91Ó°ÊÓ

The current in the wire is I(z).

The charge density is $\mathrm{λ}\left(t\right)$.

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

$\mathit{E}\mathbf{=}\frac{\mathbf{λ}}{\mathbf{2}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}\mathbf{s}}$

Here, $\mathit{λ}$ is the linear charge density, s is the distance from the wire.

The maxwell’s equation is given as follows.

$\mathbf{∇}\mathbf{×}\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{s}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{s}}\mathbf{\left(}\mathit{s}\mathit{E}\mathbf{\right)}$

(a)

The current passing through the element dz in time dt is given by,

dl = I (dz)

Calculate the current passing through the entire length,

$\begin{array}{rcl}∫dl& =& ∫I\left(dz\right)\\ I& =& {∫}_{z+dz}^{z}I\left(dz\right)\\ I& =& I{\left(z\right)}_{z+dz}^{z}I=I\left(z\right)-I(z+dz)\end{array}$

The rate of flow of the charge with respect to time, is known as current.

$I=\frac{dp}{dt}$

Substitute I (z) - I (z+dz) for I into above equation.

$\begin{array}{rcl}I\left(z\right)-I(z+dz)& =& \frac{dp}{dt}\\ I\left(z\right)-\frac{dp}{dt}dz-I\left(z\right)& =& \frac{dp}{dt}\\ -\frac{dp}{dt}dz& =& \frac{dp}{dt}..........\left(1\right)\end{array}$

The charge density in the elemental length dz is given by,

$$\begin{gathered} \mathrm{λ}=\frac{q}{dz} \\ q=\mathrm{λ}dz \end{gathered}$$

Substitute $\mathrm{λ}dz$ for q into equation (1).

role="math" localid="1658831987320" $\begin{array}{rcl}-\frac{dl}{dz}dz& =& \frac{d\left(\mathrm{λ}dz\right)}{dt}\\ -\frac{dl}{dz}dz& =& \frac{d\mathrm{λ}}{dt}dz\\ -\frac{dl}{dz}& =& \frac{d\mathrm{λ}}{dt}\\ \frac{d\mathrm{λ}}{dt}& =& -\frac{dl}{dz}\end{array}$

Hence the equation $\begin{array}{rcl}\frac{d\mathrm{λ}}{dt}& =& -\frac{dl}{dz}\end{array}$is obtained.

The equation $\begin{array}{rcl}\frac{d\mathrm{λ}}{dt}& =& -\frac{dl}{dz}\end{array}$is one dimensional equation.

$\begin{array}{rcl}\frac{d\mathrm{λ}}{dt}& =& k,-\frac{dl}{dz}=k\end{array}$

Here, k is the constant.

When $\mathrm{λ}\left(0\right)=0$

$$\begin{gathered} \frac{d\mathrm{λ}}{dt}=k \\ d\mathrm{λ}=kdt+D \end{gathered}$$ ……. (2)

Since $\mathrm{λ}\left(0\right)=0$

$$\begin{gathered} 0=k\left(0\right)+D \\ D=0 \end{gathered}$$

Substitute 0 for D into equation (2).

$$\begin{gathered} d\mathrm{λ}=kt+0 \\ d\mathrm{λ}=kt \end{gathered}$$

Integrate the above equation,

$\begin{array}{rcl}∫d\mathrm{λ}& =& ∫kdt\\ \mathrm{λ}\left(t\right)& =& kt\end{array}$

When I(0) = 0

$\begin{array}{rcl}\frac{dl}{dt}& =& -k\\ dl& =& -kdt+F\end{array}$ ……. (3)

Since I(0) = 0

0 = -k(0) +F

F = 0

Substitute 0 for F into equation (3).

$$\begin{gathered} dl=-kdt+0 \\ dl=-kdt \end{gathered}$$

Integrate the above equation.

role="math" localid="1658834851860" $\begin{array}{rcl}∫dl& =& -k∫dt\\ I\left(t\right)& =& -kz\end{array}$

Hence the expression for charge density and current when $\mathrm{λ}\left(0\right)=0,andI\left(0\right)=0are\mathrm{λ}\left(t\right)=ktandI\left(t\right)=-kz$respectively.

(b)

The electric field at point z due to infinite wire is given by,

$E=\frac{q}{2{\mathrm{Ï€Î\mu }}_0\mathrm{sl}}$

Here, l is the length of the wire.

Substitute $\mathrm{λ}$for q/I into above equation.

role="math" localid="1658834657083" $E=\frac{\mathrm{λ}}{2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}s}$

The Maxwell’s equation is given by,

role="math" localid="1658834699724" $∇·E=\frac{1}{s}\frac{∂}{∂s}\left(sE\right)$

Substitute $\frac{\mathrm{λ}}{2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}s}$for E into above equation.

$$\begin{gathered} ∇·E=\frac{1}{s}\frac{∂}{∂s}\left(s\frac{\mathrm{λ}}{2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}s}\right) \\ ∇·E=\frac{1}{s}\frac{∂}{∂s}\left(\frac{\mathrm{λ}}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right) \\ ∇·E=0 \end{gathered}$$

The curl of the electric field is always a zero.

