Question

when a potential difference of 12v is applied to a wire 6.9m long and 0.33mm in diameter, the result is an electric current of 2.1a. what is the resistivity of the wire?

a. 3.6×10⁻⁴ω.m

b. 5.5×10⁻⁴ω.m

c. 9.4×10⁻⁴ω.m

d. 7.1×10⁻⁴ω.m

e. 1.2×10⁻⁴ω.m

Explanation:

Step1: Calculate the cross - sectional area of the wire

The cross - sectional area of a circle is $A=\pi r^{2}$, where the radius $r=\frac{d}{2}$. Given $d = 0.33mm=0.33\times10^{-3}m$, so $r=\frac{0.33\times 10^{-3}}{2}m$. Then $A=\pi(\frac{0.33\times 10^{-3}}{2})^{2}=\pi\times\frac{0.33^{2}\times10^{-6}}{4}m^{2}$.

Step2: Use Ohm's law and the resistance formula

Ohm's law is $V = IR$, so the resistance $R=\frac{V}{I}$. Given $V = 12V$ and $I = 2.1A$, then $R=\frac{12}{2.1}\Omega$. The resistance formula is $R=
ho\frac{l}{A}$, where $l = 6.9m$ is the length of the wire, $
ho$ is the resistivity. We can re - arrange it to $
ho=\frac{RA}{l}$.
Substitute $R=\frac{12}{2.1}\Omega$, $A=\pi\times\frac{0.33^{2}\times10^{-6}}{4}m^{2}$ and $l = 6.9m$ into the formula for $
ho$:
\[

$$\begin{align*} ho&=\frac{\frac{12}{2.1}\times\pi\times\frac{0.33^{2}\times10^{-6}}{4}}{6.9}\\ &=\frac{12\times\pi\times0.33^{2}\times10^{-6}}{2.1\times4\times6.9}\\ &\approx 5.5\times 10^{-8}\Omega\cdot m \end{align*}$$

\]

Answer:

b. $5.5\times 10^{-8}\Omega\cdot m$