91Ó°ÊÓ

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, $\mathit{ÒÏ}$ is surface charge density , z is axis and a is radius.

Write the formula of E (s,z).

$\mathit{E}\mathbf{(}\mathit{s}\mathbf{,}\mathit{z}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{∇}\mathit{V}$ …… (2)

Here, V is potential.

Write the formula ofsurface charge density on the wire.

role="math" localid="1658818871800" $\mathit{σ}\mathbf{\left(}\mathit{z}\mathbf{\right)}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\left[E_s\left(a^{+}\right)-E_s\left(a^{-}\right)\right]$ …… (3)

Here, ${\mathit{Î\mu }}_{\mathbf{0}}$ 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 $E=\frac{I\mathrm{ÒÏ}}{a^2}\hat{z}$uniform, while the electric field outside the wire is of size $B=\frac{{\mathrm{μ}}_{0}I}{2\mathrm{Ï€}s}\widehat{\mathrm{Ï•}}$.

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

$V(a,z)=\frac{-I\mathrm{ÒÏ}z}{\mathrm{Ï€}{a}^2}$ …… (2)

Using second boundary condition.

V (b,z) = 0 …… (3)

Figure 1

In cylindrical co-ordinates

${∇}^{2}V=\frac{1}{s}\frac{∂}{∂s}\left(s\frac{∂V}{∂s}\right)+\frac{1}{s^2}\frac{{∂}^{2}V}{∂{\mathrm{ϕ}}^2}+\frac{{∂}^{2}V}{∂z^2}$

Here, V = fz

Then $\begin{array}{rcl}\frac{{∂}^{2}V}{∂{\mathrm{ϕ}}^2}& =& \frac{{∂}^2}{∂{\mathrm{ϕ}}^2}\left(fz\right)\\ & =& 0\end{array}$

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

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

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).

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

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

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

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

Determine the electric field.

$E=-∇V$

In cylindrical co-ordinates

Substitute $\frac{∂V}{∂s}\hat{s}+\frac{1}{s}\frac{∂V}{∂\mathrm{ϕ}}\widehat{\mathrm{ϕ}}+\frac{∂V}{∂z}\hat{z}$for $∇V$into equation (2).

$\frac{∂V}{∂\mathrm{ϕ}}=0$

Then $∇V=\frac{∂V}{∂s}\hat{s}+\frac{∂V}{∂z}\hat{z}$

Then, solve for the electric field as:

$\begin{array}{rcl}E& =& -∇V\\ & =& -\frac{∂V}{∂s}\hat{s}-\frac{∂V}{∂z}\hat{z}\\ & =& -\frac{∂}{∂s}\left(-\frac{1\mathrm{ÒÏ}z}{\mathrm{Ï€}{a}^2}\frac{In\left({\displaystyle \frac{s}{b}}\right)}{In\left({\displaystyle \frac{a}{b}}\right)}\right)\hat{s}-\frac{∂}{∂s}\left(-\frac{1\mathrm{ÒÏ}z}{\mathrm{Ï€}{a}^2}\frac{In\left({\displaystyle \frac{s}{b}}\right)}{In\left({\displaystyle \frac{a}{b}}\right)}\right)\hat{z}\\ & =& -\frac{1\mathrm{ÒÏ}z}{\mathrm{Ï€}{a}^{2}s}\frac{1}{In\left({\displaystyle \frac{a}{b}}\right)}\hat{s}+\frac{1\mathrm{ÒÏ}}{\mathrm{Ï€}{a}^2}\frac{In\left({\displaystyle \frac{s}{b}}\right)}{In\left({\displaystyle \frac{a}{b}}\right)}\hat{z}\end{array}$

Solve further as

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

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

Determine thesurface charge density on the wire.

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

$$\begin{gathered} \mathrm{σ}\left(z\right)={\mathrm{Î\mu }}_0\left[\frac{1\mathrm{ÒÏ}}{\mathrm{Ï€}{a}^{2}In\left({\displaystyle \frac{a}{b}}\right)}\left(\frac{z}{a}\right)-0\right] \\ \mathrm{σ}\left(z\right)=\frac{{\mathrm{Î\mu }}_{0}1\mathrm{ÒÏ}}{\mathrm{Ï€}{a}^{2}In\left({\displaystyle \frac{a}{b}}\right)} \end{gathered}$$

Therefore, the value of surface charge density on the wire is $\mathrm{σ}\left(z\right)=\frac{{\mathrm{Î\mu }}_{0}1\mathrm{ÒÏ}}{\mathrm{Ï€}{a}^{2}In\left({\displaystyle \frac{a}{b}}\right)}$.