Question

Hen a thin copper wire that is 178 m long is connected between a 1.2-v potential difference, a current of 2.0 amps flows through the wire. what is the diameter of this wire? the resistivity of copper is 1.72 × 10-8 ω • m.

Answer 1

The resistance of the cylindrical wire is $R=\frac{\rho l}{A}$.

Here $R$ is the resistance, $l$ is the length of the wire and $A$ is the area of

cross section. Since the wire is cylindrical $A=\frac{\pi d^2}{4}$ .

From the above 2 equations,

$R=\frac{\rho l}{\frac{\pi d^2}{4}} \\ d^2=\frac{4 \rho l}{\pi R} \\ d=\sqrt{\frac{4 \rho l}{\pi R} }$

The resistance of the wire is given by

$R=\frac{E}{I} \\ R=\frac{1.2}{2} \\ R=0.6$

Thus, the diameter of the wire is

$d=\sqrt{\frac{4 (1.72)10^{-8} (178)}{\pi (0.6)} }\\ d=0.00255$

The diameter of the wire is $2.55 \;mm$