Question

26.4 ohms law
how much current flows in a 2.00 mm diameter copper wire (resistivity of copper is 1.70x10^-8 ω·m) if the electric field in the wire has a magnitude of 0.100 v/cm?
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Explanation:

Step1: Convert units

The electric - field magnitude $E = 0.100\ V/cm=10\ V/m$. The diameter of the wire $d = 2.00\ mm = 2\times10^{- 3}\ m$, so the radius $r=\frac{d}{2}=1\times10^{-3}\ m$.

Step2: Use the relationship between electric field and potential difference

The potential difference $V = EL$. Also, from Ohm's law $V = IR$ and the resistance formula $R=
ho\frac{L}{A}$, where $
ho$ is the resistivity, $L$ is the length of the wire, and $A=\pi r^{2}$ is the cross - sectional area of the wire. Combining these equations, we get $I=\frac{EA}{
ho}$.

Step3: Calculate the cross - sectional area

The cross - sectional area of the wire $A=\pi r^{2}=\pi\times(1\times10^{-3})^{2}=\pi\times10^{-6}\ m^{2}$.

Step4: Calculate the current

Given $
ho = 1.70\times10^{-8}\ \Omega\cdot m$ and $E = 10\ V/m$, $A=\pi\times10^{-6}\ m^{2}$, then $I=\frac{EA}{
ho}=\frac{10\times\pi\times10^{-6}}{1.70\times10^{-8}}\ A$.
\[I=\frac{10\pi\times10^{-6}}{1.70\times10^{-8}}=\frac{10\pi}{1.70}\times10^{2}\ A\approx\frac{10\times3.14}{1.70}\times10^{2}\ A\approx1847\ A\]

Answer:

$1847\ A$