An infinite wire runs along the z axis; it carries a current I (z) that is a function ofz(but not of t ), and a charge density $\mathit{\lambda }\mathbf{\left(}\mathit{t}\mathbf{\right)}$ that is a function of t (but not of z ).

(a) By examining the charge flowing into a segment dz in a time dt, show that $\mathit{d}\mathit{\lambda }\mathbf{/}\mathit{d}\mathit{t}\mathbf{=}\mathbf{-}\mathit{d}\mathit{i}\mathbf{/}\mathit{d}\mathit{z}$. If we stipulate that $\mathit{\lambda }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$and $\mathit{I}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$, show that $\mathit{\lambda }\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{k}\mathit{t}$, $\mathit{I}\mathbf{\left(}\mathit{z}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathit{k}\mathit{z}$, where k is a constant.

(b) Assume for a moment that the process is quasistatic, so the fields are given by Eqs. 2.9 and 5.38. Show that these are in fact the exact fields, by confirming that all four of Maxwell's equations are satisfied. (First do it in differential form, for the region s > 0, then in integral form for the appropriate Gaussian cylinder/Amperian loop straddling the axis.)

The current in the wire is I(z).

The charge density is $\lambda \left(t\right)$.

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

$\mathit{E}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{s}}$

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

The maxwell’s equation is given as follows.

$\mathbf{\nabla }\mathbf{\times }\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{s}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{s}}\mathbf{\left(}\mathit{s}\mathit{E}\mathbf{\right)}$

(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,

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

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,

$$\begin{gathered} \lambda =\frac{q}{dz} \\ q=\lambda dz \end{gathered}$$

Substitute $\lambda dz$ for q into equation (1).

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

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

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

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

Here, k is the constant.

When $\lambda \left(0\right)=0$

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

Since $\lambda \left(0\right)=0$

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

Substitute 0 for D into equation (2).

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

Integrate the above equation,

$\begin{array}{rcl}\int d\lambda & =& \int kdt\\ \lambda \left(t\right)& =& kt\end{array}$

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" $\begin{array}{rcl}\int dl& =& -k\int dt\\ I\left(t\right)& =& -kz\end{array}$

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

(b)

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

$E=\frac{q}{2{\mathrm{\pi \epsilon }}_0\mathrm{sl}}$

Here, l is the length of the wire.

Substitute $\lambda$for q/I into above equation.

role="math" localid="1658834657083" $E=\frac{\lambda }{2\pi {\epsilon }_{0}s}$

The Maxwell’s equation is given by,

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

Substitute $\frac{\lambda }{2\pi {\epsilon }_{0}s}$for E into above equation.

$$\begin{gathered} \nabla ·E=\frac{1}{s}\frac{\partial }{\partial s}\left(s\frac{\lambda }{2\pi {\epsilon }_{0}s}\right) \\ \nabla ·E=\frac{1}{s}\frac{\partial }{\partial s}\left(\frac{\lambda }{2\pi {\epsilon }_0}\right) \\ \nabla ·E=0 \end{gathered}$$

The curl of the electric field is always a zero.

The integral of the electric field is given by,

$$\begin{gathered} \int E·ds=E\int ds \\ \int E·ds=E\left(2\pi sl\right) \end{gathered}$$

Substitute $E=\frac{q}{2{\mathrm{\pi \epsilon }}_{0}rl}$ for E into above equation.

$$\begin{gathered} \int E·ds=\frac{q}{2{\mathrm{\pi \epsilon }}_0\mathrm{sl}}\left(2\mathrm{\pi sl}\right) \\ \int \mathrm{E}·\mathrm{ds}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0} \end{gathered}$$

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

The divergence of the magnetic field is always zero.

$\nabla ·B=0$

The magnetic field is varying along the $\varphi$direction.

$B_{\varphi }=-\frac{{\mu }_{0}Cz}{2\pi s}$

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

$\nabla \times B=\left(\frac{1}{s}\frac{\partial B_z}{\partial \varphi }-\frac{\partial B_{\varphi }}{\partial z}\right)\hat{s}+\left(\frac{\partial B_s}{\partial z}-\frac{\partial B_{\varphi }}{\partial s}\right)\hat{\varphi }+\frac{1}{s}\left(\frac{\partial }{\partial \varphi }\left(s{B}_{\varphi }\right)-\frac{\partial B_s}{\partial \varphi }\right)\hat{z}$

Substitute $-\frac{{\mu }_{0}Cz}{2\pi s}$for $B_{\varphi }$into above equation.

$\begin{array}{rcl}\nabla \times B& =& \left(\frac{1}{s}\frac{\partial B_z}{\partial \varphi }-\frac{\partial }{\partial z}\left(-\frac{{\mu }_{0}Cz}{2\pi s}\right)\right)\hat{s}+\left(\frac{\partial B_s}{\partial z}-\frac{\partial B_z}{\partial s}\right)\hat{\varphi }+\frac{1}{s}\left(\frac{\partial }{\partial \varphi }\left(s-\frac{{\mu }_{0}Cz}{2\pi s}\right)-\frac{\partial B_s}{\partial \varphi }\right)\hat{z}\\ \nabla \times B& =& \frac{\partial }{\partial z}\left(-\frac{{\mu }_{0}Cz}{2\pi s}\right)\hat{s}-\frac{1}{s}\frac{\partial }{\partial s}\left(-\frac{{\mu }_{0}Cz}{2\pi s}\right)\hat{z}\\ \nabla \times B& =& \frac{{\mu }_{0}Cz}{2\pi s}\hat{s}\\ \nabla \times B& =& {\mu }_0{\epsilon }_0\frac{\partial E}{\partial t}\end{array}$

The electric field for the Amperian loop is given by,

$\int E·dl=-\frac{d}{dt}\int B·ds$

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

$$\begin{gathered} \int B·dl=B\int dl \\ \int B·dl=B\left(2\mathrm{\pi R}\right) \\ \int \mathrm{B}·\mathrm{dl}=2\mathrm{\pi RB} \end{gathered}$$

Substitute $\frac{{\mu }_{0}Cz}{2\pi R}$ for B into above equation.

$$\begin{gathered} \int B·dl=2\mathrm{\pi R}\left(-\frac{{\mathrm{\mu }}_0\mathrm{Cz}}{2\mathrm{\pi R}}\right) \\ \int \mathrm{B}·\mathrm{dl}={\mathrm{\mu }}_0{\mathrm{l}}_{\mathrm{endosed}}+0 \end{gathered}$$

If the electric field is perpendicular zero then value of zero is ${\mu }_0{\epsilon }_0\int \frac{\partial E}{\partial t}·ds=0$.

$\int B·dl={\mu }_0{l}_{endosed}+{\mu }_0{\epsilon }_0\int \frac{\partial E}{\partial t}·ds$

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.