Question

A current of 0.96 amps flows through a copper wire 0.44 mm in diameter when connected to a potential difference of 15 v. how long is the wire?

Answer 1

By using Ohm's law, we can calculate the resistance of the wire. Ohm's law states that:
$V=IR$
where V is the potential difference across the conductor, I is the current and R the resistance. Rearranging the equation, we get
$R= \frac{V}{I}= \frac{15 V}{0.96 A}=15.6 \Omega$

Now we can use the following equation to calculate the length of the wire:
$R= \frac{\rho L}{A}$ (1)
where
$\rho$ is the resistivity of the material
L is the length of the conductor
A is its cross-sectional area
In this problem, we have a wire of copper, with resistivity $\rho=1.68 \cdot 10^{-8} \Omega m$. The radius of the wire is half the diameter:
$r= \frac{d}{2}= \frac{0.44 mm}{2}=0.22 mm=0.22 \cdot 10^{-3} m$
And the cross-sectional area is
$A=\pi r^2=\pi (0.22 \cdot 10^{-3}m)^2=1.52 \cdot 10^{-7} m^2$

So now we can rearrange eq.(1) to calculate the length of the wire:
$L= \frac{RA}{\rho}= \frac{(15.6 \Omega)(1.52 \cdot 10^{-7} m^2)}{1.68 \cdot 10^{-8} \Omega m}=141.1 m$