91Ó°ÊÓ

The current $I$passes through the wire, a thin layer of charge appears at the boundary between the segments, in terms of $I,{\mathrm{σ}}_1,{\mathrm{σ}}_2$, and the wire's cross-section area$A$.

The current is conserved, $I_1=I_2=I$. The cross-section areas of the two wires are the identical, so the current densities are the exact:$J_1=J_2=IA$. So, the electric fields in the two segments have strengths
$E_1=\frac{J_1}{{\mathrm{σ}}_1}=\frac{I}{A{\mathrm{σ}}_1}$

$E_2=\frac{J_1}{{\mathrm{σ}}_2}=\frac{I}{A{\mathrm{σ}}_2}$

The electric field penetrates the Gaussian surface on the left and exits on the right. No flux passes via the wall of cylinder, so the net flux is${\mathrm{Φ}}_c=E_{a}a-E_a$. The Gaussian cylinder encloses charge $Q_n={\mathrm{η}}_a$on the boundary between the segments. Gauss's law is,

${\mathrm{Φ}}_e=\frac{Q_a}{{\mathrm{Î\mu }}_0}⇒E_{2}a-E_{1}a$

$=\frac{la}{A}\left(\frac{1}{{\mathrm{σ}}_2}-\frac{1}{{\mathrm{σ}}_1}\right)=\frac{n_a}{{\mathrm{Î\mu }}_0}$

Thus, an expression for the surface charge density on the boundary is

$\mathrm{η}=\frac{{\mathrm{Î\mu }}_{1}I}{A}\left(\frac{1}{{\mathrm{σ}}_2}-\frac{1}{{\mathrm{σ}}_1}\right)$

A wire that is made of two equal diameter segments with conductivities ${\mathrm{σ}}_1$and ${\mathrm{σ}}_2$. A $1.0$-mm-diameter wire made of copper and iron segments carries a $5.0\mathrm{A}$ current.

From the expression obtained in part (a)
$\mathrm{η}=\frac{Q}{s{R}^2}=\frac{l{c}_k}{\left(\mathrm{π}{R}^2\right)}\left(\frac{1}{{\mathrm{σ}}_m}-\frac{1}{{\mathrm{σ}}_{mm}}\right)$
$⇒Q=(5\mathrm{A})\left(8.85×{10}^{-12}{\mathrm{C}}^1/{\mathrm{Nm}}^2\right)\left(\frac{1}{1.0×{10}^1{\mathrm{Î\copyright }}^{-1}{\mathrm{m}}^{-4}}-\frac{1}{6.0×{10}^7{\mathrm{Î\copyright }}^{-1}{\mathrm{m}}^{-1}}\right)$
$=3.68×{10}^{-19}\mathrm{C}$
The charge accumulates at the boundary between the segments is $23$electrons.