Question

a wire of a given material has an electric field of 0.100 mv/m when a potential is applied. the wire has 8.49 x 10^28 conduction electrons per cubic meter and a mean free time between collisions of 2.50 x 10^-14 s. what is the uniform current density in the wire? 26.4 ohms law

Explanation:

Step1: Recall the formula for current density.

The formula for current density $J = n e v_d$, where $n$ is the number density of charge - carriers, $e$ is the elementary charge ($e= 1.6\times10^{-19}\ C$), and $v_d$ is the drift velocity. The drift velocity can be related to the electric field $E$ and the mean - free time $\tau$ by $v_d=\frac{eE\tau}{m}$, where $m = 9.1\times10^{-31}\ kg$ is the mass of an electron.

Step2: Substitute the expression for $v_d$ into the current - density formula.

$J=n e\times\frac{eE\tau}{m}=\frac{n e^{2}E\tau}{m}$

Step3: Identify the given values.

$n = 8.49\times10^{28}\ m^{-3}$, $e = 1.6\times10^{-19}\ C$, $E=0.100\ mV/m=0.100\times10^{-3}\ V/m$, $\tau = 2.50\times10^{-14}\ s$, $m = 9.1\times10^{-31}\ kg$

Step4: Calculate the current density.

\[

$$\begin{align*} J&=\frac{(8.49\times10^{28})\times(1.6\times10^{-19})^{2}\times(0.100\times10^{-3})\times(2.50\times10^{-14})}{9.1\times10^{-31}}\\ &=\frac{8.49\times10^{28}\times2.56\times10^{-38}\times0.100\times10^{-3}\times2.50\times10^{-14}}{9.1\times10^{-31}}\\ &=\frac{8.49\times2.56\times0.100\times2.50\times10^{28 - 38-3 - 14}}{9.1\times10^{-31}}\\ &=\frac{8.49\times2.56\times0.100\times2.50\times10^{-27}}{9.1\times10^{-31}}\\ &=\frac{8.49\times2.56\times0.100\times2.50}{9.1}\times10^{-27 + 31}\\ &=\frac{8.49\times2.56\times0.100\times2.50}{9.1}\times10^{4}\\ &\approx5.99\times10^{4}\ A/m^{2} \end{align*}$$

\]

Answer:

$5.99\times10^{4}\ A/m^{2}$