms of vector potential.

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

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

Calculate the magnetic field in terms of vector potential,

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

$\begin{array}{l}B\left(\text{r,t}\right)=-\frac{\partial }{\partial r}\left[\frac{{\mu }_0{q}_{0}c}{2\pi }\frac{1}{\sqrt{{\left(ct\right)}^2-r^2}}\widehat{z}\right]\widehat{\varphi }\\ B\left(\text{r,t}\right)=-\frac{{\mu }_0{q}_{0}c}{2\pi }\left(-\frac{1}{2}\right)\left(\frac{-2r}{{{\left(ct\right)}^2-r^2)}^{\frac{3}{2}}}\right)\widehat{\varphi }\\ B\left(\text{r,t}\right)=-\frac{{\mu }_0{q}_{0}c}{2\pi {\left({\left(ct\right)}^2-r^2\right)}^{\frac{3}{2}}}\widehat{\varphi }\end{array}$

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