Question

A copper wire has a square cross-section 2.3 mm on a side. The wire is 4.0 m long and carries a current of 3.6 A. The density of free electrons is$\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{28}}\mathbf{}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$. Find the magnitudes of (a) the current density in the wire and (b) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?

Short answer

1. Current density in the wire is$J=6.8\times {10}^5\mathrm{A}/{\mathrm{m}}^2$
2. Electric field in the wire is$E=11.69\times {10}^{-3}\mathrm{V}/\mathrm{m}$
3. Time required by the electron$\tau =22.22\mathrm{hr}$

Step-by-step solution

Step 1:

Given data:

_{$$\begin{gathered} A=l^2={\left(2.3\times {10}^{-3}\mathrm{m}\right)}^2=5.3\times {10}^{-6}{\mathrm{m}}^2 \\ L=4.0\mathrm{m} \\ l=3.6\mathrm{A} \end{gathered}$$}

Density of free electrons$=8.5\times {10}^{28}/m^3$

(a) Current density in the wire is:

_{$$\begin{gathered} J=\frac{I}{A} \\ =\frac{3.6\mathrm{A}}{5.3\times {10}^{-6}{\mathrm{m}}^2} \\ =6.8\times {10}^5\mathrm{A}/{\mathrm{m}}^2 \end{gathered}$$}

Therefore, the current density in the wire is$J=6.8\times {10}^5\mathrm{A}/{\mathrm{m}}^2$.

Step 2:

(b) Electric field in the wire is:

As the copper resistivity is$\rho =1.72\times {10}^{-8}\mathrm{\Omega }·\mathrm{m}$

So electric field is:

_{$$\begin{gathered} E=\rho J \\ =\left(1.72\times {10}^{-8}\mathrm{\Omega }·\mathrm{m}\right)\left(6.8\times {10}^5\mathrm{A}/{\mathrm{m}}^2\right) \\ =0.01169\mathrm{V}/\mathrm{m} \\ =11.69\times {10}^{-3}\mathrm{V}/\mathrm{m} \end{gathered}$$}

Hence, the electric field in the wire is$E=11.69\times {10}^{-3}\mathrm{V}/\mathrm{m}$.

Step 3:

(c) Time required to travel the length of the wire:

From the equation,

_{$\tau =\frac{L}{v_d}$}

Here $v_d$is drift speed;

$v_d=\frac{J}{n\left|q\right|}$

So, the time required is:

$$\begin{gathered} \tau =\frac{L}{v_d} \\ =\frac{L·n\left|q\right|}{J} \\ =\frac{\left(4.0\mathrm{m}\right)\times \left(8.5\times {10}^{28}\right)\times \left(\left|-1.6\times {10}^{-19}\mathrm{C}\right|\right)}{6.8\times {10}^5\mathrm{A}/{\mathrm{m}}^2} \\ =80000\mathrm{s}=1333.33\min =22.22\mathrm{hr} \end{gathered}$$

Therefore, the time required by the electron$\tau =22.22\mathrm{hr}$.