91影视

Considerthe magnetic field outside a long straight wire carrying a steady current I.

Suppose the current returns along a perfectly conducting grounded coaxial cylinder of radius b.

Write the formula of f(s).

Here, I is current, $\mathit{蚁}$ is surface charge density , z is axis and a is radius.

Write the formula of E (s,z).

$\mathit{E}\mathbf{(}\mathit{s}\mathbf{,}\mathit{z}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{鈭�}\mathit{V}$ 鈥︹�� (2)

Here, V is potential.

Write the formula ofsurface charge density on the wire.

role="math" localid="1658818871800" $\mathit{蟽}\mathbf{\left(}\mathit{z}\mathbf{\right)}\mathbf{=}{\mathit{蔚}}_{\mathbf{0}}\left[E_s\left(a^{+}\right)-E_s\left(a^{-}\right)\right]$ 鈥︹�� (3)

Here, ${\mathit{蔚}}_{\mathbf{0}}$ is relative permittivity, E_s is electric field, a⁺ radius inside the cylinder and a^- is radius outside the cylinder.

The electric field within a long, straight wire carrying a constant current I is $E=\frac{I蚁}{a^2}\hat{z}$uniform, while the electric field outside the wire is of size $B=\frac{{渭}_{0}I}{2蟺s}\widehat{蠒}$.

Given that a < s < b the potential V (s,z) satisfies equation with boundary conditions.

$V(a,z)=\frac{-I蚁z}{蟺a^2}$ 鈥︹�� (2)

Using second boundary condition.

V (b,z) = 0 鈥︹�� (3)

Figure 1

In cylindrical co-ordinates

${鈭�}^{2}V=\frac{1}{s}\frac{鈭�}{鈭�s}\left(s\frac{鈭�V}{鈭�s}\right)+\frac{1}{s^2}\frac{{鈭�}^{2}V}{鈭�{蠒}^2}+\frac{{鈭�}^{2}V}{鈭�z^2}$

Here, V = fz

Then $\begin{array}{rcl}\frac{{鈭�}^{2}V}{鈭�{蠒}^2}& =& \frac{{鈭�}^2}{鈭�{蠒}^2}\left(fz\right)\\ & =& 0\end{array}$

Then localid="1658823510111" $\begin{array}{rcl}\frac{{鈭�}^{2}V}{鈭�{蠒}^2}& =& \frac{1}{s}\frac{鈭�}{鈭�s}\left(s\frac{鈭�\left(fz\right)}{鈭�s}\right)+\frac{{鈭�}^2\left(fz\right)}{鈭�s}\\ & =& \frac{z}{s}\frac{鈭�}{鈭�s}\left(s\frac{鈭�f}{鈭�s}\right)\end{array}$

Equation in$\begin{array}{rcl}{鈭�}^{2}V& =& 0\\ \frac{z}{s}\frac{鈭�}{鈭�s}\left(s\frac{鈭�f}{鈭�s}\right)& =& 0\\ s\frac{鈭�f}{鈭�s}& =& A\left(cons\tan t\right)\\ \frac{A}{s}ds& =& dfIntegratebothsides\end{array}$

Solve further as

$f=AIn\left(\frac{s}{s_0}\right)$

Here, s₀ is another constant.

By boundary condition (3)

f (b) = 0

Then $In\left(\frac{b}{s_0}=0\right)$

Then s₀ = b

Then $V(s,z)=A_{Z}In\left(\frac{s}{b}\right)$

By boundary condition (2) we get

Determine the (s).

$\begin{array}{rcl}V(a,z)& =& \frac{-I蚁z}{蟺a^2}\\ Az\left(In\frac{a}{b}\right)& =& \frac{-I蚁z}{蟺a^2}\\ A& =& \frac{-I蚁z}{蟺a^2}\frac{1}{In\left(\frac{a}{b}\right)}\end{array}$

Then $V(s,z)=z\left(\frac{-I蚁}{蟺a^2}\frac{1}{In\left({\displaystyle \frac{a}{b}}\right)}\right)In\left(\frac{s}{b}\right)$

Then localid="1658824284124" $V(s,z)=\left(\frac{-I蚁z}{蟺a^2}\right)\frac{In\left(\frac{s}{b}\right)}{In\left(\frac{a}{b}\right)}$.

Therefore, the value of the f(s) is $V(s,z)=\left(\frac{-I蚁z}{蟺a^2}\right)\frac{In\left(\frac{s}{b}\right)}{In\left(\frac{a}{b}\right)}$.

Determine the electric field.

$E=-鈭�V$

In cylindrical co-ordinates

Substitute $\frac{鈭�V}{鈭�s}\hat{s}+\frac{1}{s}\frac{鈭�V}{鈭�蠒}\widehat{蠒}+\frac{鈭�V}{鈭�z}\hat{z}$for $鈭�V$into equation (2).

$\frac{鈭�V}{鈭�蠒}=0$

Then $鈭�V=\frac{鈭�V}{鈭�s}\hat{s}+\frac{鈭�V}{鈭�z}\hat{z}$

Then, solve for the electric field as:

$\begin{array}{rcl}E& =& -鈭�V\\ & =& -\frac{鈭�V}{鈭�s}\hat{s}-\frac{鈭�V}{鈭�z}\hat{z}\\ & =& -\frac{鈭�}{鈭�s}\left(-\frac{1蚁z}{蟺a^2}\frac{In\left({\displaystyle \frac{s}{b}}\right)}{In\left({\displaystyle \frac{a}{b}}\right)}\right)\hat{s}-\frac{鈭�}{鈭�s}\left(-\frac{1蚁z}{蟺a^2}\frac{In\left({\displaystyle \frac{s}{b}}\right)}{In\left({\displaystyle \frac{a}{b}}\right)}\right)\hat{z}\\ & =& -\frac{1蚁z}{蟺a^{2}s}\frac{1}{In\left({\displaystyle \frac{a}{b}}\right)}\hat{s}+\frac{1蚁}{蟺a^2}\frac{In\left({\displaystyle \frac{s}{b}}\right)}{In\left({\displaystyle \frac{a}{b}}\right)}\hat{z}\end{array}$

Solve further as

$E=\frac{I蚁}{蟺a^{2}In\left({\displaystyle \frac{a}{b}}\right)}\left[\frac{z}{s}\hat{s}+In\left(\frac{s}{b}\right)\hat{z}\right]$

Therefore, the value of role="math" localid="1658825058927" $E=\frac{I蚁}{蟺a^{2}In\left({\displaystyle \frac{a}{b}}\right)}\left[\frac{z}{s}\hat{s}+In\left(\frac{s}{b}\right)\hat{z}\right]$.

Determine thesurface charge density on the wire.

Substitute $\frac{I蚁}{蟺a^{2}In\left({\displaystyle \frac{a}{b}}\right)}\left(\frac{z}{a}\right)$for $E_s\left(a^{+}\right)$and 0 for $E_s\left(a^{-}\right)$into equation (3).

$$\begin{gathered} 蟽\left(z\right)={蔚}_0\left[\frac{1蚁}{蟺a^{2}In\left({\displaystyle \frac{a}{b}}\right)}\left(\frac{z}{a}\right)-0\right] \\ 蟽\left(z\right)=\frac{{蔚}_{0}1蚁}{蟺a^{2}In\left({\displaystyle \frac{a}{b}}\right)} \end{gathered}$$

Therefore, the value of surface charge density on the wire is $蟽\left(z\right)=\frac{{蔚}_{0}1蚁}{蟺a^{2}In\left({\displaystyle \frac{a}{b}}\right)}$.