91影视

The current in the wire is I(z).

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

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

$\mathit{E}\mathbf{=}\frac{\mathbf{位}}{\mathbf{2}\mathbf{蟺}{\mathbf{蔚}}_{\mathbf{0}}\mathbf{s}}$

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

The maxwell鈥檚 equation is given as follows.

$\mathbf{鈭�}\mathbf{脳}\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{s}}\frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{s}}\mathbf{\left(}\mathit{s}\mathit{E}\mathbf{\right)}$

(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,

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

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,

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

Substitute $位dz$ for q into equation (1).

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

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

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

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

Here, k is the constant.

When $位\left(0\right)=0$

$$\begin{gathered} \frac{d位}{dt}=k \\ d位=kdt+D \end{gathered}$$ 鈥︹��. (2)

Since $位\left(0\right)=0$

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

Substitute 0 for D into equation (2).

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

Integrate the above equation,

$\begin{array}{rcl}鈭�d位& =& 鈭�kdt\\ 位\left(t\right)& =& kt\end{array}$

When I(0) = 0

$\begin{array}{rcl}\frac{dl}{dt}& =& -k\\ dl& =& -kdt+F\end{array}$ 鈥︹��. (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" $\begin{array}{rcl}鈭�dl& =& -k鈭�dt\\ I\left(t\right)& =& -kz\end{array}$

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

(b)

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

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

Here, l is the length of the wire.

Substitute $位$for q/I into above equation.

role="math" localid="1658834657083" $E=\frac{位}{2蟺{蔚}_{0}s}$

The Maxwell鈥檚 equation is given by,

role="math" localid="1658834699724" $鈭�路E=\frac{1}{s}\frac{鈭�}{鈭�s}\left(sE\right)$

Substitute $\frac{位}{2蟺{蔚}_{0}s}$for E into above equation.

$$\begin{gathered} 鈭�路E=\frac{1}{s}\frac{鈭�}{鈭�s}\left(s\frac{位}{2蟺{蔚}_{0}s}\right) \\ 鈭�路E=\frac{1}{s}\frac{鈭�}{鈭�s}\left(\frac{位}{2蟺{蔚}_0}\right) \\ 鈭�路E=0 \end{gathered}$$

The curl of the electric field is always a zero.

The integral of the electric field is given by,

$$\begin{gathered} 鈭�E路ds=E鈭�ds \\ 鈭�E路ds=E\left(2蟺sl\right) \end{gathered}$$

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

$$\begin{gathered} 鈭�E路ds=\frac{q}{2{\mathrm{蟺蔚}}_0\mathrm{sl}}\left(2\mathrm{蟺蝉濒}\right) \\ 鈭�\mathrm{E}路\mathrm{ds}=\frac{\mathrm{q}}{{\mathrm{蔚}}_0} \end{gathered}$$

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

The divergence of the magnetic field is always zero.

$鈭�路B=0$

The magnetic field is varying along the $蠒$direction.

$B_{蠒}=-\frac{{渭}_{0}Cz}{2蟺s}$

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

$鈭�脳B=\left(\frac{1}{s}\frac{鈭�B_z}{鈭�蠒}-\frac{鈭�B_{蠒}}{鈭�z}\right)\hat{s}+\left(\frac{鈭�B_s}{鈭�z}-\frac{鈭�B_{蠒}}{鈭�s}\right)\widehat{蠒}+\frac{1}{s}\left(\frac{鈭�}{鈭�蠒}\left(s{B}_{蠒}\right)-\frac{鈭�B_s}{鈭�蠒}\right)\hat{z}$

Substitute $-\frac{{渭}_{0}Cz}{2蟺s}$for $B_{蠒}$into above equation.

$\begin{array}{rcl}鈭�脳B& =& \left(\frac{1}{s}\frac{鈭�B_z}{鈭�蠒}-\frac{鈭�}{鈭�z}\left(-\frac{{渭}_{0}Cz}{2蟺s}\right)\right)\hat{s}+\left(\frac{鈭�B_s}{鈭�z}-\frac{鈭�B_z}{鈭�s}\right)\widehat{蠒}+\frac{1}{s}\left(\frac{鈭�}{鈭�蠒}\left(s-\frac{{渭}_{0}Cz}{2蟺s}\right)-\frac{鈭�B_s}{鈭�蠒}\right)\hat{z}\\ 鈭�脳B& =& \frac{鈭�}{鈭�z}\left(-\frac{{渭}_{0}Cz}{2蟺s}\right)\hat{s}-\frac{1}{s}\frac{鈭�}{鈭�s}\left(-\frac{{渭}_{0}Cz}{2蟺s}\right)\hat{z}\\ 鈭�脳B& =& \frac{{渭}_{0}Cz}{2蟺s}\hat{s}\\ 鈭�脳B& =& {渭}_0{蔚}_0\frac{鈭�E}{鈭�t}\end{array}$

The electric field for the Amperian loop is given by,

$鈭�E路dl=-\frac{d}{dt}鈭�B路ds$

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

$$\begin{gathered} 鈭�B路dl=B鈭�dl \\ 鈭�B路dl=B\left(2\mathrm{蟺搁}\right) \\ 鈭�\mathrm{B}路\mathrm{dl}=2\mathrm{蟺搁B} \end{gathered}$$

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

$$\begin{gathered} 鈭�B路dl=2\mathrm{蟺搁}\left(-\frac{{\mathrm{渭}}_0\mathrm{Cz}}{2\mathrm{蟺搁}}\right) \\ 鈭�\mathrm{B}路\mathrm{dl}={\mathrm{渭}}_0{\mathrm{l}}_{\mathrm{endosed}}+0 \end{gathered}$$

If the electric field is perpendicular zero then value of zero is ${渭}_0{蔚}_0鈭�\frac{鈭�E}{鈭�t}路ds=0$.

$鈭�B路dl={渭}_0{l}_{endosed}+{渭}_0{蔚}_0鈭�\frac{鈭�E}{鈭�t}路ds$

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

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