The integral of the electric field is given by,

$$\begin{gathered} ∫E·ds=E∫ds \\ ∫E·ds=E\left(2\mathrm{π}sl\right) \end{gathered}$$

Substitute $E=\frac{q}{2{\mathrm{Ï€Î\mu }}_{0}rl}$ for E into above equation.

$$\begin{gathered} ∫E·ds=\frac{q}{2{\mathrm{Ï€Î\mu }}_0\mathrm{sl}}\left(2\mathrm{Ï€²õ\pm ô}\right) \\ ∫\mathrm{E}·\mathrm{ds}=\frac{\mathrm{q}}{{\mathrm{Î\mu }}_0} \end{gathered}$$

Hence the differential and integral Maxwell’s equation is satisfied for the Gaussian cylinder.

The divergence of the magnetic field is always zero.

$∇·B=0$

The magnetic field is varying along the $\mathrm{Ï•}$direction.

$B_{\mathrm{ϕ}}=-\frac{{\mathrm{μ}}_{0}Cz}{2\mathrm{π}s}$

The curl of magnetic field for the cylindrical coordinates is given by,

$∇×B=\left(\frac{1}{s}\frac{∂B_z}{∂\mathrm{ϕ}}-\frac{∂B_{\mathrm{ϕ}}}{∂z}\right)\hat{s}+\left(\frac{∂B_s}{∂z}-\frac{∂B_{\mathrm{ϕ}}}{∂s}\right)\widehat{\mathrm{ϕ}}+\frac{1}{s}\left(\frac{∂}{∂\mathrm{ϕ}}\left(s{B}_{\mathrm{ϕ}}\right)-\frac{∂B_s}{∂\mathrm{ϕ}}\right)\hat{z}$

Substitute $-\frac{{\mathrm{μ}}_{0}Cz}{2\mathrm{π}s}$for $B_{\mathrm{ϕ}}$into above equation.

$\begin{array}{rcl}∇×B& =& \left(\frac{1}{s}\frac{∂B_z}{∂\mathrm{Ï•}}-\frac{∂}{∂z}\left(-\frac{{\mathrm{μ}}_{0}Cz}{2\mathrm{Ï€}s}\right)\right)\hat{s}+\left(\frac{∂B_s}{∂z}-\frac{∂B_z}{∂s}\right)\widehat{\mathrm{Ï•}}+\frac{1}{s}\left(\frac{∂}{∂\mathrm{Ï•}}\left(s-\frac{{\mathrm{μ}}_{0}Cz}{2\mathrm{Ï€}s}\right)-\frac{∂B_s}{∂\mathrm{Ï•}}\right)\hat{z}\\ ∇×B& =& \frac{∂}{∂z}\left(-\frac{{\mathrm{μ}}_{0}Cz}{2\mathrm{Ï€}s}\right)\hat{s}-\frac{1}{s}\frac{∂}{∂s}\left(-\frac{{\mathrm{μ}}_{0}Cz}{2\mathrm{Ï€}s}\right)\hat{z}\\ ∇×B& =& \frac{{\mathrm{μ}}_{0}Cz}{2\mathrm{Ï€}s}\hat{s}\\ ∇×B& =& {\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{∂E}{∂t}\end{array}$

The electric field for the Amperian loop is given by,

$∫E·dl=-\frac{d}{dt}∫B·ds$

According to Ampere’s law the magnetic field around the cylinder is given by,

$$\begin{gathered} ∫B·dl=B∫dl \\ ∫B·dl=B\left(2\mathrm{Ï€¸é}\right) \\ ∫\mathrm{B}·\mathrm{dl}=2\mathrm{Ï€¸éB} \end{gathered}$$

Substitute $\frac{{\mathrm{μ}}_{0}Cz}{2\mathrm{π}R}$ for B into above equation.

$$\begin{gathered} ∫B·dl=2\mathrm{Ï€¸é}\left(-\frac{{\mathrm{μ}}_0\mathrm{Cz}}{2\mathrm{Ï€¸é}}\right) \\ ∫\mathrm{B}·\mathrm{dl}={\mathrm{μ}}_0{\mathrm{l}}_{\mathrm{endosed}}+0 \end{gathered}$$

If the electric field is perpendicular zero then value of zero is ${\mathrm{μ}}_0{\mathrm{Î\mu }}_0∫\frac{∂E}{∂t}·ds=0$.

$∫B·dl={\mathrm{μ}}_0{l}_{endosed}+{\mathrm{μ}}_0{\mathrm{Î\mu }}_0∫\frac{∂E}{∂t}·ds$

Hence the differential and integral Maxwell’s equation is satisfied for the Amperian cylinder.

Hence the maxwell’s equations of differential and integral from are satisfied.