It may have occurred to you that since parallel currents attract, the current within a single wire should contract into a tiny concentrated stream along the axis. Yet in practice the current typically distributes itself quite uniformly over the wire. How do you account for this? If the positive charges (density ${\mathit{\rho }}_{\mathbf{+}}$) are "nailed down," and the negative charges (density${\mathit{\rho }}_{-}$) move at speed $\mathit{v}$(and none of these depends on the distance from the axis), show that ${\mathit{\rho }}_{\mathbf{-}}\mathbf{=}\mathbf{-}{\mathit{\rho }}_{\mathbf{+}}{\mathit{\gamma }}^{\mathbf{2}}\mathbf{,}\mathbf{}\mathbf{Where}\mathbf{}\mathit{\gamma }\mathbf{\equiv }\mathbf{1}\mathbf{/}\sqrt{\mathbf{1}\mathbf{-}{\left(v/c\right)}^{\mathbf{2}}}\mathbf{}\mathbf{and}\mathbf{}{\mathit{c}}^{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{/}{\mathit{\mu }}_{\mathbf{0}}{\mathit{\epsilon }}_{\mathbf{0}}$. If the wire as a whole is neutral, where is the compensating charge located?${}^{\mathbf{22}}$[Notice that for typical velocities (see Prob. 5.20), the two charge densities are essentially unchanged by the current (since$\mathit{\gamma }\mathbf{\approx }\mathbf{1}$). In plasmas, however, where the positive charges are also free to move, this so-called pinch effect can be very significant.]

The Gauss’ law states that the electric flux of a closed surface is mainly equal to the division of the enclosed charge and permittivity. The law also states that the electric field is being generated with the help of an electric charge.

If the negative charges mainly start concentrating at the center of the wire, then the positive charges will be left at the surface of the sphere. Since the electric forces are much stronger than the magnetic forces, that is one of the main reasons the current typically distributes itself quite uniformly over the wire.

Thus, the electric forces are much stronger than the magnetic forces, which is one of the main reasons the current typically distributes itself quite uniformly over the wire.

The wire does not carry any amount of force. Hence, the equation of the force of the wire is expressed as:

$$\begin{gathered} F=q\left(E+v\times B\right) \\ =0 \end{gathered}$$

$E=-\left(v\times B\right)$ …(i)

Here, $q$is the charge of the wire,$E$is the electric field,$v$is the velocity and $B$is the uniform magnetic field around the wire.

The equation of the uniform magnetic field of the wire is expressed as:

$\oint B.dI=B2\mathrm{\pi s}$

Here, $dI$is the increase in the length and $s$is the increase in the length of the wire.

The above equation can also be expressed as:

localid="1657626488274" $$\begin{gathered} \oint B.dI=B2\mathrm{\pi s} \\ ={\mu }_{0}J{\mathrm{\pi s}}^2 \\ =\mathrm{B}=\frac{{\mathrm{\mu }}_0{\mathrm{\rho }}_{-}\mathrm{vs}}{2}\stackrel{⏜}{\mathrm{\varphi }} \end{gathered}$$ …(ii)

The equation of the electric field is expressed as:

$\int E.da=E2\mathrm{\pi s}I$

Here,$da$is the increase in the area of the wire and $I$is the length of the wire.

The above equation can be expressed as:

$$\begin{gathered} \int E.da=E2\mathrm{\pi s}I \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\frac{\mathit{1}}{{\mathit{\epsilon }}_{\mathit{0}}}\left({\rho }_{+}{\rho }_{-}\right){\mathrm{\pi s}}^{2}I \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}E\mathit{=}\frac{\mathit{1}}{\mathit{2}{\mathit{\epsilon }}_{\mathit{0}}}\mathit{\left(}{\mathit{\rho }}_{\mathit{+}}{\mathit{\rho }}_{\mathit{-}}\mathit{\right)}s\stackrel{\mathit{⏜}}{\mathit{s}} \end{gathered}$$ …(iii)

Substitute the value of the equation (ii) and equation (iii) in equation (i).

$$\begin{gathered} \frac{1}{2{\epsilon }_0}\left({\rho }_{+}{\rho }_{-}\right)s\stackrel{⏜}{s}=-\left(v\stackrel{⏜}{z}\times \frac{{\mu }_0{\rho }_{-}vs}{2}\stackrel{⏜}{\varphi }\right) \\ \frac{1}{2{\epsilon }_0}\left({\rho }_{+}+{\rho }_{-}\right)s\stackrel{⏜}{s}=\frac{{\mu }_0}{2}{\rho }_{-}{v}^{2}s\stackrel{⏜}{s} \end{gathered}$$

Substitute $1/{\mu }_0{\epsilon }_0\mathrm{for}{c}^2$in the above equation.

$$\begin{gathered} {\rho }_{+}+{\rho }_{-}={\rho }_{-}\left({\epsilon }_0{\mu }_0{v}^2\right) \\ ={\rho }_{-}\left(\frac{v^2}{c^2}\right) \\ {\rho }_{+}=-{\rho }_{-}\left(1-\frac{v^2}{c^2}\right) \end{gathered}$$

Substitute$\gamma$for$1/\sqrt{1-{\left(v/c\right)}^2}$in the above equation.

$$\begin{gathered} {\rho }_{+}=\frac{\rho }{{\gamma }^2} \\ {\rho }_{-}=-{\gamma }^2{\rho }_{+} \\ =-{\rho }_{+}{\gamma }^2 \end{gathered}$$

Thus, the equation ${\rho }_{-}=-{\rho }_{+}{\gamma }^2$is proved.

If the wire remains neutral, then there will not be any type of parallel currents. Hence, the compensating charge will be located at the bottom of the wire.

Thus, the compensating charge will be located at the bottom of the wire.