The magnetic field outside a long straight wire carrying a steady current I is

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{2}\mathbf{\pi }}\frac{\mathbf{I}}{\mathbf{s}}\hat{\mathbf{\varphi }}$

The electric field inside the wire is uniform:

$\mathit{E}\mathbf{=}\frac{\mathbf{I}\mathbf{\rho }}{\mathbf{\pi }{\mathbf{a}}^{\mathbf{2}}}\hat{\mathbf{z}}$,

Where $\mathit{\rho }$is the resistivity and a is the radius (see Exs. 7.1 and 7 .3). Question: What is the electric field outside the wire? 29 The answer depends on how you complete the circuit. Suppose the current returns along a perfectly conducting grounded coaxial cylinder of radius b (Fig. 7.52). In the region a < s < b, the potential V (s, z) satisfies Laplace's equation, with the boundary conditions

(i) $\mathit{V}\mathbf{(}\mathit{a}\mathbf{,}\mathit{z}\mathbf{)}\mathbf{=}\frac{\mathbf{I}\mathbf{\rho }\mathbf{z}}{\mathbf{\pi }{\mathbf{a}}^{\mathbf{2}}}$ ; (ii) $\mathit{V}\mathbf{(}\mathit{b}\mathbf{,}\mathit{z}\mathbf{)}\mathbf{=}\mathbf{0}$

Figure 7.52

This does not suffice to determine the answer-we still need to specify boundary conditions at the two ends (though for a long wire it shouldn't matter much). In the literature, it is customary to sweep this ambiguity under the rug by simply stipulating that V (s,z) is proportional to V (s,z) = zf (s) . On this assumption:

(a) Determine (s).

(b) E (s,z).

(c) Calculate the surface charge density $\mathit{\sigma }\mathbf{\left(}\mathit{z}\mathbf{\right)}$on the wire.

[Answer: $\mathit{V}\mathbf{=}\left(-Iz\rho /\pi a^2\right)$ This is a peculiar result, since E_s and $\mathit{\sigma }\mathbf{\left(}\mathit{z}\mathbf{\right)}$are not independent of localid="1658816847863" $\mathit{z}\mathbf{\to }\mathit{a}\mathit{s}$ one would certainly expect for a truly infinite wire.]

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

Write the formula of E (s,z).

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

Here, V is potential.

Write the formula ofsurface charge density on the wire.

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

Here, ${\mathit{\epsilon }}_{\mathbf{0}}$ 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 $E=\frac{I\rho }{a^2}\hat{z}$uniform, while the electric field outside the wire is of size $B=\frac{{\mu }_{0}I}{2\pi s}\hat{\varphi }$.

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

$V(a,z)=\frac{-I\rho z}{\pi a^2}$ …… (2)

Using second boundary condition.

V (b,z) = 0 …… (3)

Figure 1

In cylindrical co-ordinates

${\nabla }^{2}V=\frac{1}{s}\frac{\partial }{\partial s}\left(s\frac{\partial V}{\partial s}\right)+\frac{1}{s^2}\frac{{\partial }^{2}V}{\partial {\varphi }^2}+\frac{{\partial }^{2}V}{\partial z^2}$

Here, V = fz

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

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

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

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).

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

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

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

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

Determine the electric field.

$E=-\nabla V$

In cylindrical co-ordinates

Substitute $\frac{\partial V}{\partial s}\hat{s}+\frac{1}{s}\frac{\partial V}{\partial \varphi }\hat{\varphi }+\frac{\partial V}{\partial z}\hat{z}$for $\nabla V$into equation (2).

$\frac{\partial V}{\partial \varphi }=0$

Then $\nabla V=\frac{\partial V}{\partial s}\hat{s}+\frac{\partial V}{\partial z}\hat{z}$

Then, solve for the electric field as:

$\begin{array}{rcl}E& =& -\nabla V\\ & =& -\frac{\partial V}{\partial s}\hat{s}-\frac{\partial V}{\partial z}\hat{z}\\ & =& -\frac{\partial }{\partial s}\left(-\frac{1\rho z}{\pi a^2}\frac{In\left({\displaystyle \frac{s}{b}}\right)}{In\left({\displaystyle \frac{a}{b}}\right)}\right)\hat{s}-\frac{\partial }{\partial s}\left(-\frac{1\rho z}{\pi a^2}\frac{In\left({\displaystyle \frac{s}{b}}\right)}{In\left({\displaystyle \frac{a}{b}}\right)}\right)\hat{z}\\ & =& -\frac{1\rho z}{\pi a^{2}s}\frac{1}{In\left({\displaystyle \frac{a}{b}}\right)}\hat{s}+\frac{1\rho }{\pi a^2}\frac{In\left({\displaystyle \frac{s}{b}}\right)}{In\left({\displaystyle \frac{a}{b}}\right)}\hat{z}\end{array}$

Solve further as

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

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

Determine thesurface charge density on the wire.

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

$$\begin{gathered} \sigma \left(z\right)={\epsilon }_0\left[\frac{1\rho }{\pi a^{2}In\left({\displaystyle \frac{a}{b}}\right)}\left(\frac{z}{a}\right)-0\right] \\ \sigma \left(z\right)=\frac{{\epsilon }_{0}1\rho }{\pi a^{2}In\left({\displaystyle \frac{a}{b}}\right)} \end{gathered}$$

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