Question

What must be the diameter of a cylindrical 120-m long metal wire if its resistance is to be 6.0 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,

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

Here $l=120, R=6, \rho=1.68(10^{-8})$.

Substituting numerical values,

$d=\sqrt{\frac{4(1.68)(10^{-8}) (120)}{\pi (6)}}\\ d=0.0006541$

Te diameter of the wire is $0.6541 mm$