Question

What length of a certain metal wire of diameter 0.15 mm is needed for the wire to have a resistance of 15 ω? the resistivity of this metal is 1.68 × 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 $A=\frac{\pi d^2}{4}$ . Rearranging the above equation,

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

Here $d=0.15, R=15, \rho=1.68(10^{-8})$.

Substituting numerical values,

$l=\frac{\pi 0.15^2(10^{-6}) (15)}{4 (1.68)(10^{-8})} \\ l=15.78$

The length of the wire is $15.78 \;\;m$